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57
PROBLEM 2.55
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium and
that 500P = lb and 650Q = lb, determine the magnitudes of
the forces exerted on the rods A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
0A B= + + + =R P Q F F
Substituting components: (500 lb) [(650 lb)cos50 ]
[(650 lb)sin50 ]
( cos50 ) ( sin50 ) 0B A AF F F
= − + °
− °
+ − ° + ° =
R j i
j
i i j
In the y-direction (one unknown force):
500 lb (650 lb)sin50 sin50 0AF− − ° + ° =
Thus,
500 lb (650 lb)sin50
sin50
AF
+ °
=
°
1302.70 lb= 1303 lbAF =
In the x-direction: (650 lb)cos50 cos50 0B AF F° + − ° =
Thus, cos50 (650 lb)cos50
(1302.70 lb)cos50 (650 lb)cos50
B AF F= ° − °
= ° − °
419.55 lb= 420 lbBF =
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58
PROBLEM 2.56
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium
and that the magnitudes of the forces exerted on rods A and B
are 750AF = lb and 400BF = lb, determine the magnitudes of
P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
0A B= + + + =R P Q F F
Substituting components: cos 50 sin 50
[(750 lb)cos 50 ]
[(750 lb)sin 50 ] (400 lb)
P Q Q= − + ° − °
− °
+ ° +
R j i j
i
j i
In the x-direction (one unknown force):
cos 50 [(750 lb)cos 50 ] 400 lb 0Q ° − ° + =
(750 lb)cos 50 400 lb
cos 50
127.710 lb
Q
° −
=
°
=
In the y-direction: sin 50 (750 lb)sin 50 0P Q− − ° + ° =
sin 50 (750 lb)sin 50
(127.710 lb)sin 50 (750 lb)sin 50
476.70 lb
P Q= − ° + °
= − ° + °
= 477 lb; 127.7 lbP Q= =
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89
PROBLEM 2.87
Knowing that the tension in cable AB is 1425 N, determine the
components of the force exerted on the plate at B.
SOLUTION
2 2 2
(900 mm) (600 mm) (360 mm)
(900 mm) (600 mm) (360 mm)
1140 mm
1425 N
[ (900 mm) (600 mm) (360 mm) ]
1140 mm
(1125 N) (750 N) (450 N)
BA BA BA
BA
BA
BA
BA
T
BA
T
BA
= − + +
= + +
=
=
=
= − + +
= − + +
i j k
T λ
T i j k
i j k
( ) 1125 N, ( ) 750 N, ( ) 450 NBA x BA y BA zT T T= − = =
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90
PROBLEM 2.88
Knowing that the tension in cable AC is 2130 N, determine the
components of the force exerted on the plate at C.
SOLUTION
2 2 2
(900 mm) (600 mm) (920 mm)
(900 mm) (600 mm) (920 mm)
1420 mm
2130 N
[ (900 mm) (600 mm) (920 mm) ]
1420 mm
(1350 N) (900 N) (1380 N)
CA CA CA
CA
CA
CA
CA
T
CA
T
CA
= − + −
= + +
=
=
=
= − + −
= − + −
i j k
T λ
T i j k
i j k
( ) 1350 N, ( ) 900 N, ( ) 1380 NCA x CA y CA zT T T= − = = −
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91
PROBLEM 2.89
A frame ABC is supported in part by cable DBE that passes
through a frictionless ring at B. Knowing that the tension in the
cable is 385 N, determine the components of the force exerted by
the cable on the support at D.
SOLUTION
2 2 2
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm ) (320 mm)
770 mm
385 N
[(480 mm) (510 mm) (320 mm) ]
770 mm
(240 N) (255 N) (160 N)
DB
DB
DB
F
DB
F
DB
= − +
= + +
=
=
=
= − +
= − +
i j k
F λ
i j k
i j k
240 N, 255 N, 160.0 Nx y zF F F= + = − = +
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92
PROBLEM 2.90
For the frame and cable of Problem 2.89, determine the components
of the force exerted by the cable on the support at E.
PROBLEM 2.89 A frame ABC is supported in part by cable DBE
that passes through a frictionless ring at B. Knowing that the tension
in the cable is 385 N, determine the components of the force exerted
by the cable on the support at D.
SOLUTION
2 2 2
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm)
770 mm
385 N
[(270 mm) (400 mm) (600 mm) ]
770 mm
(135 N) (200 N) (300 N)
EB
EB
EB
F
EB
F
EB
= − +
= + +
=
=
=
= − +
= − +
i j k
F λ
i j k
F i j k
135.0 N, 200 N, 300 Nx y zF F F= + = − = +
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93
PROBLEM 2.91
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 600 N and Q = 450 N.
SOLUTION
(600 N)[sin 40 sin 25 cos40 sin 40 cos25 ]
(162.992 N) (459.63 N) (349.54 N)
(450 N)[cos55 cos30 sin55 cos55 sin30 ]
(223.53 N) (368.62 N) (129.055 N)
(386.52 N) (828.25 N) (220.49 N)
(3R
= ° ° + ° + ° °
= + +
= ° ° + ° − ° °
= + −
= +
= + +
=
P i j k
i j k
Q i j k
i j k
R P Q
i j k
2 2 2
86.52 N) (828.25 N) (220.49 N)
940.22 N
+ +
= 940 NR =
386.52 N
cos
940.22 N
x
x
R
R
θ = = 65.7xθ = °
828.25 N
cos
940.22 N
y
y
R
R
θ = = 28.2yθ = °
220.49 N
cos
940.22 N
z
z
R
R
θ = = 76.4zθ = °
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94
PROBLEM 2.92
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 450 N and Q = 600 N.
SOLUTION
(450 N)[sin 40 sin 25 cos40 sin 40 cos25 ]
(122.244 N) (344.72 N) (262.154 N)
(600 N)[cos55 cos30 sin55 cos55 sin30 ]
(298.04 N) (491.49 N) (172.073 N)
(420.28 N) (836.21 N) (90.081 N)
(R
= ° ° + ° + ° °
= + +
= ° ° + ° − ° °
= + −
= +
= + +
=
P i j k
i j k
Q i j k
i j k
R P Q
i j k
2 2 2
420.28 N) (836.21 N) (90.081 N)
940.21 N
+ +
= 940 NR =
420.28
cos
940.21
x
x
R
R
θ = = 63.4xθ = °
836.21
cos
940.21
y
y
R
R
θ = = 27.2yθ = °
90.081
cos
940.21
z
z
R
R
θ = = 84.5zθ = °
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97
PROBLEM 2.95
For the frame of Problem 2.89, determine the magnitude and
direction of the resultant of the forces exerted by the cable at B
knowing that the tension in the cable is 385 N.
PROBLEM 2.89 A frame ABC is supported in part by cable
DBE that passes through a frictionless ring at B. Knowing that
the tension in the cable is 385 N, determine the components of
the force exerted by the cable on the support at D.
SOLUTION
2 2 2
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm) (320 mm) 770 mm
BD
BD
= − + −
= + + =
i j k
(385 N)
[ (480 mm) (510 mm) (320 mm) ]
(770 mm)
(240 N) (255 N) (160 N)
BD BD BD BD
BD
T T
BD
= =
= − + −
= − + −
F λ
i j k
i j k
2 2 2
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm) 770 mm
BE
BE
= − + −
= + + =
i j k
(385 N)
[ (270 mm) (400 mm) (600 mm) ]
(770 mm)
(135 N) (200 N) (300 N)
BE BE BE BE
BE
T T
BE
= =
= − + −
= − + −
F λ
i j k
i j k
(375 N) (455 N) (460 N)BD BE= + = − + −R F F i j k
2 2 2
(375 N) (455 N) (460 N) 747.83 NR = + + = 748 NR =
375 N
cos
747.83 N
xθ
−
= 120.1xθ = °
455 N
cos
747.83 N
yθ = 52.5yθ = °
460 N
cos
747.83 N
zθ
−
= 128.0zθ = °
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112
PROBLEM 2.107
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that 0,Q = find the value of P for
which the tension in cable AD is 305 N.
SOLUTION
0: 0A AB AC ADΣ = + + + =F T T T P where P=P i
(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
(960 mm) (720 mm) (220 mm) 1220 mm
AB AB
AC AC
AD AD
= − − + =
= − − − =
= − + − =
i j k
i j k
i j k
48 12 19
53 53 53
12 3 4
13 13 13
305 N
[( 960 mm) (720 mm) (220 mm) ]
1220 mm
(240 N) (180 N) (55 N)
AB AB AB AB AB
AC AC AC AC AC
AD AD AD
AB
T T T
AB
AC
T T T
AC
T
= = = − − +
= = = − − −
= = − + −
= − + −
T λ i j k
T λ i j k
T λ i j k
i j k
Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:
48 12
: 240 N
53 13
AB ACP T T= + +i (1)
:j
12 3
180 N
53 13
AB ACT T+ = (2)
:k
19 4
55 N
53 13
AB ACT T− = (3)
Solving the system of linear equations using conventional algorithms gives:
446.71 N
341.71 N
AB
AC
T
T
=
= 960 NP =
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147
PROBLEM 2.135
Find the magnitude and direction of the resultant of the two
forces shown knowing that P = 300 N and Q = 400 N.
SOLUTION
(300 N)[ cos30 sin15 sin30 cos30 cos15 ]
(67.243 N) (150 N) (250.95 N)
(400 N)[cos50 cos20 sin50 cos50 sin 20 ]
(400 N)[0.60402 0.76604 0.21985]
(241.61 N) (306.42 N) (87.939 N)
(174.
= − ° ° + ° + ° °
= − + +
= ° ° + ° − ° °
= + −
= + −
= +
=
P i j k
i j k
Q i j k
i j
i j k
R P Q
2 2 2
367 N) (456.42 N) (163.011 N)
(174.367 N) (456.42 N) (163.011 N)
515.07 N
R
+ +
= + +
=
i j k
515 NR =
174.367 N
cos 0.33853
515.07 N
x
x
R
R
θ = = = 70.2xθ = °
456.42 N
cos 0.88613
515.07 N
y
y
R
R
θ = = = 27.6yθ = °
163.011 N
cos 0.31648
515.07 N
z
z
R
R
θ = = = 71.5zθ = °
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176
PROBLEM 3.16
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.
SOLUTION
(a) We have | |A = ×P Q
where 7 3 3= − + −P i j k
2 2 5= + +Q i j k
Then 7 3 3
2 2 5
[(15 6) ( 6 35) ( 14 6) ]
(21) (29) ( 20)
× = − −
= + + − + + − −
= + −
i j k
P Q
i j k
i j k
2 2 2
(20) (29) ( 20)A = + + − or 41.0A =
(b) We have | |A = ×P Q
where 6 5 2= − −P i j k
2 5 1= − + −Q i j k
Then 6 5 2
2 5 1
[(5 10) (4 6) (30 10) ]
(15) (10) (20)
× = − −
− −
= + + + + −
= + +
i j k
P Q
i j k
i j k
2 2 2
(15) (10) (20)A = + + or 26.9A =
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178
PROBLEM 3.18
A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from
the line to the origin O of the system of coordinates.
SOLUTION
2 2
[20 m ( 1 m)] [16 m ( 4 m)]
29 m
ABd = − − + − −
=
Assume that a force F, or magnitude F(N), acts at Point A and is
directed from A to B.
Then ABF=F λ
where
1
(21 20 )
29
B A
AB
ABd
−
=
= +
r r
i j
λ
By definition, | |O A dF= × =M r F
where (1 m) (4 m)A = − −r i j
Then [ ( 1 m) (4 m) ] [(21 m) (20 m) ]
29 m
[ (20) (84) ]
29
64
N m
29
O
F
F
F
= − − − × +
= − +
= ⋅
M i j i j
k k
k
Finally,
64
( )
29
F d F
=
64
m
29
d = 2.21 md =
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185
PROBLEM 3.25
A 200-N force is applied as shown to the bracket ABC. Determine
the moment of the force about A.
SOLUTION
We have /A C A C= ×M r F
where / (0.06 m) (0.075 m)
(200 N)cos 30 (200 N)sin 30
C A
C
= +
= − ° + °
r i j
F j k
Then 200 0.06 0.075 0
0 cos30 sin 30
200[(0.075sin 30 ) (0.06sin 30 ) (0.06cos 30 ) ]
A =
− ° °
= ° − ° − °
i j k
M
i j k
or (7.50 N m) (6.00 N m) (10.39 N m)A = ⋅ − ⋅ − ⋅M i j k
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189
PROBLEM 3.29
In Problem 3.22, determine the perpendicular distance from
point C to portion AD of the line ABAD.
PROBLEM 3.22 A small boat hangs from two davits, one of
which is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
RA exerted on the davit at A.
SOLUTION
First compute the moment about C of the force DAF exerted by the line on D:
From Problem 3.22:
/
2 2
(48 lb) (62 lb) (24 lb)
(6 ft) [ (48 lb) (62 lb) (24 lb) ]
(144 lb ft) (372 lb ft)
(144) (372)
398.90 lb ft
DA AD
C D C DA
C
= −
= − + +
= ×
= + × − + +
= − ⋅ + ⋅
= +
= ⋅
F F
i j k
M r F
i i j k
j k
M
Then C DAd=M F
Since 82 lbDAF =
398.90 lb ft
82 lb
C
DA
M
d
F
=
⋅
= 4.86 ftd =
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194
PROBLEM 3.34
Determine the value of a that minimizes the
perpendicular distance from Point C to a section
of pipeline that passes through Points A and B.
SOLUTION
Assuming a force F acts along AB,
/| | | | ( )C A C F d= × =M r F
where d = perpendicular distance from C to line AB
2 2 2
/
(24 ft) (24 ft) (28 ft)
(24) (24) (28) ft
(6) (6) (7)
11
(3 ft) (10 ft) ( 10 ft)
3 10 10
11
6 6 7
[(10 6 ) (81 6 ) 78 ]
11
AB
A C
C
F
F
F
a
F
a
F
a a
=
+ −
=
+ +
= + −
= − − −
= −
−
= + + − +
F λ
i j k
i j k
r i j k
i j k
M
i j k
Since 2 2 2
/ /| | | | or | | ( )C A C A C dF= × × =M r F r F
2 2 2 21
(10 6 ) (81 6 ) (78)
121
a a d+ + − + =
Setting 2
( ) 0d
da
d = to find a to minimize d:
1
[2(6)(10 6 ) 2( 6)(81 6 )] 0
121
a a+ + − − =
Solving 5.92 fta = or 5.92 fta =
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203
PROBLEM 3.43
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope AB is 540 N,
determine (a) the angle between rope AB and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.
SOLUTION
First note: 2 2 2
2 2 2
( 3) (3) ( 1.5) 4.5 m
( 0.08) (0.38) (0.16) 0.42 m
BA
BD
= − + + − =
= − + + =
Then ( 3 3 1.5 )
4.5
( 2 2 )
3
1
( 0.08 0.38 0.16 )
0.42
1
( 4 19 8 )
21
BA
BA
BA
BD
T
T
BD
BD
= − + −
= − + −
= = − + +
= − + +
T i j k
i j k
i j k
i j k
λ
(a) We have cosBA BD BAT θ⋅ =T λ
or
1
( 2 2 ) ( 4 19 8 ) cos
3 21
BA
BA
T
T θ− + − ⋅ − + + =i j k i j k
or
1
cos [( 2)( 4) (2)(19) ( 1)(8)]
63
0.60317
θ = − − + + −
=
or 52.9θ = °
(b) We have ( )
cos
(540 N)(0.60317)
BA BD BA BD
BA
T
T θ
= ⋅
=
=
T λ
or ( ) 326 NBA BDT =
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204
PROBLEM 3.44
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope BC is 490 N,
determine (a) the angle between rope BC and
the stake, (b) the projection on the stake of the
force exerted by rope BC at Point B.
SOLUTION
First note: 2 2 2
2 2 2
(1) (3) ( 1.5) 3.5 m
( 0.08) (0.38) (0.16) 0.42 m
BC
BD
= + + − =
= − + + =
( 3 1.5 )
3.5
(2 6 3 )
7
1
( 0.08 0.38 0.16 )
0.42
1
( 4 19 8 )
21
BC
BC
BC
BD
T
T
BD
BD
λ
= + −
= + −
= = − + +
= − + +
T i j k
i j k
i j k
i j k
(a) cosBC BD BCTλ θ⋅ =T
1
(2 6 3 ) ( 4 19 8 ) cos
7 21
BC
BC
T
T θ+ − ⋅ − + + =i j k i j k
1
cos [(2)( 4) (6)(19) ( 3)(8)]
147
0.55782
θ = − + + −
=
56.1θ = °
(b) ( )
cos
(490 N)(0.55782)
BC BD BC BD
BC
T
T
λ
θ
= ⋅
=
=
T
( ) 273 NBC BDT =
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215
PROBLEM 3.55
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable AE at A is 55 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note: AE AE
AE
T
AE
=T
2 2 2
(0.9) ( 0.6) (0.2) 1.1 mAE = + − + =
Then
55 N
(0.9 0.6 0.2 )
1.1
5[(9 N) (6 N) (2 N) ]
AE = − +
= − +
T i j k
i j k
Also, 2 2 2
(1.2) ( 0.35) (0)
1.25 m
DB = + − +
=
Then
1
(1.2 0.35 )
1.25
1
(24 7 )
25
DB
DB
DB
=
= −
= −
i j
i j
λ
Now /( )DB DB A D AEM = ⋅ ×r Tλ
where / (0.1 m) (0.2 m)A D = − +r j k
Then
24 7 0
1
(5) 0 0.1 0.2
25
9 6 2
1
( 4.8 12.6 28.8)
5
DBM
−
= −
−
= − − +
or 2.28 N mDBM = ⋅
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216
PROBLEM 3.56
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note: CF CF
CF
T
CF
=T
2 2 2
(0.6) ( 0.9) ( 0.2) 1.1 mCF = + − + − =
Then
33 N
(0.6 0.9 0.2 )
1.1
3[(6 N) (9 N) (2 N) ]
CF = − +
= − −
T i j k
i j k
Also, 2 2 2
(1.2) ( 0.35) (0)
1.25 m
DB = + − +
=
Then
1
(1.2 0.35 )
1.25
1
(24 7 )
25
DB
DB
DB
=
= −
= −
i j
i j
λ
Now /( )DB DB C D CFM = ⋅ ×r Tλ
where / (0.2 m) (0.4 m)C D = −r j k
Then
24 7 0
1
(3) 0 0.2 0.4
25
6 9 2
3
( 9.6 16.8 86.4)
25
DBM
−
= −
− −
= − + −
or 9.50 N mDBM = − ⋅
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217
PROBLEM 3.57
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 235-lb
force P.
SOLUTION
(32 in.) (30 in.) (24 in.)AB = − −i j k
2 2 2
(32) ( 30) ( 24) 50 in.AB = + − + − =
0.64 0.60 0.48AB
AB
AB
= = − −i k
λ
We shall apply the force P at Point G:
/ (5 in.) (30 in.)G B = +r i k
(21 in.) (38 in.) (18 in.)DG = − +i j k
2 2 2
(21) ( 38) (18) 47 in.DG = + − + =
21 38 18
(235 lb)
47
DG
P
DG
− +
= =
i j k
P
(105 lb) (190 lb) (90 lb)= − +P i j k
The moment of P about AB is given by Eq. (3.46):
/
0.64 0.60 0.48
( ) 5 in. 0 30 in.
105 lb 190 lb 90 lb
AB AB G B P
− −
= ⋅ × =
−
M rλ
0.64[0 (30 in.)( 190 lb)]
0.60[(30 in.)(105 lb) (5 in.)(90 lb)]
0.48[(5 in.)( 190 lb) 0]
2484 lb in.
AB = − −
− −
− − −
= + ⋅
M
207 lb ftAB = + ⋅M
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218
PROBLEM 3.58
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 174-lb
force Q.
SOLUTION
(32 in.) (30 in.) (24 in.)AB = − −i j k
2 2 2
(32) ( 30) ( 24) 50 in.AB = + − + − =
0.64 0.60 0.48AB
AB
AB
= = − −i j k
λ
We shall apply the force Q at Point H:
/ (32 in.) (17 in.)H B = − +r i j
(16 in.) (21 in.) (12 in.)DH = − − −i j k
2 2 2
(16) ( 21) ( 12) 29 in.DH = + − + − =
16 21 12
(174 lb)
29
DH
DH
− − −
= =
i j k
Q
(96 lb) (126 lb) (72 lb)Q = − − −i j k
The moment of Q about AB is given by Eq. (3.46):
/
0.64 0.60 0.48
( ) 32 in. 17 in. 0
96 lb 126 lb 72 lb
AB AB H B
− −
= ⋅ × = −
− − −
M r Qλ
0.64[(17 in.)( 72 lb) 0]
0.60[(0 ( 32 in.)( 72 lb)]
0.48[( 32 in.)( 126 lb) (17 in.)( 96 lb)]
2119.7 lb in.
AB = − −
− − − −
− − − − −
= − ⋅
M
176.6 lb ftAB = ⋅M
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219
PROBLEM 3.59
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at B and is attached to hooks
at G and H. Knowing that the tension in the cable is 450 N,
determine the moment about the diagonal AD of the force
exerted on the frame by portion BH of the cable.
SOLUTION
/( )AD AD B A BHM = ⋅ ×r Tλ
Where
/
1
(4 3 )
5
(0.5 m)
AD
B A
= −
=
i k
r i
λ
and 2 2 2
(0.375) (0.75) ( 0.75)
1.125 m
BHd = + + −
=
Then
450 N
(0.375 0.75 0.75 )
1.125
(150 N) (300 N) (300 N)
BH = + −
= + −
T i j k
i j k
Finally,
4 0 3
1
0.5 0 0
5
150 300 300
1
[( 3)(0.5)(300)]
5
ADM
−
=
−
= −
or 90.0 N mADM = − ⋅
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220
PROBLEM 3.60
In Problem 3.59, determine the moment about the diagonal AD
of the force exerted on the frame by portion BG of the cable.
PROBLEM 3.59 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal AD
of the force exerted on the frame by portion BH of the cable.
SOLUTION
/( )AD AD B A BGM = ⋅ ×r Tλ
Where
/
1
(4 3 )
5
(0.5 m)
AD
B A
= −
=
i k
r j
λ
and 2 2 2
( 0.5) (0.925) ( 0.4)
1.125 m
BG = − + + −
=
Then
450 N
( 0.5 0.925 0.4 )
1.125
(200 N) (370 N) (160 N)
BG = − + −
= − + −
T i j k
i j k
Finally,
4 0 3
1
0.5 0 0
5
200 370 160
ADM
−
=
− −
1
[( 3)(0.5)(370)]
5
= − 111.0 N mADM = − ⋅
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224
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between
cable AE and the line joining Points D and B.
PROBLEM 3.55 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable AE
at A is 55 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
From the solution to Problem 3.55: 55 N
5[(9 N) (6 N) (2 N) ]
AE
AE
=
= − +T i j k
Τ
| | 2.28 N mDBM = ⋅
1
(24 7 )
25
DB = −i jλ
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will
contribute to the moment of TAE about line .DB
Now parallel( )
1
5(9 6 2 ) (24 7 )
25
1
[(9)(24) ( 6)( 7)]
5
51.6 N
AE AE DBT = ⋅
= − + ⋅ −
= + − −
=
T
i j k i j
λ
Also, parallel perpendicular( ) ( )AE AE AE= +T T T
so that 2 2
perpendicular( ) (55) (51.6) 19.0379 NAE = + =T
Since DBλ and perpendicular( )AET are perpendicular, it follows that
perpendicular( )DB AEM d T=
or 2.28 N m (19.0379 N)d⋅ =
0.119761d = 0.1198 md =
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225
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM 3.56 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable CF at
C is 33 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
From the solution to Problem 3.56: 33 N
3[(6 N) (9 N) (2 N) ]
CF
CF
=
= − −T i j k
Τ
| | 9.50 N mDBM = ⋅
1
(24 7 )
25
DB = −i jλ
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF will
contribute to the moment of TCF about line .DB
Now parallel( )
1
3(6 9 2 ) (24 7 )
25
3
[(6)(24) ( 9)( 7)]
25
24.84 N
CF CF DB= ⋅
= − − ⋅ −
= + − −
=
T T
i j k i j
λ
Also, parallel perpendicular( ) ( )CF CF CF= +T T T
so that 2 2
perpendicular( ) (33) (24.84)
21.725 N
CF = −
=
T
Since DBλ and perpendicular( )CFT are perpendicular, it follows that
perpendicular| | ( )DB CFM d T=
or 9.50 N m 21.725 Nd⋅ = ×
or 0.437 md =
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228
PROBLEM 3.68
In Problem 3.59, determine the perpendicular distance between
portion BH of the cable and the diagonal AD.
PROBLEM 3.59 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal AD
of the force exerted on the frame by portion BH of the cable.
SOLUTION
From the solution to Problem 3.59: 450 N
(150 N) (300 N) (300 N)
BH
BH
T =
= + −T i j k
| | 90.0 N mADM = ⋅
1
(4 3 )
5
AD = −i kλ
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBH will
contribute to the moment of TBH about line .AD
Now parallel( )
1
(150 300 300 ) (4 3 )
5
1
[(150)(4) ( 300)( 3)]
5
300 N
BH BH ADT = ⋅
= + − ⋅ −
= + − −
=
T
i j k i k
λ
Also, parallel perpendicular( ) ( )BH BH BH= +T T T
so that 2 2
perpendicular( ) (450) (300) 335.41 NBHT = − =
Since ADλ and perpendicular( )BHT are perpendicular, it follows that
perpendicular( )AD BHM d T=
or 90.0 N m (335.41 N)d⋅ =
0.26833 md = 0.268 md =
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229
PROBLEM 3.69
In Problem 3.60, determine the perpendicular distance between
portion BG of the cable and the diagonal AD.
PROBLEM 3.60 In Problem 3.59, determine the moment
about the diagonal AD of the force exerted on the frame by
portion BG of the cable.
SOLUTION
From the solution to Problem 3.60: 450 N
(200 N) (370 N) (160 N)
BG
BG
=
= − + −T i j k
Τ
| | 111 N mADM = ⋅
1
(4 3 )
5
AD = −i kλ
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBG will
contribute to the moment of TBG about line .AD
Now parallel( )
1
( 200 370 160 ) (4 3 )
5
1
[( 200)(4) ( 160)( 3)]
5
64 N
BG BG ADT = ⋅
= − + − ⋅ −
= − + − −
= −
T
i j k i k
λ
Also, parallel perpendicular( ) ( )BG BG BG= +T T T
so that 2 2
perpendicular( ) (450) ( 64) 445.43 NBG = − − =T
Since ADλ and perpendicular( )BGT are perpendicular, it follows that
perpendicular( )AD BGM d T=
or 111 N m (445.43 N)d⋅ =
0.24920 md = 0.249 md =
www.elsolucionario.net
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 [(8 ) in.](25 lb)
AD AD BC BCM d F d F
d d
= +
⋅ = + + + 1.250 in.d =
www.elsolucionario.net
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Fd= −
= ° − ° −
M k
k
(6.1946 N m)= − ⋅ k or 6.19 N m= ⋅M
(c) We have / /: ( )A A B A B C A CΣ Σ × = × + × =M r F r F r F M
(0.390 m)(40 N) cos55 sin55 0
cos20 sin 20 0
(0.660 m)(40 N) cos55 sin55 0
cos20 sin 20 0
(8.9478 N m 15.1424 N m)
(6.1946 N m)
M = ° °
− ° − °
+ ° °
° °
= ⋅ − ⋅
= − ⋅
i j k
i j k
k
k or 6.19 N m= ⋅M
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239
PROBLEM 3.78
If 0,P = replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
1 2 1 2
1
/ 2 /
2 2 2
2
2
; 16 lb, 40 lb
(30 in.) [ (16 lb) ] (480 lb in.)
; (15 in.) (5 in.)
(0) (5) (10) 5 5 in.
40 lb
(5 10 )
5 5
8 5[(1 lb) (2 lb) ]
8 5 15 5 0
0 1 2
8 5[(10 lb in.)
C
E B E B
DE
F F
d
F
= + = =
= × = × − = − ⋅
= × = −
= + + =
= −
= −
= −
−
= ⋅
1
2
M M M
M r F i j k
M r F r i j
j k
j k
i j k
M
i (30 lb in.) (15 lb in.) ]
(480 lb in.) 8 5[(10 lb in.) (30 lb in.) (15 lb in.) ]
(178.885 lb in.) (536.66 lb in.) (211.67 lb in.)
+ ⋅ + ⋅
= − ⋅ + ⋅ + ⋅ + ⋅
= ⋅ + ⋅ − ⋅
j k
M k i j k
i j k
2 2 2
(178.885) (536.66) ( 211.67)
603.99 lb in
M = + + −
= ⋅ 604 lb in.M = ⋅
axis
z
0.29617 0.88852 0.35045
cos 0.29617
cos 0.88852
cos 0.35045
x
y
M
θ
θ
θ
= = + −
=
=
= −
M
i j kλ
z72.8 27.3 110.5x yθ θ θ= ° = ° = °
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you are using it without permission.
240
PROBLEM 3.79
If 20 lb,P = replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.78:
16-lb force: 1 (480 lb in.)M = − ⋅ k
40-lb force: 2 8 5[(10 lb in.) (30 lb in.) (15 lb in.) ]M = ⋅ + ⋅ + ⋅i j k
P 20 lb= 3
(30 in.) (20 lb)
(600 lb in.)
CM P= ×
= ×
= ⋅
r
i k
j
1 2 3
2 2 2
(480) 8 5 (10 30 15 ) 600
(178.885 lb in.) (1136.66 lb in.) (211.67 lb in.)
(178.885) (113.66) (211.67)
1169.96 lb in.
M
= + +
= − + + + +
= ⋅ + ⋅ − ⋅
= + +
= ⋅
M M M M
k i j k j
i j k
1170 lb in.M = ⋅
axis 0.152898 0.97154 0.180921
cos 0.152898
cos 0.97154
cos 0.180921
x
y
z
M
θ
θ
θ
= = + −
=
=
= −
M
i j kλ
81.2 13.70 100.4x y zθ θ θ= ° = ° = °
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245
PROBLEM 3.84
Solve Problem 3.83, assuming 25°.α β= =
PROBLEM 3.83 The force P has a magnitude of 250 N and is applied
at the end C of a 500-mm rod AC attached to a bracket at A and B.
Assuming α 30°= and 60°,β = replace P with (a) an equivalent force-
couple system at B, (b) an equivalent system formed by two parallel
forces applied at A and B.
SOLUTION
(a) Equivalence requires
: or 250 NB BΣ = =F F P F 25.0°
: (0.3 m)[(250 N)sin50 ] 57.453 N mB BMΣ = − ° = − ⋅M
The equivalent force-couple system at B is
250 NB =F 25.0° 57.5 N mB = ⋅M
(b) We require
Equivalence requires
(0.3 m)[(250 N)sin50 ]
[(0.2 m)sin50 ]
375 N
B AEM d Q
Q
Q
= °
= °
=
Adding the forces at B: 375 NA =F 25.0° 625 NB =F 25.0°
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 ( cos30 )(10.7 m)C AM FΣ ° = °
388.79 NAF =
or 389 NA =F 60.0°
Based on
: [(1040 N)cos30 ](6.7 m) ( cos30 )(10.7 m)A CM FΣ − ° = °
651.21 NCF =
or 651 NC =F 60.0°
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256
PROBLEM 3.95
A 110-N force acting in a vertical plane parallel to the yz-plane
is applied to the 220-mm-long horizontal handle AB of a
socket wrench. Replace the force with an equivalent force-
couple system at the origin O of the coordinate system.
SOLUTION
We have : BΣ =F P F
where 110 N[ (sin15 ) (cos15 ) ]
(28.470 N) (106.252 N)
B = − ° + °
= − +
P j k
j k
or (28.5 N) (106.3 N)= − +F j k
We have /:O B O B OMΣ × =r P M
where / [(0.22cos35 ) (0.15) (0.22sin35 ) ] m
(0.180213 m) (0.15 m) (0.126187 m)
B O = ° + − °
= + −
r i j k
i j k
0.180213 0.15 0.126187 N m
0 28.5 106.3
O⋅ =
−
i j k
M
[(12.3487) (19.1566) (5.1361) ] N mO = − − ⋅M i j k
or (12.35 N m) (19.16 N m) (5.13 N m)O = ⋅ − ⋅ − ⋅M i j k
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257
PROBLEM 3.96
An eccentric, compressive 1220-N force P is applied to the end
of a cantilever beam. Replace P with an equivalent force-couple
system at G.
SOLUTION
We have
: (1220 N)Σ − =F i F
(1220 N)= −F i
Also, we have
/:G A GΣ × =M r P M
1220 0 0.1 0.06 N m
1 0 0
− − ⋅ =
−
i j k
M
(1220 N m)[( 0.06)( 1) ( 0.1)( 1) ]= ⋅ − − − − −M j k
or (73.2 N m) (122 N m)= ⋅ − ⋅M j k
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258
PROBLEM 3.97
To keep a door closed, a wooden stick is wedged between the
floor and the doorknob. The stick exerts at B a 175-N force directed
along line AB. Replace that force with an equivalent force-couple
system at C.
SOLUTION
We have
: AB CΣ =F P F
where
(33 mm) (990 mm) (594 mm)
(175 N)
1155.00 mm
AB AB ABP=
+ −
=
P λ
i j k
or (5.00 N) (150.0 N) (90.0 N)C = + −F i j k
We have /:C B C AB CΣ × =M r P M
5 0.683 0.860 0 N m
1 30 18
(5){( 0.860)( 18) (0.683)( 18)
[(0.683)(30) (0.860)(1)] }
C = − ⋅
−
= − − − −
+ −
i j k
M
i j
k
or (77.4 N m) (61.5 N m) (106.8 N m)C = ⋅ + ⋅ + ⋅M i j k
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259
PROBLEM 3.98
A 46-lb force F and a 2120-lb⋅in. couple M are
applied to corner A of the block shown. Replace
the given force-couple system with an equivalent
force-couple system at corner H.
SOLUTION
We have 2 2 2
(18) ( 14) ( 3) 23 in.AJd = + − + − =
Then
46 lb
(18 14 3 )
23
(36 lb) (28 lb) (6 lb)
= − −
= − −
F i j k
i j k
Also 2 2 2
( 45) (0) ( 28) 53 in.ACd = − + + − =
Then
2120 lb in.
( 45 28 )
53
(1800 lb in.) (1120 lb in.)
⋅
= − −
= − ⋅ − ⋅
M i k
i k
Now /A H′ = + ×M M r F
where / (45 in.) (14 in.)A H = +r i j
Then ( 1800 1120 ) 45 14 0
36 28 6
′ = − − +
− −
i j k
M i k
( 1800 1120 ) {[(14)( 6)] [ (45)( 6)] [(45)( 28) (14)(36)] }
( 1800 84) (270) ( 1120 1764)
(1884 lb in.) (270 lb in.) (2884 lb in.)
(157 lb ft) (22.5 lb ft) (240 lb ft)
= − − + − + − − + − −
= − − + + − −
= − ⋅ + ⋅ − ⋅
= − ⋅ + ⋅ − ⋅
i k i j k
i j k
i j k
i j k
The equivalent force-couple system at H is (36.0 lb) (28.0 lb) (6.00 lb)′ = − −F i j k
(157.0 lb ft) (22.5 lb ft) (240 lb ft)′ = − ⋅ + ⋅ − ⋅M i j k
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267
PROBLEM 3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the
shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment
M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.
SOLUTION
First note that the force-couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces remain
unchanged.
: (5 lb ft) (15 lb ft) (2 ft) (10 lb)A OA = Σ = ⋅ + ⋅ + ×M M j k k i
(25 lb ft) (15 lb ft)= ⋅ + ⋅j k
: (5 lb ft) (25 lb ft)
[(4.5 ft) (1 ft) (2 ft) ] 10 lb)
(15 lb ft) (15 lb ft)
D OD = Σ = − ⋅ + ⋅
+ + + ×
= ⋅ + ⋅
M M j k
i j k i
i k
: (15 lb ft) (15 lb ft)
: (15 lb ft) (5 lb ft)
[(4.5 ft) (1 ft) ] (10 lb)
G O
I I
G
I
= Σ = ⋅ + ⋅
= Σ = ⋅ − ⋅
+ + ×
M M i j
M M j k
i j j
(15 lb ft) (15 lb ft)= ⋅ − ⋅j k
The equivalent force-couple system is the system at corner D. www.elsolucionario.net
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 ( in.)(11.7 lb)
DF
d L
Σ − −
− + = −
or 375.4 3.5 11.7 ( , in in.)d L d L+ =
(a) 25 in.d =
We have 375.4 3.5(25) 11.7 or 39.6 in.L L+ = =
The resultant passes through a point 39.6 in. to the right of D.
(b) 42 in.L =
We have 375.4 3.5 11.7(42)d+ = or 33.1 in.d =
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273
PROBLEM 3.109
A couple M and the three forces shown are applied to an angle bracket.
Find the moment of the couple if the line of action of the resultant of the
force system is to pass through (a) Point A, (b) Point B, (c) Point C.
SOLUTION
In each case, we must have 1 0R
=M
(a) (12 in.)[(30 lb)sin 60 ] (8 in.)(45 lb) 0B
A AM M M= Σ = + ° − =
48.231 lb in.M = + ⋅ 48.2 lb in.= ⋅M
(b) (12 in.)(10 lb) (8 in.)(45 lb) 0R
B BM M M= Σ = + − =
240 lb in.M = + ⋅ 240 lb in.= ⋅M
(c) (12 in.)(10 lb) (8 in.)[(30 lb)cos60 ] 0R
C CM M M= Σ = + − ° =
0M = 0=M
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274
PROBLEM 3.110
A 32-lb motor is mounted on the floor. Find the resultant of the
weight and the forces exerted on the belt, and determine where
the line of action of the resultant intersects the floor.
SOLUTION
We have
: (60 lb) (32 lb) (140 lb)(cos30 sin30 )Σ − + ° + ° =F i j i j R
(181.244 lb) (38.0 lb)= +R i j
or 185.2 lb=R 11.84°
We have :O O yM M xRΣ Σ =
[(140 lb)cos30 ][(4 2cos30 )in.] [(140 lb)sin30 ][(2 in.)sin30 ]− ° + ° − ° °
(60 lb)(2 in.) (38.0 lb)x− =
1
( 694.97 70.0 120) in.
38.0
x = − − −
and 23.289 in.x = −
Or resultant intersects the base (x-axis) 23.3 in. to the left of
the vertical centerline (y-axis) of the motor.
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276
PROBLEM 3.112
Solve Problem 3.111, assuming that P = 60 N.
PROBLEM 3.111 A machine component is subjected to the
forces and couples shown. The component is to be held in
place by a single rivet that can resist a force but not a couple.
For P = 0, determine the location of the rivet hole if it is to
be located (a) on line FG, (b) on line GH.
SOLUTION
See the solution to Problem 3.111 leading to the development of Equation (1):
and
(55.544 0.13 ) N m
(152.142 ) N
G
x
M P
R P
= − + ⋅
= +
For 60 NP =
we have (152.142 60)
212.14 N
[55.544 0.13(60)]
63.344 N m
x
G
R
M
= +
=
= − +
= − ⋅
Then with R at I, : 63.344 N m (244.53 N)GM aΣ − ⋅ = −
or 0.259 ma =
and with R at J, : 63.344 N m (212.14 N)GM bΣ − ⋅ = −
or 0.299 mb =
(a) The rivet hole is 0.299 m above G.
(b) The rivet hole is 0.259 m to the right of G.
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 [240 lbsin 70 ](4 ft)
(160 lb)(12 ft) [300 lbcos40 ](6 ft)
[300 lbsin 40 ](20 ft) (180 lb)(8 ft)
7232.5 lb ft
AMΣ = − ° − °
− + °
− ° −
= − ⋅
7232.5 lb ft
758.36 lb
9.5370 ft
d
− ⋅
=
−
= or 9.54 ft to the right ofd A=
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281
PROBLEM 3.117
Solve Problem 3.116, assuming that the 760-N force is directed
to the right.
PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as
shown. (a) Find the resultant of these forces. (b) Locate the two
points where the line of action of the resultant intersects the edge
of the plate.
SOLUTION
(a)
( 400N 160 N 760 N)
(600 N 300 N 300 N)
(520 N) (1200 N)
= Σ
= − + +
+ + +
= +
R F
i
j
i j
2 2
(520 N) (1200 N) 1307.82 N
1200 N
tan 2.3077
520 N
66.5714
R
θ
θ
= + =
= =
= ° 1308 N=R 66.6°
(b)
(0.5 m) (300 N 300 N)
(300 N m)
R
C = Σ ×
= × +
= ⋅
M r F
i j
k
(300 N m) (1200 N)
0.25000 m
x
x
⋅ = ×
=
k i j
or 0.250 mmx =
(300 N m) [ (0.375 m) ] [(520 N) (1200 N) ]
(1200 195)
x
x
′⋅ = + × +
′= −
k i j i j
k
0.41250 mx′ =
or 412.5 mmx′ =
Intersection 412 mm to the right of A and 250 mm to the right of C www.elsolucionario.net
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284
PROBLEM 3.119
As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet
metal enclosure, the insertion tools exert the forces shown on the enclosure.
Each of the forces is parallel to one of the coordinate axes. Replace these
forces with an equivalent force-couple system at C.
SOLUTION
For equivalence, :ΣF
(17 N) (12 N) (16 N) (21 N)
(21 N) (29 N) (16 N)
A B C D= + + +
= − − − −
= − − −
R F F F F
j j k i
i j k
/ / /:C A C A B C B D C DMΣ = × + × + ×M r F r F r F
[(0.11 m) (0.03 m) ] [ (17 N)]
[(0.02 m) (0.11 m) (0.03 m) ] [ (12 N)]
[(0.03 m) (0.03 m) (0.03 m) ] [ (21 N)]
(0.51 N m) [ (0.24 N m) (0.36 N m) ]
[(0.63 N m) (0.63 N m) ]
M = − × −
+ + − × −
+ + − × −
= − ⋅ + − ⋅ − ⋅
+ ⋅ + ⋅
j k j
i j k j
i j k i
i k i
k j
∴ The equivalent force-couple system at C is
(21.0 N) (29.0 N) (16.00 N)= − − −R i j k
(0.870 N m) (0.630 N m) (0.390 N m)= − ⋅ + ⋅ + ⋅M i j k
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286
PROBLEM 3.121
Four forces are applied to the machine component
ABDE as shown. Replace these forces with an
equivalent force-couple system at A.
SOLUTION
(50 N) (300 N) (120 N) (250 N)
(420 N) (50 N) (250 N)
(0.2 m)
(0.2 m) (0.16 m)
(0.2 m) (0.1 m) (0.16 m)
B
D
E
= − − − −
= − − −
=
= +
= − +
R j i i k
R i j k
r i
r ki
r i j k
[ (300 N) (50 N) ]
( 250 N) ( 120 N)
0.2 m 0 0 0.2 m 0 0.16 m
300 N 50 N 0 0 0 250 N
0.2 m 0.1 m 0.16 m
120 N 0 0
(10 N m) (50 N m) (19.2 N m) (12 N m)
R
A B
D
= × − −
+ × − + × −
= +
− − −
+ −
−
= − ⋅ + ⋅ − ⋅ − ⋅
M r i j
r k r i
i j k i j k
i j k
k j j k
Force-couple system at A is
(420 N) (50 N) (250 N) (30.8 N m) (220 N m)R
A= − − − = ⋅ − ⋅R i j k M j k
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288
PROBLEM 3.123
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to a force-couple
system at A consisting of (30 N) + +y zR R= −R i j k
and (12 N · m) .R
A = −M i (b) Find the corresponding
values of yR and .zR (c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(a) Equivalence requires :Σ = +F R B C
or (30 N) ( )y z x y zR R B C C C− + + = − + − + +i j k k i j k
Equating the i coefficients: : 30 N or 30 Nx xC C− = − =i
Also, / /: R
A A B A C AΣ = × + ×M M r B r C
or (12 N m) [(0.2 m) (0.15 m) ] ( )
(0.4 m) [ (30 N) ]y z
B
C C
− ⋅ = + × −
+ × − + +
i i j k
i i j k
Equating coefficients: : 12 N m (0.15 m) or 80 N
: 0 (0.4 m) or 0
: 0 (0.2 m)(80 N) (0.4 m) or 40 N
y y
z z
B B
C C
C C
− ⋅ = − =
= =
= − =
i
k
j
(80.0 N) (30.0 N) (40.0 N)= − = − +B k C i k
(b) Now we have for the equivalence of forces
(30 N) (80 N) [( 30 N) (40 N) ]y zR R− + + = − + − +i j k k i k
Equating coefficients: : 0yR =j 0yR =
: 80 40zR = − +k or 40.0 NzR = −
(c) First note that (30 N) (40 N) .= − −R i k Thus, the screw is best able to resist the lateral force zR
when the slot in the head of the screw is vertical. www.elsolucionario.net
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289
PROBLEM 3.124
In order to unscrew the tapped faucet A, a plumber uses two
pipe wrenches as shown. By exerting a 40-lb force on each
wrench, at a distance of 10 in. from the axis of the pipe and in a
direction perpendicular to the pipe and to the wrench, he
prevents the pipe from rotating, and thus avoids loosening or
further tightening the joint between the pipe and the tapped
elbow C. Determine (a) the angle θ that the wrench at A should
form with the vertical if elbow C is not to rotate about the
vertical, (b) the force-couple system at C equivalent to the two
40-lb forces when this condition is satisfied.
SOLUTION
We first reduce the given forces to force-couple systems at A and B, noting that
| | | | (40 lb)(10 in.)
400 lb in.
A B= =
= ⋅
M M
We now determine the equivalent force-couple system at C.
(40 lb)(1 cos ) (40 lb)sinθ θ= − −R i j (1)
(15 in.) [ (40 lb)cos (40 lb)sin ]
(7.5 in.) (40 lb)
400 400 600cos 600sin 300
(600 lb in.)sin (300 lb in.)(1 2cos )
R
C A B θ θ
θ θ
θ θ
= + + × − −
+ ×
= + − − + +
= ⋅ + ⋅ −
M M M k i j
k i
j i j
i j (2)
(a) For no rotation about vertical, y component of R
CM must be zero.
1 2cos 0
cos 1/2
θ
θ
− =
=
60.0θ = °
(b) For 60.0θ = ° in Eqs. (1) and (2),
(20.0 lb) (34.641 lb) ; (519.62 lb in.)R
C= − = ⋅R i j M i
(20.0 lb) (34.6 lb) ; (520 lb in.)R
C= − = ⋅R i j M i
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290
PROBLEM 3.125
Assuming θ = 60° in Prob. 3.124, replace the two 40-lb forces
with an equivalent force-couple system at D and determine
whether the plumber’s action tends to tighten or loosen the
joint between (a) pipe CD and elbow D, (b) elbow D and
pipe DE. Assume all threads to be right-handed.
PROBLEM 3.124 In order to unscrew the tapped faucet A,
a plumber uses two pipe wrenches as shown. By exerting a
40-lb force on each wrench, at a distance of 10 in. from the
axis of the pipe and in a direction perpendicular to the pipe
and to the wrench, he prevents the pipe from rotating, and
thus avoids loosening or further tightening the joint between
the pipe and the tapped elbow C. Determine (a) the angle θ
that the wrench at A should form with the vertical if elbow C
is not to rotate about the vertical, (b) the force-couple system
at C equivalent to the two 40-lb forces when this condition
is satisfied.
SOLUTION
The equivalent force-couple system at C for 60θ = ° was obtained in the solution to Prob. 3.124:
(20.0 lb) (34.641 lb)
(519.62 lb in.)R
C
= −
= ⋅
R i j
M i
The equivalent force-couple system at D is made of R and R
DM where
/
(519.62 lb in.) (25.0 in.) [(20.0 lb) (34.641 lb) ]
(519.62 lb in.) (500 lb in.)
R R
D C C D= + ×
= ⋅ + × −
= ⋅ − ⋅
M M r R
i j i j
i k
Equivalent force-couple at D:
(20.0 lb) (34.6 lb) ; (520 lb in.) (500 lb in.)R
C= − = ⋅ − ⋅R i j M i k
(a) Since R
DM has no component along the y-axis, the plumber’s action will neither loosen
nor tighten the joint between pipe CD and elbow.
(b) Since the x component of R
DM is , the plumber’s action will tend to tighten
the joint between elbow and pipe DE. www.elsolucionario.net
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303
PROBLEM 3.135*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have : (10 lb) (11lb)Σ − − =F j j R
(21lb)= −R j
We have : ( ) R
O O C OΣ Σ × + Σ =M r F M M
0 0 20 lb in. 0 0 15 lb in. (12 lb in)
0 10 0 0 11 0
(35 lb in.) (12 lb in.)
R
O = ⋅ + − ⋅ − ⋅
− −
= ⋅ − ⋅
i j k i j k
M j
i j
(a) (21lb)= −R j or (21.0 lb)= −R j
(b) We have 1
1
( ) [(35 lb in.) (12 lb in.) ]
12 lb in. and (12 lb in.)
R
R O RM
R
= ⋅ =
= − ⋅ ⋅ − ⋅
= ⋅ = − ⋅
R
λ M λ
j i j
M j
and pitch 1 12 lb in.
0.57143 in.
21lb
M
p
R
⋅
= = = or 0.571in.p =
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304
PROBLEM 3.135* (Continued)
(c) We have 1 2
2 1 (35 lb in.)
R
O
R
O
= +
= − = ⋅
M M M
M M M i
We require 2 /
(35 lb in.) ( ) [ (21lb) ]
35 (21 ) (21 )
Q O
x z
x z
= ×
⋅ = + × −
= − +
M r R
i i k j
i k i
From i: 35 21
1.66667 in.
z
z
=
=
From k: 0 21
0
x
z
= −
=
The axis of the wrench is parallel to the y-axis and intersects the xz-plane at 0, 1.667 in.x z= =
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305
PROBLEM 3.136*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple system.
We have : (20 N) (15 N) 25 NRΣ − − = =F i j R
We have : ( ) R
O O C OΣ Σ × + Σ =M r F M M
20 N(0.1 m) (4 N m) (1 N m)
(4 N m) (3 N m)
R
O = − − ⋅ − ⋅
= − ⋅ − ⋅
M j i j
i j
(a) (20.0 N) (15.00 N)= − −R i j
(b) We have 1
( 0.8 0.6 ) [ (4 N m) (3 N m) ]
5 N m
R
R OM
R
= ⋅ =
= − − ⋅ − ⋅ − ⋅
= ⋅
R
M
i j i j
λ λ
Pitch: 1 5 N m
0.200 m
25 N
M
p
R
⋅
= = =
or 0.200 mp =
(c) From above, note that
1
R
O=M M
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the
xy-plane with a slope of
3
4
y x=
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306
PROBLEM 3.137*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have : (84 N) (80 N) 116 NRΣ − − = =F j k R
and : ( ) R
O O C OΣ Σ × + Σ =M r F M M
0.6 0 0.1 0.4 0.3 0 ( 30 32 ) N m
0 84 0 0 0 80
R
O+ + − − ⋅ =
i j k i j k
j k M
(15.6 N m) (2 N m) (82.4 N m)R
O = − ⋅ + ⋅ − ⋅M i j k
(a) (84.0 N) (80.0 N)= − −R j k
(b) We have 1
84 80
[ (15.6 N m) (2 N m) (82.4 N m) ]
116
55.379 N m
R
R O RM
R
= ⋅ =
− −
= − ⋅ − ⋅ + ⋅ − ⋅
= ⋅
R
λ M λ
j k
i j k
and 1 1 (40.102 N m) (38.192 N m)RM λ= = − ⋅ − ⋅M j k
Then pitch 1 55.379 N m
0.47741m
116 N
M
p
R
⋅
= = = or 0.477 mp =
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307
PROBLEM 3.137* (Continued)
(c) We have 1 2
2 1 [( 15.6 2 82.4 ) (40.102 38.192 )] N m
(15.6 N m) (42.102 N m) (44.208 N m)
R
O
R
O
= +
= − = − + − − − ⋅
= − ⋅ + ⋅ − ⋅
M M M
M M M i j k j k
i j k
We require 2 /
( 15.6 42.102 44.208 ) ( ) (84 80 )
(84 ) (80 ) (84 )
Q O
x z
z x x
= ×
− + − = + × −
= + −
M r R
i j k i k j k
i j k
From i: 15.6 84
0.185714 m
z
z
− =
= −
or 0.1857 mz = −
From k: 44.208 84
0.52629 m
x
x
− = −
=
or 0.526 mx =
The axis of the wrench intersects the xz-plane at
0.526 m 0 0.1857 mx y z= = = −
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308
PROBLEM 3.138*
Two bolts at A and B are tightened by applying the forces and
couples shown. Replace the two wrenches with a single equivalent
wrench and determine (a) the resultant R, (b) the pitch of the
single equivalent wrench, (c) the point where the axis of the
wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin at B.
(a) We have
8 15
: (26.4 lb) (17 lb)
17 17
Σ − − + =
F k i j R
(8.00 lb) (15.00 lb) (26.4 lb)= − − −R i j k
and 31.4 lbR =
We have /: R
B A B A A B BΣ × + + =M r F M M M
8 15
0 10 0 220 238 264 220 14(8 15 )
17 17
0 0 26.4
(152 lb in.) (210 lb in.) (220 lb in.)
R
B
R
B
= − − − + = − − +
−
= ⋅ − ⋅ − ⋅
i j k
M k i j i k i j
M i j k
(b) We have 1
8.00 15.00 26.4
[(152 lb in.) (210 lb in.) (220 lb in.) ]
31.4
246.56 lb in.
R
R O RM
R
= ⋅ =
− − −
= ⋅ ⋅ − ⋅ − ⋅
= ⋅
R
λ M λ
i j k
i j k
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311
PROBLEM 3.139* (Continued)
(b) We have 1
R
R OM = ⋅Mλ
where
3 25 1
(2 20 ) (2 20 )
25 27 5 9 5
R
P
R P
= = − − = − −
R
i j k i j kλ
Then 1
1 24 8
(2 20 ) ( )
59 5 15 5
Pa Pa
M
−
= − − ⋅ − − =i j k i k
and pitch 1 8 25 8
8115 5 27 5
M Pa a
p
R P
− −
= = =
or 0.0988p a= −
(c) 1 1
8 1 8
(2 20 ) ( 2 20 )
67515 5 9 5
R
Pa Pa
M
−
= = − − = − + +
M i j k i j kλ
Then 2 1
24 8 8
( ) ( 2 20 ) ( 430 20 406 )
5 675 675
R
O
Pa Pa Pa
= − = − − − − + + = − − −M M M i k i j k i j k
We require 2 /Q O= ×M r R
8 3
( 403 20 406 ) ( ) (2 20 )
675 25
3
[20 ( 2 ) 20 ]
25
Pa P
x z
P
z x z x
− − − = + × − −
= + + −
i j k i k i j k
i j k
From i:
3
8( 403) 20 1.99012
675 25
Pa P
z z a
− = = −
From k:
3
8( 406) 20 2.0049
675 25
Pa P
x x a
− = − =
The axis of the wrench intersects the xz-plane at
2.00 , 1.990x a z a= = −
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312
PROBLEM 3.140*
Two ropes attached at A and B are used to move the trunk of a
fallen tree. Replace the forces exerted by the ropes with an
equivalent wrench and determine (a) the resultant force R,
(b) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the yz-plane.
SOLUTION
(a) First replace the given forces with an equivalent force-couple system ( ), R
OR M at the origin.
We have
2 2 2
2 2 2
(6) (2) (9) 11 m
(14) (2) (5) 15 m
AC
BD
d
d
= + + =
= + + =
Then
1650 N
(6 2 9 )
11
(900 N) (300 N) (1350 N)
ACT = = + +
= + +
i j k
i j k
and
1500 N
(14 2 5 )
15
(1400 N) (200 N) (500 N)
BDT = = + +
= + +
i j k
i j k
Equivalence then requires
:
(900 300 1350 )
(1400 200 500 )
(2300 N) (500 N) (1850 N)
AC BDΣ = +
= + +
+ + +
= + +
F R T T
i j k
i j k
i j k
: R
O O A AC B BDΣ = × + ×M M r T r T
(12 m) [(900 N) (300 N) (1350 N) ]
(9 m) [(1400 N) (200 N) (500 N) ]
(3600) (10,800 4500) (1800)
(3600 N m) (6300 N m) (1800 N m)
= × + +
+ × + +
= − + − +
= − ⋅ + ⋅ + ⋅
k i j k
i i j k
i j k
i j k
The components of the wrench are 1( , ),R M where
(2300 N) (500 N) (1850 N)= + +R i j k
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313
PROBLEM 3.140* (Continued)
(b) We have
2 2 2
100 (23) (5) (18.5) 2993.7 NR = + + =
Let
axis
1
(23 5 18.5 )
29.937R
= = + +
R
i j kλ
Then 1 axis
1
(23 5 18.5 ) ( 3600 6300 1800 )
29.937
1
[(23)( 36) (5)(63) (18.5)(18)]
0.29937
601.26 N m
R
OM = ⋅
= + + ⋅ − + +
= − + +
= − ⋅
M
i j k i j k
λ
Finally, 1 601.26 N m
2993.7 N
M
P
R
− ⋅
= =
or 0.201 mP = −
(c) We have 1 1 axis
1
( 601.26 N m) (23 5 18.5 )
29.937
M M=
= − ⋅ × + +
λ
i j k
or 1 (461.93 N m) (100.421 N m) (371.56 N m)= − ⋅ − ⋅ − ⋅M i j k
Now 2 1
( 3600 6300 1800 )
( 461.93 100.421 371.56 )
(3138.1 N m) (6400.4 N m) (2171.6 N m)
R
O= −
= − + +
− − − −
= − ⋅ + ⋅ + ⋅
M M M
i j k
i j k
i j k
For equivalence:
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315
PROBLEM 3.141*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the
yz-plane. If it cannot be so reduced, replace the given system
with an equivalent wrench and determine its resultant, its pitch,
and the point where its axis intersects the yz-plane.
SOLUTION
First determine the resultant of the forces at D. We have
2 2 2
2 2 2
( 12) (9) (8) 17 in.
( 6) (0) ( 8) 10 in.
DA
ED
d
d
= − + + =
= − + + − =
Then
34 lb
( 12 9 8 )
17
(24 lb) (18 lb) (16 lb)
DA = = − + +
= − + +
F i j k
i j k
and
30 lb
( 6 8 )
10
(18 lb) (24 lb)
ED = = − −
= − −
F i k
i k
Then
: DA EDΣ = +F R F F
( 24 18 16 ( 18 24 )
(42 lb) (18 lb) (8 lb)
= − + + + − −
= − + −
i j k i k
i j k
For the applied couple
2 2 2
( 6) ( 6) (18) 6 11 in.AKd = − + − + =
Then
160 lb in.
( 6 6 18 )
6 11
160
[ (1 lb in.) (1 lb in.) (3 lb in.) ]
11
⋅
= − − +
= − ⋅ − ⋅ + ⋅
M i j k
i j k
To be able to reduce the original forces and couple to a single equivalent force, R and M
must be perpendicular. Thus
?
0⋅ =R M
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316
PROBLEM 3.141* (Continued)
Substituting
?160
( 42 18 8 ) ( 3 ) 0
11
− + − ⋅ − − + =i j k i j k
or
?160
[( 42)( 1) (18)( 1) ( 8)(3)] 0
11
− − + − + − =
or 0 0=
R and M are perpendicular so that the given system can be reduced to the single equivalent force.
(42.0 lb) (18.00 lb) (8.00 lb)= − + −R i j k
Then for equivalence,
Thus, we require /P D= ×M r R
where / (12 in.) [( 3)in.] ( in.)P D y z= − + − +r i j k
Substituting:
160
( 3 ) 12 ( 3)
11
42 18 8
[( 3)( 8) ( )(18)]
[( )( 42) ( 12)( 8)]
[( 12)(18) ( 3)( 42)]
y z
y z
z
y
− − + = − −
− −
= − − −
+ − − − −
+ − − − −
i j k
i j k
i
j
k
Equating coefficients:
160
: 42 96 or 1.137 in.
11
480
: 216 42( 3) or 11.59 in.
11
z z
y y
− = − − = −
= − + − =
j
k
The line of action of R intersects the yz-plane at 0 11.59 in. 1.137 in.x y z= = = −
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317
PROBLEM 3.142*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the
yz-plane. If it cannot be so reduced, replace the given system
with an equivalent wrench and determine its resultant, its
pitch, and the point where its axis intersects the yz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have : A GΣ + =F F F R
(40 mm) (60 mm) (120 mm)
(50 N) 70 N
140 mm
(20 N) (30 N) (10 N)
+ −
= +
= + −
i j k
R k
i j k
and 37.417 NR =
We have : ( ) R
O O C OΣ Σ × + Σ =M r F M M
0
[(0.12 m) (50 N) ] {(0.16 m) [(20 N) (30 N) (60 N) ]}
(160 mm) (120 mm)
(10 N m)
200 mm
(40 mm) (120 mm) (60 mm)
(14 N m)
140 mm
(18 N m) (8.4 N m) (10.8 N m)
R
O
R
= × + × + −
−
+ ⋅
− +
+ ⋅
= ⋅ − ⋅ + ⋅
M j k i i j k
i j
i j k
M i j k
To be able to reduce the original forces and couples to a single equivalent force, R and M must be
perpendicular. Thus, 0.⋅ =R M
Substituting
?
(20 30 10 ) (18 8.4 10.8 ) 0+ − ⋅ − + =i j k i j k
or
?
(20)(18) (30)( 8.4) ( 10)(10.8) 0+ − + − =
or 0 0=
R and M are perpendicular so that the given system can be reduced to the single equivalent force.
(20.0 N) (30.0 N) (10.00 N)= + −R i j k
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328
PROBLEM 3.147
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a) (300 N)cos 25
271.89 N
(300 N)sin 25
126.785 N
(271.89 N) (126.785 N)
x
y
F
F
= °
=
= °
=
= +F i j
(0.1m) (0.2 m)
[ (0.1m) (0.2 m) ] [(271.89 N) (126.785 N) ]
(12.6785 N m) (54.378 N m)
(41.700 N m)
D
D
DA= = − −
= ×
= − − × +
= − ⋅ + ⋅
= ⋅
r i j
M r F
M i j i j
k k
k
41.7 N mD = ⋅M
(b) The smallest force Q at B must be perpendicular to
DB
at 45°
( )
41.700 N m (0.28284 m)
D Q DB
Q
=
⋅ =
M
147.4 NQ = 45.0°
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329
PROBLEM 3.148
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
SOLUTION
First note 2 2
(12.0 in.) (2.33 in.)
12.2241in.
CBd = +
=
Then
12.0 in.
cos
12.2241in.
2.33 in.
sin
12.2241in.
θ
θ
=
=
and cos sin
125 lb
[(12.0 in.) (2.33 in.) ]
12.2241in.
CB CB CBF Fθ θ= −
= −
F i j
i j
Now /A B A CB= ×M r F
where / (15.3 in.) (12.0 in. 2.33 in.)
(15.3 in.) (14.33 in.)
B A = − +
= −
r i j
i j
Then
125 lb
[(15.3 in.) (14.33 in.) ] (12.0 2.33 )
12.2241in.
(1393.87 lb in.)
A = − × −
= ⋅
M i j i j
k
(116.156 lb ft)= ⋅ k or 116.2 lb ftA = ⋅M
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332
PROBLEM 3.151
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the moments about the y
and the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a.
SOLUTION
First note (2.2 m) (3.2 m) ( m)BA a= − −i j k
Now /D A D BA= ×M r T
where / (2.2 m) (1.6 m)
(2.2 3.2 ) (N)
A D
BA
BA
BA
T
a
d
= +
= − −
r i j
T i j k
Then 2.2 1.6 0
2.2 3.2
{ 1.6 2.2 [(2.2)( 3.2) (1.6)(2.2)] }
BA
D
BA
BA
BA
T
d
a
T
a a
d
=
− −
= − + + − −
i j k
M
i j k
Thus 2.2 (N m)
10.56 (N m)
BA
y
BA
BA
z
BA
T
M a
d
T
M
d
= ⋅
= − ⋅
Then forming the ratio
y
z
M
M
2.2 (N m)120 N m
460 N m 10.56 (N m)
BA
BA
BA
BA
T
d
T
d
⋅⋅
=
− ⋅ − ⋅
or 1.252 ma =
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333
PROBLEM 3.152
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that 25 ,θ = °
Mx 61lb ft,= − ⋅ and 43 lb ft,zM = − ⋅ determine φ and d.
SOLUTION
We have /:O A O OΣ × =M r F M
where / (4 in.) (11in.) ( )
(cos cos sin cos sin )
A O d
F θ φ θ θ φ
= − + −
= − +
r i j k
F i j k
For 70 lb, 25F θ= = °
(70 lb)[(0.90631cos ) 0.42262 (0.90631sin ) ]
(70 lb) 4 11 in.
0.90631cos 0.42262 0.90631sin
(70 lb)[(9.9694sin 0.42262 ) ( 0.90631 cos 3.6252sin )
(1.69048 9.9694cos ) ] in.
O d
d d
φ φ
φ φ
φ φ φ
φ
= − +
= − −
− −
= − + − +
+ −
F i j k
i j k
M
i j
k
and (70 lb)(9.9694sin 0.42262 ) in. (61lb ft)(12 in./ft)xM dφ= − = − ⋅ (1)
(70 lb)( 0.90631 cos 3.6252sin ) in.yM d φ φ= − + (2)
(70 lb)(1.69048 9.9694cos ) in. 43 lb ft(12 in./ft)zM φ= − = − ⋅ (3)
From Equation (3): 1 634.33
cos 24.636
697.86
φ −
= = °
or 24.6φ = °
From Equation (1):
1022.90
34.577 in.
29.583
d
= =
or 34.6 in.d =
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335
PROBLEM 3.154
While tapping a hole, a machinist applies the horizontal forces
shown to the handle of the tap wrench. Show that these forces
are equivalent to a single force, and specify, if possible, the
point of application of the single force on the handle.
SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.
We have 2.9 lb 2.65 lb 0.25 lb,BF = − = where the 2.65-lb force is part of the couple. Combining the two
parallel forces,
couple (2.65 lb)[(3.2 in. 2.8 in.)cos25 ]
14.4103 lb in.
M = + °
= ⋅
and couple 14.4103 lb in.= ⋅M
A single equivalent force will be located in the negative z direction.
Based on : 14.4103 lb in. [(0.25 lb)cos 25 ]( )BM aΣ − ⋅ = °
63.600 in.a =
F′ (0.25 lb)(cos 25 sin 25 )= ° + °i k
F′ (0.227 lb) (0.1057 lb)= +i k and is applied on an extension of handle BD at a
distance of 63.6 in. to the right of B.
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336
PROBLEM 3.155
Replace the 150-N force with an equivalent force-couple
system at A.
SOLUTION
Equivalence requires : (150 N)( cos 35 sin 35 )
(122.873 N) (86.036 N)
Σ = − ° − °
= − −
F F j k
j k
/:A D AΣ = ×M M r F
where / (0.18 m) (0.12 m) (0.1 m)D A = − +r i j k
Then 0.18 0.12 0.1 N m
0 122.873 86.036
[( 0.12)( 86.036) (0.1)( 122.873)]
[ (0.18)( 86.036)]
[(0.18)( 122.873)]
(22.6 N m) (15.49 N m) (22.1 N m)
= − ⋅
− −
= − − − −
+ − −
+ −
= ⋅ + ⋅ − ⋅
i j k
M
i
j
k
i j k
The equivalent force-couple system at A is
(122.9 N) (86.0 N)= − −F j k
(22.6 N m) (15.49 N m) (22.1 N m)= ⋅ + ⋅ − ⋅M i j k
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353
PROBLEM 4.9
For the beam and loading shown, determine the range of the
distance a for which the reaction at B does not exceed 100 lb
downward or 200 lb upward.
SOLUTION
Assume B is positive when directed .
Sketch showing distance from D to forces.
0: (300 lb)(8 in. ) (300 lb)( 2 in.) (50 lb)(4 in.) 16 0DM a a BΣ = − − − − + =
600 2800 16 0a B− + + =
(2800 16 )
600
B
a
+
= (1)
For 100 lbB = 100 lb,= − Eq. (1) yields:
[2800 16( 100)] 1200
2 in.
600 600
a
+ −
≥ = = 2.00 in.a ≥
For 200B = 200 lb,= + Eq. (1) yields:
[2800 16(200)] 6000
10 in.
600 600
a
+
≤ = = 10.00 in.a ≤
Required range: 2.00 in. 10.00 in.a≤ ≤
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370
PROBLEM 4.24
A lever AB is hinged at C and attached to a control cable at A. If the
lever is subjected to a 75-lb vertical force at B, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Geometry:
(10 in.)cos20 9.3969 in.
(10 in.)sin 20 3.4202 in.
AC
AC
x
y
= ° =
= ° =
12 in. 3.4202 in. 8.5798 in.DAy = − =
1 1 8.5798
tan tan 42.397
9.3969
DA
AC
y
x
α − −
= = = °
90 20 42.397 27.603β = ° − ° − ° = °
Equilibrium for lever:
(a) 0: cos27.603 (10 in.) (75 lb)[(15 in.)cos 20°] 0C ADM TΣ = ° − =
119.293 lbADT = 119.3 lbADT =
(b) 0: (119.293 lb)cos42.397 0x xF CΣ = + ° =
88.097 lbxC = −
0: 75 lb lb)sin 42.397 0y yF CΣ = − − (119.293 ° =
155.435yC =
Thus, 2 2 2 2
( 88.097) (155.435) 178.665 lbx yC C C= + = − + =
and 1 1 155.435
tan tan 60.456
88.097
y
x
C
C
θ − −
= = = ° 178.7 lb=C 60.5°
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you are using it without permission.
458
PROBLEM 4.99
The 45-lb square plate shown is supported by three vertical wires.
Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
/ / /0: ( 45 lb) 0B C B C A B A G BM T TΣ = × + × + × − =r j r j r j
[ (20 in.) (15 in.) ] (20 in.)C AT T− + × + ×i k j k j
[ (10 in.) (10 in.) ] [ (45 lb) ] 0+ − + × − =i k j
20 15 20 450 450 0C C AT T T− − − + + =k i i k i
Equating to zero the coefficients of the unit vectors,
:k 20 450 0CT− + = 22.5 lbCT =
:i 15(22.5) 20 450 0AT− − + = 5.625 lbAT =
0:yFΣ = 45 lb 0A B CT T T+ + − =
5.625 lb 22.5 lb 45 lb 0BT+ + − = 16.875 lbBT =
5.63 lb; 16.88 lb; 22.5 lbA B CT T T= = =
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459
PROBLEM 4.100
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by
three legs equally spaced around the edge. A vertical load P of magnitude
100 lb is applied to the top of the table at D. Determine the maximum
value of a if the table is not to tip over. Show, on a sketch, the area of the
table over which P can act without tipping the table.
SOLUTION
2 ft sin30 1ftr b r= = ° =
We shall sum moments about AB.
( ) ( ) 0b r C a b P bW+ + − − =
(1 2) ( 1)100 (1)30 0C a+ + − − =
1
[30 ( 1)100]
3
C a= − −
If table is not to tip, 0.C ≥
[30 ( 1)100] 0
30 ( 1)100
a
a
− − ≥
≥ −
1 0.3 1.3 ft 1.300 fta a a− ≤ ≤ =
Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located
in shaded area for no tipping.
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you are using it without permission.
462
PROBLEM 4.103
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
/ / /0: ( 80 lb) 0B A B A C B C G BT TΣ = × + × + × − =M r j r j r j
(60 in.) [(60 in.) (15 in.) ] [(30 in.) (30 in.) ] ( 80 lb) 0A CT T× + + × + + × − =k j i k j i k j
60 60 15 2400 2400 0A C CT T T− + − − + =i k i k i
Equating to zero the coefficients of the unit vectors,
:i 60 15(40) 2400 0AT − + = 30.0 lbAT =
:k 60 2400 0CT − = 40.0 lbCT =
0:yFΣ = 80 lb 0A B CT T T+ + − =
30 lb 40 lb 80 lb 0BT+ + − = 10.00 lbBT = www.elsolucionario.net
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495
PROBLEM 4.121
The assembly shown is welded to collar A that fits on the vertical
pin shown. The pin can exert couples about the x and z axes but
does not prevent motion about or along the y-axis. For the loading
shown, determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
First note:
2 2
2 2
(0.08 m) (0.06 m)
(0.08) (0.06) m
( 0.8 0.6 )
(0.12 m) (0.09 m)
(0.12) (0.09) m
(0.8 0.6 )
− +
= =
+
= − +
−
= =
+
= −
CF CF CF CF
CF
DE DE DE DE
DE
T T
T
T T
T
i j
T λ
i j
j k
T λ
j k
(a) From F.B.D. of assembly:
0: 0.6 0.8 480 N 0Σ = + − =y CF DEF T T
or 0.6 0.8 480 NCF DET T+ = (1)
0: (0.8 )(0.135 m) (0.6 )(0.08 m) 0Σ = − + =y CF DEM T T
or 2.25DE CFT T= (2)
Substituting Equation (2) into Equation (1),
0.6 0.8[(2.25) ] 480 NCF CFT T+ =
200.00 NCFT =
or 200 NCFT =
and from Equation (2): 2.25(200.00 N) 450.00DET = =
or 450 NDET =
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[(450 N)(0.8)](0.09 m) 0
Σ = + −
− =
xx AM M
16.2000 N mxAM = − ⋅
0: (480 N)(0.08 m) [(200.00 N)(0.6)](0.08 m)
[(450 N)(0.8)](0.08 m) 0
zz AM MΣ = − +
+ =
0zAM =
or (16.20 N m)A = − ⋅M i
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576
PROBLEM 5.23
The first moment of the shaded area with respect to the x-axis is denoted by Qx.
(a) Express Qx in terms of b, c, and the distance y from the base of the shaded
area to the x-axis. (b) For what value of y is xQ maximum, and what is that
maximum value?
SOLUTION
Shaded area:
( )
1
( )[ ( )]
2
x
A b c y
Q yA
c y b c y
= −
=
= + −
(a) 2 21
( )
2
xQ b c y= −
(b) For max,Q
1
0 or ( 2 ) 0
2
dQ
b y
dy
= − = 0y =
For 0,y = 21
( )
2
xQ bc= 21
( )
2
xQ bc=
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590
PROBLEM 5.36
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
At ( , ),a h 2
1 :y h ka=
or 2
h
k
a
=
2 :y h ma=
or
h
m
a
=
Now
1 2
1
( )
2
EL
EL
x x
y y y
=
= +
and 2
2 1 2
2
2
( )
( )
h h
dA y y dx x x dx
a a
h
ax x dx
a
= − = −
= −
Then 2 2 3
2 20
0
1 1
( )
2 3 6
a
a h h a
A dA ax x dx x x ah
a a
= = − = − =
and
( )
2 3 4 2
2 20
0
2 2
1 2 2 1 2 1
1 1
( )
3 4 12
1 1
( )[( ) ]
2 2
a
a
EL
EL
h h a
x dA x ax x dx x x a h
a a
y dA y y y y dx y y dx
= − = − =
= + − = −
2 2
2 4
2 40
2 2
3 5
4
0
2
1
2
1 1
2 3 5
1
15
a
a
h h
x x dx
a a
h a
x x
a
ah
= −
= −
=
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604
PROBLEM 5.45
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line.
Now cos andELx r dL rdθ θ= =
Then
7 /4
7 /4
/4
/4
3
[ ]
2
L dL r d r r
π
π
π
π
θ θ π= = = =
and
7 /4
/4
7 /42
/4
2
2
cos ( )
[sin ]
1 1
2 2
2
ELx dL r rd
r
r
r
π
π
π
π
θ θ
θ
=
=
= − −
= −
Thus 23
: 2
2
xL xdL x r rπ
= = −
2 2
3
x r
π
= −
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605
PROBLEM 5.46
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line.
Now 3 2 2
cos andELx a dL dx dyθ= = +
where 3 2
3 2
cos : 3 cos sin
sin : 3 sin cos
x a dx a d
y a dy a d
θ θ θ θ
θ θ θ θ
= = −
= =
Then 2 2 2 2 1/2
2 2 1/2
/2
/2
2
0
0
[( 3 cos sin ) (3 sin cos ) ]
3 cos sin (cos sin )
3 cos sin
1
3 cos sin 3 sin
2
3
2
dL a d a d
a d
a d
L dL a d a
a
π
π
θ θ θ θ θ θ
θ θ θ θ θ
θ θ θ
θ θ θ θ
= − +
= +
=
= = =
=
and
/2
3
0
/2
2 5 2
0
cos (3 cos sin )
1 3
3 cos
5 5
ELx dL a a d
a a
π
π
θ θ θ θ
θ
=
= − =
Hence, 23 3
:
2 5
ELxL x dL x a a
= =
2
5
x a=
Alternative Solution:
2/3
3 2
2/3
3 2
cos cos
sin sin
x
x a
a
y
y a
a
θ θ
θ θ
= =
= =
2/3 2/3
2/3 2/3 3/2
1 or ( )
x y
y a x
a a
+ = = −
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607
PROBLEM 5.47*
A homogeneous wire is bent into the shape shown. Determine by
direct integration the x coordinate of its centroid. Express your
answer in terms of a.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
We have at ,x a= 3/2 1
, , ory a a ka k
a
= = =
Then 3/21
y x
a
=
and 1/23
2
dy
x
dx a
=
Now ELx x=
and
2
1/22
1/2
1
3
1
2
1
4 9
2
dy
dL dx
dx
x dx
a
a x dx
a
= +
= +
= +
Then
0
3/2
0
3/2
1
4 9
2
1 2 1
(4 9 )
3 92
[(13) 8]
27
1.43971
a
a
L dL a x dx
a
a x
a
a
a
= = +
= × +
= −
=
and
0
1
4 9
2
a
ELx dL x a x dx
a
= +
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608
PROBLEM 5.47* (Continued)
Use integration by parts with
3/2
4 9
2
(4 9 )
27
u x dv a x dx
du dx v a x
= = +
= = +
Then 3/2 3/2
0
0
3/2
2 5/2
0
2
3/2 5/2
2
1 2 2
(4 9 ) (4 9 )
27 272
(13) 1 2
(4 9 )
27 4527
2
(13) [(13) 32]
27 45
0.78566
a
a
EL
a
x dL x a x a x dx
a
a a x
a
a
a
= × + − +
= − +
= − −
=
2
: (1.43971 ) 0.78566ELxL x dL x a a= = or 0.546x a=
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609
PROBLEM 5.48
Determine by direct integration the centroid of the area shown.
SOLUTION
/2 /2
0 0
/2
0
sin
1
, ,
2
sin
cos
EL EL
L L
L
x
y a
L
x x y y dA y dx
x
A y dx a dx
L
L x aL
A a
L
π
π
π
π π
=
= = =
= =
= − =
/2 /2
0 0
sin
L L
EL
x
x dA xydx xa dx
L
π
= =
Setting ,
x
u
L
π
= we have ,
L
x u
π
= ,
L
dx du
π
=
2
/2 /2
0 0
sin sinEL
L L L
x dA u a u du a u xdu
π π
π π π
= =
Integrating by parts,
2 2/2
/2
0 20
2/2 /2 /2
2 2 2 2
0 0 0
/22 2/2
2
2 0
0
[ cos ] cos
1 1
sin sin
2 2 2
1 1 1
(1 cos2 ) sin 2
2 4 2 82
EL
L L
EL
L aL
x dA a u u u du
x a L
y dA y dx a dx u du
L
a L a L
u du u u a L
π
π
π
π
π
π π
π
π
ππ
= − + =
= = =
= − = − =
2
2
:
π π
= =
EL
aL aL
xA x dA x
π
=
L
x
21
:
8π
= =
EL
aL
yA y dA y a L
8
π
=y a
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612
PROBLEM 5.50
Determine the centroid of the area shown when 2a = in.
SOLUTION
We have
1 1 1
1
2 2
EL
EL
x x
y y
x
=
= = −
and
1
1dA ydx dx
x
= = −
Then 2
1
1
1
1 [ ln ] ( ln 1) in
2
a
adx
A dA x x a a
x
= = − = − = − −
and
2 2
3
1
1
1 1
1 in
2 2 2
a
a
EL
x a
x dA x dx x a
x
= − = − = − +
21 1
3
1
1 1 1 1 2 1
1 1 1
2 2
1 1 1 1
2ln 2ln in
2 2
a a
EL
a
y dA dx dx
x x x x
x x a a
x a
= − − = − +
= − − = − −
2
1
2 2
1
: in.
ln 1
2ln
: in.
2( ln 1)
a
EL
a
EL
a
xA x dA x
a a
a a
yA y dA y
a a
− +
= =
− −
− −
= =
− −
Find x and y when 2 in.=a
We have
21 1
2 2
(2) 2
2 ln 2 1
x
− +
=
− −
or 1.629 in.x =
and
1
2
2 2ln 2
2(2 ln 2 1)
y
− −
=
− −
or 0.1853 in.y =
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613
PROBLEM 5.51
Determine the value of a for which the ratio /x y is 9.
SOLUTION
We have
1 1 1
1
2 2
EL
EL
x x
y y
x
=
= = −
and
1
1dA y dx dx
x
= = −
Then 1
1
2
1
1 [ ln ]
2
( ln 1) in
a
adx
A dA x x
x
a a
= = − = −
= − −
and
2
1
1
2
3
1
1
2
1
in
2 2
a
a
EL
x
x dA x dx x
x
a
a
= − = −
= − +
21 1
1
3
1 1 1 1 2 1
1 1 1
2 2
1 1
2ln
2
1 1
2ln in
2
a a
EL
a
y dA dx dx
x x x x
x x
x
a a
a
= − − = − +
= − −
= − −
2
1
2 2
1
: in.
ln 1
2ln
: in.
2( ln 1)
a
EL
a
EL
a
xA x dA x
a a
a a
yA y dA y
a a
− +
= =
− −
− −
= =
− −
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you are using it without permission.
615
PROBLEM 5.52
Determine the volume and the surface area of the solid obtained by rotating
the area of Problem 5.1 about (a) the x-axis, (b) the y-axis.
SOLUTION
From the solution of Problem 5.1, we have
2
3
3
11in
11.5 in
39.5 in
A
xA
yA
=
Σ =
Σ =
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about the x-axis:
area
3
Volume 2 2
2 (39.5 in )
y A yAπ π
π
= = Σ
= or 3
Volume 248 in=
line line
2 2 3 3 4 4 5 5 6 6 7 7 8 8
Area 2 2 ( )
2 ( )
2 [(1)(2) (2)(3) (2.5)(1) (3)(3) (5.5)(5) (8)(1) (4)(8)]
π π
π
π
= = Σ
= + + + + + +
= + + + + + +
y L y L
y L y L y L y L y L y L y L
or 2
Area 547 in=
(b) Rotation about the y-axis:
area
3
Volume 2 2
2 (11.5 in )
x A xAπ π
π
= = Σ
= or 3
Volume 72.3 in=
line line
1 1 2 2 3 3 4 4 5 5 6 6 7 7
Area 2 2 ( )
2 ( )
2 [(0.5)(1) (1)(2) (2.5)(3) (4)(1) (2.5)(3) (1)(5) (0.5)(1)]
x L x L
x L x L x L x L x L x L x L
π π
π
π
= = Σ
= + + + + + +
= + + + + + +
or 2
Area 169.6 in= www.elsolucionario.net
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617
PROBLEM 5.54
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.8 about (a) the line x = −60 mm, (b) the line y = 120 mm.
SOLUTION
From the solution of Problem 5.8, we have
2
3 3
3 3
7200 mm
72 10 mm
629.83 10 mm
A
x A
y A
=
Σ = − ×
Σ = ×
Applying the theorems of Pappus-Guldinus, we have
(a) Rotation about line 60 mm:x = −
3
Volume 2 ( 60) 2 ( 60 )
2 [ 72 10 60(7200)]
x A xA Aπ π
π
= + = Σ +
= − × + 6 3
Volume 2.26 10 mm= ×
line line
1 1 2 2 3 3
Area 2 2 ( )
2 ( )
2(60) (60) 2(60) (60)
2 60 60 (60)(120)
2 2
x L x L
x L x L x L
π π
π
π π
π
π π
= = Σ
= + +
= − + + +
where 1 2 3, ,x x x are measured with respect to line 60 mm.x = − 3 2
Area 116.3 10 mm= ×
(b) Rotation about line 120 mm:y =
3
Volume 2 (120 ) 2 (120 )
2 [120(7200) 629.83 10 ]
y A A yAπ π
π
= − = −Σ
= − × 6 3
Volume 1.471 10 mm= ×
line line
1 1 2 2 4 4
Area 2 2 ( )
2 ( )
y L y L
y L y L y L
π π
π
= = Σ
= + +
where 1 2 4, ,y y y are measured with respect to line 120 mm.y =
2(60) (60) 2(60) (60)
Area 2 120 (60)(120)
2 2
π π
π
π π
= − + +
3 2
Area 116.3 10 mm= ×
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you are using it without permission.
619
PROBLEM 5.56
Determine the volume and the surface area of the chain link
shown, which is made from a 6-mm-diameter bar, if R = 10 mm
and L = 30 mm.
SOLUTION
The area A and circumference C of the cross section of the bar are
2
and .
4
A d C d
π
π= =
Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V,
side end2( ) 2( )
2( ) 2( )
2( )
V V V
AL RA
L R A
π
π
= +
= +
= +
or 2
2[30 mm (10 mm)] (6 mm)
4
V
π
π
= +
3
3470 mm= 3
or 3470 mmV =
For the area A, side end2( ) 2( )
2( ) 2( )
2( )
A A A
CL RC
L R C
π
π
= +
= +
= +
or 2[30 mm (10 mm)][ (6 mm)]A π π= +
2
2320 mm= 2
or 2320 mmA =
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629
PROBLEM 5.65*
The shade for a wall-mounted light is formed from a thin sheet of
translucent plastic. Determine the surface area of the outside of
the shade, knowing that it has the parabolic cross section shown.
SOLUTION
First note that the required surface area A can be generated by rotating the parabolic cross section through π
radians about the y-axis. Applying the first theorem of Pappus-Guldinus, we have
A xLπ=
Now at
2 1
100 mm, 250 mm
250 (100) or 0.025 mm
x y
k k −
= =
= =
and
2
1
ELx x
dy
dL dx
dx
=
= +
where 2
dy
kx
dx
=
Then 2 2
1 4dL k x dx= +
We have ( )100
2 2
0
1 4ELxL x dL x k x dx= = +
{ }
100
2 2 3/2
2
0
2 2 3/2 3/2
2
2
1 1
(1 4 )
3 4
1 1
[1 4(0.025) (100) ] (1)
12 (0.025)
17,543.3 mm
xL k x
k
= +
= + −
=
Finally, 2
(17,543.3 mm )A π= or 3 2
55.1 10 mmA = × www.elsolucionario.net
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 300(4 ) N
2
R a a
R a a
= =
= − = −
Then 0: 900 300(4 ) 0y y yF A a a BΣ = − − − + =
or 1200 600y yA B a+ = +
Now 600 300 (N)y y y yA B A B a= = = + (1)
Also, 0: (4 m) 4 m [(900 ) N]
3
B y
a
M A a
Σ = − + −
1
(4 ) m [300(4 ) N] 0
3
a a
+ − − =
or 2
400 700 50yA a a= + − (2)
Equating Eqs. (1) and (2), 2
600 300 400 700 50a a a+ = + −
or 2
8 4 0a a− + =
Then
2
8 ( 8) 4(1)(4)
2
a
± − −
=
or 0.53590 ma = 7.4641ma =
Now 4 ma ≤ 0.536 ma =
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 300(4 ) N
2
R a a
R a a
= =
= − = −
Then
8
0: m (900 N) m [300(4 )N] (4 m) 0
3 3
A y
a a
M a a B
+
Σ = − − − + =
or 2
50 100 800yB a a= − + (1)
Then 100 100 0
ydB
a
da
= − = or 1.000 ma =
(b) From Eq. (1): 2
50(1) 100(1) 800 750 NyB = − + = 750 N=B
and 0: 0x xF AΣ = =
0: 900(1) N 300(4 1) N 750 N 0y yF AΣ = − − − + =
or 1050 NyA = 1050 N=A
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652
PROBLEM 5.85
The 3 × 4-m side of an open tank is hinged at its bottom A and is held in
place by a thin rod BC. The tank is to be filled with glycerine, whose density
is 1263 kg/m3
. Determine the force T in the rod and the reactions at the hinge
after the tank is filled to a depth of 2.9 m.
SOLUTION
Consider the free-body diagram of the side.
We have
3 2
1 1
( )
2 2
1
[(2.9 m)(4 m)] [(1263 kg/m )(9.81 m/s )(2.9 m)]
2
= 208.40 kN
P Ap A gdρ= =
=
Then 0: 0y yF AΣ = =
2.9
0: (3 m) m (208.4 kN) 0
3
AM T
Σ = − =
or 67.151 kNT = 67.2 kN=T
0: 208.40 kN 67.151 kN 0x xF AΣ = + − =
or 141.249 kNxA = − 141.2 kN=A
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653
PROBLEM 5.86
The friction force between a 6 × 6-ft square sluice gate AB and its guides is equal
to 10 percent of the resultant of the pressure forces exerted by the water on the face
of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb.
SOLUTION
Consider the free-body diagram of the gate.
Now 2 3
I I
1 1
[(6 6) ft ][(62.4 lb/ft )(9 ft)]
2 2
10,108.8 lb
P Ap= = ×
=
2 3
II II
1 1
[(6 6) ft ][(62.4 lb/ft )(15 ft)]
2 2
16,848 lb
P Ap= = ×
=
Then I II0.1 0.1( )
0.1(10,108.8 16,848) lb
2695.7 lb
F P P P= = +
= +
=
Finally 0: 2695.7 lb 1000 lb 0yF TΣ = − − = or 3.70 kips=T
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654
PROBLEM 5.87
A tank is divided into two sections by a 1 × 1-m square gate that is
hinged at A. A couple of magnitude 490 N · m is required for the
gate to rotate. If one side of the tank is filled with water at the rate
of 0.1 m3
/min and the other side is filled simultaneously with
methyl alcohol (density ρma = 789 kg/m3
) at the rate of 0.2 m3
/min,
determine at what time and in which direction the gate will rotate.
SOLUTION
Consider the free-body diagram of the gate.
First note baseV A d= and .V rt=
Then
3
3
0.1 m / min (min)
0.25 (m)
(0.4 m)(1m)
0.2 m / min (min)
(m)
(0.2 m)(1m)
W
MA
t
d t
t
d t
×
= =
×
= =
Now
1 1
( )
2 2
P Ap A ghρ= = so that
3 3 2
2
3 2
2
1
[(0.25 ) m (1 m)][(10 kg/m )(9.81 m/s )(0.25 ) m]
2
306.56 N
1
[( ) m (1 m)][(789 kg/m )(9.81m/s )( ) m]
2
3870 N
W
MA
P t t
t
P t t
t
= ×
=
= ×
=
Now assume that the gate will rotate clockwise and when 0.6 m.≤MAd When rotation of the gate is
impending, we require
1 1
: 0.6 m 0.6 m
3 3
Σ = − − −
A R MA MA W WM M d P d P
Substituting
2 21 1
490 N m 0.6 m (3870 ) N 0.6 0.25 m (306.56 ) N
3 3
t t t t
⋅ = − × − − × ×
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662
PROBLEM 5.92
A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled
with water. The gate is hinged along its top edge A and rests on a
frictionless stop at B. Determine the reactions at A and B when
cable BCD is slack.
SOLUTION
First consider the force of the water on the gate.
We have
1 1
( )
2 2
ρ= =P Ap A gh
so that 3 3 2
I
1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.45 m)]
2
882.9 N
P = ×
=
3 3 2
II
1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.93 m)]
2
1824.66 N
P = ×
=
Reactions at A and B when T = 0:
We have
1 2
0: (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) (0.8 m) 0
3 3
AM BΣ = − =
or 1510.74 NB =
or 1511 N=B 53.1°
0: 1510.74 N 882.9 N 1824.66 N 0F AΣ = + − − =
or 1197 N=A 53.1°
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663
PROBLEM 5.93
A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with
water. The gate is hinged along its top edge A and rests on a frictionless
stop at B. Determine the minimum tension required in cable BCD to
open the gate.
SOLUTION
First consider the force of the water on the gate.
We have
1 1
( )
2 2
ρ= =P Ap A gh
so that 3 3 2
I
1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.45 m)]
2
882.9 N
P = ×
=
3 3 2
II
1
[(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.93 m)]
2
1824.66 N
P = ×
=
T to open gate:
First note that when the gate begins to open, the reaction at B 0.
Then
1 2
0: (0.8 m)(882.9 N)+ (0.8 m)(1824.66 N)
3 3
8
(0.45 0.27)m 0
17
AM
T
Σ =
− + × =
or 235.44 973.152 0.33882 0T+ − =
or 3570 N=T
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664
PROBLEM 5.94
A 4 2-ft× gate is hinged at A and is held in position by rod CD.
End D rests against a spring whose constant is 828 lb/ft. The spring
is undeformed when the gate is vertical. Assuming that the force
exerted by rod CD on the gate remains horizontal, determine the
minimum depth of water d for which the bottom B of the gate will
move to the end of the cylindrical portion of the floor.
SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor.
We have
2
sin 30
4
θ θ= = °
Then (3 ft)tan30SPx = °
and
828 lb/ft 3 ft tan30°
1434.14 lb
SP SPF kx=
= × ×
=
Assume 4 ftd ≥
We have
1 1
( )
2 2
γ= =P Ap A h
Then 3
I
1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4) ft]
2
249.6( 4) lb
P d
d
= × −
= −
3
II
1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4 4cos30 )]
2
249.6( 0.53590 ) lb
P d
d
= × − + °
= − °
For mind so that the gate opens, 0W =
Using the above free-body diagrams of the gate, we have
4 8
0: ft [249.6( 4) lb] ft [249.6( 0.53590) lb]
3 3
AM d d
Σ = − + −
(3 ft)(1434.14 lb) 0− =
or (332.8 1331.2) (665.6 356.70) 4302.4 0d d− + − − =
or 6.00 ftd =
4 ftd ≥ assumption correct 6.00 ftd =
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665
PROBLEM 5.95
Solve Problem 5.94 if the gate weighs 1000 lb.
PROBLEM 5.94 A 4 2-ft× gate is hinged at A and is held in
position by rod CD. End D rests against a spring whose constant is
828 lb/ft. The spring is undeformed when the gate is vertical.
Assuming that the force exerted by rod CD on the gate remains
horizontal, determine the minimum depth of water d for which the
bottom B of the gate will move to the end of the cylindrical portion
of the floor.
SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor.
We have
2
sin 30
4
θ θ= = °
Then (3 ft)tan30SPx = °
and 828 lb/ft 3 ft tan30°
1434.14 lb
SP SPF kx= = × ×
=
Assume 4 ftd ≥
We have
1 1
( )
2 2
γ= =P Ap A h
Then 3
I
1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4) ft]
2
249.6( 4) lb
P d
d
= × −
= −
3
II
1
[(4 ft)(2 ft)] [(62.4 lb/ft )( 4 4cos30 )]
2
249.6( 0.53590 ) lb
P d
d
= × − + °
= − °
For mind so that the gate opens, 1000 lb=W
Using the above free-body diagrams of the gate, we have
4 8
0: ft [249.6( 4) lb] ft [249.6( 0.53590) lb]
3 3
(3 ft)(1434.14 lb) (1ft)(1000 lb) 0
AM d d
Σ = − + −
− − =
or (332.8 1331.2) (665.6 356.70) 4302.4 1000 0− + − − − =d d
or 7.00 ftd =
4 ftd ≥ assumption correct 7.00 ftd =
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666
PROBLEM 5.96
A hemisphere and a cone are attached as shown. Determine the location of the
centroid of the composite body when (a) 1.5 ,=h a (b) 2 .=h a
SOLUTION
V y yV
Cone I 21
3
πa h
4
h 2 21
12
πa h
Hemisphere II 32
3
πa
3
8
−
a 41
4
π− a
2
2 2 2
1
( 2 )
3
1
( 3 )
12
π
π
= +
Σ = −
V a h a
yV a h a
(a) For 1.5 ,h a= 2 21 7
(1.5 2 )
3 6
π π= + =V a a a a
2 2 2 41 1
[(1.5 ) 3 ]
12 16
π πΣ = − = −yV a a a a
3 47 1 3
:
6 16 56
π π
=Σ = − = −
Y V yV Y a a Y a
Centroid is 0.0536a below base of cone.
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
667
PROBLEM 5.96 (Continued)
(b) For 2 ,h a= 2 31 4
(2 2 )
3 3
π π= + =V a a a a
2 2 2 41 1
[(2 ) 3 ]
12 12
yV a a a aπ πΣ = − =
3 44 1 1
:
3 12 16
π π
= Σ = =
Y V yV Y a a Y a
Centroid is 0.0625a above base of cone.
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4
270.22 10 m
(7860 kg/m )(270.22 10 m )
2.1239 kg
V
m V
π
ρ
−
−
= −
= ×
= = ×
=
Each brass plate: 6 3
3 6 3
1
(0.096 m)(0.192 m)(0.006 m) 55.296 10 m
2
(8470 kg/m )(55.296 10 m ) 0.46836 kgρ
−
−
= = ×
= = × =
V
m V
Flagpole base:
2.1239 kg 3(0.46836 kg) 3.5290 kg
(0.096 m)(2.1239 kg) 3[(0.064 m)(0.46836 kg)] 0.29382 kg m
: (3.5290 kg) 0.29382 kg m
m
ym
Y m ym Y
Σ = + =
Σ = + = ⋅
Σ = Σ = ⋅
0.083259 m=Y 83.3 mm=Y above the base
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4
3.8693 lb
W Vγ
π
=
= −
=
Component W(lb) (in.)y (lb in.)yW ⋅
Rod 0.81229 2 1.62458
Collar 3.8693 1.25 4.8366
Σ 4.6816 6.4612
: (4.6816 lb) 6.4612 lb in.Y W yW YΣ = Σ = ⋅
1.38013 in.Y = 1.380 in.Y =
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 0.23889 lb
4
0.90 in. (0.284 lb/in ) [(1.125 0.75 ) in ](1in.) 0.156834 lb
4
0.70 in. (0.318 lb/in ) [(0.75 0
4
y W
y W
y W
π
π
π
= = − =
= = − =
= = −
2 2
.5 ) in ](1.4 in.) 0.109269 lb
=
We have
(0.20 in.)(0.23889 lb) (0.90 in.)(0.156834 lb) (0.70 in.)(0.109269 lb)
0.23889 lb 0.156834 lb 0.109269 lb
Y W yW
Y
Σ = Σ
+ +
=
+ +
or 0.526 in.Y =
(above base)
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you are using it without permission.
711
PROBLEM 5.129*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the y-axis. (Hint: Use a
thin cylindrical shell of radius r and thickness dr as the element
of volume.)
SOLUTION
First note that symmetry implies 0x =
0z =
Choose as the element of volume a cylindrical shell of radius r and thickness dr.
Then
1
(2 )( )( ),
2
ELdV r y dr y yπ= =
Now sin
2
r
y b
a
π
=
so that 2 sin
2
r
dV br dr
a
π
π=
Then
2
2 sin
2
a
a
r
V br dr
a
π
π=
Use integration by parts with
sin
2
2
cos
2
r
u rd dv dr
a
a r
du dr v
a
π
π
π
= =
= = −
Then
[ ]
2
2
2
2
2
2 2
2 ( ) cos cos
2 2
2 4
2 ( )( ) sin2 1
2
a
a
a
a
a
a
a r a r
V b r dr
a a
a a r
b a
a
π π
π
π π
π
π
π π
= − −
= − + −
2 2
2
2
2
4 4
2
1
8 1
5.4535
a a
V b
a b
a b
π
π π
π
= −
= −
=
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you are using it without permission.
714
PROBLEM 5.131
Determine by direct integration the location of the centroid of one-half of a
thin, uniform hemispherical shell of radius R.
SOLUTION
First note that symmetry implies 0x =
The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy
plane. Now
( )( )
2
EL
dA r Rd
r
y
π θ
π
=
= −
where sinr R θ=
so that 2
sin
2
sinEL
dA R d
R
y
π θ θ
θ
π
=
= −
Then
/2
2 2 /2
0
0
2
sin [ cos ]A R d R
R
π
π
π θ θ π θ
π
= = −
=
and
/2
2
0
/2
3
0
3
2
sin ( sin )
sin 2
2
2 4
2
EL
R
y dA R d
R
R
π
π
θ π θ θ
π
θ θ
π
= −
= − −
= −
Now 2 3
: ( )
2
ELyA y dA y R R
π
π= = − or
1
2
y R= −
Symmetry implies z y=
1
2
z R= − www.elsolucionario.net
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715
PROBLEM 5.132
The sides and the base of a punch bowl are of
uniform thickness t. If t R<< and R = 250 mm,
determine the location of the center of gravity of
(a) the bowl, (b) the punch.
SOLUTION
(a) Bowl:
First note that symmetry implies 0x =
0z =
for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of
gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of
area is obtained by rotating the arc ds about the y-axis. Then
wall (2 sin )( )dA R Rdπ θ θ=
and wall( ) cosELy R θ= −
Then
/2
2
wall
/6
/22
/6
2
2 sin
2 [ cos ]
3
A R d
R
R
π
π
π
π
π θ θ
π θ
π
=
= −
=
and wall wall wall
/2
2
/6
/23 2
/6
3
( )
( cos )(2 sin )
[cos ]
3
4
ELy A y dA
R R d
R
R
π
π
π
π
θ π θ θ
π θ
π
=
= −
=
= −
By observation, 2
base base
3
,
4 2
A R y R
π
= = −
Now y A yAΣ = Σ
or 2 2 3 23 3
3
4 4 4 2
y R R R R R
π π
π π
+ = − + −
or 0.48763 250 mmy R R= − = 121.9 mmy = −
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718
PROBLEM 5.133 (Continued)
Then 0
0
2 (1 cos ) [ sin ]
2
aht
V d aht
aht
π
π
θ θ θ θ
π
= − = −
=
and
0
2
2
0
2
0
2
2 (5 cos ) (1 cos )
12 2
(5 6cos cos )
12
sin 2
5 6sin
12 2 4
11
24
EL
h aht
y dV d
ah t
d
ah t
ah t
π
π
π
θ θ θ
θ θ θ
θ θ
θ θ
π
= − −
= − +
= − + +
=
0
2
0
2
2 cos (1 cos )
2
sin 2
sin
2 4
1
2
EL
aht
z dV a d
a ht
a ht
π
π
θ θ θ
θ θ
θ
π
= −
= − −
= −
Now 211
: ( )
24
ELyV y dV y aht ah tπ π= = or
11
24
y h=
and 21
: ( )
2
π π= = − ELzV z dV z aht a ht or
1
2
z a= −
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719
PROBLEM 5.134*
Locate the centroid of the section shown, which was cut from an elliptical
cylinder by an oblique plane.
SOLUTION
First note that symmetry implies 0x =
Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then
1
2 , ,
24
EL ELdV xy dz y z z= = =
Now 2 2a
x b z
b
= −
and
/2
( )
2 2
= − + = −
h h h
y z b z
b b
Then 2 2
2 ( )
2−
= − −
b
b
a h
V b z b z dz
b b
Let sin cosz b dz b dθ θ θ= =
Then
/2
2 /2
/2
2 2
/2
/2
3
/2
( cos )[ (1 sin )] cos
(cos sin cos )
sin 2 1
cos
2 4 3
1
2
ah
V b b b d
b
abh d
abh
V abh
π
π
π
π
π
π
θ θ θ θ
θ θ θ θ
θ θ
θ
π
−
−
= −
= −
= + +
=
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720
PROBLEM 5.134* (Continued)
and 2 2
2
2 2 2
3
1
( ) 2 ( )
2 2 2
1
( )
4
−
−
= × − − −
= − −
b
EL
b
b
b
h a h
y dV b z b z b z dz
b b b
ah
b z b z dz
b
Let sin cosz b dz b dθ θ θ= =
Then
2 /2
2
3 /2
/2
2 2 2 2 2
/2
1
[ (1 sin )] ( cos ) ( cos )
4
1
(cos 2sin cos sin cos )
4
π
π
π
π
θ θ θ θ
θ θ θ θ θ θ
−
−
= − ×
= − +
EL
ah
y dV b b b d
b
abh d
Now 2 21 1
sin (1 cos2 ) cos (1 cos2 )
2 2
θ θ θ θ= − = +
so that 2 2 21
sin cos (1 cos 2 )
4
θ θ θ= −
Then
/2
2 2 2 2
/2
/2
2 3
/2
2
1 1
cos 2sin cos (1 cos 2 )
4 4
1 sin 2 1 1 1 sin 4
cos
4 2 4 3 4 4 2 8
5
32
ELy dV abh d
abh
abh
π
π
π
π
θ θ θ θ θ
θ θ θ θ
θ θ
π
−
−
= − + −
= + + + − +
=
Also, 2 2
2 2
2
2 ( )
2
( )
−
−
= − −
= − −
b
EL
b
b
b
a h
z dV z a z b z dz
b b
ah
z b z b z dz
b
Let sin cosz b dz b dθ θ θ= =
Then
/2
2 /2
/2
2 2 2 2
/2
( sin )[ (1 sin )]( cos ) ( cos )
(sin cos sin cos )
EL
ah
z dV b b b b d
b
ab h d
π
π
π
π
θ θ θ θ θ
θ θ θ θ θ
−
−
= − ×
= −
Using 2 2 21
sin cos (1 cos 2 )
4
θ θ θ= − from above,
/2
2 2 2
/2
1
sin cos (1 cos 2 )
4
ELz dV ab h d
π
π
θ θ θ θ
−
= − −
/2
2 3 2
/2
1 1 1 sin 4 1
cos
3 4 4 2 8 8
ab h ab h
π
π
θ θ
θ θ π
−
= − − + + = −
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732
PROBLEM 5.142
Three different drive belt profiles are to be
studied. If at any given time each belt makes
contact with one-half of the circumference of
its pulley, determine the contact area between
the belt and the pulley for each design.
SOLUTION SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area CA of a belt
is given by
CA yL yLπ π= = Σ
where the individual lengths are the lengths of the belt cross section that are
in contact with the pulley.
(a) 1 1 2 2[2( ) ]
0.125 0.125 in.
2 3 in. [(3 0.125) in.](0.625 in.)
2 cos20
CA y L y Lπ
π
= +
= − + − °
or 2
8.10 inCA =
(b) 1 1[2( )]
0.375 0.375 in.
2 3 0.08 in.
2 cos20
CA y Lπ
π
=
= − − °
or 2
6.85 inCA =
(c) 1 1[2( )]
2(0.25)
3 in. [ (0.25 in.)]
CA y Lπ
π π
π
=
= −
or 2
7.01inCA = www.elsolucionario.net
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735
PROBLEM 5.145
The base of a dam for a lake is designed to resist up to 120 percent of
the horizontal force of the water. After construction, it is found that
silt (that is equivalent to a liquid of density 3 3
1.76 10 kg/m )sρ = × is
settling on the lake bottom at the rate of 12 mm/year. Considering a
1-m-wide section of dam, determine the number of years until the
dam becomes unsafe.
SOLUTION
First determine force on dam without the silt,
3 3 2
allow
1 1
( )
2 2
1
[(6.6 m)(1m)][(10 kg/m )(9.81m/s )(6.6 m)]
2
213.66 kN
1.2 (1.5)(213.66 kN) 256.39 kN
ww p
w
P A A gh
P P
ρ= =
=
=
= = =
Next determine the force P′on the dam face after a depth d of silt has settled.
We have 3 3 2
2
3 3 2
I
2
3 3 2
II
2
1
[(6.6 ) m (1 m)][(10 kg/m )(9.81m/s )(6.6 ) m]
2
4.905(6.6 ) kN
( ) [ (1 m)][(10 kg/m )(9.81m/s )(6.6 ) m]
9.81(6.6 ) kN
1
( ) [ (1m)][(1.76 10 kg/m )(9.81 m/s )( ) m]
2
8.6328 kN
w
s
s
P d d
d
P d d
d d
P d d
d
′ = − × −
= −
= −
= −
= ×
=
2
I II
2 2
2
( ) ( ) [4.905(43.560 13.2000 )
9.81(6.6 ) 8.6328 ] kN
[3.7278 213.66] kN
w s sP P P P d d
d d d
d
′ ′= + + = − +
+ − +
= +
Now it’s required that allowP P′ = to determine the maximum value of d.
2
(3.7278 213.66) kN 256.39 kNd + =
or 3.3856 md =
Finally, 3 m
3.3856 m 12 10 N
year
−
= × × or 282 yearsN =
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737
PROBLEM 5.146 (Continued)
(b) For 2.5 ,h b= 2 21
(2.5 ) 1.8333
3
V a b b a bπ π
= + =
2 2 2
2 2
2 2
1 1 1
(2.5 ) (2.5 )
2 3 12
[0.5 0.8333 0.52083]
1.85416
xV a b b b b
a b
a b
π
π
π
Σ = + +
= + +
=
2 2 2
: (1.8333 ) 1.85416 1.01136XV xV X a b a b X bπ π= Σ = =
Centroid is 0.01136b to right of base of cone.
Note: Centroid is at base of cone for 6 2.449 .h b b= =
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you are using it without permission.
739
PROBLEM 5.148
Locate the centroid of the volume obtained by rotating the shaded area
about the x-axis.
SOLUTION
First note that symmetry implies 0y =
and 0z =
We have 2
( )y k X h= −
At 0, ,x y a= = 2
( )a k h= −
or 2
a
k
h
=
Choose as the element of volume a disk of radius r and thickness dx. Then
2
, ELdV r dx X xπ= =
Now 2
2
( )
a
r x h
h
= −
so that
2
4
4
( )
a
dV x h dx
h
π= −
Then
2 2
4 5
04 40
2
( ) [( ) ]
5
1
5
h
ha a
V x h dx x h
h h
a h
π
π
π
= − = −
=
and
2
4
40
2
5 4 2 3 3 2 4
4 0
2
6 5 2 4 3 3 4 2
4
0
2 2
( )
( 4 6 4 )
1 4 3 4 1
6 5 2 3 2
1
30
h
EL
h
h
a
x dV x x h dx
h
a
x hx h x h x h x dx
h
a
x hx h x h x h x
h
a h
π
π
π
π
= −
= − + − +
= − + − +
=
Now 2 2 2
:
5 30
ELxV x dV x a h a h
π π
= =
1
or
6
x h=
www.elsolucionario.net
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Beer vector mechanics for engineers statics 10th solutions
- 5. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 PROBLEM 2.1 Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 3.30 kN, 66.6R α= = ° 3.30 kN=R 66.6° www.elsolucionario.net www.elsolucionario.net
- 6. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. SOLUTION We measure: 51.3 59.0 α β = ° = ° (a) Parallelogram law: (b) Triangle rule: We measure: 139.1 lb,R = 67.0γ = ° 139.1lbR = 67.0° www.elsolucionario.net www.elsolucionario.net
- 7. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5 PROBLEM 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 20.1 kN,R = 21.2α = ° 20.1kN=R 21.2° www.elsolucionario.net www.elsolucionario.net
- 8. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6 PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: 8.03 kips, 3.8R α= = ° 8.03 kips=R 3.8° www.elsolucionario.net www.elsolucionario.net
- 9. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 7 PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the triangle rule and the law of sines: (a) 120 N sin30 sin 25 P = ° ° 101.4 NP = (b) 30 25 180 180 25 30 125 β β ° + + ° = ° = ° − ° − ° = ° 120 N sin30 sin125 = ° ° R 196.6 N=R www.elsolucionario.net www.elsolucionario.net
- 10. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8 PROBLEM 2.6 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant? SOLUTION Using the triangle rule and the law of sines: (a) 1600 N sin 25° sin 75 P = ° 3660 NP = (b) 25 75 180 180 25 75 80 β β ° + + ° = ° = ° − ° − ° = ° 1600 N sin 25° sin80 R = ° 3730 NR = www.elsolucionario.net www.elsolucionario.net
- 11. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 9 PROBLEM 2.7 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N. SOLUTION Using the law of cosines: 2 2 2 (1600 N) (2500 N) 2(1600 N)(2500 N)cos 75° 2596 N P P = + − = Using the law of sines: sin sin75 1600 N 2596 N 36.5 α α ° = = ° P is directed 90 36.5 or 53.5°° − ° below the horizontal. 2600 N=P 53.5° www.elsolucionario.net www.elsolucionario.net
- 12. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10 PROBLEM 2.8 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75 40 180 180 75 40 65 α α ° + ° + = ° = ° − ° − ° = ° 2800 lb sin 65 sin 75 T = ° ° 2 853 lbT = (b) 800 lb sin 65 sin 40 R = ° ° 567 lbR = www.elsolucionario.net www.elsolucionario.net
- 13. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 11 PROBLEM 2.9 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75 40 180 180 75 40 65 β β ° + ° + = ° = ° − ° − ° = ° 11000 lb sin75° sin 65 T = ° 1 938 lbT = (b) 1000 lb sin75° sin 40 R = ° 665 lbR = www.elsolucionario.net www.elsolucionario.net
- 14. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 12 PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and law of sines: (a) sin sin 25 50 N 35 N sin 0.60374 α α ° = = 37.138α = ° 37.1α = ° (b) 25 180 180 25 37.138 117.862 α β β + + ° = ° = ° − ° − ° = ° 35 N sin117.862 sin 25 R = ° ° 73.2 NR = www.elsolucionario.net www.elsolucionario.net
- 15. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 13 PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 50 60 180 180 50 60 70 β β + ° + ° = ° = ° − ° − ° = ° 425 lb sin 70 sin 60 P = ° ° 392 lbP = (b) 425 lb sin 70 sin50 R = ° ° 346 lbR = www.elsolucionario.net www.elsolucionario.net
- 16. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 14 PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) ( 30 ) 60 180 180 ( 30 ) 60 90 sin (90 ) sin60 425 lb 500 lb α β β α β α α + ° + ° + = ° = ° − + ° − ° = ° − ° − ° = 90 47.402α° − = ° 42.6α = ° (b) 500 lb sin (42.598 30 ) sin 60 R = ° + ° ° 551 lbR = www.elsolucionario.net www.elsolucionario.net
- 17. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 15 PROBLEM 2.13 A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) (425 lb)cos30P = ° 368 lb=P (b) (425 lb)sin30R = ° 213 lbR = www.elsolucionario.net www.elsolucionario.net
- 18. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 16 PROBLEM 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) (50 N)sin 25P = ° 21.1 N=P (b) (50 N)cos 25R = ° 45.3 NR = www.elsolucionario.net www.elsolucionario.net
- 19. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 17 PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. SOLUTION 8 tan 10 38.66 6 tan 10 30.96 α α β β = = ° = = ° Using the triangle rule: 180 38.66 30.96 180 110.38 α β ψ ψ ψ + + = ° ° + ° + = ° = ° Using the law of cosines: 2 22 (120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38 139.08 lb R R = + − ° = Using the law of sines: sin sin110.38 40 lb 139.08 lb γ ° = 15.64 (90 ) (90 38.66 ) 15.64 66.98 γ φ α γ φ φ = ° = ° − + = ° − ° + ° = ° 139.1 lb=R 67.0° www.elsolucionario.net www.elsolucionario.net
- 20. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 18 PROBLEM 2.16 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the force triangle and the laws of cosines and sines: We have: 180 (50 25 ) 105 γ = ° − ° + ° = ° Then 2 2 2 2 (4 kips) (6 kips) 2(4 kips)(6 kips)cos105 64.423 kips 8.0264 kips R R = + − ° = = And 4 kips 8.0264 kips sin(25 ) sin105 sin(25 ) 0.48137 25 28.775 3.775 α α α α = ° + ° ° + = ° + = ° = ° 8.03 kips=R 3.8° www.elsolucionario.net www.elsolucionario.net
- 21. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 19 PROBLEM 2.17 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the laws of cosines and sines: 2 2 2 (120 N) (160 N) 2(120 N)(160 N)cos25 72.096 N P P = + − ° = And sin sin 25 120 N 72.096 N sin 0.70343 44.703 α α α ° = = = ° 72.1 N=P 44.7° www.elsolucionario.net www.elsolucionario.net
- 22. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 20 PROBLEM 2.18 For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the force triangle and the laws of cosines and sines: We have 180 (50 25 ) 105 β = ° − ° + ° = ° Then 2 2 2 2 2 (75 N) (50 N) 2(75 N)(50 N)cos105 10,066.1 N 100.330 N R R R = + − ° = = and sin sin105 75 N 100.330 N sin 0.72206 46.225 γ γ γ ° = = = ° Hence: 25 46.225 25 21.225γ − ° = ° − ° = ° 100.3 N=R 21.2° www.elsolucionario.net www.elsolucionario.net
- 23. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 21 PROBLEM 2.19 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have 180 (20 10 ) 150 γ = ° − ° + ° = ° Then 2 2 2 (48 N) (60 N) 2(48 N)(60 N)cos150 104.366 N R R = + − ° = and 48 N 104.366 N sin sin150 sin 0.22996 13.2947 α α α = ° = = ° Hence: 180 80 180 13.2947 80 86.705 φ α= ° − − ° = ° − ° − ° = ° 104.4 N=R 86.7° www.elsolucionario.net www.elsolucionario.net
- 24. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 22 PROBLEM 2.20 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have 180 (20 10 ) 150 γ = ° − ° + ° = ° Then 2 2 2 (60 N) (48 N) 2(60 N)(48 N)cos150 104.366 N R R = + − ° = and 60 N 104.366 N sin sin150 sin 0.28745 16.7054 α α α = ° = = ° Hence: 180 180 180 16.7054 80 83.295 φ α= ° − − ° = ° − ° − ° = ° 104.4 N=R 83.3° www.elsolucionario.net www.elsolucionario.net
- 25. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 23 PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION 80-N Force: (80 N)cos40xF = + ° 61.3 NxF = (80 N)sin 40yF = + ° 51.4 NyF = 120-N Force: (120 N)cos70xF = + ° 41.0 NxF = (120 N)sin 70yF = + ° 112.8 NyF = 150-N Force: (150 N)cos35xF = − ° 122. 9 NxF = − (150 N)sin35yF = + ° 86.0 NyF = www.elsolucionario.net www.elsolucionario.net
- 26. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 24 PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION 40-lb Force: (40 lb)cos60xF = + ° 20.0 lbxF = (40 lb)sin60yF = − ° 34.6 lbyF = − 50-lb Force: (50 lb)sin50xF = − ° 38.3 lbxF = − (50 lb)cos50yF = − ° 32.1 lbyF = − 60-lb Force: (60 lb)cos25xF = + ° 54.4 lbxF = (60 lb)sin 25yF = + ° 25.4 lbyF = www.elsolucionario.net www.elsolucionario.net
- 27. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 25 PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: 2 2 2 2 2 2 (600) (800) 1000 mm (560) (900) 1060 mm (480) (900) 1020 mm OA OB OC = + = = + = = + = 800-N Force: 800 (800 N) 1000 xF = + 640 NxF = + 600 (800 N) 1000 yF = + 480 NyF = + 424-N Force: 560 (424 N) 1060 xF = − 224 NxF = − 900 (424 N) 1060 yF = − 360 NyF = − 408-N Force: 480 (408 N) 1020 xF = + 192.0 NxF = + 900 (408 N) 1020 yF = − 360 NyF = − www.elsolucionario.net www.elsolucionario.net
- 28. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 26 PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: 2 2 2 2 2 2 (24 in.) (45 in.) 51.0 in. (28 in.) (45 in.) 53.0 in. (40 in.) (30 in.) 50.0 in. OA OB OC = + = = + = = + = 102-lb Force: 24 in. 102 lb 51.0 in. xF = − 48.0 lbxF = − 45 in. 102 lb 51.0 in. yF = + 90.0 lbyF = + 106-lb Force: 28 in. 106 lb 53.0 in. = +xF 56.0 lbxF = + 45 in. 106 lb 53.0 in. yF = + 90.0 lbyF = + 200-lb Force: 40 in. 200 lb 50.0 in. xF = − 160.0 lbxF = − 30 in. 200 lb 50.0 in. yF = − 120.0 lbyF = − www.elsolucionario.net www.elsolucionario.net
- 29. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 27 PROBLEM 2.25 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC. SOLUTION (a) 750 N sin 20P= ° 2192.9 NP = 2190 NP = (b) cos20ABCP P= ° (2192.9 N)cos20= ° 2060 NABCP = www.elsolucionario.net www.elsolucionario.net
- 30. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 28 PROBLEM 2.26 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) cos 55 yP P = ° 350 lb cos 55 610.21 lb = ° = 610 lbP = (b) sin 55xP P= ° (610.21 lb)sin 55 499.85 lb = ° = 500 lbxP = www.elsolucionario.net www.elsolucionario.net
- 31. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 29 PROBLEM 2.27 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION 2 2 (650 mm) (720 mm) 970 mm BC = + = (a) 650 970 xP P = or 970 650 970 325 N 650 485 N xP P = = = 485 NP = (b) 720 970 720 485 N 970 360 N yP P = = = 970 NyP = www.elsolucionario.net www.elsolucionario.net
- 32. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 30 PROBLEM 2.28 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) 240 lb sin 40 sin 40° = = ° yP P or 373 lbP = (b) 240 lb tan 40 tan 40° y x P P = = ° or 286 lbxP = www.elsolucionario.net www.elsolucionario.net
- 33. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 31 PROBLEM 2.29 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) 37 12 37 (720 N) 12 2220 N = = = xP P 2.22 kNP = (b) 35 12 35 (720 N) 12 2100 N y xP P= = = 2.10 kN=yP www.elsolucionario.net www.elsolucionario.net
- 34. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 32 PROBLEM 2.30 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB. SOLUTION 180 45 90 30 180 45 90 30 15 α α ° = ° + + ° + ° = ° − ° − ° − ° = ° (a) cos cos 600 N cos15 621.17 N x x P P P P α α = = = ° = 621 NP = (b) tan tan (600 N)tan15 160.770 N y x y x P P P P α α = = = ° = 160.8 NyP = www.elsolucionario.net www.elsolucionario.net
- 35. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 33 PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.23: Force x Comp. (N) y Comp. (N) 800 lb +640 +480 424 lb –224 –360 408 lb +192 –360 608xR = + 240yR = − (608 lb) ( 240 lb) tan 240 608 21.541 240 N sin(21.541°) 653.65 N x y y x R R R R R α α = + = + − = = = ° = = R i j i j 654 N=R 21.5° www.elsolucionario.net www.elsolucionario.net
- 36. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 34 PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.21: Force x Comp. (N) y Comp. (N) 80 N +61.3 +51.4 120 N +41.0 +112.8 150 N –122.9 +86.0 20.6xR = − 250.2yR = + ( 20.6 N) (250.2 N) tan 250.2 N tan 20.6 N tan 12.1456 85.293 250.2 N sin85.293 x y y x R R R R R α α α α = + = − + = = = = ° = ° R i j i j 251 N=R 85.3° www.elsolucionario.net www.elsolucionario.net
- 37. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 35 PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Force x Comp. (lb) y Comp. (lb) 40 lb +20.00 –34.64 50 lb –38.30 –32.14 60 lb +54.38 +25.36 36.08xR = + 41.42yR = − ( 36.08 lb) ( 41.42 lb) tan 41.42 lb tan 36.08 lb tan 1.14800 48.942 41.42 lb sin 48.942 x y y x R R R R R α α α α = + = + + − = = = = ° = ° R i j i j 54.9 lb=R 48.9° www.elsolucionario.net www.elsolucionario.net
- 38. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36 PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.24: Force x Comp. (lb) y Comp. (lb) 102 lb −48.0 +90.0 106 lb +56.0 +90.0 200 lb −160.0 −120.0 152.0xR = − 60.0yR = ( 152 lb) (60.0 lb) tan 60.0 lb tan 152.0 lb tan 0.39474 21.541 α α α α = + = − + = = = = ° x y y x R R R R R i j i j 60.0 lb sin 21.541 R = ° 163.4 lb=R 21.5° www.elsolucionario.net www.elsolucionario.net
- 39. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 37 PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown. SOLUTION 100-N Force: (100 N)cos35 81.915 N (100 N)sin35 57.358 N x y F F = + ° = + = − ° = − 150-N Force: (150 N)cos65 63.393 N (150 N)sin 65 135.946 N x y F F = + ° = + = − ° = − 200-N Force: (200 N)cos35 163.830 N (200 N)sin35 114.715 N x y F F = − ° = − = − ° = − Force x Comp. (N) y Comp. (N) 100 N +81.915 −57.358 150 N +63.393 −135.946 200 N −163.830 −114.715 18.522xR = − 308.02yR = − ( 18.522 N) ( 308.02 N) tan 308.02 18.522 86.559 x y y x R R R R α α = + = − + − = = = ° R i j i j 308.02 N sin86.559 R = 309 N=R 86.6° www.elsolucionario.net www.elsolucionario.net
- 40. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38 PROBLEM 2.36 Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC. SOLUTION Determine force components: Cable force AC: 960 (365 N) 240 N 1460 1100 (365 N) 275 N 1460 = − = − = − = − x y F F 500-N Force: 24 (500 N) 480 N 25 7 (500 N) 140 N 25 x y F F = = = = 200-N Force: 4 (200 N) 160 N 5 3 (200 N) 120 N 5 x y F F = = = − = − and 2 2 2 2 240 N 480 N 160 N 400 N 275 N 140 N 120 N 255 N (400 N) ( 255 N) 474.37 N = Σ = − + + = = Σ = − + − = − = + = + − = x x y y x y R F R F R R R Further: 255 tan 400 32.5 α α = = ° 474 N=R 32.5° www.elsolucionario.net www.elsolucionario.net
- 41. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 39 PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos20 56.382 lb (60 lb)sin 20 20.521lb x y F F = ° = = ° = 80-lb Force: (80 lb)cos60 40.000 lb (80 lb)sin 60 69.282 lb x y F F = ° = = ° = 120-lb Force: (120 lb)cos30 103.923 lb (120 lb)sin30 60.000 lb x y F F = ° = = − ° = − and 2 2 200.305 lb 29.803 lb (200.305 lb) (29.803 lb) 202.510 lb x x y y R F R F R = Σ = = Σ = = + = Further: 29.803 tan 200.305 α = 1 29.803 tan 200.305 8.46 α − = = ° 203 lb=R 8.46° www.elsolucionario.net www.elsolucionario.net
- 42. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 40 PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos 20 56.382 lb (60 lb)sin 20 20.521 lb x y F F = ° = = ° = 80-lb Force: (80 lb)cos 95 6.9725 lb (80 lb)sin 95 79.696 lb x y F F = ° = − = ° = 120-lb Force: (120 lb)cos 5 119.543 lb (120 lb)sin 5 10.459 lb x y F F = ° = = ° = Then 168.953 lb 110.676 lb x x y y R F R F = Σ = = Σ = and 2 2 (168.953 lb) (110.676 lb) 201.976 lb R = + = 110.676 tan 168.953 tan 0.65507 33.228 α α α = = = ° 202 lb=R 33.2° www.elsolucionario.net www.elsolucionario.net
- 43. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 41 PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION (100 N)cos (150 N)cos( 30 ) (200 N)cos (100 N)cos (150 N)cos( 30 ) x x x R F R α α α α α = Σ = + + ° − = − + + ° (1) (100 N)sin (150 N)sin( 30 ) (200 N)sin (300 N)sin (150 N)sin( 30 ) y y y R F R α α α α α = Σ = − − + ° − = − − + ° (2) (a) For R to be vertical, we must have 0.xR = We make 0xR = in Eq. (1): 100cos 150cos( 30 ) 0 100cos 150(cos cos 30 sin sin 30 ) 0 29.904cos 75sin α α α α α α α − + + ° = − + ° − ° = = 29.904 tan 75 0.39872 21.738 α α = = = ° 21.7α = ° (b) Substituting for α in Eq. (2): 300sin 21.738 150sin51.738 228.89 N yR = − ° − ° = − | | 228.89 NyR R= = 229 NR = www.elsolucionario.net www.elsolucionario.net
- 44. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 42 PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. SOLUTION 960 24 4 (500 N) (200 N) 1460 25 5 48 640 N 73 x x AC x AC R F T R T = Σ = − + + = − + (1) 1100 7 3 (500 N) (200 N) 1460 25 5 55 20 N 73 y y AC y AC R F T R T = Σ = − + − = − + (2) (a) For R to be horizontal, we must have 0.yR = Set 0yR = in Eq. (2): 55 20 N 0 73 ACT− + = 26.545 NACT = 26.5 NACT = (b) Substituting for ACT into Eq. (1) gives 48 (26.545 N) 640 N 73 622.55 N 623 N = − + = = = x x x R R R R 623 NR = www.elsolucionario.net www.elsolucionario.net
- 45. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 43 PROBLEM 2.41 A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION xR = (200 lb)sin 40 (400 lb)cos40xF PΣ = + ° − ° 177.860 lbxR P= − (1) yR = (200 lb)cos40 (400 lb)sin 40yFΣ = ° + ° 410.32 lbyR = (2) (a) For R to be vertical, we must have 0.xR = Set 0xR = in Eq. (1) 0 177.860 lb 177.860 lb P P = − = 177.9 lbP = (b) Since R is to be vertical: 410 lb= =yR R 410 lbR = www.elsolucionario.net www.elsolucionario.net
- 46. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 44 PROBLEM 2.42 A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine (a) the required value of α if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION xR = 250 lb (200 lb)sin (400 lb)cosxF α αΣ = + − 250 lb (200 lb)sin (400 lb)cosxR α α= + − (1) yR = (200 lb)cos (400 lb)sinyF α αΣ = + (a) For R to be vertical, we must have 0.xR = Set 0xR = in Eq. (1) 0 250 lb (200 lb)sin (400 lb)cosα α= + − 2 2 2 2 2 (400 lb)cos (200 lb)sin 250 lb 2cos sin 1.25 4cos sin 2.5sin 1.5625 4(1 sin ) sin 2.5sin 1.5625 0 5sin 2.5sin 2.4375 α α α α α α α α α α α α = + = + = + + − = + + = + − Using the quadratic formula to solve for the roots gives sin 0.49162α = or 29.447α = ° 29.4α = ° (b) Since R is to be vertical: (200 lb)cos29.447 (400 lb)sin 29.447yR R= = ° + ° 371 lb=R www.elsolucionario.net www.elsolucionario.net
- 47. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 45 PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 1100 tan 960 48.888 400 tan 960 22.620 α α β β = = ° = = ° Force Triangle Law of sines: 15.696 kN sin 22.620 sin 48.888 sin108.492 AC BCT T = = ° ° ° (a) 15.696 kN (sin 22.620 ) sin108.492 ACT = ° ° 6.37 kNACT = (b) 15.696 kN (sin 48.888 ) sin108.492 BCT = ° ° 12.47 kNBCT = www.elsolucionario.net www.elsolucionario.net
- 48. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 46 PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION 3 tan 2.25 53.130 1.4 tan 2.25 31.891 α α β β = = ° = = ° Free-Body Diagram Law of sines: Force-Triangle 660 N sin31.891 sin53.130 sin94.979 AC BCT T = = ° ° ° (a) 660 N (sin31.891 ) sin94.979 ACT = ° ° 350 NACT = (b) 660 N (sin53.130 ) sin94.979 BCT = ° ° 530 NBCT = www.elsolucionario.net www.elsolucionario.net
- 49. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 47 PROBLEM 2.45 Knowing that 20 ,α = ° determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 1200 lb sin 110 sin 5 sin 65 AC BCT T = = ° ° ° (a) 1200 lb sin 110 sin 65 ACT = ° ° 1244 lbACT = (b) 1200 lb sin 5 sin 65 BCT = ° ° 115.4 lbBCT = www.elsolucionario.net www.elsolucionario.net
- 50. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 48 PROBLEM 2.46 Knowing that 55α = ° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 300 lb sin 35 sin 50 sin 95 AC BCF T = = ° ° ° (a) 300 lb sin 35 sin 95 ACF = ° ° 172.7 lbACF = (b) 300 lb sin 50 sin 95 BCT = ° ° 231 lbBCT = www.elsolucionario.net www.elsolucionario.net
- 51. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 49 PROBLEM 2.47 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 1.4 tan 4.8 16.2602 1.6 tan 3 28.073 α α β β = = ° = = ° Force Triangle Law of sines: 1.98 kN sin 61.927 sin 73.740 sin 44.333 AC BCT T = = ° ° ° (a) 1.98 kN sin 61.927 sin 44.333 ACT = ° ° 2.50 kNACT = (b) 1.98 kN sin 73.740 sin 44.333 BCT = ° ° 2.72 kNBCT = www.elsolucionario.net www.elsolucionario.net
- 52. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 50 PROBLEM 2.48 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 500 N sin35 sin 75 sin70° AC BCT T = = ° ° (a) 500 N sin35 sin 70 ACT = ° ° 305 NACT = (b) 500 N sin 75 sin 70 BCT = ° ° 514 NBCT = www.elsolucionario.net www.elsolucionario.net
- 53. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 51 PROBLEM 2.49 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TC and TD. SOLUTION Free-Body Diagram 0 15 kips 8 kips cos40 0x DF TΣ = − − ° = 9.1379 kipsDT = 0yFΣ = sin 40 0D CT T° − = (9.1379 kips)sin 40 0 5.8737 kips ° − = = C C T T 5.87 kips=CT 9.14 kipsDT = www.elsolucionario.net www.elsolucionario.net
- 54. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 52 PROBLEM 2.50 Two forces of magnitude TA = 6 kips and TC = 9 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TB and TD. SOLUTION Free-Body Diagram 0xFΣ = 6 kips cos40 0B DT T− − ° = (1) 0yFΣ = sin 40 9 kips 0 9 kips sin 40 14.0015 kips D D D T T T ° − = = ° = Substituting for TD into Eq. (1) gives: 6 kips (14.0015 kips)cos40 0 16.7258 kips B B T T − − ° = = 16.73 kipsBT = 14.00 kipsDT = www.elsolucionario.net www.elsolucionario.net
- 55. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 53 PROBLEM 2.51 Two cables are tied together at C and loaded as shown. Knowing that 360 N,P = determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free Body: C (a) 12 4 0: (360 N) 0 13 5 x ACTΣ = − + =F 312 NACT = (b) 5 3 0: (312 N) (360 N) 480 N 0 13 5 y BCTΣ = + + − =F 480 N 120 N 216 NBCT = − − 144 NBCT = www.elsolucionario.net www.elsolucionario.net
- 56. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 54 PROBLEM 2.52 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free Body: C 12 4 0: 0 13 5 x ACTΣ = − + =F P 13 15 ACT P= (1) 5 3 0: 480 N 0 13 5 y AC BCT T PΣ = + + − =F Substitute for ACT from (1): 5 13 3 480 N 0 13 15 5 BCP T P + + − = 14 480 N 15 BCT P= − (2) From (1), 0ACT Ͼ requires 0.P Ͼ From (2), 0BCT Ͼ requires 14 480 N, 514.29 N 15 P PϽ Ͻ Allowable range: 0 514 NPϽ Ͻ www.elsolucionario.net www.elsolucionario.net
- 57. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 55 PROBLEM 2.53 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that 30α = ° and 10β = ° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. SOLUTION Free-Body Diagram 0: cos 10 cos 30 cos 30 0x ACB ACB CDF T T TΣ = ° − ° − ° = 0.137158CD ACBT T= (1) 0: sin 10 sin 30 sin 30 900 0y ACB ACB CDF T T TΣ = ° + ° + ° − = 0.67365 0.5 900ACB CDT T+ = (2) (a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 900ACB ACBT T+ = 1212.56 NACBT = 1213 NACBT = (b) From (1): 0.137158(1212.56 N)CDT = 166.3 NCDT = www.elsolucionario.net www.elsolucionario.net
- 58. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 56 PROBLEM 2.54 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that 25α = ° and 15β = ° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB. SOLUTION Free-Body Diagram 0: cos 15 cos 25 (80 N)cos 25 0x ACB ACBF T TΣ = ° − ° − ° = 1216.15 NACBT = 0: (1216.15 N)sin 15 (1216.15 N)sin 25yFΣ = ° + ° (80 N)sin 25 0 862.54 N W W + ° − = = (a) 863 NW = (b) 1216 NACBT = www.elsolucionario.net www.elsolucionario.net
- 59. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 57 PROBLEM 2.55 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that 500P = lb and 650Q = lb, determine the magnitudes of the forces exerted on the rods A and B. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: 0A B= + + + =R P Q F F Substituting components: (500 lb) [(650 lb)cos50 ] [(650 lb)sin50 ] ( cos50 ) ( sin50 ) 0B A AF F F = − + ° − ° + − ° + ° = R j i j i i j In the y-direction (one unknown force): 500 lb (650 lb)sin50 sin50 0AF− − ° + ° = Thus, 500 lb (650 lb)sin50 sin50 AF + ° = ° 1302.70 lb= 1303 lbAF = In the x-direction: (650 lb)cos50 cos50 0B AF F° + − ° = Thus, cos50 (650 lb)cos50 (1302.70 lb)cos50 (650 lb)cos50 B AF F= ° − ° = ° − ° 419.55 lb= 420 lbBF = www.elsolucionario.net www.elsolucionario.net
- 60. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 58 PROBLEM 2.56 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are 750AF = lb and 400BF = lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: 0A B= + + + =R P Q F F Substituting components: cos 50 sin 50 [(750 lb)cos 50 ] [(750 lb)sin 50 ] (400 lb) P Q Q= − + ° − ° − ° + ° + R j i j i j i In the x-direction (one unknown force): cos 50 [(750 lb)cos 50 ] 400 lb 0Q ° − ° + = (750 lb)cos 50 400 lb cos 50 127.710 lb Q ° − = ° = In the y-direction: sin 50 (750 lb)sin 50 0P Q− − ° + ° = sin 50 (750 lb)sin 50 (127.710 lb)sin 50 (750 lb)sin 50 476.70 lb P Q= − ° + ° = − ° + ° = 477 lb; 127.7 lbP Q= = www.elsolucionario.net www.elsolucionario.net
- 61. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 59 PROBLEM 2.57 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram: C Force Triangle Force triangle is isosceles with 2 180 85 47.5 β β = ° − ° = ° (a) 2(800 N)cos 47.5° 1081 NP = = Since 0,P Ͼ the solution is correct. 1081 NP = (b) 180 50 47.5 82.5α = ° − ° − ° = ° 82.5α = ° www.elsolucionario.net www.elsolucionario.net
- 62. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 60 PROBLEM 2.58 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 1200 N in cable AC and 600 N in cable BC, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram Force Triangle (a) Law of cosines: 2 2 2 (1200 N) (600 N) 2(1200 N)(600 N)cos 85 1294.02 N P P = + − ° = Since 1200 N,P Ͼ the solution is correct. 1294 NP = (b) Law of sines: sin sin 85 1200 N 1294.02 N 67.5 180 50 67.5 β β α ° = = ° = ° − ° − ° 62.5α = ° www.elsolucionario.net www.elsolucionario.net
- 63. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 61 PROBLEM 2.59 For the situation described in Figure P2.45, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.45 Knowing that 20 ,α = ° determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle To be smallest, BCT must be perpendicular to the direction of .ACT (a) Thus, 5α = ° 5.00α = ° (b) (1200 lb)sin 5BCT = ° 104.6 lbBCT = www.elsolucionario.net www.elsolucionario.net
- 64. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 62 PROBLEM 2.60 For the structure and loading of Problem 2.46, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension. SOLUTION BCT must be perpendicular to ACF to be as small as possible. Free-Body Diagram: C Force Triangle is a right triangle To be a minimum, BCT must be perpendicular to .ACF (a) We observe: 90 30α = ° − ° 60.0α = ° (b) (300 lb)sin 50BCT = ° or 229.81lbBCT = 230 lbBCT = www.elsolucionario.net www.elsolucionario.net
- 65. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 63 PROBLEM 2.61 For the cables of Problem 2.48, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α. SOLUTION Free-Body Diagram Force Triangle (a) Law of cosines 2 2 2 (600) (750) 2(600)(750)cos(25 45 )P = + − ° + ° 784.02 NP = 784 NP = (b) Law of sines sin sin (25 45 ) 600 N 784.02 N β ° + ° = 46.0β = ° 46.0 25α∴ = ° + ° 71.0α = ° www.elsolucionario.net www.elsolucionario.net
- 66. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 64 PROBLEM 2.62 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN. SOLUTION Free-Body Diagram tan 0.6 m α = h (1) Isosceles Force Triangle Law of sines: 1 2 1 2 (2.8 kN) sin 5 kN (2.8 kN) sin 5 kN 16.2602 AC AC T T α α α = = = = ° From Eq. (1): tan16.2602 0.175000 m 0.6 m h h° = ∴ = Half length of chain 2 2 (0.6 m) (0.175 m) 0.625 m AC= = + = Total length: 2 0.625 m= × 1.250 m www.elsolucionario.net www.elsolucionario.net
- 67. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 65 PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) 4.5 in.,x = (b) 15 in.x = SOLUTION (a) Free Body: Collar A Force Triangle 50 lb 4.5 20.5 P = 10.98 lbP = (b) Free Body: Collar A Force Triangle 50 lb 15 25 P = 30.0 lbP = www.elsolucionario.net www.elsolucionario.net
- 68. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 66 PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb. SOLUTION Free Body: Collar A Force Triangle 2 2 2 (50) (48) 196 14.00 lb N N = − = = Similar Triangles 48 lb 20 in. 14 lb x = 68.6 in.x = www.elsolucionario.net www.elsolucionario.net
- 69. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 67 PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N. SOLUTION Combine the two 150-N forces into a resultant force Q: 2(150 N)cos25 271.89 N Q = ° = Equivalent loading at A: Using the law of cosines: 2 2 2 (600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 ) cos(55 ) 0.132685 α α = + + ° + ° + = Two values for :α 55 82.375 27.4 α α ° + = = ° or 55 82.375 55 360 82.375 222.6 α α α ° + = − ° ° + = ° − ° = ° For 600 lb:R < 27.4 222.6α° < < www.elsolucionario.net www.elsolucionario.net
- 70. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 68 PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.) SOLUTION Free-Body Diagram: Pulley A 5 0: 2 cos 0 281 cos 0.59655 53.377 xF P P α α α Σ = − + = = = ± ° For 53.377 :α = + ° 16 0: 2 sin53.377 1962 N 0 281 yF P P Σ = + ° − = 724 N=P 53.4° For 53.377 :α = − ° 16 0: 2 sin( 53.377 ) 1962 N 0 281 yF P P Σ = + − ° − = 1773=P 53.4° www.elsolucionario.net www.elsolucionario.net
- 71. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 69 PROBLEM 2.67 A 600-lb crate is supported by several rope-and- pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley (a) 0: 2 (600 lb) 0 1 (600 lb) 2 yF T T Σ = − = = 300 lbT = (b) 0: 2 (600 lb) 0 1 (600 lb) 2 yF T T Σ = − = = 300 lbT = (c) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T Σ = − = = 200 lbT = (d) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T Σ = − = = 200 lbT = (e) 0: 4 (600 lb) 0 1 (600 lb) 4 yF T T Σ = − = = 150.0 lbT = www.elsolucionario.net www.elsolucionario.net
- 72. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 70 PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley and Crate (b) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T Σ = − = = 200 lbT = (d) 0: 4 (600 lb) 0 1 (600 lb) 4 yF T T Σ = − = = 150.0 lbT = www.elsolucionario.net www.elsolucionario.net
- 73. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 71 PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that 750 N,P = determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C (a) 0: (cos25 cos55 ) (750 N)cos55° 0x ACBF TΣ = ° − ° − = Hence: 1292.88 NACBT = 1293 NACBT = (b) 0: (sin 25 sin55 ) (750 N)sin55 0 (1292.88 N)(sin 25 sin55 ) (750 N)sin55 0 y ACBF T Q Q Σ = ° + ° + ° − = ° + ° + ° − = or 2219.8 NQ = 2220 NQ = www.elsolucionario.net www.elsolucionario.net
- 74. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 72 PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P. SOLUTION Free-Body Diagram: Pulley C 0: (cos25 cos55 ) cos55 0x ACBF T PΣ = ° − ° − ° = or 0.58010 ACBP T= (1) 0: (sin 25 sin55 ) sin55 1800 N 0y ACBF T PΣ = ° + ° + ° − = or 1.24177 0.81915 1800 NACBT P+ = (2) (a) Substitute Equation (1) into Equation (2): 1.24177 0.81915(0.58010 ) 1800 NACB ACBT T+ = Hence: 1048.37 NACBT = 1048 NACBT = (b) Using (1), 0.58010(1048.37 N) 608.16 NP = = 608 NP = www.elsolucionario.net www.elsolucionario.net
- 75. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 73 PROBLEM 2.71 Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION cos 65 (900 N)cos 65 380.36 N h h F F F = ° = ° = (a) sin 20 (380.36 N)sin 20° x hF F= ° = 130.091 N,= −xF 130.1 NxF = − sin65 (900 N)sin 65° 815.68 N, y y F F F = ° = = + 816 NyF = + cos20 (380.36 N)cos20 357.42 N = ° = ° = + z h z F F F 357 NzF = + (b) 130.091 N cos 900 N x x F F θ − = = 98.3xθ = ° 815.68 N cos 900 N y y F F θ + = = 25.0yθ = ° 357.42 N cos 900 N z z F F θ + = = 66.6zθ = ° www.elsolucionario.net www.elsolucionario.net
- 76. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 74 PROBLEM 2.72 Determine (a) the x, y, and z components of the 750-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION sin 35 (750 N)sin 35 430.18 N h h F F F = ° = ° = (a) cos 25 (430.18 N)cos 25° = ° = x hF F 389.88 N,= +xF 390 NxF = + cos35 (750 N)cos 35° 614.36 N, y y F F F = ° = = + 614 NyF = + sin 25 (430.18 N)sin 25 181.802 N z h z F F F = ° = ° = + 181.8 NzF = + (b) 389.88 N cos 750 N x x F F θ + = = 58.7xθ = ° 614.36 N cos 750 N y y F F θ + = = 35.0yθ = ° 181.802 N cos 750 N z z F F θ + = = 76.0zθ = ° www.elsolucionario.net www.elsolucionario.net
- 77. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 75 PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force 400 NF = (400 N)cos40 306.42 N HF∴ = ° = (a) sin35 (306.42 N)sin35 x HF F= − ° = − ° 175.755 N= − 175.8 NxF = − sin 40 (400 N)sin 40 257.12 N yF F= − ° = − ° = − 257 NyF = − cos35 (306.42 N)cos35 251.00 N z HF F= + ° = + ° = + 251 NzF = + (b) 175.755 N cos 400 N x x F F θ − = = 116.1xθ = ° 257.12 N cos 400 N y y F F θ − = = 130.0yθ = ° 251.00 N cos 400 N z z F F θ = = 51.1zθ = ° www.elsolucionario.net www.elsolucionario.net
- 78. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 76 PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force 400 NF = (400 N)cos25 362.52 N HF∴ = ° = (a) cos15 (362.52 N)cos15 x HF F= + ° = + ° 350.17 N= + 350 NxF = + sin 25 (400 N)sin 25 169.047 N yF F= − ° = − ° = − 169.0 NyF = − sin15 (362.52 N)sin15 93.827 N z HF F= + ° = + ° = + 93.8 NzF = + (b) 350.17 N cos 400 N x x F F θ + = = 28.9xθ = ° 169.047 N cos 400 N y y F F θ − = = 115.0yθ = ° 93.827 N cos 400 N z z F F θ + = = 76.4zθ = ° www.elsolucionario.net www.elsolucionario.net
- 79. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 77 PROBLEM 2.75 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles ,xθ ,yθ and zθ defining the direction of that force. SOLUTION From triangle AOB: 56 ft cos 65 ft 0.86154 30.51 y y θ θ = = = ° (a) sin cos20 (3900 lb)sin30.51 cos20 x yF F θ= − ° = − ° ° 1861 lbxF = − cos (3900 lb)(0.86154)y yF F θ= + = 3360 lbyF = + (3900 lb)sin 30.51° sin 20°zF = + 677 lbzF = + (b) 1861 lb cos 0.4771 3900 lb x x F F θ = = − = − 118.5xθ = ° From above: 30.51yθ = ° 30.5yθ = ° 677 lb cos 0.1736 3900 lb z z F F θ = = + = + 80.0zθ = ° www.elsolucionario.net www.elsolucionario.net
- 80. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 78 PROBLEM 2.76 Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force. SOLUTION In triangle AOB: 70 ft 56 ft 5250 lb AC OA F = = = 56 ft cos 70 ft 36.870 sin (5250 lb)sin36.870 3150.0 lb y y H yF F θ θ θ = = ° = = ° = (a) sin50 (3150.0 lb)sin50 2413.04 lbx HF F= − ° = − ° = − 2413 lbxF = − cos (5250 lb)cos36.870 4200.0 lby yF F θ= + = + ° = + 4200 lbyF = + cos50 3150cos50 2024.8 lbz HF F= − ° = − ° = − 2025 lbzF = − (b) 2413.04 lb cos 5250 lb x x F F θ − = = 117.4xθ = ° From above: 36.870yθ = ° 36.9yθ = ° 2024.8 lb 5250 lb z z F F θ − = = 112.7zθ = ° www.elsolucionario.net www.elsolucionario.net
- 81. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 79 PROBLEM 2.77 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) (120 lb)cos 60 cos 20xF = ° ° 56.382 lbxF = 56.4 lbxF = + (120 lb)sin 60 103.923 lb y y F F = − ° = − 103.9 lbyF = − (120 lb)cos 60 sin 20 20.521 lb z z F F = − ° ° = − 20.5 lbzF = − (b) 56.382 lb cos 120 lb x x F F θ = = 62.0xθ = ° 103.923 lb cos 120 lb y y F F θ − = = 150.0yθ = ° 20.52 lb cos 120 lb z z F F θ − = = 99.8zθ = ° www.elsolucionario.net www.elsolucionario.net
- 82. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 80 PROBLEM 2.78 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION (a) (85 lb)sin 36 sin 48xF = ° ° 37.129 lb= 37.1 lbxF = (85 lb)cos 36 68.766 lb yF = − ° = − 68.8 lbyF = − (85 lb)sin 36 cos 48 33.431 lb zF = ° ° = 33.4 lbzF = (b) 37.129 lb cos 85 lb x x F F θ = = 64.1xθ = ° 68.766 lb cos 85 lb y y F F θ − = = 144.0yθ = ° 33.431 lb cos 85 lb z z F F θ = = 66.8zθ = ° www.elsolucionario.net www.elsolucionario.net
- 83. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 81 PROBLEM 2.79 Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k. SOLUTION 2 2 2 2 2 2 (690 N) (300 N) (580 N) (690 N) (300 N) ( 580 N) 950 N x y zF F F F = + − = + + = + + − = F i j k 950 NF = 690 N cos 950 N x x F F θ = = 43.4xθ = ° 300 N cos 950 N y y F F θ = = 71.6yθ = ° 580 N cos 950 N z z F F θ − = = 127.6zθ = ° www.elsolucionario.net www.elsolucionario.net
- 84. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 82 PROBLEM 2.80 Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k. SOLUTION 2 2 2 2 2 2 (650 N) (320 N) (760 N) (650 N) ( 320 N) (760 N) x y zF F F F = − + = + + = + − + F i j k 1050 NF = 650 N cos 1050 N x x F F θ = = 51.8xθ = ° 320 N cos 1050 N y y F F θ − = = 107.7yθ = ° 760 N cos 1050 N z z F F θ = = 43.6zθ = ° www.elsolucionario.net www.elsolucionario.net
- 85. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83 PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 75° and θz = 130°. Knowing that the y component of the force is +300 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force. SOLUTION 2 2 2 2 2 2 cos cos cos 1 cos (75 ) cos cos (130 ) 1 cos 0.72100 x y z y y θ θ θ θ θ + + = ° + + ° = = ± (a) Since 0,yF Ͼ we choose cos 0.72100yθ ϭϩ 43.9yθ∴ = ° (b) cos 300 lb (0.72100) y yF F F θ= = 416.09 lbF = 416 lbF = cos 416.09 lbcos75x xF F θ= = ° 107.7 lbxF = + cos 416.09 lbcos130z zF F θ= = ° 267 lbzF = − www.elsolucionario.net www.elsolucionario.net
- 86. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 84 PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is −500 N, determine (a) the angle θx, (b) the other components and the magnitude of the force. SOLUTION 2 2 2 2 2 2 cos cos cos 1 cos cos 55 cos 45 1 cos 0.41353 x y z x x θ θ θ θ θ + + = + ° + ° = = ± (a) Since 0,yF Ͻ we choose cos 0.41353xθ ϭϪ 114.4xθ∴ = ° (b) cos 500 N ( 0.41353) x xF F F θ= − = − 1209.10 NF = 1209.1 NF = cos 1209.10 Ncos55y yF F θ= = ° 694 NyF = + cos 1209.10 Ncos45z zF F θ= = ° 855 NzF = + www.elsolucionario.net www.elsolucionario.net
- 87. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 85 PROBLEM 2.83 A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that θx = 32.5°, Fy = −60 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz. SOLUTION (a) We have cos (230 N)cos32.5x xF F θ= = ° 194.0 NxF = − Then: 193.980 NxF = 2 2 2 2 x y zF F F F= + + So: 2 2 2 2 (230 N) (193.980 N) ( 60 N) zF= + − + Hence: 2 2 2 (230 N) (193.980 N) ( 60 N)zF = + − − − 108.0 NzF = (b) 108.036 NzF = 60 N cos 0.26087 230 N y y F F θ − = = = − 105.1yθ = ° 108.036 N cos 0.46972 230 N z z F F θ = = = 62.0zθ = ° www.elsolucionario.net www.elsolucionario.net
- 88. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 86 PROBLEM 2.84 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy. SOLUTION (a) cos (210 N)cos151.2z zF F θ= = ° 184.024 N= − 184.0 NzF = − Then: 2 2 2 2 x y zF F F F= + + So: 2 2 2 2 (210 N) (80 N) ( ) (184.024 N)yF= + + Hence: 2 2 2 (210 N) (80 N) (184.024 N)yF = − − − 61.929 N= − 62.0 lbyF = − (b) 80 N cos 0.38095 210 N x x F F θ = = = 67.6xθ = ° 61.929 N cos 0.29490 210 N y y F F θ = = = − 107.2yθ = ° www.elsolucionario.net www.elsolucionario.net
- 89. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 87 PROBLEM 2.85 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: ( 46.765 ft) (45 ft) (36 ft) 74.216 ft 46.765 45 36 74.216 AB AB AB AB AB AB T − + + = = − + + = = i j k λ i j k T λ ( ) 1.260 kipsAB xT = − ( ) 1.213 kipsAB yT = + ( ) 0.970 kipsAB zT = + www.elsolucionario.net www.elsolucionario.net
- 90. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 88 PROBLEM 2.86 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: ( 46.765 ft) (55.8 ft) ( 45 ft) 85.590 ft 46.765 55.8 45 (1.5 kips) 85.590 AC AC AC AC AC AC T − + + − = = − + − = = i j k λ i j k T λ ( ) 0.820 kipsAC xT = − ( ) 0.978 kipsAC yT = + ( ) 0.789 kips= −AC zT www.elsolucionario.net www.elsolucionario.net
- 91. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 89 PROBLEM 2.87 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B. SOLUTION 2 2 2 (900 mm) (600 mm) (360 mm) (900 mm) (600 mm) (360 mm) 1140 mm 1425 N [ (900 mm) (600 mm) (360 mm) ] 1140 mm (1125 N) (750 N) (450 N) BA BA BA BA BA BA BA T BA T BA = − + + = + + = = = = − + + = − + + i j k T λ T i j k i j k ( ) 1125 N, ( ) 750 N, ( ) 450 NBA x BA y BA zT T T= − = = www.elsolucionario.net www.elsolucionario.net
- 92. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 90 PROBLEM 2.88 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION 2 2 2 (900 mm) (600 mm) (920 mm) (900 mm) (600 mm) (920 mm) 1420 mm 2130 N [ (900 mm) (600 mm) (920 mm) ] 1420 mm (1350 N) (900 N) (1380 N) CA CA CA CA CA CA CA T CA T CA = − + − = + + = = = = − + − = − + − i j k T λ T i j k i j k ( ) 1350 N, ( ) 900 N, ( ) 1380 NCA x CA y CA zT T T= − = = − www.elsolucionario.net www.elsolucionario.net
- 93. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 91 PROBLEM 2.89 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION 2 2 2 (480 mm) (510 mm) (320 mm) (480 mm) (510 mm ) (320 mm) 770 mm 385 N [(480 mm) (510 mm) (320 mm) ] 770 mm (240 N) (255 N) (160 N) DB DB DB F DB F DB = − + = + + = = = = − + = − + i j k F λ i j k i j k 240 N, 255 N, 160.0 Nx y zF F F= + = − = + www.elsolucionario.net www.elsolucionario.net
- 94. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 92 PROBLEM 2.90 For the frame and cable of Problem 2.89, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.89 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION 2 2 2 (270 mm) (400 mm) (600 mm) (270 mm) (400 mm) (600 mm) 770 mm 385 N [(270 mm) (400 mm) (600 mm) ] 770 mm (135 N) (200 N) (300 N) EB EB EB F EB F EB = − + = + + = = = = − + = − + i j k F λ i j k F i j k 135.0 N, 200 N, 300 Nx y zF F F= + = − = + www.elsolucionario.net www.elsolucionario.net
- 95. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 93 PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N. SOLUTION (600 N)[sin 40 sin 25 cos40 sin 40 cos25 ] (162.992 N) (459.63 N) (349.54 N) (450 N)[cos55 cos30 sin55 cos55 sin30 ] (223.53 N) (368.62 N) (129.055 N) (386.52 N) (828.25 N) (220.49 N) (3R = ° ° + ° + ° ° = + + = ° ° + ° − ° ° = + − = + = + + = P i j k i j k Q i j k i j k R P Q i j k 2 2 2 86.52 N) (828.25 N) (220.49 N) 940.22 N + + = 940 NR = 386.52 N cos 940.22 N x x R R θ = = 65.7xθ = ° 828.25 N cos 940.22 N y y R R θ = = 28.2yθ = ° 220.49 N cos 940.22 N z z R R θ = = 76.4zθ = ° www.elsolucionario.net www.elsolucionario.net
- 96. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 94 PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 450 N and Q = 600 N. SOLUTION (450 N)[sin 40 sin 25 cos40 sin 40 cos25 ] (122.244 N) (344.72 N) (262.154 N) (600 N)[cos55 cos30 sin55 cos55 sin30 ] (298.04 N) (491.49 N) (172.073 N) (420.28 N) (836.21 N) (90.081 N) (R = ° ° + ° + ° ° = + + = ° ° + ° − ° ° = + − = + = + + = P i j k i j k Q i j k i j k R P Q i j k 2 2 2 420.28 N) (836.21 N) (90.081 N) 940.21 N + + = 940 NR = 420.28 cos 940.21 x x R R θ = = 63.4xθ = ° 836.21 cos 940.21 y y R R θ = = 27.2yθ = ° 90.081 cos 940.21 z z R R θ = = 84.5zθ = ° www.elsolucionario.net www.elsolucionario.net
- 97. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 95 PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION 2 2 2 2 2 2 (40 in.) (45 in.) (60 in.) (40 in.) (45 in.) (60 in.) 85 in. (100 in.) (45 in.) (60 in.) (100 in.) (45 in.) (60 in.) 125 in. (40 in.) (45 in.) (60 in.) (425 lb) 85 in. AB AB AB AB AB AB AC AC AB T T AB = − + = + + = = − + = + + = − + = = = i j k i j k i j k T λ (200 lb) (225 lb) (300 lb) (100 in.) (45 in.) (60 in.) (510 lb) 125 in. (408 lb) (183.6 lb) (244.8 lb) (608) (408.6 lb) (544.8 lb) AB AC AC AC AC AC AB AC AC T T AC = − + − + = = = = − + = + = − + T i j k i j k T λ T i j k R T T i j k Then: 912.92 lbR = 913 lbR = and 608 lb cos 0.66599 912.92 lb xθ = = 48.2xθ = ° 408.6 lb cos 0.44757 912.92 lb yθ = = − 116.6yθ = ° 544.8 lb cos 0.59677 912.92 lb zθ = = 53.4zθ = ° www.elsolucionario.net www.elsolucionario.net
- 98. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 96 PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION 2 2 2 2 2 2 (40 in.) (45 in.) (60 in.) (40 in.) (45 in.) (60 in.) 85 in. (100 in.) (45 in.) (60 in.) (100 in.) (45 in.) (60 in.) 125 in. (40 in.) (45 in.) (60 in.) (510 lb) 85 in. AB AB AB AB AB AB AC AC AB T T AB = − + = + + = = − + = + + = − + = = = i j k i j k i j k T λ (240 lb) (270 lb) (360 lb) (100 in.) (45 in.) (60 in.) (425 lb) 125 in. (340 lb) (153 lb) (204 lb) (580 lb) (423 lb) (564 lb) AB AC AC AC AC AC AB AC AC T T AC = − + − + = = = = − + = + = − + T i j k i j k T λ T i j k R T T i j k Then: 912.92 lbR = 913 lbR = and 580 lb cos 0.63532 912.92 lb xθ = = 50.6xθ = ° 423 lb cos 0.46335 912.92 lb yθ − = = − 117.6yθ = ° 564 lb cos 0.61780 912.92 lb zθ = = 51.8zθ = ° www.elsolucionario.net www.elsolucionario.net
- 99. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 97 PROBLEM 2.95 For the frame of Problem 2.89, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.89 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION 2 2 2 (480 mm) (510 mm) (320 mm) (480 mm) (510 mm) (320 mm) 770 mm BD BD = − + − = + + = i j k (385 N) [ (480 mm) (510 mm) (320 mm) ] (770 mm) (240 N) (255 N) (160 N) BD BD BD BD BD T T BD = = = − + − = − + − F λ i j k i j k 2 2 2 (270 mm) (400 mm) (600 mm) (270 mm) (400 mm) (600 mm) 770 mm BE BE = − + − = + + = i j k (385 N) [ (270 mm) (400 mm) (600 mm) ] (770 mm) (135 N) (200 N) (300 N) BE BE BE BE BE T T BE = = = − + − = − + − F λ i j k i j k (375 N) (455 N) (460 N)BD BE= + = − + −R F F i j k 2 2 2 (375 N) (455 N) (460 N) 747.83 NR = + + = 748 NR = 375 N cos 747.83 N xθ − = 120.1xθ = ° 455 N cos 747.83 N yθ = 52.5yθ = ° 460 N cos 747.83 N zθ − = 128.0zθ = ° www.elsolucionario.net www.elsolucionario.net
- 100. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 98 PROBLEM 2.96 For the cables of Problem 2.87, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION (use results of Problem 2.87) ( ) 1125 N ( ) 750 N ( ) 450 N (use results of Problem 2.88) ( ) 1350 N ( ) 900 N ( ) 1380 N AB BA AB x AB y AB z AC CA AC x AC y AC z T T T T T T T T T T = − = + = − = − = − = + = − = + Resultant: 2 2 2 2 2 2 1125 1350 2475 N 750 900 1650 N 450 1380 930 N ( 2475) ( 1650) ( 930) x x y y z z x y z R F R F R F R R R R = Σ = + + = + = Σ = − − = − = Σ = − + = + = + + = + + − + + 3116.6 N= 3120 NR = 2475 cos 3116.6 x x R R θ + = = 37.4xθ = ° 1650 cos 3116.6 y y R R θ − = = 122.0yθ = ° 930 cos 3116.6 z z R R θ + = = 72.6zθ = ° www.elsolucionario.net www.elsolucionario.net
- 101. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 99 PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION Cable AB: 183 lbABT = ( 48 in.) (29 in.) (24 in.) (183lb) 61in. (144 lb) (87 lb) (72 lb) AB AB AB AB AB AB T T AB − + + = = = = − + + i j k T T i j k λ Cable AC: ( 48 in.) (25 in.) ( 36 in.) 65 in. 48 25 36 65 65 65 AC AC AC AC AC AC AC AC AC AC T T T AC T T T − + + − = = = = − + − i j k T T i j k λ Load P: P=P j For resultant to be directed along OA, i.e., x-axis 36 0: (72 lb) 0 65 z z ACR F T′= Σ = − = 130.0 lbACT = www.elsolucionario.net www.elsolucionario.net
- 102. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 100 PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write 25 0: (87 lb) 0 65 y y ACR F T P= Σ = + − = 130.0 lbACT = from Problem 2.97. Then 25 (87 lb) (130.0 lb) 0 65 P+ − = 137.0 lbP = www.elsolucionario.net www.elsolucionario.net
- 103. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 101 PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. SOLUTION Free-Body Diagram at A: The forces applied at A are: , , , andAB AC ADT T T W where .W=W j To express the other forces in terms of the unit vectors i, j, k, we write (450 mm) (600 mm) 750 mm (600 mm) (320 mm) 680 mm (500 mm) (600 mm) (360 mm) 860 mm AB AB AC AC AD AD = − + = = + − = = + + + = i j j k i j k and ( 450 mm) (600 mm) 750 mm AB AB AB AB AB AB T T T AB − + = = = i j T λ 45 60 75 75 ABT = − + i j (600 mm) (320 mm) 680 mm 60 32 68 68 (500 mm) (600 mm) (360 mm) 860 mm 50 60 36 86 86 86 − = = = = − + + = = = = + + AC AC AC AC AC AC AD AD AD AD AD AD AC T T T AC T AD T T T AD T i j T λ j k i j k T λ i j k www.elsolucionario.net www.elsolucionario.net
- 104. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 102 PROBLEM 2.99 (Continued) Equilibrium condition: 0: 0AB AC ADFΣ = ∴ + + + =T T T W Substituting the expressions obtained for , , and ;AB AC ADT T T factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: 45 50 0 75 86 AB ADT T− + = (1) From j: 60 60 60 0 75 68 86 AB AC ADT T T W+ + − = (2) From k: 32 36 0 68 86 AC ADT T− + = (3) Setting 6 kNABT = in (1) and (2), and solving the resulting set of equations gives 6.1920 kN 5.5080 kN AC AC T T = = 13.98 kNW = www.elsolucionario.net www.elsolucionario.net
- 105. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 103 PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN. SOLUTION See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: 45 50 0 75 86 AB ADT T− + = (1) 60 60 60 0 75 68 86 AB AC ADT T T W+ + − = (2) 32 36 0 68 86 AC ADT T− + = (3) Setting 4.3 kNADT = into the above equations gives 4.1667 kN 3.8250 kN AB AC T T = = 9.71kNW = www.elsolucionario.net www.elsolucionario.net
- 106. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 104 PROBLEM 2.101 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N. SOLUTION The forces applied at A are: , , , andAB AC ADT T T P where .P=P j To express the other forces in terms of the unit vectors i, j, k, we write (4.20 m) (5.60 m) 7.00 m (2.40 m) (5.60 m) (4.20 m) 7.40 m (5.60 m) (3.30 m) 6.50 m AB AB AC AC AD AD = − − = = − + = = − − = i j i j k j k and ( 0.6 0.8 ) (0.32432 0.75676 0.56757 ) ( 0.86154 0.50769 ) AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T T AB AC T T T AC AD T T T AD = = = − − = = = − + = = = − − T λ i j T λ i j k T λ j k www.elsolucionario.net www.elsolucionario.net
- 107. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 105 PROBLEM 2.101 (Continued) Equilibrium condition: 0: 0AB AC ADF PΣ = + + + =T T T j Substituting the expressions obtained for , , andAB AC ADT T T and factoring i, j, and k: ( 0.6 0.32432 ) ( 0.8 0.75676 0.86154 ) (0.56757 0.50769 ) 0 AB AC AB AC AD AC AD T T T T T P T T − + + − − − + + − = i j k Equating to zero the coefficients of i, j, k: 0.6 0.32432 0AB ACT T− + = (1) 0.8 0.75676 0.86154 0AB AC ADT T T P− − − + = (2) 0.56757 0.50769 0AC ADT T− = (3) Setting 481 NADT = in (2) and (3), and solving the resulting set of equations gives 430.26 N 232.57 N AC AD T T = = 926 N=P www.elsolucionario.net www.elsolucionario.net
- 108. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 106 PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). 0.6 0.32432 0AB ACT T− + = (1) 0.8 0.75676 0.86154 0AB AC ADT T T P− − − + = (2) 0.56757 0.50769 0AC ADT T− = (3) From Eq. (1): 0.54053AB ACT T= From Eq. (3): 1.11795AD ACT T= Substituting for ABT and ADT in terms of ACT into Eq. (2) gives 0.8(0.54053 ) 0.75676 0.86154(1.11795 ) 0AC AC ACT T T P− − − + = 2.1523 ; 800 N 800 N 2.1523 371.69 N AC AC T P P T = = = = Substituting into expressions for ABT and ADT gives 0.54053(371.69 N) 1.11795(371.69 N) AB AD T T = = 201 N, 372 N, 416 NAB AC ADT T T= = = www.elsolucionario.net www.elsolucionario.net
- 109. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 107 PROBLEM 2.103 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb. SOLUTION The forces applied at A are: , , andAB AC ADT T T W where .P=P j To express the other forces in terms of the unit vectors i, j, k, we write (36 in.) (60 in.) (27 in.) 75 in. (60 in.) (32 in.) 68 in. (40 in.) (60 in.) (27 in.) 77 in. AB AB AC AC AD AD = − + − = = + = = + − = i j k j k i j k and ( 0.48 0.8 0.36 ) (0.88235 0.47059 ) (0.51948 0.77922 0.35065 ) AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T AB T AC T T AC T AD T T AD T = = = − + − = = = + = = = + − T λ i j k T λ j k T λ i j k Equilibrium Condition with W= −W j 0: 0AB AC ADF WΣ = + + − =T T T j www.elsolucionario.net www.elsolucionario.net
- 110. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 108 PROBLEM 2.103 (Continued) Substituting the expressions obtained for , , andAB AC ADT T T and factoring i, j, and k: ( 0.48 0.51948 ) (0.8 0.88235 0.77922 ) ( 0.36 0.47059 0.35065 ) 0 AB AD AB AC AD AB AC AD T T T T T W T T T − + + + + − + − + − = i j k Equating to zero the coefficients of i, j, k: 0.48 0.51948 0 0.8 0.88235 0.77922 0 0.36 0.47059 0.35065 0 AB AD AB AC AD AB AC AD T T T T T W T T T − + = + + − = − + − = Substituting 750 lbABT = in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: 1090.1lb 693 lb AC AD T T = = 2100 lbW = www.elsolucionario.net www.elsolucionario.net
- 111. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 109 PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AD is 616 lb. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48 0.51948 0AB ADT T− + = (1) 0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = (2) 0.36 0.47059 0.35065 0AB AC ADT T T− + − = (3) Substituting 616 lbADT = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: 667.67 lb 969.00 lb AB AC T T = = 1868 lbW = www.elsolucionario.net www.elsolucionario.net
- 112. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 110 PROBLEM 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48 0.51948 0AB ADT T− + = (1) 0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = (2) 0.36 0.47059 0.35065 0AB AC ADT T T− + − = (3) Substituting 544 lbACT = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: 374.27 lb 345.82 lb AB AD T T = = 1049 lbW = www.elsolucionario.net www.elsolucionario.net
- 113. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 111 PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable. SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48 0.51948 0AB ADT T− + = (1) 0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = (2) 0.36 0.47059 0.35065 0AB AC ADT T T− + − = (3) Substituting 1600 lbW = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives 571 lbABT = 830 lbACT = 528 lbADT = www.elsolucionario.net www.elsolucionario.net
- 114. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 112 PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 0,Q = find the value of P for which the tension in cable AD is 305 N. SOLUTION 0: 0A AB AC ADΣ = + + + =F T T T P where P=P i (960 mm) (240 mm) (380 mm) 1060 mm (960 mm) (240 mm) (320 mm) 1040 mm (960 mm) (720 mm) (220 mm) 1220 mm AB AB AC AC AD AD = − − + = = − − − = = − + − = i j k i j k i j k 48 12 19 53 53 53 12 3 4 13 13 13 305 N [( 960 mm) (720 mm) (220 mm) ] 1220 mm (240 N) (180 N) (55 N) AB AB AB AB AB AC AC AC AC AC AD AD AD AB T T T AB AC T T T AC T = = = − − + = = = − − − = = − + − = − + − T λ i j k T λ i j k T λ i j k i j k Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives: 48 12 : 240 N 53 13 AB ACP T T= + +i (1) :j 12 3 180 N 53 13 AB ACT T+ = (2) :k 19 4 55 N 53 13 AB ACT T− = (3) Solving the system of linear equations using conventional algorithms gives: 446.71 N 341.71 N AB AC T T = = 960 NP = www.elsolucionario.net www.elsolucionario.net
- 115. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 113 PROBLEM 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that 1200 N,P = determine the values of Q for which cable AD is taut. SOLUTION We assume that 0ADT = and write 0: (1200 N) 0A AB AC QΣ = + + + =F T T j i (960 mm) (240 mm) (380 mm) 1060 mm (960 mm) (240 mm) (320 mm) 1040 mm AB AB AC AC = − − + = = − − − = i j k i j k 48 12 19 53 53 53 12 3 4 13 13 13 AB AB AB AB AB AC AC AC AC AC AB T T T AB AC T T T AC = = = − − + = = = − − − T λ i j k T λ i j k Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives: 48 12 : 1200 N 0 53 13 AB ACT T− − + =i (1) 12 3 : 0 53 13 AB ACT T Q− − + =j (2) 19 4 : 0 53 13 AB ACT T− =k (3) Solving the resulting system of linear equations using conventional algorithms gives: 605.71 N 705.71 N 300.00 N AB AC T T Q = = = 0 300 NQՅ Ͻ Note: This solution assumes that Q is directed upward as shown ( 0),Q Ն if negative values of Q are considered, cable AD remains taut, but AC becomes slack for 460 N.Q = − www.elsolucionario.net www.elsolucionario.net
- 116. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 114 PROBLEM 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. SOLUTION We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. Free Body A: 0: 0AB AC ADF PΣ = + + + =T T T j We have: ( ) (320 mm) (480 mm) (360 mm) 680 mm (450 mm) (480 mm) (360 mm) 750 mm (250 mm) (480 mm) 360 mm 650 mm AB AB AC AC AD AD = − − + = = − + = = − − = i j k i j k i j k Thus: ( ) 8 12 9 17 17 17 0.6 0.64 0.48 5 9.6 7.2 13 13 13 AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T T AB AC T T T AC AD T T T AD = = = − − + = = = − + = = = − − T λ i j k T λ i j k T λ i j k Substituting into the Eq. 0FΣ = and factoring , , :i j k 8 5 0.6 17 13 12 9.6 0.64 17 13 9 7.2 0.48 0 17 13 AB AC AD AB AC AD AB AC AD T T T T T T P T T T − + + + − − − + + + − = i j k Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 117. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 115 PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: :i 8 5 0.6 0 17 13 AB AC ADT T T− + + = (1) :j 12 9.6 0.64 0 7 13 AB AC ADT T T P− − − + = (2) :k 9 7.2 0.48 0 17 13 AB AC ADT T T+ − = (3) Making 60 NACT = in (1) and (3): 8 5 36 N 0 17 13 AB ADT T− + + = (1′) 9 7.2 28.8 N 0 17 13 AB ADT T+ − = (3′) Multiply (1′) by 9, (3′) by 8, and add: 12.6 554.4 N 0 572.0 N 13 AD ADT T− = = Substitute into (1′) and solve for :ABT 17 5 36 572 544.0 N 8 13 AB ABT T = + × = Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) 0.64(60 N) (572 N) 17 13 844.8 N P = + + = Weight of plate 845 NP= = www.elsolucionario.net www.elsolucionario.net
- 118. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 116 PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 0.6 0 17 13 AB AC ADT T T− + + = (1) 12 9.6 0.64 0 17 13 AB AC ADT T T P− + − + = (2) 9 7.2 0.48 0 17 13 AB AC ADT T T+ − = (3) Making 520 NADT = in Eqs. (1) and (3): 8 0.6 200 N 0 17 AB ACT T− + + = (1′) 9 0.48 288 N 0 17 AB ACT T+ − = (3′) Multiply (1′) by 9, (3′) by 8, and add: 9.24 504 N 0 54.5455 NAC ACT T− = = Substitute into (1′) and solve for :ABT 17 (0.6 54.5455 200) 494.545 N 8 AB ABT T= × + = Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) 0.64(54.5455 N) (520 N) 17 13 768.00 N P = + + = Weight of plate 768 NP= = www.elsolucionario.net www.elsolucionario.net
- 119. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 117 PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 630 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION Free Body A: 0: 0AB AC AD PΣ = + + + =F T T T j = 45 90 30 105 ft 30 90 65 115 ft 20 90 60 110 ft AB AB AC AC AD AD − − + = = − + = = − − = i j k i j k i j k We write 3 6 2 7 7 7 AB AB AB AB AB AB T T AB T = = = − − + T λ i j k 6 18 13 23 23 23 AC AC AC AC AC AC T T AC T = = = − + T λ i j k 2 9 6 11 11 11 AD AD AD AD AD AD T T AD T = = = − − T λ i j k Substituting into the Eq. 0Σ =F and factoring , , :i j k 3 6 2 7 23 11 6 18 9 7 23 11 2 13 6 0 7 23 11 AB AC AD AB AC AD AB AC AD T T T T T T P T T T − + + + − − − + + + − = i j k www.elsolucionario.net www.elsolucionario.net
- 120. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 118 PROBLEM 2.111 (Continued) Setting the coefficients of , , ,i j k equal to zero: :i 3 6 2 0 7 23 11 AB AC ADT T T− + + = (1) :j 6 18 9 0 7 23 11 AB AC ADT T T P− − − + = (2) :k 2 13 6 0 7 23 11 AB AC ADT T T+ − = (3) Set 630 lbABT = in Eqs. (1) – (3): 6 2 270 lb 0 23 11 AC ADT T− + + = (1′) 18 9 540 lb 0 23 11 AC ADT T P− − − + = (2′) 13 6 180 lb 0 23 11 AC ADT T+ − = (3′) Solving, 467.42 lb 814.35 lb 1572.10 lbAC ADT T P= = = 1572 lbP = www.elsolucionario.net www.elsolucionario.net
- 121. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 119 PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 920 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 0 7 23 11 AB AC ADT T T− + + = (1) 6 18 9 0 7 23 11 AB AC ADT T T P− − − + = (2) 2 13 6 0 7 23 11 AB AC ADT T T+ − = (3) Substituting for 920 lbACT = in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 2 240 lb 0 7 11 AB ADT T− + + = (1′) 6 9 720 lb 0 7 11 AB ADT T P− − − + = (2′) 2 6 520 lb 0 7 11 AB ADT T+ − = (3′) Solving, 1240.00 lb 1602.86 lb AB AD T T = = 3094.3 lbP = 3090 lbP = www.elsolucionario.net www.elsolucionario.net
- 122. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 120 PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION Free-Body Diagram at A 16 30 34 34 N = + N i j and (180 lb)W= = −W j j ( 30 ft) (20 ft) (12 ft) 38 ft 15 10 6 19 19 19 AC AC AC AC AC AC AC T T T AC T − + − = = = = − + − i j k T λ i j k ( 30 ft) (24 ft) (32 ft) 50 ft 15 12 16 25 25 25 AB AB AB AB AB AB AB T T T AB T − + + = = = = − + + i j k T λ i j k Equilibrium condition: 0Σ =F 0AB AC+ + + =T T N W www.elsolucionario.net www.elsolucionario.net
- 123. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 121 PROBLEM 2.113 (Continued) Substituting the expressions obtained for , , ,AB ACT T N and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: 15 15 16 0 25 19 34 AB ACT T N− − + = (1) From j: 12 10 30 (180 lb) 0 25 19 34 AB ACT T N+ + − = (2) From k: 16 6 0 25 19 AB ACT T− = (3) Solving the resulting set of equations gives: 31.7 lb; 64.3 lbAB ACT T= = www.elsolucionario.net www.elsolucionario.net
- 124. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 122 PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(60 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force ( 60 lb) .= −P k 15 15 16 0 25 19 34 AB ACT T N− − + = (1) 12 10 30 (180 lb) 0 25 19 34 AB ACT T N+ + − = (2) 16 6 (60 lb) 0 25 19 AB ACT T− − = (3) Solving the resulting set of equations simultaneously gives: 99.0 lbABT = 10.55 lbACT = www.elsolucionario.net www.elsolucionario.net
- 125. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 123 PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting 792 NP = gives: 8 5 0.6 0 17 13 AB AC ADT T T− + + = (1) 12 9.6 0.64 792 N 0 17 13 AB AC ADT T T− − − + = (2) 9 7.2 0.48 0 17 13 AB AC ADT T T+ − = (3) Solving Equations (1), (2), and (3) by conventional algorithms gives 510.00 NABT = 510 NABT = 56.250 NACT = 56.2 NACT = 536.25 NADT = 536 NADT = www.elsolucionario.net www.elsolucionario.net
- 126. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 124 PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that 2880 NP = and 0.Q = SOLUTION 0: 0A AB AC ADΣ = + + + + =F T T T P Q Where P=P i and Q=Q j (960 mm) (240 mm) (380 mm) 1060 mm (960 mm) (240 mm) (320 mm) 1040 mm (960 mm) (720 mm) (220 mm) 1220 mm AB AB AC AC AD AD = − − + = = − − − = = − + − = i j k i j k i j k 48 12 19 53 53 53 12 3 4 13 13 13 48 36 11 61 61 61 AB AB AB AB AB AC AC AC AC AC AD AD AD AD AD AB T T T AB AC T T T AC AD T T T AD = = = − − + = = = − − − = = = − + − T λ i j k T λ i j k T λ i j k Substituting into 0,AΣ =F setting (2880 N)P = i and 0,Q = and setting the coefficients of , ,i j k equal to 0, we obtain the following three equilibrium equations: 48 12 48 : 2880 N 0 53 13 61 AB AC ADT T T− − − + =i (1) 12 3 36 : 0 53 13 61 AB AC ADT T T− − + =j (2) 19 4 11 : 0 53 13 61 AB AC ADT T T− − =k (3) www.elsolucionario.net www.elsolucionario.net
- 127. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 125 PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: 1340.14 N 1025.12 N 915.03 N AB AC AD T T T = = = 1340 NABT = 1025 NACT = 915 NADT = www.elsolucionario.net www.elsolucionario.net
- 128. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 126 PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that 2880 NP = and 576 N.Q = SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 48 12 48 0 53 13 61 AB AC ADT T T P− − − + = (1) 12 3 36 0 53 13 61 AB AC ADT T T Q− − + + = (2) 19 4 11 0 53 13 61 AB AC ADT T T− − = (3) Setting 2880 NP = and 576 NQ = gives: 48 12 48 2880 N 0 53 13 61 AB AC ADT T T− − − + = (1′) 12 3 36 576 N 0 53 13 61 AB AC ADT T T− − + + = (2′) 19 4 11 0 53 13 61 AB AC ADT T T− − = (3′) Solving the resulting set of equations using conventional algorithms gives: 1431.00 N 1560.00 N 183.010 N AB AC AD T T T = = = 1431 NABT = 1560 NACT = 183.0 NADT = www.elsolucionario.net www.elsolucionario.net
- 129. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 127 PROBLEM 2.118 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that 2880 NP = and 576Q = − N. (Q is directed downward). SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 48 12 48 0 53 13 61 AB AC ADT T T P− − − + = (1) 12 3 36 0 53 13 61 AB AC ADT T T Q− − + + = (2) 19 4 11 0 53 13 61 AB AC ADT T T− − = (3) Setting 2880 NP = and 576 NQ = − gives: 48 12 48 2880 N 0 53 13 61 AB AC ADT T T− − − + = (1′) 12 3 36 576 N 0 53 13 61 AB AC ADT T T− − + − = (2′) 19 4 11 0 53 13 61 AB AC ADT T T− − = (3′) Solving the resulting set of equations using conventional algorithms gives: 1249.29 N 490.31 N 1646.97 N AB AC AD T T T = = = 1249 NABT = 490 NACT = 1647 NADT = www.elsolucionario.net www.elsolucionario.net
- 130. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 128 PROBLEM 2.119 For the transmission tower of Problems 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 2100 lb. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 0 7 23 11 AB AC ADT T T− + + = (1) 6 18 9 0 7 23 11 AB AC ADT T T P− − − + = (2) 2 13 6 0 7 23 11 AB AC ADT T T+ − = (3) Substituting for 2100 lbP = in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 6 2 0 7 23 11 AB AC ADT T T− + + = (1′) 6 18 9 2100 lb 0 7 23 11 AB AC ADT T T− − − + = (2′) 2 13 6 0 7 23 11 AB AC ADT T T+ − = (3′) 841.55 lb 624.38 lb 1087.81 lb AB AC AD T T T = = = 842 lbABT = 624 lbACT = 1088 lbADT = www.elsolucionario.net www.elsolucionario.net
- 131. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 129 PROBLEM 2.120 A horizontal circular plate weighing 60 lb is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire. SOLUTION 0:xFΣ = (sin30 )(sin50 ) (sin30 )(cos40 ) (sin30 )(cos60 ) 0AD BD CDT T T− ° ° + ° ° + ° ° = Dividing through by sin30° and evaluating: 0.76604 0.76604 0.5 0AD BD CDT T T− + + = (1) 0: (cos30 ) (cos30 ) (cos30 ) 60 lb 0y AD BD CDF T T TΣ = − ° − ° − ° + = or 69.282 lbAD BD CDT T T+ + = (2) 0: sin30 cos50 sin30 sin 40 sin30 sin 60 0z AD BD CDF T T TΣ = ° ° + ° ° − ° ° = or 0.64279 0.64279 0.86603 0AD BD CDT T T+ − = (3) Solving Equations (1), (2), and (3) simultaneously: 29.5 lbADT = 10.25 lbBDT = 29.5 lbCDT = www.elsolucionario.net www.elsolucionario.net
- 132. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 130 PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Free Body Diagram at A: Since tension inBACT = cable BAC, it follows that AB AC BACT T T= = ( 17.5 in.) (60 in.) 17.5 60 62.5 in. 62.5 62.5 (60 in.) (25 in.) 60 25 65 in. 65 65 (80 in.) (60 in.) 4 3 100 in. 5 5 AB BAC AB BAC BAC AC BAC AC BAC BAC AD AD AD AD AD AE AE AE AE T T T T T T T T T T T − + − = = = + + = = = + + = = = + = = i j T λ i j i k T λ j k i j T λ i j T λ (60 in.) (45 in.) 4 3 75 in. 5 5 AET − = − j k j k www.elsolucionario.net www.elsolucionario.net
- 133. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 131 PROBLEM 2.121 (Continued) Substituting into 0,AΣ =F setting ( 200 lb) ,= −P j and setting the coefficients of i, j, k equal to ,φ we obtain the following three equilibrium equations: From 17.5 4 : 0 62.5 5 BAC ADT T− + =i (1) From 60 60 3 4 : 200 lb 0 62.5 65 5 5 BAC AD AET T T + + + − = j (2) From 25 3 : 0 65 5 BAC AET T− =k (3) Solving the system of linear equations using convential acgorithms gives: 76.7 lb; 26.9 lb; 49.2 lbBAC AD AET T T= = = www.elsolucionario.net www.elsolucionario.net
- 134. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 132 PROBLEM 2.122 Knowing that the tension in cable AE of Prob. 2.121 is 75 lb, determine (a) the magnitude of the load P, (b) the tension in cables BAC and AD. PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Refer to the solution to Problem 2.121 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity: 17.5 4 0 62.5 5 BAC ADT T− + = (1) 60 60 3 4 0 62.5 65 5 5 BAC AD AET T T P + + + − = (2) 25 3 0 65 5 BAC AET T− = (3) Substituting for 75 lbAET = and solving simultaneously gives: 305 lb; 117.0 lb; 40.9 lbBAC ADP T T= = = www.elsolucionario.net www.elsolucionario.net
- 135. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 133 PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P=P i and Q Q= k are applied to the ring to maintain the container in the position shown. Knowing that W 376 N,= determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION ( 130 mm) (400 mm) (160 mm) 450 mm 13 40 16 45 45 45 AB ABT AB T AB T T = = − + + = = − + + T λ i j k i j k Free-Body A: ( 150 mm) (400 mm) ( 240 mm) 490 mm 15 40 24 49 49 49 0: 0 AC AC AB AC T AC T AC T T F = = − + + − = = − + − Σ = + + + + = T λ i j k i j k T T Q P W Setting coefficients of i, j, k equal to zero: 13 15 : 0 0.59501 45 49 T T P T P− − + = =i (1) 40 40 : 0 1.70521 45 49 T T W T W+ + − = =j (2) 16 24 : 0 0.134240 45 49 T T Q T Q+ − + = =k (3) www.elsolucionario.net www.elsolucionario.net
- 136. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 134 PROBLEM 2.123 (Continued) Data: 376 N 1.70521 376 N 220.50 NW T T= = = 0.59501(220.50 N) P= 131.2 NP = 0.134240(220.50 N) Q= 29.6 NQ = www.elsolucionario.net www.elsolucionario.net
- 137. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 135 PROBLEM 2.124 For the system of Problem 2.123, determine W and Q knowing that 164 N.P = PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P=P i and Q=Q k are applied to the ring to maintain the container in the position shown. Knowing that W 376 N,= determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting 164 NP = we have: Eq. (1): 0.59501 164 NT = 275.63 NT = Eq. (2): 1.70521(275.63 N) W= 470 NW = Eq. (3): 0.134240(275.63 N) Q= 37.0 NQ = www.elsolucionario.net www.elsolucionario.net
- 138. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 136 PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force (341 N)=P j is applied to collar A, determine (a) the tension in the wire when y 155 mm,= (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION For both Problems 2.125 and 2.126: Free-Body Diagrams of Collars: 2 2 2 2 ( )AB x y z= + + Here 2 2 2 2 (0.525 m) (0.20 m) y z= + + or 2 2 2 0.23563 my z+ = Thus, when y given, z is determined, Now 1 (0.20 )m 0.525 m 0.38095 1.90476 1.90476 AB AB AB y z y z = = − + = − + λ i j k i j k Where y and z are in units of meters, m. From the F.B. Diagram of collar A: 0: 0x z AB ABN N P T λΣ = + + + =F i k j Setting the j coefficient to zero gives (1.90476 ) 0ABP y T− = With 341 N 341 N 1.90476 AB P T y = = Now, from the free body diagram of collar B: 0: 0x y AB ABN N Q TΣ = + + − =F i j k λ Setting the k coefficient to zero gives (1.90476 ) 0ABQ T z− = And using the above result for ,ABT we have 341 N (341 N)( ) (1.90476 ) (1.90476) AB z Q T z z y y = = = www.elsolucionario.net www.elsolucionario.net
- 139. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 137 PROBLEM 2.125 (Continued) Then from the specifications of the problem, 155 mm 0.155 my = = 2 2 2 0.23563 m (0.155 m) 0.46 m z z = − = and (a) 341 N 0.155(1.90476) 1155.00 N ABT = = or 1155 N=ABT and (b) 341 N(0.46 m)(0.866) (0.155 m) (1012.00 N) Q = = or 1012 N=Q www.elsolucionario.net www.elsolucionario.net
- 140. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 138 PROBLEM 2.126 Solve Problem 2.125 assuming that 275 mm.y = PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force (341 N)=P j is applied to collar A, determine (a) the tension in the wire when y 155 mm,= (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION From the analysis of Problem 2.125, particularly the results: 2 2 2 0.23563 m 341 N 1.90476 341 N AB y z T y Q z y + = = = With 275 mm 0.275 m,y = = we obtain: 2 2 2 0.23563 m (0.275 m) 0.40 m z z = − = and (a) 341 N 651.00 (1.90476)(0.275 m) ABT = = or 651 NABT = and (b) 341 N(0.40 m) (0.275 m) Q = or 496 NQ = www.elsolucionario.net www.elsolucionario.net
- 141. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 139 PROBLEM 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines, we have 180 (40 20 ) 120 γ = ° − ° + ° = ° Then 2 2 2 2 (15 kN) (10 kN) 2(15 kN)(10 kN)cos120 475 kN 21.794 kN R R = + − ° = = and 10 kN 21.794 kN sin sin120 10 kN sin sin120 21.794 kN 0.39737 23.414 α α α = ° = ° = = Hence: 50 73.414φ α= + ° = 21.8 kN=R 73.4° www.elsolucionario.net www.elsolucionario.net
- 142. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 140 PROBLEM 2.128 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION (a) sin35 300 lbP ° = 300 lb sin35 P = ° 523 lbP = (b) Vertical component cos35vP P= ° (523 lb)cos35= ° 428 lbvP = www.elsolucionario.net www.elsolucionario.net
- 143. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 141 PROBLEM 2.129 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. SOLUTION Using the x and y axes shown: sin10 (50 lb)cos35 (75 lb)cos60 sin10 78.458 lb x x AC AC R F T T = Σ = ° + ° + ° = ° + (1) (50 lb)sin35 (75 lb)sin 60 cos10 93.631 lb cos10 y y AC y AC R F T R T = Σ = ° + ° − ° = − ° (2) (a) Set 0yR = in Eq. (2): 93.631 lb cos10 0 95.075 lb AC AC T T − ° = = 95.1 lbACT = (b) Substituting for ACT in Eq. (1): (95.075 lb)sin10 78.458 lb 94.968 lb x x R R R = ° + = = 95.0 lbR = www.elsolucionario.net www.elsolucionario.net
- 144. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 142 PROBLEM 2.130 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle 2 mg (200 kg)(9.81m/s ) 1962 N W = = = Law of sines: 1962 N sin 15 sin 105 sin 60 AC BCT T = = ° ° ° (a) (1962 N) sin 15 sin 60 ACT ° = ° 586 NACT = (b) (1962 N)sin 105 sin 60 BCT ° = ° 2190 NBCT = www.elsolucionario.net www.elsolucionario.net
- 145. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 143 PROBLEM 2.131 A welded connection is in equilibrium under the action of the four forces shown. Knowing that 8AF = kN and 16BF = kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 0: 0 5 5 x B C AF F F FΣ = − − = With 8 kN 16 kN A B F F = = 4 4 (16 kN) (8 kN) 5 5 CF = − 6.40 kNCF = 3 3 0: 0 5 5 y D B AF F F FΣ = − + − = With FA and FB as above: 3 3 (16 kN) (8 kN) 5 5 DF = − 4.80 kNDF = www.elsolucionario.net www.elsolucionario.net
- 146. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 144 PROBLEM 2.132 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable. SOLUTION Free-Body Diagram 0: cos60 75 lb 0x BCF T QΣ = − − ° + = 75 lb cos60BCT Q= − ° (1) 0: sin 60 0y ACF T QΣ = − ° = sin 60ACT Q= ° (2) Requirement: 60 lb:ACT Յ From Eq. (2): sin60 60 lbQ °Յ 69.3 lbQ Յ Requirement: 60 lb:BCT Յ From Eq. (1): 75 lb sin 60 60 lbQ− °Յ 30.0 lbQ Ն 30.0 lb 69.3 lbQՅ Յ www.elsolucionario.net www.elsolucionario.net
- 147. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 145 PROBLEM 2.133 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles θx, θy, and θz that the force exerted at A forms with the coordinate axes. SOLUTION (a) sin30 sin50 110.3 N (Given)xF F= ° ° = 110.3 N 287.97 N sin 30° sin 50° F = = 288 NF = (b) 110.3 N cos 0.38303 287.97 N x x F F θ = = = 67.5xθ = ° cos30 249.39 249.39 N cos 0.86603 287.97 N y y y F F F F θ = ° = = = = 30.0yθ = ° sin30 cos50 (287.97 N)sin 30°cos 50° 92.552 N zF F= − ° ° = − = − 92.552 N cos 0.32139 287.97 N z z F F θ − = = = − 108.7zθ = ° www.elsolucionario.net www.elsolucionario.net
- 148. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 146 PROBLEM 2.134 A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is −500 lb, determine (a) the angle θx, (b) the other components and the magnitude of the force. SOLUTION (a) We have 2 2 2 2 2 2 (cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )x y z y y zθ θ θ θ θ θ+ + = = − − Since 0,xF Ͻ we must have cos 0.xθ Ͻ Thus, taking the negative square root, from above, we have 2 2 cos 1 (cos55) (cos45) 0.41353xθ = − − − = 114.4xθ = ° (b) Then 500 lb 1209.10 lb cos 0.41353 x x F F θ = = = 1209 lbF = and cos (1209.10 lb)cos55y yF F θ= = ° 694 lbyF = cos (1209.10 lb)cos45z zF F θ= = ° 855 lbzF = www.elsolucionario.net www.elsolucionario.net
- 149. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 147 PROBLEM 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION (300 N)[ cos30 sin15 sin30 cos30 cos15 ] (67.243 N) (150 N) (250.95 N) (400 N)[cos50 cos20 sin50 cos50 sin 20 ] (400 N)[0.60402 0.76604 0.21985] (241.61 N) (306.42 N) (87.939 N) (174. = − ° ° + ° + ° ° = − + + = ° ° + ° − ° ° = + − = + − = + = P i j k i j k Q i j k i j i j k R P Q 2 2 2 367 N) (456.42 N) (163.011 N) (174.367 N) (456.42 N) (163.011 N) 515.07 N R + + = + + = i j k 515 NR = 174.367 N cos 0.33853 515.07 N x x R R θ = = = 70.2xθ = ° 456.42 N cos 0.88613 515.07 N y y R R θ = = = 27.6yθ = ° 163.011 N cos 0.31648 515.07 N z z R R θ = = = 71.5zθ = ° www.elsolucionario.net www.elsolucionario.net
- 150. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 148 PROBLEM 2.136 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AC is 444 N. SOLUTION See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.6 0.32432 0AB ACT T− + = (1) 0.8 0.75676 0.86154 0AB AC ADT T T P− − − + = (2) 0.56757 0.50769 0AC ADT T− = (3) Substituting 444 NACT = in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives 240 N 496.36 N AB AD T T = = 956 N=P www.elsolucionario.net www.elsolucionario.net
- 151. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 149 PROBLEM 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when 9 in.,x = (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system. SOLUTION Free-Body Diagrams of Collars: A: B: (20 in.) 25 in. AB AB x z AB − − + = = i j k λ Collar A: 0: 0y z AB ABP N N TΣ = + + + =F i j k λ Substitute for ABλ and set coefficient of i equal to zero: 0 25 in. ABT x P − = (1) Collar B: 0: (60 lb) 0x y AB ABN N T′ ′Σ = + + − =F k i j λ Substitute for ABλ and set coefficient of k equal to zero: 60 lb 0 25 in. ABT z − = (2) (a) 2 2 2 2 9 in. (9 in.) (20 in.) (25 in.) 12 in. x z z = + + = = From Eq. (2): 60 lb (12 in.) 25 in. ABT− 125.0 lbABT = (b) From Eq. (1): (125.0 lb)(9 in.) 25 in. P = 45.0 lbP = www.elsolucionario.net www.elsolucionario.net
- 152. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 150 PROBLEM 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when 120 lbP = and 60 lb.Q = SOLUTION See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below: 0 25 in. ABT x P = = (1) 60 lb 0 25 in. ABT z − = (2) For 120 lb,P = Eq. (1) yields (25 in.)(20 lb)ABT x = (1′) From Eq. (2): (25 in.)(60 lb)ABT z = (2′) Dividing Eq. (1′) by (2′), 2 x z = (3) Now write 2 2 2 2 (20 in.) (25 in.)x z+ + = (4) Solving (3) and (4) simultaneously, 2 2 2 4 400 625 45 6.7082 in. z z z z + + = = = From Eq. (3): 2 2(6.7082 in.) 13.4164 in. x z= = = 13.42 in., 6.71 in.x z= = www.elsolucionario.net www.elsolucionario.net
- 153. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 151 PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the free- body diagram needed to determine the tension in AC and BC. SOLUTION Free-Body Diagram of Point C: www.elsolucionario.net www.elsolucionario.net
- 154. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 152 PROBLEM 2.F2 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F. SOLUTION Free-Body Diagram of Point B: 1 1 250 N 765 N 1015 N 8.25 tan 30.510 14 10 tan 22.620 24 E AB BC W θ θ − − = + = = = ° = = ° Use this free body to determine TAB and TBC. Free-Body Diagram of Point C: 1 1.1 tan 10.3889 6 CDθ − = = ° Use this free body to determine TCD and WF. Then weight of skier WS is found by 250 NS FW W= − www.elsolucionario.net www.elsolucionario.net
- 155. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 153 PROBLEM 2.F3 Two cables are tied together at A and loaded as shown. Draw the free- body diagram needed to determine the tension in each cable. SOLUTION Free-Body Diagram of Point A: www.elsolucionario.net www.elsolucionario.net
- 156. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 154 PROBLEM 2.F4 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. SOLUTION Free-Body Diagram of Point A: www.elsolucionario.net www.elsolucionario.net
- 157. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 155 PROBLEM 2.F5 A 36-lb triangular plate is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each wire. SOLUTION Free-Body Diagram of Point D: www.elsolucionario.net www.elsolucionario.net
- 158. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 156 PROBLEM 2.F6 A 70-kg cylinder is supported by two cables AC and BC, which are attached to the top of vertical posts. A horizontal force P, perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable. SOLUTION Free-Body Diagram of Point C: 2 (70 kg)(9.81 m/s ) 686.7 N (686.7 N) W = = = −W j www.elsolucionario.net www.elsolucionario.net
- 159. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 157 PROBLEM 2.F7 Three cables are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. The cables support a 180-lb cylinder as shown. Draw the free-body diagram needed to determine the tension in each cable. SOLUTION Free-Body Diagram of Point D: www.elsolucionario.net www.elsolucionario.net
- 160. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 158 PROBLEM 2.F8 A 100-kg container is suspended from ring A, to which cables AC and AE are attached. A force P is applied to end F of a third cable that passes over a pulley at B and through ring A and then is attached to a support at D. Draw the free-body diagram needed to determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION Free-Body Diagram of Ring A: 2 (100 kg)(9.81 m/s ) 981 N (681 N) W = = = −W j www.elsolucionario.net www.elsolucionario.net
- 163. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 161 PROBLEM 3.1 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components. SOLUTION Free-Body Diagram of Rod AB: (9 in.)cos65 3.8036 in. (9 in.)sin65 8.1568 in. x y = ° = = ° = (20 lb cos25 ) ( 20 lb sin 25 ) (18.1262 lb) (8.4524 lb) x yF F= + = ° + − ° = − F i j i j i j / ( 3.8036 in.) (8.1568 in.)A B BA= = − +r i j / ( 3.8036 8.1568 ) (18.1262 8.4524 ) 32.150 147.852 115.702 lb-in. B A B= × = − + × − = − = − M r F i j i j k k 115.7 lb-in.B =M www.elsolucionario.net www.elsolucionario.net
- 164. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 162 PROBLEM 3.2 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB. SOLUTION Free-Body Diagram of Rod AB: 90 (65 25 ) 50 θ = ° − ° − ° = ° (20 lb)cos50 12.8558 lb (9 in.) (12.8558 lb)(9 in.) 115.702 lb-in. B Q M Q = ° = = = = 115.7 lb-in.B =M www.elsolucionario.net www.elsolucionario.net
- 165. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 163 PROBLEM 3.3 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb·in. clockwise, determine the value of α. SOLUTION Free-Body Diagram of Rod AB: 25α θ= − ° (20 lb)cosQ θ= and ( )(9 in.)BM Q= Therefore, 120 lb-in. (20 lb)(cos )(9 in.) 120 lb-in. cos 180 lb-in. θ θ = = or 48.190θ = ° Therefore, 48.190 25α = ° − ° 23.2α = ° www.elsolucionario.net www.elsolucionario.net
- 166. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 164 PROBLEM 3.4 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E. SOLUTION (a) By definition, 2 80 kg(9.81m/s ) 784.8 NW mg= = = We have : (784.8 N)(0.25 m)E EM MΣ = 196.2 N mE = ⋅M (b) For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: 2 2 (0.85 m) (0.5 m) 0.98615 md = + = We have : 196.2 N m (0.98615 m)E BM FΣ ⋅ = 198.954 NBF = and 1 0.85 m tan 59.534 0.5 m θ − = = ° or 199.0 NB =F 59.5° www.elsolucionario.net www.elsolucionario.net
- 167. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165 PROBLEM 3.5 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E. SOLUTION First note. . . 2 mg (80 kg)(9.81 m/s ) 784.8 NW = = = (a) We have / (0.25 m)(784.8 N) 196.2 N mE H EM r W= = = ⋅ or 196.2 N mE = ⋅M (b) For FA to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have / min( ) ( ) .E A E AM r F− = Where 2 2 / (0.35 m) (0.5 m) 0.61033 mA Er = + = then min196.2 N m (0.61033 m)( )AF⋅ = or min( ) 321 NAF = Also 0.35 m tan 0.5 m φ = or 35.0φ = ° min( ) 321 NA =F 35.0° (c) For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then / min min ( ) ( ) 196.2 N m (0.85 m)( ) E D E vertical vertical M r F F − = ⋅ = or min( ) 231 Nvertical =F at Point D www.elsolucionario.net www.elsolucionario.net
- 168. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 166 PROBLEM 3.6 A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P. SOLUTION (0.2 m)cos40 0.153209 m (0.2 m)sin 40 0.128558 m x y = ° = = ° = / (0.153209 m) (0.128558 m)A O∴ = +r i j (a) (300 N)sin30 150 N (300 N)cos30 259.81 N x y F F = ° = = ° = (150 N) (259.81 N)= +F i j / (0.153209 0.128558 ) m (150 259.81 ) N (39.805 19.2837 ) N m (20.521 N m) O A O= × = + × + = − ⋅ = ⋅ M r F i j i j k k k 20.5 N mO = ⋅M (b) OM Fd= 20.521 N m (300 N)( ) 0.068403 m d d ⋅ = = 68.4 mmd = www.elsolucionario.net www.elsolucionario.net
- 169. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 167 PROBLEM 3.7 A 400-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into components along line OA and in a direction perpendicular to that line. (b) Determine the magnitude and direction of the smallest force Q applied at B that has the same moment as P about O. SOLUTION (a) Portion OA of crank: 90 30 40 20 θ θ = ° − ° − ° = ° / sin (400 N)sin 20 136.81 N M (0.2 m)(136.81 N) 27.362 N m O O A S P r S θ= = ° = = = = ⋅ 27.4 N mO = ⋅M (b) Smallest force Q must be perpendicular to OB. Portion OB of crank: / (0.120 m) O O B O M r Q M Q = = 27.362 N m (0.120 m)Q⋅ = 228 N=Q 42.0° www.elsolucionario.net www.elsolucionario.net
- 170. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 168 PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α = 10°, (c) the smallest force P that creates the same moment about B. SOLUTION (a) We have / (4 in.)(200 lb) 800 lb in. B C B NM r F= = = ⋅ or 800 lb in.BM = ⋅ (b) By definition, / sin 10 (180 70 ) 120 B A BM r P θ θ = = ° + ° − ° = ° Then 800 lb in. (18 in.) sin120P⋅ = × ° or 51.3 lbP = (c) For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus min / B A B M dP d r = = or min800 lb in. (18 in.)P⋅ = or min 44.4 lbP = min 44.4 lb=P 20° www.elsolucionario.net www.elsolucionario.net
- 171. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169 PROBLEM 3.9 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION Using (a): 1 1( ) ( ) 7 24 (2.24 in.) 500 lb (1.68 in.) 500 lb 25 25 1120 lb in. C AB x AB yM y F x F= + = × + × = ⋅ (a) 1.120 kip in.C = ⋅M Using (b): 2 ( ) 7 (8 in.) 500 lb 25 1120 lb in. C AB xM y F= = × = ⋅ (b) 1.120 kip in.C = ⋅M www.elsolucionario.net www.elsolucionario.net
- 172. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 170 PROBLEM 3.10 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION Using (a): 1 1( ) ( ) 7 24 (2.24 in.) 500 lb (1.68 in.) 500 lb 25 25 492.8 lb in. C AB x AB yM y F x F= − + = − × + × = + ⋅ (a) 493 lb in.C = ⋅M Using (b): 2 ( ) 7 (3.52 in.) 500 lb 25 492.8 lb in. C AB xM y F= = × = + ⋅ (b) 493 lb in.C = ⋅M www.elsolucionario.net www.elsolucionario.net
- 173. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 171 PROBLEM 3.11 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E. SOLUTION (a) Slope of line: 0.875 m 5 1.90 m 0.2 m 12 EC = = + Then 12 ( ) 13 ABx ABT T= 12 (1040 N) 13 960 N = = (a) and 5 (1040 N) 13 400 N AByT = = Then (0.875 m) (0.2 m) (960 N)(0.875 m) (400 N)(0.2 m) D ABx AByM T T= − = − 760 N m= ⋅ or 760 N mD = ⋅M (b) We have ( ) ( )D ABx ABxM T y T x= + (960 N)(0) (400 N)(1.90 m) 760 N m = + = ⋅ (b) or 760 N mD = ⋅M www.elsolucionario.net www.elsolucionario.net
- 174. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 172 PROBLEM 3.12 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D. SOLUTION Slope of line: 0.875 m 7 2.80 m 0.2 m 24 EC = = + Then 24 25 ABx ABT T= and 7 25 ABy ABT T= We have ( ) ( )D ABx AByM T y T x= + 24 7 960 N m (0) (2.80 m) 25 25 1224 N AB AB AB T T T ⋅ = + = or 1224 NABT = www.elsolucionario.net www.elsolucionario.net
- 175. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173 PROBLEM 3.13 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D. SOLUTION The minimum value of d can be found based on the equation relating the moment of the force ABT about D: max( ) ( )D AB yM T d= where 960 N mDM = ⋅ max max( ) sin (2400 N)sinAB y ABT T θ θ= = Now 2 2 2 2 0.875 m sin ( 0.20) (0.875) m 0.875 960 N m 2400 N ( ) ( 0.20) (0.875) d d d θ = + + ⋅ = + + or 2 2 ( 0.20) (0.875) 2.1875d d+ + = or 2 2 2 ( 0.20) (0.875) 4.7852d d+ + = or 2 3.7852 0.40 0.8056 0d d− − = Using the quadratic equation, the minimum values of d are 0.51719 m and 0.41151 m.− Since only the positive value applies here, 0.51719 md = or 517 mmd = www.elsolucionario.net www.elsolucionario.net
- 176. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 174 PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O. SOLUTION We have /C B C B= ×M r F Noting the direction of the moment of each force component about C is clockwise, C By BxM xF yF= + where 120 mm 65 mm 55 mm 72 mm 90 mm 162 mm x y = − = = + = and 2 2 2 2 65 (485 N) 325 N (65) (72) 72 (485 N) 360 N (65) (72) Bx By F F = = + = = + (55 mm)(360 N) (162)(325 N) 72450 N m 72.450 N m CM = + = ⋅ = ⋅ or 72.5 N mC = ⋅M www.elsolucionario.net www.elsolucionario.net
- 177. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 175 PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity 1 1 sin cos sin ( ) sin ( ). 2 2 α β α β α β= + + − SOLUTION Note: (cos sin ) (cos sin ) (cos sin ) B B C β β β β α α = + ′ = − = + B i j B i j C i j By definition, | | sin ( )BC α β× = −B C (1) | | sin ( )BC α β′× = +B C (2) Now (cos sin ) (cos sin )B Cβ β α α× = + × +B C i j i j (cos sin sin cos )BC β α β α= − k (3) and (cos sin ) (cos sin )B Cβ β α α′× = − × +B C i j i j (cos sin sin cos )BC β α β α= + k (4) Equating the magnitudes of ×B C from Equations (1) and (3) yields: sin( ) (cos sin sin cos )BC BCα β β α β α− = − (5) Similarly, equating the magnitudes of ′×B C from Equations (2) and (4) yields: sin( ) (cos sin sin cos )BC BCα β β α β α+ = + (6) Adding Equations (5) and (6) gives: sin( ) sin( ) 2cos sinα β α β β α− + + = or 1 1 sin cos sin( ) sin( ) 2 2 α β α β α β= + + − www.elsolucionario.net www.elsolucionario.net
- 178. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 176 PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k. SOLUTION (a) We have | |A = ×P Q where 7 3 3= − + −P i j k 2 2 5= + +Q i j k Then 7 3 3 2 2 5 [(15 6) ( 6 35) ( 14 6) ] (21) (29) ( 20) × = − − = + + − + + − − = + − i j k P Q i j k i j k 2 2 2 (20) (29) ( 20)A = + + − or 41.0A = (b) We have | |A = ×P Q where 6 5 2= − −P i j k 2 5 1= − + −Q i j k Then 6 5 2 2 5 1 [(5 10) (4 6) (30 10) ] (15) (10) (20) × = − − − − = + + + + − = + + i j k P Q i j k i j k 2 2 2 (15) (10) (20)A = + + or 26.9A = www.elsolucionario.net www.elsolucionario.net
- 179. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 177 PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k. SOLUTION (a) We have | | × = × A B λ A B where 1 2 5= + −A i j k 4 7 5= − −B i j k Then 1 2 5 4 7 5 ( 10 35) (20 5) ( 7 8) 15(3 1 1 ) × = + − − − = − − + + + − − = − − i j k A B i j k i j k and 2 2 2 | | 15 ( 3) ( 1) ( 1) 15 11× = − + − + − =A B 15( 3 1 1 ) 15 11 − − − = i j k λ or 1 ( 3 ) 11 = − − −λ i j k (b) We have | | × = × A B λ A B where 3 3 2= − +A i j k 2 6 4= − + −B i j k Then 3 3 2 2 6 4 (12 12) ( 4 12) (18 6) (8 12 ) × = − − − = − + − + + − = + i j k A B i j k j k and 2 2 | | 4 (2) (3) 4 13× = + =A B 4(2 3 ) 4 13 + = j k λ or 1 (2 3 ) 13 = +λ j k www.elsolucionario.net www.elsolucionario.net
- 180. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178 PROBLEM 3.18 A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates. SOLUTION 2 2 [20 m ( 1 m)] [16 m ( 4 m)] 29 m ABd = − − + − − = Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B. Then ABF=F λ where 1 (21 20 ) 29 B A AB ABd − = = + r r i j λ By definition, | |O A dF= × =M r F where (1 m) (4 m)A = − −r i j Then [ ( 1 m) (4 m) ] [(21 m) (20 m) ] 29 m [ (20) (84) ] 29 64 N m 29 O F F F = − − − × + = − + = ⋅ M i j i j k k k Finally, 64 ( ) 29 F d F = 64 m 29 d = 2.21 md = www.elsolucionario.net www.elsolucionario.net
- 181. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179 PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k. SOLUTION O = ×M r F (a) 2 3 4 4 3 5 O = − − i j k M (15 12) ( 16 10) ( 6 12)= − + − − + − −i j k 3 26 18O = − −M i j k (b) 8 6 10 4 3 5 O = − − − i j k M (30 30) ( 40 40) (24 24)= − + − + + −i j k 0O =M (c) 8 6 5 4 3 5 O = − − i j k M ( 30 15) (20 40) ( 24 24)= − + + − + − +i j k 15 20O = − −M i j Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional. www.elsolucionario.net www.elsolucionario.net
- 182. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180 PROBLEM 3.20 Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k. SOLUTION O = ×M r F (a) 3 6 5 2 3 4 O = − − i j k M (24 15) (10 12) (9 12)= − + + + +i j k 9 22 21O = + +M i j k (b) 1 4 2 2 3 4 O = − − − i j k M (16 6) ( 4 4) (3 8)= + + − + + +i j k 22 11O = +M i k (c) 4 6 8 2 3 4 O = − − i j k M ( 24 24) ( 16 16) (12 12)= − + + − + + −i j k 0O =M Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional. www.elsolucionario.net www.elsolucionario.net
- 183. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 181 PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION 2 2 2 (0.21 m) (0.16 m) (0.12 m) (0.21 m) ( 0.16 m) (0.12 m) 0.29 m AE AE = − + = + − + = i j k (a) 0.21 0.16 0.12 (435 N) 0.29 (315 N) (240 N) (180 N) A A AE AE F F AE = = − + = = − + F i j k i j k λ / (0.09 m) (0.16 m)A O = − +r i j 0.09 0.16 0 315 240 180 O = − − i j k M 28.8 16.20 (21.6 50.4)= + + −i j k (28.8 N m) (16.20 N m) (28.8 N m)O = ⋅ + ⋅ − ⋅M i j k (b) (315 N) (240 N) (180 N)E A= − = − + −F F i j k / (0.12 m) (0.12 m)E O = +r i k 0.12 0 0.12 315 240 180 O = − − i j k M 28.8 ( 37.8 21.6) 28.8= − + − + +i j k (28.8 N m) (16.20 N m) (28.8 N m)O = − ⋅ − ⋅ + ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 184. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 182 PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION We have 2A AB AD= +R F F where (82 lb)AB = −F j and 6 7.75 3 (82 lb) 10.25 (48 lb) (62 lb) (24 lb) AD AD AD AD AD − − = = = − − i j k F F F i j k Thus 2 (48 lb) (226 lb) (24 lb)A AB AD= + = − −R F F i j k Also / (7.75 ft) (3 ft)A C = +r j k Using Eq. (3.21): 0 7.75 3 48 226 24 (492 lb ft) (144.0 lb ft) (372 lb ft) C = − − = ⋅ + ⋅ − ⋅ i j k M i j k (492 lb ft) (144.0 lb ft) (372 lb ft)C = ⋅ + ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 185. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183 PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION We have (6 lb)cos 8 5.9416 lbxzT = ° = Then sin 30 2.9708 lb sin 8 0.83504 lb cos 30 5.1456 lb x xz y BC z xz T T T T T T = ° = = ° = − = ° = − Now /A B A BC= ×M r T where / (6sin 45 ) (6cos 45 ) 6 ft ( ) 2 B A = ° − ° = − r j k j k Then 6 0 1 1 2 2.9708 0.83504 5.1456 6 6 6 ( 5.1456 0.83504) (2.9708) (2.9708) 2 2 2 A = − − − = − − − − i j k M i j k or (25.4 lb ft) (12.60 lb ft) (12.60 lb ft)A = − ⋅ − ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 186. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 184 PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION where 900 N BD F F BD = =F 2 2 2 (1 m) (2 m) (2 m) ( 1 m) ( 2 m) (2 m) 3 m BD BD = − − + = − + − + = i j k / 2 2 (900 N) 3 (300 N) (600 N) (600 N) (2.5 m) (2 m)B O − − + = = − − + = + i j k F i j k r i j /O B O= ×M r F 2.5 2 0 300 600 600 1200 1500 ( 1500 600) = − − = − + − + i j k i j k (1200 N m) (1500 N m) (900 N m)O = ⋅ − ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 187. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 185 PROBLEM 3.25 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. SOLUTION We have /A C A C= ×M r F where / (0.06 m) (0.075 m) (200 N)cos 30 (200 N)sin 30 C A C = + = − ° + ° r i j F j k Then 200 0.06 0.075 0 0 cos30 sin 30 200[(0.075sin 30 ) (0.06sin 30 ) (0.06cos 30 ) ] A = − ° ° = ° − ° − ° i j k M i j k or (7.50 N m) (6.00 N m) (10.39 N m)A = ⋅ − ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 188. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 186 PROBLEM 3.26 The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B. SOLUTION First note 2 2 2 ( 6) (2.4) ( 4) 7.6 m BCd = − + + − = Then 2.5 kN ( 6 2.4 4 ) 7.6 BC = − + −T i j k We have /A B A BC= ×M r T where / (6 m)B A =r i Then 2.5 kN (6 m) ( 6 2.4 4 ) 7.6 A = × − + −M i i j k or (7.89 kN m) (4.74 kN m)A = ⋅ + ⋅M j k www.elsolucionario.net www.elsolucionario.net
- 189. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187 PROBLEM 3.27 In Prob. 3.21, determine the perpendicular distance from point O to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION From the solution to Prob. 3.21 2 2 2 (28.8 N m) (16.20 N m) (28.8 N m) (28.8) (16.20) (28.8) 43.8329 N m O OM = ⋅ + ⋅ − ⋅ = + + = ⋅ M i j k But or O O A A M M F d d F = = 43.8329 N m 435 N 0.100765 m d ⋅ = = 100.8 mmd = www.elsolucionario.net www.elsolucionario.net
- 190. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 188 PROBLEM 3.28 In Prob. 3.21, determine the perpendicular distance from point B to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION From the solution to Prob. 3.21 / / (315 N) (240 N) (180 N) (0.210 m) 0.21 (315 240 180 ) A A B B A B A = − + = − = × = − × − + F i j k r i M r F i i j k 50.4 37.8= +k j 2 2 (50.4) (37.8) 63.0 N m BM = + = ⋅ or B B A A M M F d d F = = 63.0 N m 435 N 0.144829 m d ⋅ = = 144.8 mmd = www.elsolucionario.net www.elsolucionario.net
- 191. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189 PROBLEM 3.29 In Problem 3.22, determine the perpendicular distance from point C to portion AD of the line ABAD. PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION First compute the moment about C of the force DAF exerted by the line on D: From Problem 3.22: / 2 2 (48 lb) (62 lb) (24 lb) (6 ft) [ (48 lb) (62 lb) (24 lb) ] (144 lb ft) (372 lb ft) (144) (372) 398.90 lb ft DA AD C D C DA C = − = − + + = × = + × − + + = − ⋅ + ⋅ = + = ⋅ F F i j k M r F i i j k j k M Then C DAd=M F Since 82 lbDAF = 398.90 lb ft 82 lb C DA M d F = ⋅ = 4.86 ftd = www.elsolucionario.net www.elsolucionario.net
- 192. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190 PROBLEM 3.30 In Prob. 3.23, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION From the solution to Prob. 3.23: 2 2 2 (25.4 lb ft) (12.60 lb ft) (12.60 lb ft) ( 25.4) ( 12.60) ( 12.60) 31.027 lb ft A AM = − ⋅ − ⋅ − ⋅ = − + − + − = ⋅ M i j k or A A BC BC M M T d d T = = 31.027 lb ft 6 lb 5.1712 ft ⋅ = = 5.17 ftd = www.elsolucionario.net www.elsolucionario.net
- 193. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 191 PROBLEM 3.31 In Prob. 3.23, determine the perpendicular distance from point D to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION 6 ft 6 lbBC AB T = = We have | |D BCT d=M where d = perpendicular distance from D to line BC. / / (6sin 45 ft) (4.2426 ft)D B D BC B D= × = ° =M r T r j : ( ) (6 lb)cos8 sin30 2.9708 lbBC BC xT = ° ° =T ( ) (6 lb)sin8 0.83504 lb ( ) (6 lb)cos8 cos30 5.1456 lb BC y BC z T T = − ° = − = − ° ° = − (2.9708 lb) (0.83504 lb) (5.1456 lb) 0 4.2426 0 2.9708 0.83504 5.1456 (21.831 lb ft) (12.6039 lb ft) BC D = − − = − − = − ⋅ − ⋅ T i j k i j k M i 2 2 | | ( 21.831) ( 12.6039) 25.208 lb ftDM = − + − = ⋅ 25.208 lb ft (6 lb)d⋅ = 4.20 ftd = www.elsolucionario.net www.elsolucionario.net
- 194. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 192 PROBLEM 3.32 In Prob. 3.24, determine the perpendicular distance from point O to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION From the solution to Prob. 3.24 we have 2 2 2 (1200 N m) (1500 N m) (900 N m) (1200) ( 1500) ( 900) 2121.3 N m O O O O M M M Fd d F = ⋅ − ⋅ − ⋅ = + − + − = ⋅ = = M i j k 2121.3 N m 900 N ⋅ = 2.36 md = www.elsolucionario.net www.elsolucionario.net
- 195. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 193 PROBLEM 3.33 In Prob. 3.24, determine the perpendicular distance from point C to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION From the solution to Prob. 3.24 we have / / (300 N) (600 N) (600 N) (2 m) (2 m) ( 300 N 600 N 600 N ) B C C B C = − − + = = × = × − − + F i j k r j M r F j i j k (600 N m) (1200 N m)= ⋅ + ⋅k i 2 2 (600) (1200) 1341.64 N mCM = + = ⋅ C C M M Fd d F = = 1341.64 N m 900 N ⋅ = 1.491 md = www.elsolucionario.net www.elsolucionario.net
- 196. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 194 PROBLEM 3.34 Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B. SOLUTION Assuming a force F acts along AB, /| | | | ( )C A C F d= × =M r F where d = perpendicular distance from C to line AB 2 2 2 / (24 ft) (24 ft) (28 ft) (24) (24) (28) ft (6) (6) (7) 11 (3 ft) (10 ft) ( 10 ft) 3 10 10 11 6 6 7 [(10 6 ) (81 6 ) 78 ] 11 AB A C C F F F a F a F a a = + − = + + = + − = − − − = − − = + + − + F λ i j k i j k r i j k i j k M i j k Since 2 2 2 / /| | | | or | | ( )C A C A C dF= × × =M r F r F 2 2 2 21 (10 6 ) (81 6 ) (78) 121 a a d+ + − + = Setting 2 ( ) 0d da d = to find a to minimize d: 1 [2(6)(10 6 ) 2( 6)(81 6 )] 0 121 a a+ + − − = Solving 5.92 fta = or 5.92 fta = www.elsolucionario.net www.elsolucionario.net
- 197. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 195 PROBLEM 3.35 Given the vectors P = 3i − j + 2k, Q = 4i + 5j − 3k, and S = −2i + 3j − k, compute the scalar products P · Q, P · S, and Q · S. SOLUTION (3 2 ) (4 5 3 ) (3)(4) ( 1)(5) (2)( 3) 12 5 6 ⋅ = − + ⋅ + − = + − + − = − − P Q i j k i j k 1⋅ = +P Q (3 2 ) ( 2 3 ) (3)( 2) ( 1)(3) (2)( 1) 6 3 2 ⋅ = − + ⋅ − + − = − + − + − = − − − P S i j k i j k 11⋅ = −P S (4 5 3 ) ( 2 3 ) (4)( 2) (5)(3) ( 3)( 1) 8 15 3 ⋅ = + − ⋅ − + − = − + + − − = − + + Q S i j k i j k 10⋅ = +Q S www.elsolucionario.net www.elsolucionario.net
- 198. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 196 PROBLEM 3.36 Form the scalar product B · C and use the result obtained to prove the identity cos (α − β) = cos α cos β + sin α sin β . SOLUTION cos sinB Bα α= +B i j (1) cos sinC Cβ β= +C i j (2) By definition: cos( )BC α β⋅ = −B C (3) From (1) and (2): ( cos sin ) ( cos sin ) (cos cos sin sin ) B B C C BC α α β β α β α β ⋅ = + ⋅ + = + B C i j i j (4) Equating the right-hand members of (3) and (4), cos( ) cos cos sin sinα β α β α β− = + www.elsolucionario.net www.elsolucionario.net
- 199. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 197 PROBLEM 3.37 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC. SOLUTION First note: 2 2 2 2 2 2 ( 6.5) ( 8) (2) 10.5 ft (0) ( 8) (6) 10 ft AB AC = − + − + = = + − + = and (6.5 ft) (8 ft) (2 ft) (8 ft) (6 ft) AB AC = − − + = − + i j k j k By definition, ( )( )cosAB AC AB AC θ⋅ = or ( 6.5 8 2 ) ( 8 6 ) (10.5)(10)cos ( 6.5)(0) ( 8)( 8) (2)(6) 105cos θ θ − − + ⋅ − + = − + − − + = i j k j k or cos 0.72381θ = or 43.6θ = ° www.elsolucionario.net www.elsolucionario.net
- 200. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 198 PROBLEM 3.38 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD. SOLUTION First note: 2 2 2 2 2 2 (0) ( 8) (6) 10 ft (4) ( 8) (1) 9 ft AC AD = + − + = = + − + = and (8 ft) (6 ft) (4 ft) (8 ft) (1 ft) AC AD = − + = − + j k i j k By definition, ( )( )cosAC AD AC AD θ⋅ = or ( 8 6 ) (4 8 ) (10)(9)cosθ− + ⋅ − + =j k i j k (0)(4) ( 8)( 8) (6)(1) 90cosθ+ − − + = or cos 0.77778θ = or 38.9θ = ° www.elsolucionario.net www.elsolucionario.net
- 201. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 199 PROBLEM 3.39 Three cables are used to support a container as shown. Determine the angle formed by cables AB and AD. SOLUTION First note: 2 2 2 2 2 (450 mm) (600 mm) 750 mm ( 500 mm) (600 mm) (360 mm) 860 mm AB AD = + = = − + + = and (450 mm) (600 mm) ( 500 mm) (600 mm) (360 mm) AB AD = + = − + + i j i j k By definition, ( )( )cosAB AD AB AD θ⋅ = (450 600 ) ( 500 600 360 ) (750)(860)cosθ+ ⋅ − − + =i j i j k (450)( 500) (600)(600) (0)(360) (750)(860)cosθ− + + = or cos 0.20930θ = 77.9θ = ° www.elsolucionario.net www.elsolucionario.net
- 202. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 200 PROBLEM 3.40 Three cables are used to support a container as shown. Determine the angle formed by cables AC and AD. SOLUTION First note: 2 2 2 2 2 (600 mm) ( 320 mm) 680 mm ( 500 mm) (600 mm) (360 mm) 860 mm AC AD = + − = = − + + = and (600 mm) ( 320 mm) ( 500 mm) (600 mm) (360 mm) AC AD = + − = − + + j k i j k By definition, ( )( )cosAC AD AC AD θ⋅ = (600 320 ) ( 500 600 360 ) (680)(860)cosθ− ⋅ − + + =j k i j k 0( 500) (600)(600) ( 320)(360) (680)(860)cosθ− + + − = cos 0.41860θ = 65.3θ = ° www.elsolucionario.net www.elsolucionario.net
- 203. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 201 PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC. SOLUTION (a) Using Eq. (3.32): 2 2 2 2 2 2 11 12 24 (11) ( 12) (24) 29 in. 31 12 24 (31) ( 12) (24) 41 in. CA CA CB CB = − + = + − + = = − + = + − + = i j k i j k cos ( )( ) (11 12 24 ) (31 12 24 ) (29)(41) (11)(31) ( 12)( 12) (24)(24) (29)(41) 0.89235 CA CB CA CB θ ⋅ = − + ⋅ − + = + − − + = = i j k i j k 26.8θ = ° (b) Law of cosines: 2 2 2 2 2 2 ( ) ( ) ( ) 2( )( )cos (20) (29) (41) 2(29)(41)cos cos 0.89235 AB CA CB CA CB θ θ θ = + − = + − = 26.8θ = ° www.elsolucionario.net www.elsolucionario.net
- 204. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 202 PROBLEM 3.42 Solve Prob. 3.41 for the position corresponding to x = 4 in. PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC. SOLUTION (a) Using Eq. (3.32): 2 2 2 2 2 2 4 12 24 (4) ( 12) (24) 27.129 in. 24 12 24 (24) ( 12) (24) 36 in. CA CA CB CB = − + = + − + = = − + = + − + = i j k i j k cos ( )( ) (4 12 24 ) (24 12 24 ) (27.129)(36) 0.83551 CA CB CA CB θ ⋅ = − + ⋅ − + = = i j k i j k 33.3θ = ° (b) Law of cosines: 2 2 2 2 2 2 ( ) ( ) ( ) 2( )( )cos (20) (27.129) (36) 2(27.129)(36)cos cos 0.83551 AB CA CB CA CB θ θ θ = + − = + − = 33.3θ = ° www.elsolucionario.net www.elsolucionario.net
- 205. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 203 PROBLEM 3.43 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B. SOLUTION First note: 2 2 2 2 2 2 ( 3) (3) ( 1.5) 4.5 m ( 0.08) (0.38) (0.16) 0.42 m BA BD = − + + − = = − + + = Then ( 3 3 1.5 ) 4.5 ( 2 2 ) 3 1 ( 0.08 0.38 0.16 ) 0.42 1 ( 4 19 8 ) 21 BA BA BA BD T T BD BD = − + − = − + − = = − + + = − + + T i j k i j k i j k i j k λ (a) We have cosBA BD BAT θ⋅ =T λ or 1 ( 2 2 ) ( 4 19 8 ) cos 3 21 BA BA T T θ− + − ⋅ − + + =i j k i j k or 1 cos [( 2)( 4) (2)(19) ( 1)(8)] 63 0.60317 θ = − − + + − = or 52.9θ = ° (b) We have ( ) cos (540 N)(0.60317) BA BD BA BD BA T T θ = ⋅ = = T λ or ( ) 326 NBA BDT = www.elsolucionario.net www.elsolucionario.net
- 206. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 204 PROBLEM 3.44 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at Point B. SOLUTION First note: 2 2 2 2 2 2 (1) (3) ( 1.5) 3.5 m ( 0.08) (0.38) (0.16) 0.42 m BC BD = + + − = = − + + = ( 3 1.5 ) 3.5 (2 6 3 ) 7 1 ( 0.08 0.38 0.16 ) 0.42 1 ( 4 19 8 ) 21 BC BC BC BD T T BD BD λ = + − = + − = = − + + = − + + T i j k i j k i j k i j k (a) cosBC BD BCTλ θ⋅ =T 1 (2 6 3 ) ( 4 19 8 ) cos 7 21 BC BC T T θ+ − ⋅ − + + =i j k i j k 1 cos [(2)( 4) (6)(19) ( 3)(8)] 147 0.55782 θ = − + + − = 56.1θ = ° (b) ( ) cos (490 N)(0.55782) BC BD BC BD BC T T λ θ = ⋅ = = T ( ) 273 NBC BDT = www.elsolucionario.net www.elsolucionario.net
- 207. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 205 PROBLEM 3.45 Given the vectors 4 2 3 , 2 4 5 ,= − + = + −P i j k Q i j k and 2 ,xS= − +S i j k determine the value of xS for which the three vectors are coplanar. SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to ( ).×Q S ( ) 0⋅ × =P Q S (or, the volume of a parallelepiped defined by P, Q, and S is zero). Then 4 2 3 2 4 5 0 1 2xS − − = − or 32 10 6 20 8 12 0x xS S+ − − + − = 7xS = www.elsolucionario.net www.elsolucionario.net
- 208. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 206 PROBLEM 3.46 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k, (b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k. SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a) 3 Vol. ( ) 4 3 2 2 5 1 in. 7 1 1 (20 21 4 70 6 4) 67 = ⋅ × − = − − − = − − + + − = P Q S or Volume 67.0= (b) 3 Vol. ( ) 5 1 6 2 3 1 in. 3 2 4 (60 3 24 54 8 10) 111 = ⋅ × − = − − = + − + + + = P Q S or Volume 111.0= www.elsolucionario.net www.elsolucionario.net
- 209. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 207 PROBLEM 3.47 Knowing that the tension in cable AB is 570 N, determine the moment about each of the coordinate axes of the force exerted on the plate at B. SOLUTION 2 2 2 ( 900 mm) (600 mm) (360 mm) ( 900) (600) (360) 1140 mm 900 600 360 (570 N) 1140 (450 N) (300 N) (180 N) (0.9 m) B B B BA BA BA F BA = − + + = − + + = = − + + = = − + + = i j k F i j k i j k r i 0.9 ( 450 300 180 )O B B= × = × − + +M r F i i j k 270 162= −k j (162 N m) (270 N m) O x y zM M M= + + = − ⋅ + ⋅ M i j k j k Therefore, 0, 162.0 N m, 270 N mx y zM M M= = − ⋅ = + ⋅ www.elsolucionario.net www.elsolucionario.net
- 210. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 208 PROBLEM 3.48 Knowing that the tension in cable AC is 1065 N, determine the moment about each of the coordinate axes of the force exerted on the plate at C. SOLUTION 2 2 2 ( 900 mm) (600 mm) ( 920 mm) ( 900) (600) ( 920) 1420 mm 900 600 920 (1065 N) 1420 (675 N) (450 N) (690 N) (0.9 m) (1.28 m) C C C CA CA CA F CA = − + + − = − + + − = = − + − = = − + − = + i j k F i j k i j k r i k Using Eq. (3.19): 0.9 0 1.28 675 450 690 (576 N m) (243 N m) (405 N m) O C C O = × = − − = − ⋅ − ⋅ + ⋅ i j k M r F M i j k But O x y zM M M= + +M i j k Therefore, 576 N m, 243 N m, 405 N mx y zM M M= − ⋅ = − ⋅ = + ⋅ www.elsolucionario.net www.elsolucionario.net
- 211. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 209 PROBLEM 3.49 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z-axis of the resultant force AR exerted on the davit at A must not exceed 279 lb⋅ft in absolute value. Determine the largest allowable tension in line ABAD when 6x = ft. SOLUTION First note: 2A AB AD= +R T T Also note that only TAD will contribute to the moment about the z-axis. Now 2 2 2 (6) ( 7.75) ( 3) 10.25 ft AD = + − + − = Then (6 7.75 3 ) 10.25 AD AD T AD T = = − − T i j k Now /( )z A C ADM = ⋅ ×k r T where / (7.75 ft) (3 ft)A C = +r j k Then for max,T max max 0 0 1 279 0 7.75 3 10.25 6 7.75 3 | (1)(7.75)(6)| 10.25 T T = − − = − or max 61.5 lbT = www.elsolucionario.net www.elsolucionario.net
- 212. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 210 PROBLEM 3.50 For the davit of Problem 3.49, determine the largest allowable distance x when the tension in line ABAD is 60 lb. SOLUTION From the solution of Problem 3.49, ADT is now 2 2 2 60 lb ( 7.75 3 ) ( 7.75) ( 3) AD AD T AD x x = = − − + − + − T i j k Then /( )z A C ADM = ⋅ ×k r T becomes 2 2 2 2 2 2 0 0 1 60 279 0 7.75 3 ( 7.75) ( 3) 7.75 3 60 279 | (1)(7.75)( ) | 69.0625 279 69.0625 465 0.6 69.0625 x x x x x x x x = + − + − − − = − + + = + = Squaring both sides: 2 2 0.36 24.8625x x+ = 2 38.848x = 6.23 ftx = www.elsolucionario.net www.elsolucionario.net
- 213. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 211 PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have : ( ) ( ) ( ) ( )x AB z A DE z D xM T y T y MΣ + = where ( ) ( ) 12 12 255 lb 17 180 lb AB z AB AB AB T T λ = ⋅ = ⋅ − − + = ⋅ = k T k i j k k ( ) ( ) 1.5 14 12 18.5 0.64865 12 ft 14 ft 4728 lb ft DE z DE DE DE DE DE A D x T T T T y y M λ = ⋅ = ⋅ − + = ⋅ = = = = ⋅ k T k i j k k (180 lb)(12 ft) (0.64865 )(14 ft) 4728 lb ftDET+ = ⋅ and 282.79 lbDET = or 283 lbDET = www.elsolucionario.net www.elsolucionario.net
- 214. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212 PROBLEM 3.52 Solve Problem 3.51 when the tension in cable AB is 306 lb. PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have : ( ) ( ) ( ) ( )x AB z A DE z D xM T y T y MΣ + = Where ( ) ( ) 12 12 306 lb 17 216 lb AB z AB AB AB T T = ⋅ = ⋅ − − + = ⋅ = k T k λ i j k k ( ) ( ) 1.5 14 12 18.5 0.64865 12 ft 14 ft 4728 lb ft DE z DE DE DE DE DE A D x T T T T y y M = ⋅ = ⋅ − + = ⋅ = = = = ⋅ k T k λ i j k k (216 lb)(12 ft) (0.64865 )(14 ft) 4728 lb ftDET+ = ⋅ and 235.21 lbDET = or 235 lbDET = www.elsolucionario.net www.elsolucionario.net
- 215. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 213 PROBLEM 3.53 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Knowing that Mx = +20 N · m and My = −8.75 N · m, and Mz = −30 N · m, determine the magnitude of P and the values of φ and θ. SOLUTION (0.25 m) (0.2 m)sin (0.2 m)cos sin cos 0.25 0.2sin 0.2cos 0 sin cos C O C P P P P θ θ φ φ θ θ φ φ = + + = − + = × = − r i j k P j k i j k M r P Expanding the determinant, we find (0.2) (sin cos cos sin )xM P θ φ θ φ= + (0.2) sin( )xM P θ φ= + (1) (0.25) cosyM P φ= − (2) (0.25) sinzM P φ= − (3) Dividing Eq. (3) by Eq. (2) gives: tan z y M M φ = (4) 30 N m tan 8.75 N m φ − ⋅ = − ⋅ 73.740φ = 73.7φ = ° Squaring Eqs. (2) and (3) and adding gives: 2 2 2 2 2 2 (0.25) or 4y z y zM M P P M M+ = = + (5) 2 2 4 (8.75) (30) 125.0 N P = + = 125.0 NP = Substituting data into Eq. (1): ( 20 N m) 0.2 m(125.0 N)sin( ) ( ) 53.130 and ( ) 126.87 20.6 and 53.1 θ φ θ φ θ φ θ θ + ⋅ = + + = ° + = ° = − ° = ° 53.1Q = ° www.elsolucionario.net www.elsolucionario.net
- 216. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 214 PROBLEM 3.54 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Determine the moment Mx of P about the x-axis when θ = 65°, knowing that My = −15 N · m and Mz = −36 N · m. SOLUTION See the solution to Prob. 3.53 for the derivation of the following equations: (0.2) sin( )xM P θ φ= + (1) tan z y M M φ = (4) 2 2 4 y zP M M= + (5) Substituting for known data gives: 2 2 36 N m tan 15 N m 67.380 4 ( 15) ( 36) 156.0 N 0.2 m(156.0 N)sin(65 67.380 ) 23.047 N m x P P M φ φ − ⋅ = − ⋅ = ° = − + − = = ° + ° = ⋅ 23.0 N mxM = ⋅ www.elsolucionario.net www.elsolucionario.net
- 217. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215 PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note: AE AE AE T AE =T 2 2 2 (0.9) ( 0.6) (0.2) 1.1 mAE = + − + = Then 55 N (0.9 0.6 0.2 ) 1.1 5[(9 N) (6 N) (2 N) ] AE = − + = − + T i j k i j k Also, 2 2 2 (1.2) ( 0.35) (0) 1.25 m DB = + − + = Then 1 (1.2 0.35 ) 1.25 1 (24 7 ) 25 DB DB DB = = − = − i j i j λ Now /( )DB DB A D AEM = ⋅ ×r Tλ where / (0.1 m) (0.2 m)A D = − +r j k Then 24 7 0 1 (5) 0 0.1 0.2 25 9 6 2 1 ( 4.8 12.6 28.8) 5 DBM − = − − = − − + or 2.28 N mDBM = ⋅ www.elsolucionario.net www.elsolucionario.net
- 218. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 216 PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note: CF CF CF T CF =T 2 2 2 (0.6) ( 0.9) ( 0.2) 1.1 mCF = + − + − = Then 33 N (0.6 0.9 0.2 ) 1.1 3[(6 N) (9 N) (2 N) ] CF = − + = − − T i j k i j k Also, 2 2 2 (1.2) ( 0.35) (0) 1.25 m DB = + − + = Then 1 (1.2 0.35 ) 1.25 1 (24 7 ) 25 DB DB DB = = − = − i j i j λ Now /( )DB DB C D CFM = ⋅ ×r Tλ where / (0.2 m) (0.4 m)C D = −r j k Then 24 7 0 1 (3) 0 0.2 0.4 25 6 9 2 3 ( 9.6 16.8 86.4) 25 DBM − = − − − = − + − or 9.50 N mDBM = − ⋅ www.elsolucionario.net www.elsolucionario.net
- 219. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217 PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P. SOLUTION (32 in.) (30 in.) (24 in.)AB = − −i j k 2 2 2 (32) ( 30) ( 24) 50 in.AB = + − + − = 0.64 0.60 0.48AB AB AB = = − −i k λ We shall apply the force P at Point G: / (5 in.) (30 in.)G B = +r i k (21 in.) (38 in.) (18 in.)DG = − +i j k 2 2 2 (21) ( 38) (18) 47 in.DG = + − + = 21 38 18 (235 lb) 47 DG P DG − + = = i j k P (105 lb) (190 lb) (90 lb)= − +P i j k The moment of P about AB is given by Eq. (3.46): / 0.64 0.60 0.48 ( ) 5 in. 0 30 in. 105 lb 190 lb 90 lb AB AB G B P − − = ⋅ × = − M rλ 0.64[0 (30 in.)( 190 lb)] 0.60[(30 in.)(105 lb) (5 in.)(90 lb)] 0.48[(5 in.)( 190 lb) 0] 2484 lb in. AB = − − − − − − − = + ⋅ M 207 lb ftAB = + ⋅M www.elsolucionario.net www.elsolucionario.net
- 220. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218 PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. SOLUTION (32 in.) (30 in.) (24 in.)AB = − −i j k 2 2 2 (32) ( 30) ( 24) 50 in.AB = + − + − = 0.64 0.60 0.48AB AB AB = = − −i j k λ We shall apply the force Q at Point H: / (32 in.) (17 in.)H B = − +r i j (16 in.) (21 in.) (12 in.)DH = − − −i j k 2 2 2 (16) ( 21) ( 12) 29 in.DH = + − + − = 16 21 12 (174 lb) 29 DH DH − − − = = i j k Q (96 lb) (126 lb) (72 lb)Q = − − −i j k The moment of Q about AB is given by Eq. (3.46): / 0.64 0.60 0.48 ( ) 32 in. 17 in. 0 96 lb 126 lb 72 lb AB AB H B − − = ⋅ × = − − − − M r Qλ 0.64[(17 in.)( 72 lb) 0] 0.60[(0 ( 32 in.)( 72 lb)] 0.48[( 32 in.)( 126 lb) (17 in.)( 96 lb)] 2119.7 lb in. AB = − − − − − − − − − − − = − ⋅ M 176.6 lb ftAB = ⋅M www.elsolucionario.net www.elsolucionario.net
- 221. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 219 PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION /( )AD AD B A BHM = ⋅ ×r Tλ Where / 1 (4 3 ) 5 (0.5 m) AD B A = − = i k r i λ and 2 2 2 (0.375) (0.75) ( 0.75) 1.125 m BHd = + + − = Then 450 N (0.375 0.75 0.75 ) 1.125 (150 N) (300 N) (300 N) BH = + − = + − T i j k i j k Finally, 4 0 3 1 0.5 0 0 5 150 300 300 1 [( 3)(0.5)(300)] 5 ADM − = − = − or 90.0 N mADM = − ⋅ www.elsolucionario.net www.elsolucionario.net
- 222. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 220 PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION /( )AD AD B A BGM = ⋅ ×r Tλ Where / 1 (4 3 ) 5 (0.5 m) AD B A = − = i k r j λ and 2 2 2 ( 0.5) (0.925) ( 0.4) 1.125 m BG = − + + − = Then 450 N ( 0.5 0.925 0.4 ) 1.125 (200 N) (370 N) (160 N) BG = − + − = − + − T i j k i j k Finally, 4 0 3 1 0.5 0 0 5 200 370 160 ADM − = − − 1 [( 3)(0.5)(370)] 5 = − 111.0 N mADM = − ⋅ www.elsolucionario.net www.elsolucionario.net
- 223. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 221 PROBLEM 3.61 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA. SOLUTION We have /( )OA OA C OM = ⋅ ×r Pλ From triangle OBC: ( ) 2 1 ( ) ( ) tan30 2 3 2 3 x z x a OA a a OA OA = = ° = = Since 2 2 2 2 ( ) ( ) ( ) ( )x y zOA OA OA OA= + + or 22 2 2 2 2 2 ( ) 2 2 3 2 ( ) 4 12 3 y y a a a OA a a OA a a = + + = − − = Then / 2 2 3 2 3 A O a a a= + +r i j k and 1 2 1 2 3 2 3 OA = + +i j kλ / ( sin30 ) ( cos30 ) ( ) ( 3 ) 2 BC C O a a P P P a a ° − ° = = = − = i k P i k r i λ 1 2 1 2 3 2 3 ( ) 1 0 0 2 1 0 3 2 (1)( 3) 2 3 2 OA P M a aP aP = − = − − = 2 OA aP M = www.elsolucionario.net www.elsolucionario.net
- 224. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 222 PROBLEM 3.62 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.61 to determine the perpendicular distance between edges OA and BC. SOLUTION (a) For edge OA to be perpendicular to edge BC, 0OA BC⋅ = From triangle OBC: ( ) 2 1 ( ) ( ) tan30 2 3 2 3 ( ) 2 2 3 x z x y a OA a a OA OA a a OA OA = = ° = = = + + i j k and ( sin30 ) ( cos30 ) 3 ( 3 ) 2 2 2 BC a a a a a = ° − ° = − = − i k i k i k Then ( ) ( 3 ) 0 2 22 3 y a a a OA + + ⋅ − = i j k i k or 2 2 ( ) (0) 0 4 4 0 y a a OA OA BC + − = ⋅ = so that OA is perpendicular to .BC (b) We have ,OAM Pd= with P acting along BC and d the perpendicular distance from OA to .BC From the results of Problem 3.57, 2 2 OA Pa M Pa Pd = = or 2 a d = www.elsolucionario.net www.elsolucionario.net
- 225. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 223 PROBLEM 3.63 Two forces 1F and 2F in space have the same magnitude F. Prove that the moment of 1F about the line of action of 2F is equal to the moment of 2F about the line of action of 1F . SOLUTION First note that 1 1 1 2 2 2andF F= =F Fλ λ Let 1 2moment ofM = F about the line of action of 1F and 2 momentM = of 1F about the line of action of 2F . Now, by definition, 1 1 / 2 1 / 2 2 2 2 / 1 2 / 1 1 ( ) ( ) ( ) ( ) B A B A A B A B M F M F = ⋅ × = ⋅ × = ⋅ × = ⋅ × r F r r F r λ λ λ λ λ λ Since 1 2 / / 1 1 / 2 2 2 / 1 and ( ) ( ) A B B A B A B A F F F M F M F = = = − = ⋅ × = ⋅ − × r r r r λ λ λ λ Using Equation (3.39): 1 / 2 2 / 1( ) ( )B A B A⋅ × = ⋅ − ×r rλ λ λ λ so that 2 1 / 2( )B AM F= ⋅ ×rλ λ 12 21M M= www.elsolucionario.net www.elsolucionario.net
- 226. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 224 PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION From the solution to Problem 3.55: 55 N 5[(9 N) (6 N) (2 N) ] AE AE = = − +T i j k Τ | | 2.28 N mDBM = ⋅ 1 (24 7 ) 25 DB = −i jλ Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will contribute to the moment of TAE about line .DB Now parallel( ) 1 5(9 6 2 ) (24 7 ) 25 1 [(9)(24) ( 6)( 7)] 5 51.6 N AE AE DBT = ⋅ = − + ⋅ − = + − − = T i j k i j λ Also, parallel perpendicular( ) ( )AE AE AE= +T T T so that 2 2 perpendicular( ) (55) (51.6) 19.0379 NAE = + =T Since DBλ and perpendicular( )AET are perpendicular, it follows that perpendicular( )DB AEM d T= or 2.28 N m (19.0379 N)d⋅ = 0.119761d = 0.1198 md = www.elsolucionario.net www.elsolucionario.net
- 227. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 225 PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION From the solution to Problem 3.56: 33 N 3[(6 N) (9 N) (2 N) ] CF CF = = − −T i j k Τ | | 9.50 N mDBM = ⋅ 1 (24 7 ) 25 DB = −i jλ Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF will contribute to the moment of TCF about line .DB Now parallel( ) 1 3(6 9 2 ) (24 7 ) 25 3 [(6)(24) ( 9)( 7)] 25 24.84 N CF CF DB= ⋅ = − − ⋅ − = + − − = T T i j k i j λ Also, parallel perpendicular( ) ( )CF CF CF= +T T T so that 2 2 perpendicular( ) (33) (24.84) 21.725 N CF = − = T Since DBλ and perpendicular( )CFT are perpendicular, it follows that perpendicular| | ( )DB CFM d T= or 9.50 N m 21.725 Nd⋅ = × or 0.437 md = www.elsolucionario.net www.elsolucionario.net
- 228. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 226 PROBLEM 3.66 In Prob. 3.57, determine the perpendicular distance between rod AB and the line of action of P. PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P. SOLUTION (32 in.) (30 in.) (24 in.)AB = − −i j k 2 2 2 (32) ( 30) ( 24) 50 in.AB = + − + − = 0.64 0.60 0.48AB AB AB = = − −i j k λ 105 190 90 235 P P − + = = P i j k λ Angle θ between AB and P: cos 105 190 90 (0.64 0.60 0.48 ) 235 0.58723 AB Pθ = ⋅ − + = − − ⋅ = i j k i j k λ λ 54.039θ∴ = ° The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to AB by the perpendicular distance d from AB to P: ( sin )AB P dθ=M From the solution to Prob. 3.57: 207 lb ft 2484 lb in.AB = ⋅ = ⋅M We have 2484 lb in. (235 lb)(sin54.039)d⋅ = 13.06 in.d = www.elsolucionario.net www.elsolucionario.net
- 229. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 227 PROBLEM 3.67 In Prob. 3.58, determine the perpendicular distance between rod AB and the line of action of Q. PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. SOLUTION (32 in.) (30 in.) (24 in.)AB = − −i j k 2 2 2 (32) ( 30) ( 24) 50 in.AB = + − + − = 0.64 0.60 0.48AB AB AB = = − −i j k λ 96 126 72 174 Q Q − − − = = Q i j k λ Angle θ between AB and Q: cos ( 96 126 72 ) (0.64 0.60 0.48 ) 174 0.28000 AB Qθ = ⋅ − − − = − − ⋅ = i j k i j k λ λ 73.740θ∴ = ° The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to AB by the perpendicular distance d from AB to Q: ( sin )AB Q dθ=M From the solution to Prob. 3.58: 176.6 lb ft 2119.2 lb in.AB = ⋅ = ⋅M 2119.2 lb in. (174 lb)(sin 73.740 )d⋅ = ° 12.69 in.d = www.elsolucionario.net www.elsolucionario.net
- 230. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 228 PROBLEM 3.68 In Problem 3.59, determine the perpendicular distance between portion BH of the cable and the diagonal AD. PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION From the solution to Problem 3.59: 450 N (150 N) (300 N) (300 N) BH BH T = = + −T i j k | | 90.0 N mADM = ⋅ 1 (4 3 ) 5 AD = −i kλ Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBH will contribute to the moment of TBH about line .AD Now parallel( ) 1 (150 300 300 ) (4 3 ) 5 1 [(150)(4) ( 300)( 3)] 5 300 N BH BH ADT = ⋅ = + − ⋅ − = + − − = T i j k i k λ Also, parallel perpendicular( ) ( )BH BH BH= +T T T so that 2 2 perpendicular( ) (450) (300) 335.41 NBHT = − = Since ADλ and perpendicular( )BHT are perpendicular, it follows that perpendicular( )AD BHM d T= or 90.0 N m (335.41 N)d⋅ = 0.26833 md = 0.268 md = www.elsolucionario.net www.elsolucionario.net
- 231. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 229 PROBLEM 3.69 In Problem 3.60, determine the perpendicular distance between portion BG of the cable and the diagonal AD. PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. SOLUTION From the solution to Problem 3.60: 450 N (200 N) (370 N) (160 N) BG BG = = − + −T i j k Τ | | 111 N mADM = ⋅ 1 (4 3 ) 5 AD = −i kλ Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBG will contribute to the moment of TBG about line .AD Now parallel( ) 1 ( 200 370 160 ) (4 3 ) 5 1 [( 200)(4) ( 160)( 3)] 5 64 N BG BG ADT = ⋅ = − + − ⋅ − = − + − − = − T i j k i k λ Also, parallel perpendicular( ) ( )BG BG BG= +T T T so that 2 2 perpendicular( ) (450) ( 64) 445.43 NBG = − − =T Since ADλ and perpendicular( )BGT are perpendicular, it follows that perpendicular( )AD BGM d T= or 111 N m (445.43 N)d⋅ = 0.24920 md = 0.249 md = www.elsolucionario.net www.elsolucionario.net
- 232. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 230 PROBLEM 3.70 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-lb forces, (b) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 72 lb in.⋅ clockwise and d is 42 in. SOLUTION (a) We have 1 1 1M d F= where 1 1 16 in. 21 lb d F = = 1 (16 in.)(21 lb) 336 lb in. M = = ⋅ or 1 336 lb in.= ⋅M (b) We have 1 2 0+ =M M or 2336 lb in. (12 lb) 0d⋅ − = 2 28.0 in.d = (c) We have total 1 2= +M M M or 72 lb in. 336 lb in. (42 in.)(sin )(12 lb)α− ⋅ = ⋅ − sin 0.80952α = and 54.049α = ° or 54.0α = ° www.elsolucionario.net www.elsolucionario.net
- 233. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 231 PROBLEM 3.71 Four 1-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension? SOLUTION (a) (35 lb)(7 in.) (25 lb)(9 in.) 245 lb in. 225 lb in. M = + = ⋅ + ⋅ 470 lb in.M = ⋅ (b) With only one string, pegs A and D, or B and C should be used. We have 6 tan 36.9 90 53.1 8 θ θ θ= = ° ° − = ° Direction of forces: With pegs A and D: 53.1θ = ° With pegs B and C: 53.1θ = ° (c) The distance between the centers of the two pegs is 2 2 8 6 10 in.+ = Therefore, the perpendicular distance d between the forces is 1 10 in. 2 in. 2 11 in. d = + = We must have 470 lb in. (11 in.)M Fd F= ⋅ = 42.7 lbF = www.elsolucionario.net www.elsolucionario.net
- 234. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 232 PROBLEM 3.72 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb·in. counterclockwise. SOLUTION 485 lb in. [(6 ) in.](35 lb) [(8 ) in.](25 lb) AD AD BC BCM d F d F d d = + ⋅ = + + + 1.250 in.d = www.elsolucionario.net www.elsolucionario.net
- 235. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 233 PROBLEM 3.73 A piece of plywood in which several holes are being drilled successively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N·m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C. SOLUTION (a) 12 N m (0.45 m) M Fd F = ⋅ = 26.7 NF = (b) 12 N m (0.24 m) M Fd F = ⋅ = 50.0 NF = (c) 2 2 (0.45 m) (0.24 m) 0.510 m M Fd d= = + = 12 N m (0.510 m)F⋅ = 23.5 NF = www.elsolucionario.net www.elsolucionario.net
- 236. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 234 PROBLEM 3.74 Two parallel 40-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A. SOLUTION (a) We have 1 2:B x yd C d CΣ − + =M M where 1 2 (0.270 m)sin55° 0.22117 m (0.270 m)cos55 0.154866 m d d = = = ° = (40 N)cos20 37.588 N (40 N)sin 20 13.6808 N x y C C = ° = = ° = (0.22117 m)(37.588 N) (0.154866 m)(13.6808 N) (6.1946 N m) = − + = − ⋅ M k k k or 6.19 N m= ⋅M (b) We have ( ) 40 N[(0.270 m)sin(55 20 )]( ) Fd= − = ° − ° − M k k (6.1946 N m)= − ⋅ k or 6.19 N m= ⋅M (c) We have / /: ( )A A B A B C A CΣ Σ × = × + × =M r F r F r F M (0.390 m)(40 N) cos55 sin55 0 cos20 sin 20 0 (0.660 m)(40 N) cos55 sin55 0 cos20 sin 20 0 (8.9478 N m 15.1424 N m) (6.1946 N m) M = ° ° − ° − ° + ° ° ° ° = ⋅ − ⋅ = − ⋅ i j k i j k k k or 6.19 N m= ⋅M www.elsolucionario.net www.elsolucionario.net
- 237. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 235 PROBLEM 3.75 The two shafts of a speed-reducer unit are subjected to couples of magnitude M1 = 15 lb·ft and M2 = 3 lb·ft, respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION 1 (15 lb ft)M = ⋅ k 2 (3 lb ft)M = ⋅ i 2 2 1 2 2 2 (15) (3) 15.30 lb ft M M M= + = + = ⋅ 15 tan 5 3 xθ = = 78.7xθ = ° 90yθ = ° 90 78.7 11.30 zθ = ° − ° = ° 15.30 lb ft; 78.7 , 90.0 , 11.30x y zM θ θ θ= ⋅ = ° = ° = ° www.elsolucionario.net www.elsolucionario.net
- 238. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 236 PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Replace the couple in the ABCD plane with two couples P and Q shown: 160 mm (50 N) (50 N) 40 N 200 mm CD P CG = = = 120 mm (50 N) (50 N) 30 N 200 mm CF Q CG = = = Couple vector M1 perpendicular to plane ABCD: 1 (40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM = − = ⋅ Couple vector M2 in the xy plane: 2 (12.5 N)(0.192 m) 2.40 N mM = − = − ⋅ 144 mm tan 36.870 192 mm θ θ= = ° 1 (4.80 cos36.870 ) (4.80 sin36.870 ) 3.84 2.88 = ° + ° = + M j k j k 2 2.40= −M j 1 2 1.44 2.88= + = +M M M j k 3.22 N m; 90.0 , 53.1 , 36.9x y zθ θ θ= ⋅ = ° = ° = °M www.elsolucionario.net www.elsolucionario.net
- 239. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 237 PROBLEM 3.77 Solve Prob. 3.76, assuming that two 10-N vertical forces have been added, one acting upward at C and the other downward at B. PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Replace the couple in the ABCD plane with two couples P and Q shown. 160 mm (50 N) (50 N) 40 N 200 mm CD P CG = = = 120 mm (50 N) (50 N) 30 N 200 mm CF Q CG = = = Couple vector M1 perpendicular to plane ABCD. 1 (40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM = − = ⋅ 144 mm tan 36.870 192 mm θ θ= = ° 1 (4.80cos36.870 ) (4.80sin36.870 ) 3.84 2.88 = ° + ° = + M j k j k 2 (12.5 N)(0.192 m) 2.40 N m 2.40 M = − = − ⋅ = − j 3 / 3 /; (0.16 m) (0.144 m) (0.192 m) (0.16 m) (0.144 m) (0.192 m) ( 10 N) 1.92 1.6 B C B CM= × = + − = + − × − = − − M r r i j k i j k j i k www.elsolucionario.net www.elsolucionario.net
- 240. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 238 PROBLEM 3.77 (Continued) 1 2 3 (3.84 2.88 ) 2.40 ( 1.92 1.6 ) (1.92 N m) (1.44 N m) (1.28 N m) M M M M= + + = + − + − − = − ⋅ + ⋅ + ⋅ j k j i k i j k 2 2 2 ( 1.92) (1.44) (1.28) 2.72 N mM = − + + = ⋅ 2.72 N mM = ⋅ cos 1.92/2.72 cos 1.44/2.72 cos 1.28/2.72 x y z θ θ θ = − = = 134.9 58.0 61.9x y zθ θ θ= ° = ° = ° www.elsolucionario.net www.elsolucionario.net
- 241. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 239 PROBLEM 3.78 If 0,P = replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION 1 2 1 2 1 / 2 / 2 2 2 2 2 ; 16 lb, 40 lb (30 in.) [ (16 lb) ] (480 lb in.) ; (15 in.) (5 in.) (0) (5) (10) 5 5 in. 40 lb (5 10 ) 5 5 8 5[(1 lb) (2 lb) ] 8 5 15 5 0 0 1 2 8 5[(10 lb in.) C E B E B DE F F d F = + = = = × = × − = − ⋅ = × = − = + + = = − = − = − − = ⋅ 1 2 M M M M r F i j k M r F r i j j k j k i j k M i (30 lb in.) (15 lb in.) ] (480 lb in.) 8 5[(10 lb in.) (30 lb in.) (15 lb in.) ] (178.885 lb in.) (536.66 lb in.) (211.67 lb in.) + ⋅ + ⋅ = − ⋅ + ⋅ + ⋅ + ⋅ = ⋅ + ⋅ − ⋅ j k M k i j k i j k 2 2 2 (178.885) (536.66) ( 211.67) 603.99 lb in M = + + − = ⋅ 604 lb in.M = ⋅ axis z 0.29617 0.88852 0.35045 cos 0.29617 cos 0.88852 cos 0.35045 x y M θ θ θ = = + − = = = − M i j kλ z72.8 27.3 110.5x yθ θ θ= ° = ° = ° www.elsolucionario.net www.elsolucionario.net
- 242. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 240 PROBLEM 3.79 If 20 lb,P = replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION From the solution to Problem. 3.78: 16-lb force: 1 (480 lb in.)M = − ⋅ k 40-lb force: 2 8 5[(10 lb in.) (30 lb in.) (15 lb in.) ]M = ⋅ + ⋅ + ⋅i j k P 20 lb= 3 (30 in.) (20 lb) (600 lb in.) CM P= × = × = ⋅ r i k j 1 2 3 2 2 2 (480) 8 5 (10 30 15 ) 600 (178.885 lb in.) (1136.66 lb in.) (211.67 lb in.) (178.885) (113.66) (211.67) 1169.96 lb in. M = + + = − + + + + = ⋅ + ⋅ − ⋅ = + + = ⋅ M M M M k i j k j i j k 1170 lb in.M = ⋅ axis 0.152898 0.97154 0.180921 cos 0.152898 cos 0.97154 cos 0.180921 x y z M θ θ θ = = + − = = = − M i j kλ 81.2 13.70 100.4x y zθ θ θ= ° = ° = ° www.elsolucionario.net www.elsolucionario.net
- 243. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 241 PROBLEM 3.80 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. If the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION 1 2 3 2 2 2 (1.5 N m)( cos20 sin 20 ) (1.5 N m) (1.75 N m)( cos25 sin 25 ) (4.4956 N m) (0.22655 N m) (0) ( 4.4956) (0.22655) 4.5013 N m M = + + = ⋅ − ° + ° − ⋅ + ⋅ − ° + ° = − ⋅ + ⋅ = + − + = ⋅ M M M M j k j j k j k 4.50 N mM = ⋅ axis (0.99873 0.050330 ) cos 0 cos 0.99873 cos 0.050330 x y z M θ θ θ = = − + = = − = M j kλ 90.0 , 177.1 , 87.1x y zθ θ θ= ° = ° = ° www.elsolucionario.net www.elsolucionario.net
- 244. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 242 PROBLEM 3.81 A 260-lb force is applied at A to the rolled-steel section shown. Replace that force with an equivalent force-couple system at the center C of the section. SOLUTION 2 2 (2.5 in.) (6.0 in.) 6.50 in.AB = + = 2.5 in. 5 sin 6.5 in. 13 6.0 in. 12 cos 22.6 6.5 in. 13 α α α = = = = = ° sin cos 5 12 (260 lb) (260 lb) 13 13 (100.0 lb) (240 lb) F Fα α= − − = − − = − − F i j i j i j / (2.5 4.0 ) ( 100.0 240 ) 400 600 (200 lb in.) C A C= × = + × − − = − = − ⋅ M r F i j i j k k k 260 lb=F 67.4°; 200 lb in.C = ⋅M www.elsolucionario.net www.elsolucionario.net
- 245. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 243 PROBLEM 3.82 A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a. SOLUTION (a) (30 lb)(5 in.) 150.0 lb in. BM = = ⋅ 30.0 lb=F , 150.0 lb in.B = ⋅M (b) 150 lb in. 50.0 lb 3.0 in. B C ⋅ = = = 50.0 lb=B ; 50.0 lb=C www.elsolucionario.net www.elsolucionario.net
- 246. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244 PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α 30°= and 60°,β = replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. SOLUTION (a) Equivalence requires : or 250 NΣ = =F F P F 60° : (0.3 m)(250 N) 75 N mB MΣ = − = − ⋅M The equivalent force-couple system at B is 250 NB =F 60° 75.0 N mB = ⋅M (b) We require Equivalence then requires : 0 cos cosx A BF F Fφ φΣ = + or cos 0A BF F φ= − = : 250 sin siny A BF F Fφ φΣ − = − − Now if 250 0,A BF F= − − = reject. cos 0φ = or 90φ = ° and 250A BF F+ = Also, : (0.3 m)(250 N) (0.2m)B AM FΣ − = or 375 NAF = − and 625 NBF = 375 NA =F 60° 625 NB =F 60.0° www.elsolucionario.net www.elsolucionario.net
- 247. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 245 PROBLEM 3.84 Solve Problem 3.83, assuming 25°.α β= = PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α 30°= and 60°,β = replace P with (a) an equivalent force- couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. SOLUTION (a) Equivalence requires : or 250 NB BΣ = =F F P F 25.0° : (0.3 m)[(250 N)sin50 ] 57.453 N mB BMΣ = − ° = − ⋅M The equivalent force-couple system at B is 250 NB =F 25.0° 57.5 N mB = ⋅M (b) We require Equivalence requires (0.3 m)[(250 N)sin50 ] [(0.2 m)sin50 ] 375 N B AEM d Q Q Q = ° = ° = Adding the forces at B: 375 NA =F 25.0° 625 NB =F 25.0° www.elsolucionario.net www.elsolucionario.net
- 248. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 246 PROBLEM 3.85 The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in part a. SOLUTION (a) Based on : 80 NBF F FΣ = = or 80.0 NB =F : B BM M FdΣ = 80 N (0.05 m) 4.0000 N m = = ⋅ or 4.00 N mB = ⋅M (b) If the two vertical forces are to be equivalent to MB, they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then with CF and DF acting as shown, : D CM M F dΣ = 4.0000 N m (0.04 m) 100.000 N C C F F ⋅ = = or 100.0 NC =F : 0y D CF F FΣ = − 100.000 NDF = or 100.0 ND =F www.elsolucionario.net www.elsolucionario.net
- 249. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 247 PROBLEM 3.86 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C. SOLUTION Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical. Then for equivalence, : (1040 N)sin30 sin sinx A BF F Fα αΣ ° = + (1) : (1040 N)cos30 cos cosy A BF F Fα αΣ − ° = − − (2) Dividing Equation (1) by Equation (2), ( )sin(1040 N)sin30 (1040 N)cos30 ( )cos A B A B F F F F α α +° = − ° − + Simplifying yields 30 .α = ° Based on : [(1040 N)cos30 ](4 m) ( cos30 )(10.7 m)C AM FΣ ° = ° 388.79 NAF = or 389 NA =F 60.0° Based on : [(1040 N)cos30 ](6.7 m) ( cos30 )(10.7 m)A CM FΣ − ° = ° 651.21 NCF = or 651 NC =F 60.0° www.elsolucionario.net www.elsolucionario.net
- 250. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 248 PROBLEM 3.87 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in part a, and specify its point of application on the lever. SOLUTION (a) First note that the two 90-N forces form a couple. Then 216 N=F θ where 180 (60 55 ) 65θ = ° − ° + ° = ° and (0.450 m)(216 N)cos55 (1.050 m)(90 N)cos20 33.049 N m BM M= Σ = ° − ° = − ⋅ The equivalent force-couple system at B is 216 N=F 65.0 ;° 33.0 N m= ⋅M (b) The single equivalent force F′ is equal to F. Further, since the sense of M is clockwise, F′ must be applied between A and B. For equivalence, : cos55BM M aF′Σ = ° where a is the distance from B to the point of application of F′. Then 33.049 N m (216 N)cos55a− ⋅ = − ° 0.26676 ma = or 216 N′ =F 65.0° applied to the lever 267 mm to the left of B www.elsolucionario.net www.elsolucionario.net
- 251. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 249 PROBLEM 3.88 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E. SOLUTION From the statement of the problem, it follows that 0EMΣ = for the given force-couple system. Further, for Pmin, we must require that P be perpendicular to / .B Er Then min : (0.2 sin 30 0.2)m 300 N (0.2 m)sin 30 300 N (0.4 m) 0 EM P Σ ° + × + °× − = or min 300 NP = min 300 N=P 30.0° www.elsolucionario.net www.elsolucionario.net
- 252. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 250 PROBLEM 3.89 A force and couple act as shown on a square plate of side a = 25 in. Knowing that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by a single force applied at a point located (a) on line AB, (b) on line AC. In each case determine the distance from A to the point of application of the force. SOLUTION Replace the given force-couple system with an equivalent force- couple system at A. (60 lb)(cos50 ) 38.567 lbxP = ° = (60 lb)(sin50 ) 45.963 lbyP = ° = (45.963 lb)(25 in.) (40 lb)(25 in.) A yM P a Qa= − = − 149.075 lb in.= ⋅ (a) Equating moments about A gives: 149.075 lb in. (45.963 lb) 3.24 in. x x ⋅ = = 60.0 lb=P 50.0°; 3.24 in. from A (b) 149.075 lb in. (38.567 lb) 3.87 in. y y ⋅ = = 60.0 lb=P 50.0°; 3.87 in. below A www.elsolucionario.net www.elsolucionario.net
- 253. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 251 PROBLEM 3.90 The force and couple shown are to be replaced by an equivalent single force. Knowing that P = 2Q, determine the required value of α if the line of action of the single equivalent force is to pass through (a) Point A, (b) Point C. SOLUTION (a) We must have 0AM = ( sin ) ( ) 0P a Q aα − = 1 sin 2 2 Q Q P Q α = = = 30.0α = ° (b) We must have 0CM = ( sin ) ( cos ) ( ) 0P a P a Q aα α− − = 1 sin cos 2 2 Q Q P Q α α− = = = 1 sin cos 2 α α= + (1) 2 2 1 sin cos cos 4 α α α= + + 2 2 1 1 cos cos cos 4 α α α− = + + 2 2cos cos 0.75 0α α+ − = (2) Solving the quadratic in cos :α 1 7 cos 65.7 or 155.7 4 α α − ± = = ° ° Only the first value of α satisfies Eq. (1), therefore 65.7α = ° www.elsolucionario.net www.elsolucionario.net
- 254. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 252 PROBLEM 3.91 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.) SOLUTION Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple (0.18)(250 N) 45 N m HM = = ⋅ Then (900 N)HM x= or 45 N m (900 N) 0.05 m x x ⋅ = = 900 N=F 50.0 mmx = www.elsolucionario.net www.elsolucionario.net
- 255. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 253 PROBLEM 3.92 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed. SOLUTION (a) We have : (360 N) (360 N) (600 N)Σ = − −F F j j k or (600 N)= −F k and : (360 N)(0.15 m) (600 N)( )DM dΣ = 0.09 md = or 90.0 mm belowd ED= (b) We have from part a: (600 N)= −F k and : (360 N)(0.15 m) (600 N)( )DM dΣ − = − 0.09 md = or 90.0 mm aboved ED= www.elsolucionario.net www.elsolucionario.net
- 256. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 254 PROBLEM 3.93 An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna. SOLUTION We have 2 2 2 ( 64) ( 128) (16) 144 ftABd = − + − + = Then 288 lb ( 64 128 16 ) 144 (32 lb)( 4 8 ) AB = − − + = − − + T i j k i j k Now / 128 32( 4 8 ) (4096 lb ft) (16,384 lb ft) O A O AB= = × = × − − + = ⋅ + ⋅ M M r T j i j k i k The equivalent force-couple system at O is (128.0 lb) (256 lb) (32.0 lb)= − − +F i j k (4.10 kip ft) (16.38 kip ft)= ⋅ + ⋅M i k www.elsolucionario.net www.elsolucionario.net
- 257. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 255 PROBLEM 3.94 An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna. SOLUTION We have 2 2 2 ( 64) ( 128) ( 128) 192 ft ADd = − + − + − = Then 270 lb ( 64 128 128 ) 192 (90 lb)( 2 2 ) AD = − − + = − − − T i j k i j k Now / 128 90( 2 2 ) (23,040 lb ft) (11,520 lb ft) O A O AD= = × = × − − − = − ⋅ + ⋅ M M r T j i j k i k The equivalent force-couple system at O is (90.0 lb) (180.0 lb) (180.0 lb)= − − −F i j k (23.0 kip ft) (11.52 kip ft)= − ⋅ + ⋅M i k www.elsolucionario.net www.elsolucionario.net
- 258. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 256 PROBLEM 3.95 A 110-N force acting in a vertical plane parallel to the yz-plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent force- couple system at the origin O of the coordinate system. SOLUTION We have : BΣ =F P F where 110 N[ (sin15 ) (cos15 ) ] (28.470 N) (106.252 N) B = − ° + ° = − + P j k j k or (28.5 N) (106.3 N)= − +F j k We have /:O B O B OMΣ × =r P M where / [(0.22cos35 ) (0.15) (0.22sin35 ) ] m (0.180213 m) (0.15 m) (0.126187 m) B O = ° + − ° = + − r i j k i j k 0.180213 0.15 0.126187 N m 0 28.5 106.3 O⋅ = − i j k M [(12.3487) (19.1566) (5.1361) ] N mO = − − ⋅M i j k or (12.35 N m) (19.16 N m) (5.13 N m)O = ⋅ − ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 259. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 257 PROBLEM 3.96 An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G. SOLUTION We have : (1220 N)Σ − =F i F (1220 N)= −F i Also, we have /:G A GΣ × =M r P M 1220 0 0.1 0.06 N m 1 0 0 − − ⋅ = − i j k M (1220 N m)[( 0.06)( 1) ( 0.1)( 1) ]= ⋅ − − − − −M j k or (73.2 N m) (122 N m)= ⋅ − ⋅M j k www.elsolucionario.net www.elsolucionario.net
- 260. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 258 PROBLEM 3.97 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C. SOLUTION We have : AB CΣ =F P F where (33 mm) (990 mm) (594 mm) (175 N) 1155.00 mm AB AB ABP= + − = P λ i j k or (5.00 N) (150.0 N) (90.0 N)C = + −F i j k We have /:C B C AB CΣ × =M r P M 5 0.683 0.860 0 N m 1 30 18 (5){( 0.860)( 18) (0.683)( 18) [(0.683)(30) (0.860)(1)] } C = − ⋅ − = − − − − + − i j k M i j k or (77.4 N m) (61.5 N m) (106.8 N m)C = ⋅ + ⋅ + ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 261. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 259 PROBLEM 3.98 A 46-lb force F and a 2120-lb⋅in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H. SOLUTION We have 2 2 2 (18) ( 14) ( 3) 23 in.AJd = + − + − = Then 46 lb (18 14 3 ) 23 (36 lb) (28 lb) (6 lb) = − − = − − F i j k i j k Also 2 2 2 ( 45) (0) ( 28) 53 in.ACd = − + + − = Then 2120 lb in. ( 45 28 ) 53 (1800 lb in.) (1120 lb in.) ⋅ = − − = − ⋅ − ⋅ M i k i k Now /A H′ = + ×M M r F where / (45 in.) (14 in.)A H = +r i j Then ( 1800 1120 ) 45 14 0 36 28 6 ′ = − − + − − i j k M i k ( 1800 1120 ) {[(14)( 6)] [ (45)( 6)] [(45)( 28) (14)(36)] } ( 1800 84) (270) ( 1120 1764) (1884 lb in.) (270 lb in.) (2884 lb in.) (157 lb ft) (22.5 lb ft) (240 lb ft) = − − + − + − − + − − = − − + + − − = − ⋅ + ⋅ − ⋅ = − ⋅ + ⋅ − ⋅ i k i j k i j k i j k i j k The equivalent force-couple system at H is (36.0 lb) (28.0 lb) (6.00 lb)′ = − −F i j k (157.0 lb ft) (22.5 lb ft) (240 lb ft)′ = − ⋅ + ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 262. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 260 PROBLEM 3.99 A 77-N force F1 and a 31-N ⋅ m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system (F2, M2) at corner B and if (M2)z = 0, determine (a) the distance d, (b) F2 and M2. SOLUTION (a) We have 2: 0Bz zM MΣ = / 1 1( ) 0H B zM⋅ × + =k r F (1) where / (0.31m) (0.0233)H B = −r i j 1 1 1 1 1 1 2 (0.06 m) (0.06 m) (0.07 m) (77 N) 0.11 m (42 N) (42 N) (49 N) (0.03 m) (0.07 m) (31 N m) 0.0058 m EH z EJ F M M d d = + − = = + − = ⋅ = − + − = ⋅ + F λ i j k i j k k M M λ i j k Then from Equation (1), 2 0 0 1 ( 0.07 m)(31 N m) 0.31 0.0233 0 0 0.005842 42 49 d − ⋅ − + = +− Solving for d, Equation (1) reduces to 2 2.17 N m (13.0200 0.9786) 0 0.0058d ⋅ + − = + from which 0.1350 md = or 135.0 mmd = www.elsolucionario.net www.elsolucionario.net
- 263. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 261 PROBLEM 3.99 (Continued) (b) 2 1 (42 42 49 ) N= = + −F F i j k or 2 (42.0 N) (42.0 N) (49.0 N)= + −F i j k 2 / 1 1 2 (0.1350) 0.03 0.07 0.31 0.0233 0 (31 N m) 0.155000 42 42 49 (1.14170 15.1900 13.9986 ) N m ( 27.000 6.0000 14.0000 ) N m (25.858 N m) (21.190 N m) H B= × + + − = − + ⋅ − = + + ⋅ + − + − ⋅ = − ⋅ + ⋅ M r F M i j k i j k i j k i j k M i j or 2 (25.9 N m) (21.2 N m)= − ⋅ + ⋅M i j www.elsolucionario.net www.elsolucionario.net
- 264. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 262 PROBLEM 3.100 A 2.6-kip force is applied at Point D of the cast iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section. SOLUTION (12 in.) (5 in.) ; 13.00 in.DE DE= − − =j k (2.6 kips) DE DE =F 12 5 (2.6 kips) 13 − − = j k F (2.40 kips) (1.000 kip)= − −F j k /A D A= ×M r F where / (6 in.) (12 in.)D A = +r i j 6 in. 12 in. 0 0 2.4 kips 1.0 kips A = − − i j k M (12.00 kip in.) (6.00 kip in.) (14.40 kip in.)A = − ⋅ + ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 265. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 263 PROBLEM 3.101 A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force- couple system at end A of the beam. (b) Which of the loadings are equivalent? SOLUTION (a) (a) We have : 300 N 200 NY aF RΣ − − = or 500 Na =R and : 400 N m (200 N)(3 m)A aM MΣ − ⋅ − = or 1000 N ma = ⋅M (b) We have : 200 N 300 NY bF RΣ + = or 500 Nb =R and : 400 N m (300 N)(3 m)A bM MΣ − ⋅ + = or 500 N mb = ⋅M (c) We have : 200 N 300 NY cF RΣ − − = or 500 Nc =R and : 400 N m (300 N)(3 m)A cM MΣ ⋅ − = or 500 N mc = ⋅M www.elsolucionario.net www.elsolucionario.net
- 266. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 264 PROBLEM 3.101 (Continued) (d) We have : 500 NY dF RΣ − = or 500 Nd =R and : 400 N m (500 N)(3 m)A dM MΣ ⋅ − = or 1100 N md = ⋅M (e) We have : 300 N 800 NY eF RΣ − = or 500 Ne =R and : 400 N m 1000 N m (800 N)(3 m)A eM MΣ ⋅ + ⋅ − = or 1000 N me = ⋅M (f ) We have : 300 N 200 NY fF RΣ − − = or 500 Nf =R and : 400 N m (200 N)(3 m)A fM MΣ ⋅ − = or 200 N mf = ⋅M (g) We have : 800 N 300 NY gF RΣ − + = or 500 Ng =R and : 1000 N m 400 N m (300 N)(3 m)A gM MΣ ⋅ + ⋅ + = or 2300 N mg = ⋅M (h) We have : 250 N 250 NY hF RΣ − − = or 500 Nh =R and : 1000 N m 400 N m (250 N)(3 m)A hM MΣ ⋅ + ⋅ − = or 650 N mh = ⋅M (b) Therefore, loadings (a) and (e) are equivalent. www.elsolucionario.net www.elsolucionario.net
- 267. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 265 PROBLEM 3.102 A 3-m-long beam is loaded as shown. Determine the loading of Prob. 3.101 that is equivalent to this loading. SOLUTION We have : 200 N 300 NYF RΣ − − = or 500 N=R and : 500 N m 200 N m (300 N)(3 m)AM MΣ ⋅ + ⋅ − = or 200 N m= ⋅M Problem 3.101 equivalent force-couples at A: Case R M (a) 500 N 1000 N⋅m (b) 500 N 500 N⋅m (c) 500 N 500 N⋅m (d) 500 N 1100 N⋅m (e) 500 N 1000 N⋅m (f ) 500 N 200 N⋅m (g) 500 N 2300 N⋅m (h) 500 N 650 N⋅m Equivalent to case (f ) of Problem 3.101 www.elsolucionario.net www.elsolucionario.net
- 268. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 266 PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102. SOLUTION For equivalent single force at distance d from A: (a) We have : 300 N 200 NYF RΣ − − = or 500 N=R and : 400 N m (300 N)( ) (200 N)(3 ) 0 CM d d Σ − ⋅ + − − = or 2.00 md = (b) We have : 200 N 300 NYF RΣ + = or 500 N=R and : 400 N m (200 N)( ) (300 N)(3 ) 0 CM d d Σ − ⋅ − + − = or 1.000 md = (c) We have : 200 N 300 NYF RΣ − − = or 500 N=R and : 500 N m 200 N m (200 N)( ) (300 N)(3 ) 0 CM d d Σ ⋅ + ⋅ + − − = or 0.400 md = www.elsolucionario.net www.elsolucionario.net
- 269. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 267 PROBLEM 3.104 Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin. SOLUTION First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces remain unchanged. : (5 lb ft) (15 lb ft) (2 ft) (10 lb)A OA = Σ = ⋅ + ⋅ + ×M M j k k i (25 lb ft) (15 lb ft)= ⋅ + ⋅j k : (5 lb ft) (25 lb ft) [(4.5 ft) (1 ft) (2 ft) ] 10 lb) (15 lb ft) (15 lb ft) D OD = Σ = − ⋅ + ⋅ + + + × = ⋅ + ⋅ M M j k i j k i i k : (15 lb ft) (15 lb ft) : (15 lb ft) (5 lb ft) [(4.5 ft) (1 ft) ] (10 lb) G O I I G I = Σ = ⋅ + ⋅ = Σ = ⋅ − ⋅ + + × M M i j M M j k i j j (15 lb ft) (15 lb ft)= ⋅ − ⋅j k The equivalent force-couple system is the system at corner D. www.elsolucionario.net www.elsolucionario.net
- 270. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 268 PROBLEM 3.105 Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when (a) P = 200 N, (b) P = 2400 N, (c) P = 1000 N. SOLUTION (a) 200 N 600 N 400 N 800 NDR = + − − = − (200 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m) 150.0 N m DM = − + + = + ⋅ 150 N m 0.1875 m 800 N DM y R ⋅ = = = 800 N=R ; 187.5 mmy = (b) 2400 N 600 N 400 N 1400 NDR = + − − = + (2400 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m) 840 N m DM = − + + = − ⋅ 840 N m 0.600 m 1400 N DM y R ⋅ = = = 1400 N=R ; 600 mmy = www.elsolucionario.net www.elsolucionario.net
- 271. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 269 PROBLEM 3.105 (Continued) (c) 1000 600 400 0DR = + − − = (1000 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m) 210 N m DM = − + + = − ⋅ y∴ = ∞ System reduces to a couple. 210 N mD = ⋅M www.elsolucionario.net www.elsolucionario.net
- 272. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 270 PROBLEM 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d = 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe. SOLUTION For equivalence, : 4.1 4.1 3.5 or 11.7 lbyF RΣ − − − = − =R : (10 in.)(4.1 lb) (44 in.)(4.1 lb) [(4.4 ) in.](3.5 lb) ( in.)(11.7 lb) DF d L Σ − − − + = − or 375.4 3.5 11.7 ( , in in.)d L d L+ = (a) 25 in.d = We have 375.4 3.5(25) 11.7 or 39.6 in.L L+ = = The resultant passes through a point 39.6 in. to the right of D. (b) 42 in.L = We have 375.4 3.5 11.7(42)d+ = or 33.1 in.d = www.elsolucionario.net www.elsolucionario.net
- 273. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 271 PROBLEM 3.107 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb. SOLUTION (a) For the resultant weight to act at C, 0 60 lbC CM WΣ = = Then (84 lb)(6 ft) 60 lb( ) 64 lb(6 ft) 0d− − = 2.00 ft to the right ofd C= (b) For the resultant weight to act at C, 0 52 lbC CM WΣ = = Then (84 lb)(6 ft) 52 lb( ) 64 lb(6 ft) 0d− − = 2.31 ft to the right ofd C= www.elsolucionario.net www.elsolucionario.net
- 274. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 272 PROBLEM 3.108 A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. SOLUTION (a) We have : ( 10 ) (30 cos 60 ) 30 sin 60 ( 45 ) (30 lb) (15.9808 lb) Σ = − + ° + ° + − = − + F R j i j i i j or 34.0 lb=R 28.0° (b) First reduce the given forces and couple to an equivalent force-couple system ( , )BR M at B. We have : (54 lb in) (12 in.)(10 lb) (8 in.)(45 lb) 186 lb in. B BM MΣ = ⋅ + − = − ⋅ Then with R at D, : 186 lb in (15.9808 lb)BM aΣ − ⋅ = or 11.64 in.a = and with R at E, : 186 lb in (30 lb)BM CΣ − ⋅ = or 6.2 in.C = The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B. www.elsolucionario.net www.elsolucionario.net
- 275. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 273 PROBLEM 3.109 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C. SOLUTION In each case, we must have 1 0R =M (a) (12 in.)[(30 lb)sin 60 ] (8 in.)(45 lb) 0B A AM M M= Σ = + ° − = 48.231 lb in.M = + ⋅ 48.2 lb in.= ⋅M (b) (12 in.)(10 lb) (8 in.)(45 lb) 0R B BM M M= Σ = + − = 240 lb in.M = + ⋅ 240 lb in.= ⋅M (c) (12 in.)(10 lb) (8 in.)[(30 lb)cos60 ] 0R C CM M M= Σ = + − ° = 0M = 0=M www.elsolucionario.net www.elsolucionario.net
- 276. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 274 PROBLEM 3.110 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor. SOLUTION We have : (60 lb) (32 lb) (140 lb)(cos30 sin30 )Σ − + ° + ° =F i j i j R (181.244 lb) (38.0 lb)= +R i j or 185.2 lb=R 11.84° We have :O O yM M xRΣ Σ = [(140 lb)cos30 ][(4 2cos30 )in.] [(140 lb)sin30 ][(2 in.)sin30 ]− ° + ° − ° ° (60 lb)(2 in.) (38.0 lb)x− = 1 ( 694.97 70.0 120) in. 38.0 x = − − − and 23.289 in.x = − Or resultant intersects the base (x-axis) 23.3 in. to the left of the vertical centerline (y-axis) of the motor. www.elsolucionario.net www.elsolucionario.net
- 277. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 275 PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION We have First replace the applied forces and couples with an equivalent force-couple system at G. Thus, : 200cos15 120cos 70x xF P RΣ ° − ° + = or (152.142 ) NxR P= + : 200sin 15 120sin 70 80y yF RΣ − ° − ° − = or 244.53 NyR = − : (0.47 m)(200 N)cos15 (0.05 m)(200 N)sin15 (0.47 m)(120 N)cos70 (0.19 m)(120 N)sin 70 (0.13 m)( N) (0.59 m)(80 N) 42 N m 40 N m G G M P M Σ − ° + ° + ° − ° − − + ⋅ + ⋅ = or (55.544 0.13 ) N mGM P= − + ⋅ (1) Setting 0P = in Eq. (1): Now with R at I, : 55.544 N m (244.53 N)GM aΣ − ⋅ = − or 0.227 ma = and with R at J, : 55.544 N m (152.142 N)GM bΣ − ⋅ = − or 0.365 mb = (a) The rivet hole is 0.365 m above G. (b) The rivet hole is 0.227 m to the right of G. www.elsolucionario.net www.elsolucionario.net
- 278. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 276 PROBLEM 3.112 Solve Problem 3.111, assuming that P = 60 N. PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION See the solution to Problem 3.111 leading to the development of Equation (1): and (55.544 0.13 ) N m (152.142 ) N G x M P R P = − + ⋅ = + For 60 NP = we have (152.142 60) 212.14 N [55.544 0.13(60)] 63.344 N m x G R M = + = = − + = − ⋅ Then with R at I, : 63.344 N m (244.53 N)GM aΣ − ⋅ = − or 0.259 ma = and with R at J, : 63.344 N m (212.14 N)GM bΣ − ⋅ = − or 0.299 mb = (a) The rivet hole is 0.299 m above G. (b) The rivet hole is 0.259 m to the right of G. www.elsolucionario.net www.elsolucionario.net
- 279. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 277 PROBLEM 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G. SOLUTION We have (240 lb)(cos70 sin 70 ) (160 lb) (300 lb)( cos40 sin 40 ) (180 lb) = Σ = ° − ° − + − ° − ° − R F R i j j i j j 2 2 2 2 1 1 (147.728 lb) (758.36 lb) (147.728) (758.36) 772.62 lb tan 758.36 tan 147.728 78.977 x y y x R R R R R θ − − = − − = + = + = = − = − = ° R i j or 773 lb=R 79.0° We have A yM dRΣ = where [240 lbcos70 ](6 ft) [240 lbsin 70 ](4 ft) (160 lb)(12 ft) [300 lbcos40 ](6 ft) [300 lbsin 40 ](20 ft) (180 lb)(8 ft) 7232.5 lb ft AMΣ = − ° − ° − + ° − ° − = − ⋅ 7232.5 lb ft 758.36 lb 9.5370 ft d − ⋅ = − = or 9.54 ft to the right ofd A= www.elsolucionario.net www.elsolucionario.net
- 280. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 278 PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when 30 ,α = ° (b) the value of α so that the single equivalent force is applied at Point B. SOLUTION We have (a) For equivalence, : 100 cos30 400 cos65 90 cos65x xF RΣ − ° + ° + ° = or 120.480 lbxR = : 100 sin 160 400 sin65 90 sin65y yF RαΣ + + ° + ° = or (604.09 100sin ) lbyR α= + (1) With 30 ,α = ° 654.09 lbyR = Then 2 2 654.09 (120.480) (654.09) tan 120.480 665 lb or 79.6 R θ θ = + = = = ° Also : (46 in.)(160 lb) (66 in.)(400 lb)sin65 (26 in.)(400 lb)cos65 (66 in.)(90 lb)sin65 (36 in.)(90 lb)cos65 (654.09 lb) AM d Σ + ° + ° + ° + ° = or 42,435 lb in. and 64.9 in.AM dΣ = ⋅ = 665 lbR = 79.6° and R is applied 64.9 in. to the right of A. (b) We have 66 in.d = Then : 42,435 lb in (66 in.)A yM RΣ ⋅ = or 642.95 lbyR = Using Eq. (1): 642.95 604.09 100sinα= + or 22.9α = ° www.elsolucionario.net www.elsolucionario.net
- 281. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 279 PROBLEM 3.115 Solve Prob. 3.114, assuming that the 90-lb force is removed. PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when 30 ,α = ° (b) the value of α so that the single equivalent force is applied at Point B. SOLUTION (a) For equivalence, : (100 lb)cos30 (400 lb)sin 25x xF RΣ − ° + ° = or 82.445 lbxR = : 160 lb (100 lb)sin30 (400 lb)cos25y yF RΣ + ° + ° = or 572.52 lbyR = 2 (82.445) (572.52) 578.43 lbR = + = 572.52 tan 82.445 θ = or 81.806θ = ° : (46 in.)(160 lb) (66 in.)(400 lb)cos25 (26 in.)(400 lb)sin 25 (527.52 lb) AM d Σ + ° + ° = 62.3 in.d = 578 lb=R 81.8° and is applied 62.3 in. to the right of A. (b) We have 66.0 in.d = For R applied at B, : (66 in.) (160 lb)(46 in.) (66 in.)(400 lb)cos25 (26 in.)(400 lb)sin 25A yM RΣ = + ° + ° 540.64 lbyR = : 160 lb (100 lb)sin (400 lb)cos25 540.64 lbYF αΣ + + ° = 10.44α = ° www.elsolucionario.net www.elsolucionario.net
- 282. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 280 PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION (a) ( 400 N 160 N 760 N) (600 N 300 N 300 N) (1000 N) (1200 N) = Σ = − + − + + + = − + R F i j i j 2 2 (1000 N) (1200 N) 1562.09 N 1200 N tan 1000 N 1.20000 50.194 R θ θ = + = = − = − = − ° 1562 N=R 50.2° (b) (0.5 m) (300 N 300 N) (300 N m) R C = Σ × = × + = ⋅ M r F i j k (300 N m) (1200 N) 0.25000 m 250 mm (300 N m) ( 1000 N) 0.30000 m 300 mm x x x y y y ⋅ = × = = ⋅ = × − = = k i j j i Intersection 250 mm to right of C and 300 mm above C www.elsolucionario.net www.elsolucionario.net
- 283. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 281 PROBLEM 3.117 Solve Problem 3.116, assuming that the 760-N force is directed to the right. PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION (a) ( 400N 160 N 760 N) (600 N 300 N 300 N) (520 N) (1200 N) = Σ = − + + + + + = + R F i j i j 2 2 (520 N) (1200 N) 1307.82 N 1200 N tan 2.3077 520 N 66.5714 R θ θ = + = = = = ° 1308 N=R 66.6° (b) (0.5 m) (300 N 300 N) (300 N m) R C = Σ × = × + = ⋅ M r F i j k (300 N m) (1200 N) 0.25000 m x x ⋅ = × = k i j or 0.250 mmx = (300 N m) [ (0.375 m) ] [(520 N) (1200 N) ] (1200 195) x x ′⋅ = + × + ′= − k i j i j k 0.41250 mx′ = or 412.5 mmx′ = Intersection 412 mm to the right of A and 250 mm to the right of C www.elsolucionario.net www.elsolucionario.net
- 284. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 282 PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at Point D obtained by drawing the perpendicular from the point of contact to the x-axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent force-couple system at D is maximum. SOLUTION (a) The slope of any tangent to the surface of member C is 2 2 2 2 1 dy d x b b x dx dx a a − = − = Since the force F is perpendicular to the surface, 1 2 1 tan 2 dy a dx b x α − = − = For equivalence, :FΣ =F R : ( cos )( )D A DM F y MαΣ = where 2 2 2 2 2 3 2 2 4 2 2 2 cos ( ) (2 ) 1 2 4 A D bx a bx x y b a x Fb x a M a b x α = + = − − = + Therefore, the equivalent force-couple system at D is F=R 2 1 tan 2 a bx − 3 2 2 4 2 2 2 4 x Fb x a a b x − = + M www.elsolucionario.net www.elsolucionario.net
- 285. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 283 PROBLEM 3.118 (Continued) (b) To maximize M, the value of x must satisfy 0 dM dx = where for 1 m, 2 ma b= = 3 2 2 2 3 2 1/2 2 8 ( ) 1 16 1 1 16 (1 3 ) ( ) (32 )(1 16 ) 2 8 0 (1 16 ) F x x M x x x x x x x dM F dx x − − = + + − − − + = = + 2 2 3 (1 16 )(1 3 ) 16 ( ) 0x x x x x+ − − − = or 4 2 32 3 1 0x x+ − = 2 2 23 9 4(32)( 1) 0.136011m and 0.22976 m 2(32) x − ± − − = = − Using the positive value of x2 : 0.36880 mx = or 369 mmx = www.elsolucionario.net www.elsolucionario.net
- 286. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 284 PROBLEM 3.119 As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet metal enclosure, the insertion tools exert the forces shown on the enclosure. Each of the forces is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at C. SOLUTION For equivalence, :ΣF (17 N) (12 N) (16 N) (21 N) (21 N) (29 N) (16 N) A B C D= + + + = − − − − = − − − R F F F F j j k i i j k / / /:C A C A B C B D C DMΣ = × + × + ×M r F r F r F [(0.11 m) (0.03 m) ] [ (17 N)] [(0.02 m) (0.11 m) (0.03 m) ] [ (12 N)] [(0.03 m) (0.03 m) (0.03 m) ] [ (21 N)] (0.51 N m) [ (0.24 N m) (0.36 N m) ] [(0.63 N m) (0.63 N m) ] M = − × − + + − × − + + − × − = − ⋅ + − ⋅ − ⋅ + ⋅ + ⋅ j k j i j k j i j k i i k i k j ∴ The equivalent force-couple system at C is (21.0 N) (29.0 N) (16.00 N)= − − −R i j k (0.870 N m) (0.630 N m) (0.390 N m)= − ⋅ + ⋅ + ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 287. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 285 PROBLEM 3.120 Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz-plane. Replace the belt forces shown with an equivalent force- couple system at A. SOLUTION Equivalent force-couple at each pulley: Pulley B: (145 N)( cos20 sin 20 ) 215 N (351.26 N) (49.593 N) (215 N 145 N)(0.075 m) (5.25 N m) B B = − ° + ° − = − + = − − = − ⋅ R j k j j k M i i Pulley C: (155 N 240 N)( sin10 cos10 ) (68.591 N) (389.00 N) (240 N 155 N)(0.075 m) (6.3750 N m) C C = + − ° − ° = − − = − = ⋅ R j k j k M i i Then (419.85 N) (339.41)B C= + = − −R R R j k or (420 N) (339 N)= −R j k / / (5.25 N m) (6.3750 N m) 0.225 0 0 N m 0 351.26 49.593 + 0.45 0 0 N m 0 68.591 389.00 (1.12500 N m) (163.892 N m) (109.899 N m) A B C B A B C A C= + + × + × = − ⋅ + ⋅ + ⋅ − ⋅ − − = ⋅ + ⋅ − ⋅ M M M r R r R i j k i i i j k i j k or (1.125 N m) (163.9 N m) (109.9 N m)A = ⋅ + ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 288. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 286 PROBLEM 3.121 Four forces are applied to the machine component ABDE as shown. Replace these forces with an equivalent force-couple system at A. SOLUTION (50 N) (300 N) (120 N) (250 N) (420 N) (50 N) (250 N) (0.2 m) (0.2 m) (0.16 m) (0.2 m) (0.1 m) (0.16 m) B D E = − − − − = − − − = = + = − + R j i i k R i j k r i r ki r i j k [ (300 N) (50 N) ] ( 250 N) ( 120 N) 0.2 m 0 0 0.2 m 0 0.16 m 300 N 50 N 0 0 0 250 N 0.2 m 0.1 m 0.16 m 120 N 0 0 (10 N m) (50 N m) (19.2 N m) (12 N m) R A B D = × − − + × − + × − = + − − − + − − = − ⋅ + ⋅ − ⋅ − ⋅ M r i j r k r i i j k i j k i j k k j j k Force-couple system at A is (420 N) (50 N) (250 N) (30.8 N m) (220 N m)R A= − − − = ⋅ − ⋅R i j k M j k www.elsolucionario.net www.elsolucionario.net
- 289. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 287 PROBLEM 3.122 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force- couple system at A consisting of the force (2.6 lb) +=R i (0.7yR −j lb)k and the couple R A xM= +M i (1.0 lb · ft) −j (0.72 lb · ft) .k (b) Find the corresponding values of yR and .xM SOLUTION (a) From the statement of the problem, equivalence requires :Σ + =F B C R or : 2.6 lbx x xF B CΣ + = (1) :y y yF C RΣ − = (2) : 0.7 lb or 0.7 lbz z zF C CΣ − = − = and / /: ( ) R A B A B C A AMΣ × + + × =M r B M r C or 1.75 : (1 lb ft) ft ( ) 12 x y xC M Σ ⋅ + = M (3) 3.75 1.75 3.5 : ft ( ) ft ( ) ft (0.7 lb) 1 lb ft 12 12 12 y x xM B C Σ + + = ⋅ or 3.75 1.75 9.55x xB C+ = Using Eq. (1): 3.75 1.75(2.6 ) 9.55x xB B+ = or 2.5 lbxB = and 0.1 lbxC = 3.5 : ft ( ) 0.72 lb ft 12 z yM C Σ − = − ⋅ or 2.4686 lbyC = (2.50 lb) (0.1000 lb) (2.47 lb) (0.700 lb)= = − −B i C i j k (b) Eq. (2) 2.47 lbyR = − Using Eq. (3): 1.75 1 (2.4686) 12 xM + = or 1.360 lb ftxM = ⋅ www.elsolucionario.net www.elsolucionario.net
- 290. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 288 PROBLEM 3.123 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of (30 N) + +y zR R= −R i j k and (12 N · m) .R A = −M i (b) Find the corresponding values of yR and .zR (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown? SOLUTION (a) Equivalence requires :Σ = +F R B C or (30 N) ( )y z x y zR R B C C C− + + = − + − + +i j k k i j k Equating the i coefficients: : 30 N or 30 Nx xC C− = − =i Also, / /: R A A B A C AΣ = × + ×M M r B r C or (12 N m) [(0.2 m) (0.15 m) ] ( ) (0.4 m) [ (30 N) ]y z B C C − ⋅ = + × − + × − + + i i j k i i j k Equating coefficients: : 12 N m (0.15 m) or 80 N : 0 (0.4 m) or 0 : 0 (0.2 m)(80 N) (0.4 m) or 40 N y y z z B B C C C C − ⋅ = − = = = = − = i k j (80.0 N) (30.0 N) (40.0 N)= − = − +B k C i k (b) Now we have for the equivalence of forces (30 N) (80 N) [( 30 N) (40 N) ]y zR R− + + = − + − +i j k k i k Equating coefficients: : 0yR =j 0yR = : 80 40zR = − +k or 40.0 NzR = − (c) First note that (30 N) (40 N) .= − −R i k Thus, the screw is best able to resist the lateral force zR when the slot in the head of the screw is vertical. www.elsolucionario.net www.elsolucionario.net
- 291. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 289 PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied. SOLUTION We first reduce the given forces to force-couple systems at A and B, noting that | | | | (40 lb)(10 in.) 400 lb in. A B= = = ⋅ M M We now determine the equivalent force-couple system at C. (40 lb)(1 cos ) (40 lb)sinθ θ= − −R i j (1) (15 in.) [ (40 lb)cos (40 lb)sin ] (7.5 in.) (40 lb) 400 400 600cos 600sin 300 (600 lb in.)sin (300 lb in.)(1 2cos ) R C A B θ θ θ θ θ θ = + + × − − + × = + − − + + = ⋅ + ⋅ − M M M k i j k i j i j i j (2) (a) For no rotation about vertical, y component of R CM must be zero. 1 2cos 0 cos 1/2 θ θ − = = 60.0θ = ° (b) For 60.0θ = ° in Eqs. (1) and (2), (20.0 lb) (34.641 lb) ; (519.62 lb in.)R C= − = ⋅R i j M i (20.0 lb) (34.6 lb) ; (520 lb in.)R C= − = ⋅R i j M i www.elsolucionario.net www.elsolucionario.net
- 292. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 290 PROBLEM 3.125 Assuming θ = 60° in Prob. 3.124, replace the two 40-lb forces with an equivalent force-couple system at D and determine whether the plumber’s action tends to tighten or loosen the joint between (a) pipe CD and elbow D, (b) elbow D and pipe DE. Assume all threads to be right-handed. PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied. SOLUTION The equivalent force-couple system at C for 60θ = ° was obtained in the solution to Prob. 3.124: (20.0 lb) (34.641 lb) (519.62 lb in.)R C = − = ⋅ R i j M i The equivalent force-couple system at D is made of R and R DM where / (519.62 lb in.) (25.0 in.) [(20.0 lb) (34.641 lb) ] (519.62 lb in.) (500 lb in.) R R D C C D= + × = ⋅ + × − = ⋅ − ⋅ M M r R i j i j i k Equivalent force-couple at D: (20.0 lb) (34.6 lb) ; (520 lb in.) (500 lb in.)R C= − = ⋅ − ⋅R i j M i k (a) Since R DM has no component along the y-axis, the plumber’s action will neither loosen nor tighten the joint between pipe CD and elbow. (b) Since the x component of R DM is , the plumber’s action will tend to tighten the joint between elbow and pipe DE. www.elsolucionario.net www.elsolucionario.net
- 293. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 291 PROBLEM 3.126 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if 21.2 lbR = and 13.25 lb · ft.M = SOLUTION We have : A BCλΣ = =F R R R where (42 in.) (96 in.) (16 in.) 106 in. BC − − = i j k λ 21.2 lb (42 96 16 ) 106 A = − −R i j k or (8.40 lb) (19.20 lb) (3.20 lb)A = − −R i j k We have /:A C A AΣ × + =M r R M M where / 1 (42 in.) (48 in.) (42 48 )ft 12 (3.5 ft) (4.0 ft) C A = + = + = + r i k i k i k (8.40 lb) (19.50 lb) (3.20 lb) 42 96 16 (13.25 lb ft) 106 (5.25 lb ft) (12 lb ft) (2 lb ft) BC Mλ = − − = − − + + = ⋅ = − ⋅ + ⋅ + ⋅ R i j k M i j k i j k Then 3.5 0 4.0 lb ft ( 5.25 12 2 ) lb ft 8.40 19.20 3.20 A⋅ + − + + ⋅ = − − i j k i j k M (71.55 lb ft) (56.80 lb ft) (65.20 lb ft)A = ⋅ + ⋅ − ⋅M i j k or (71.6 lb ft) (56.8 lb ft) (65.2 lb ft)A = ⋅ + ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 294. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 292 PROBLEM 3.127 Three children are standing on a 5 5-m× raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights. SOLUTION We have : A B CΣ + + =F F F F R (375 N) (260 N) (400 N) (1035 N) − − − = − = j j j R j R or 1035 NR = We have : ( ) ( ) ( ) ( )x A A B B C C DM F z F z F z R zΣ + + = (375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (1035 N)( )Dz+ + = 3.0483 mDz = or 3.05 mDz = We have : ( ) ( ) ( ) ( )z A A B B C C DM F x F x F x R xΣ + + = 375 N(1 m) (260 N)(1.5 m) (400 N)(4.75 m) (1035 N)( )Dx+ + = 2.5749 mDx = or 2.57 mDx = www.elsolucionario.net www.elsolucionario.net
- 295. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 293 PROBLEM 3.128 Three children are standing on a 5 5-m raft.× The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft. SOLUTION We have : A B CΣ + + =F F F F R (375 N) (260 N) (400 N) (425 N)− − − − =j j j j R (1460 N)= −R j We have : ( ) ( ) ( ) ( ) ( )x A A B B C C D D HM F z F z F z F z R zΣ + + + = (375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (425 N)( ) (1460 N)(2.5 m)Dz + + + = 1.16471mDz = or 1.165 mDz = We have : ( ) ( ) ( ) ( ) ( )z A A B B C C D D HM F x F x F x F x R xΣ + + + = (375 N)(1m) (260 N)(1.5 m) (400 N)(4.75 m) (425 N)( ) (1460 N)(2.5 m)Dx + + + = 2.3235 mDx = or 2.32 mDx = www.elsolucionario.net www.elsolucionario.net
- 296. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 294 PROBLEM 3.129 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when 1 fta = and 12 ft.b = SOLUTION We have Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0). Equivalence then requires : 105 90 160 50zF RΣ − − − − = − or 405 lbR = : (5 ft)(105 lb) (1 ft)(90 lb) (3 ft)(160 lb) (5.5 ft)(50 lb) (405 lb) xM y Σ − + + = − or 2.94 fty = − : (5.5 ft)(105 lb) (12 ft)(90 lb) (14.5 ft)(160 lb) (22.5 ft)(50 lb) (405 lb) yM x Σ + + + = − or 12.60 ftx = R acts 12.60 ft to the right of member AB and 2.94 ft below member BC. www.elsolucionario.net www.elsolucionario.net
- 297. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 295 PROBLEM 3.130 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G. SOLUTION Since R acts at G, equivalence then requires that GΣM of the applied system of forces also be zero. Then at : : ( 3) ft (90 lb) (2 ft)(105 lb) (2.5 ft)(50 lb) 0 xG M aΣ − + × + + = or 0.722 fta = : (9 ft)(105 ft) (14.5 ) ft (90 lb) (8 ft)(50 lb) 0 yM bΣ − − − × + = or 20.6 ftb = www.elsolucionario.net www.elsolucionario.net
- 298. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 296 PROBLEM 3.131* A group of students loads a 2 3.3-m× flatbed trailer with two 0.66 0.66 0.66-m× × boxes and one0.66 0.66 1.2-m× × box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second0.66 0.66× × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.66 0.66 1.2-m× × box should be placed adjacent to one of the edges of the trailer with the 0.66 0.66-m× side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. We have : (224 N) (392 N) (176 N)Σ − − − =F j j j R (792 N)= −R j We have : (224 N)(0.33 m) (392 N)(1.67 m) (176 N)(1.67 m) ( 792 N)( )zM xΣ − − − = − 1.29101mRx = We have : (224 N)(0.33 m) (392 N)(0.6 m) (176 N)(2.0 m) (792 N)( )xM zΣ + + = 0.83475 mRz = From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, 0.GΣ =M Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply (0.33 m 1 m) (1.5 m 2.97 m)x z≤ ≤ ≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 299. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 297 PROBLEM 3.131* (Continued) With 0.33 mLx = at : : (1 0.33) m (1.29101 1) m (792 N) 0z LG M WΣ − × − − × = or 344.00 NLW = Now we must check if this is physically possible, at : : ( 1.5) m 344 N) (1.5 0.83475) m (792 N) 0x LG M zΣ − × − − × = or 3.032 mLz = which is not acceptable. With 2.97 m:Lz = at : : (2.97 1.5) m (1.5 0.83475) m (792 N) 0x LG M WΣ − × − − × = or 358.42 NLW = Now check if this is physically possible, at : : (1 ) m (358.42 N) (1.29101 1) m (792 N) 0z LG M xΣ − × − − × = or 0.357 m ok!Lx = The minimum weight of the fourth box is 358 NLW = And it is placed on end A (0.66 0.66-m× side down) along side AB with the center of the box 0.357 m from side AD. www.elsolucionario.net www.elsolucionario.net
- 300. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 298 PROBLEM 3.132* Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer. PROBLEM 3.131* A group of students loads a 2 3.3-m× flatbed trailer with two 0.66 0.66 0.66-m× × boxes and one 0.66 0.66 1.2-m× × box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 0.66 1.2-m× × box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION First replace the three known loads with a single equivalent force R applied at coordinate ( , 0, ).R Rx z Equivalence requires : 224 392 176yF RΣ − − − = − or 792 N=R : (0.33 m)(224 N) (0.6 m)(392 N) (2 m)(176 N) (792 N) x R M z Σ + + = or 0.83475 mRz = : (0.33 m)(224 N) (1.67 m)(392 N) (1.67 m)(176 N) (792 N) z R M x Σ − − − = or 1.29101 mRx = From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes through G, the point of intersection of the two center lines. Thus, 0GΣ =M Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are 0.6 m or 2.7 mH Hx z= = www.elsolucionario.net www.elsolucionario.net
- 301. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 299 PROBLEM 3.132* (Continued) Now consider these two possibilities. With 0.6 mHx = at : : (1 0.6) m (1.29101 1) m (792 N) 0z HG M WΣ − × − − × = or 576.20 NHW = Checking if this is physically possible at : : ( 1.5) m (576.20 N) (1.5 0.83475) m (792 N) 0x HG M zΣ − × − − × = or 2.414 mHz = which is acceptable. With 2.7 mHz = at : : (2.7 1.5) m (1.5 0.83475) m (792 N) 0x HG M WΣ − × − − × = or 439 NHW = Since this is less than the first case, the maximum weight of the fourth box is 576 NHW = and it is placed with a 0.66 1.2-m× side down, a 0.66-m edge along side AD, and the center 2.41 m from side DC. www.elsolucionario.net www.elsolucionario.net
- 302. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 300 PROBLEM 3.133* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION First reduce the given forces to an equivalent force-couple system ( ), R OR M at the origin. We have : P P P RΣ − + + =F j j k or P=R k 5 : ( ) ( ) 2 R O OaP aP aP M Σ − + − + = M j i k or 5 2 R O aP = − − + M i j k (a) Then for the wrench, R P= and axis R = = R λ k cos 0 cos 0 cos 1x y zθ θ θ= = = or 90 90 0x y zθ θ θ= ° = ° = ° (b) Now 1 axis 5 2 5 2 R OM aP aP = ⋅ = ⋅ − − + = M k i j k λ Then 5 1 2 aP P R P = = M or 5 2 P a= www.elsolucionario.net www.elsolucionario.net
- 303. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 301 PROBLEM 3.133* (Continued) (c) The components of the wrench are 1( , ),R M where 1 1 axis ,M=M λ and the axis of the wrench is assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require z Q R= ×M r R where 1z O= ×M M M Then 5 5 ( ) 2 2 aP aP x y P − − + − = + + i j k k i j k Equating coefficients: : or : or aP yP y a aP xP x a − = = − − = − = i j The axis of the wrench is parallel to the z-axis and intersects the xy-plane at , .x a y a= = − www.elsolucionario.net www.elsolucionario.net
- 304. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 302 PROBLEM 3.134* Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION Force-couple system at O: ( )P P P P= + + = + +R i j k i j k R O a P a P a P Pa Pa Pa = × + × + × = − − − M j i k j i k k i j ( )R O Pa= − + +M i j k Since R and R OM have the same direction, they form a wrench with 1 .R O=M M Thus, the axis of the wrench is the diagonal OA. We note that 1 cos cos cos 3 3 x y z a a θ θ θ= = = = 1 3 54.7 3 x y z R O R P M M Pa θ θ θ= = = = ° = = − 1 3 Pitch 3 M Pa p a R P − = = = = − (a) 3 54.7x y zR P θ θ θ= = = = ° (b) – a (c) Axis of the wrench is diagonal OA. www.elsolucionario.net www.elsolucionario.net
- 305. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 303 PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have : (10 lb) (11lb)Σ − − =F j j R (21lb)= −R j We have : ( ) R O O C OΣ Σ × + Σ =M r F M M 0 0 20 lb in. 0 0 15 lb in. (12 lb in) 0 10 0 0 11 0 (35 lb in.) (12 lb in.) R O = ⋅ + − ⋅ − ⋅ − − = ⋅ − ⋅ i j k i j k M j i j (a) (21lb)= −R j or (21.0 lb)= −R j (b) We have 1 1 ( ) [(35 lb in.) (12 lb in.) ] 12 lb in. and (12 lb in.) R R O RM R = ⋅ = = − ⋅ ⋅ − ⋅ = ⋅ = − ⋅ R λ M λ j i j M j and pitch 1 12 lb in. 0.57143 in. 21lb M p R ⋅ = = = or 0.571in.p = www.elsolucionario.net www.elsolucionario.net
- 306. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 304 PROBLEM 3.135* (Continued) (c) We have 1 2 2 1 (35 lb in.) R O R O = + = − = ⋅ M M M M M M i We require 2 / (35 lb in.) ( ) [ (21lb) ] 35 (21 ) (21 ) Q O x z x z = × ⋅ = + × − = − + M r R i i k j i k i From i: 35 21 1.66667 in. z z = = From k: 0 21 0 x z = − = The axis of the wrench is parallel to the y-axis and intersects the xz-plane at 0, 1.667 in.x z= = www.elsolucionario.net www.elsolucionario.net
- 307. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 305 PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple system. We have : (20 N) (15 N) 25 NRΣ − − = =F i j R We have : ( ) R O O C OΣ Σ × + Σ =M r F M M 20 N(0.1 m) (4 N m) (1 N m) (4 N m) (3 N m) R O = − − ⋅ − ⋅ = − ⋅ − ⋅ M j i j i j (a) (20.0 N) (15.00 N)= − −R i j (b) We have 1 ( 0.8 0.6 ) [ (4 N m) (3 N m) ] 5 N m R R OM R = ⋅ = = − − ⋅ − ⋅ − ⋅ = ⋅ R M i j i j λ λ Pitch: 1 5 N m 0.200 m 25 N M p R ⋅ = = = or 0.200 mp = (c) From above, note that 1 R O=M M Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy-plane with a slope of 3 4 y x= www.elsolucionario.net www.elsolucionario.net
- 308. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 306 PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have : (84 N) (80 N) 116 NRΣ − − = =F j k R and : ( ) R O O C OΣ Σ × + Σ =M r F M M 0.6 0 0.1 0.4 0.3 0 ( 30 32 ) N m 0 84 0 0 0 80 R O+ + − − ⋅ = i j k i j k j k M (15.6 N m) (2 N m) (82.4 N m)R O = − ⋅ + ⋅ − ⋅M i j k (a) (84.0 N) (80.0 N)= − −R j k (b) We have 1 84 80 [ (15.6 N m) (2 N m) (82.4 N m) ] 116 55.379 N m R R O RM R = ⋅ = − − = − ⋅ − ⋅ + ⋅ − ⋅ = ⋅ R λ M λ j k i j k and 1 1 (40.102 N m) (38.192 N m)RM λ= = − ⋅ − ⋅M j k Then pitch 1 55.379 N m 0.47741m 116 N M p R ⋅ = = = or 0.477 mp = www.elsolucionario.net www.elsolucionario.net
- 309. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 307 PROBLEM 3.137* (Continued) (c) We have 1 2 2 1 [( 15.6 2 82.4 ) (40.102 38.192 )] N m (15.6 N m) (42.102 N m) (44.208 N m) R O R O = + = − = − + − − − ⋅ = − ⋅ + ⋅ − ⋅ M M M M M M i j k j k i j k We require 2 / ( 15.6 42.102 44.208 ) ( ) (84 80 ) (84 ) (80 ) (84 ) Q O x z z x x = × − + − = + × − = + − M r R i j k i k j k i j k From i: 15.6 84 0.185714 m z z − = = − or 0.1857 mz = − From k: 44.208 84 0.52629 m x x − = − = or 0.526 mx = The axis of the wrench intersects the xz-plane at 0.526 m 0 0.1857 mx y z= = = − www.elsolucionario.net www.elsolucionario.net
- 310. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 308 PROBLEM 3.138* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin at B. (a) We have 8 15 : (26.4 lb) (17 lb) 17 17 Σ − − + = F k i j R (8.00 lb) (15.00 lb) (26.4 lb)= − − −R i j k and 31.4 lbR = We have /: R B A B A A B BΣ × + + =M r F M M M 8 15 0 10 0 220 238 264 220 14(8 15 ) 17 17 0 0 26.4 (152 lb in.) (210 lb in.) (220 lb in.) R B R B = − − − + = − − + − = ⋅ − ⋅ − ⋅ i j k M k i j i k i j M i j k (b) We have 1 8.00 15.00 26.4 [(152 lb in.) (210 lb in.) (220 lb in.) ] 31.4 246.56 lb in. R R O RM R = ⋅ = − − − = ⋅ ⋅ − ⋅ − ⋅ = ⋅ R λ M λ i j k i j k www.elsolucionario.net www.elsolucionario.net
- 311. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 309 PROBLEM 3.138* (Continued) and 1 1 (62.818 lb in.) (117.783 lb in.) (207.30 lb in.)RM λ= = − ⋅ − ⋅ − ⋅M i j k Then pitch 1 246.56 lb in. 7.8522 in. 31.4 lb M p R ⋅ = = = or 7.85 in.p = (c) We have 1 2 2 1 (152 210 220 ) ( 62.818 117.783 207.30 ) (214.82 lb in.) (92.217 lb in.) (12.7000 lb in.) R B R B = + = − = − − − − − − = ⋅ − ⋅ − ⋅ M M M M M M i j k i j k i j k We require 2 /Q B= ×M r R 214.82 92.217 12.7000 0 8 15 26.4 (15 ) (8 ) (26.4 ) (15 ) x z z z x x − − = − − − = − + − i j k i j k i j j k From i: 214.82 15 14.3213 in.z z= = From k: 12.7000 15 0.84667 in.x x− = − = The axis of the wrench intersects the xz-plane at 0.847 in. 0 14.32 in.x y z= = = www.elsolucionario.net www.elsolucionario.net
- 312. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 310 PROBLEM 3.139* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION (a) First reduce the given force system to a force-couple at the origin. We have : BA DC DEP P PΣ + + =F λ λ λ R 4 3 3 4 9 4 12 5 5 5 5 25 5 25 P − = − + − + − + R j k i j i j k 3 (2 20 ) 25 P = − −R i j k 2 2 23 27 5 (2) (20) (1) 25 25 P R P= + + = We have : ( ) R O OPΣ Σ × =M r M 4 3 3 4 9 4 12 (24 ) (20 ) (20 ) 5 5 5 5 25 5 25 R O P P P P P P P a a a − − × − + × − + × − + = j j k j i j j i j k M 24 ( ) 5 R O Pa = − −M i k www.elsolucionario.net www.elsolucionario.net
- 313. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 311 PROBLEM 3.139* (Continued) (b) We have 1 R R OM = ⋅Mλ where 3 25 1 (2 20 ) (2 20 ) 25 27 5 9 5 R P R P = = − − = − − R i j k i j kλ Then 1 1 24 8 (2 20 ) ( ) 59 5 15 5 Pa Pa M − = − − ⋅ − − =i j k i k and pitch 1 8 25 8 8115 5 27 5 M Pa a p R P − − = = = or 0.0988p a= − (c) 1 1 8 1 8 (2 20 ) ( 2 20 ) 67515 5 9 5 R Pa Pa M − = = − − = − + + M i j k i j kλ Then 2 1 24 8 8 ( ) ( 2 20 ) ( 430 20 406 ) 5 675 675 R O Pa Pa Pa = − = − − − − + + = − − −M M M i k i j k i j k We require 2 /Q O= ×M r R 8 3 ( 403 20 406 ) ( ) (2 20 ) 675 25 3 [20 ( 2 ) 20 ] 25 Pa P x z P z x z x − − − = + × − − = + + − i j k i k i j k i j k From i: 3 8( 403) 20 1.99012 675 25 Pa P z z a − = = − From k: 3 8( 406) 20 2.0049 675 25 Pa P x x a − = − = The axis of the wrench intersects the xz-plane at 2.00 , 1.990x a z a= = − www.elsolucionario.net www.elsolucionario.net
- 314. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 312 PROBLEM 3.140* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz-plane. SOLUTION (a) First replace the given forces with an equivalent force-couple system ( ), R OR M at the origin. We have 2 2 2 2 2 2 (6) (2) (9) 11 m (14) (2) (5) 15 m AC BD d d = + + = = + + = Then 1650 N (6 2 9 ) 11 (900 N) (300 N) (1350 N) ACT = = + + = + + i j k i j k and 1500 N (14 2 5 ) 15 (1400 N) (200 N) (500 N) BDT = = + + = + + i j k i j k Equivalence then requires : (900 300 1350 ) (1400 200 500 ) (2300 N) (500 N) (1850 N) AC BDΣ = + = + + + + + = + + F R T T i j k i j k i j k : R O O A AC B BDΣ = × + ×M M r T r T (12 m) [(900 N) (300 N) (1350 N) ] (9 m) [(1400 N) (200 N) (500 N) ] (3600) (10,800 4500) (1800) (3600 N m) (6300 N m) (1800 N m) = × + + + × + + = − + − + = − ⋅ + ⋅ + ⋅ k i j k i i j k i j k i j k The components of the wrench are 1( , ),R M where (2300 N) (500 N) (1850 N)= + +R i j k www.elsolucionario.net www.elsolucionario.net
- 315. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 313 PROBLEM 3.140* (Continued) (b) We have 2 2 2 100 (23) (5) (18.5) 2993.7 NR = + + = Let axis 1 (23 5 18.5 ) 29.937R = = + + R i j kλ Then 1 axis 1 (23 5 18.5 ) ( 3600 6300 1800 ) 29.937 1 [(23)( 36) (5)(63) (18.5)(18)] 0.29937 601.26 N m R OM = ⋅ = + + ⋅ − + + = − + + = − ⋅ M i j k i j k λ Finally, 1 601.26 N m 2993.7 N M P R − ⋅ = = or 0.201 mP = − (c) We have 1 1 axis 1 ( 601.26 N m) (23 5 18.5 ) 29.937 M M= = − ⋅ × + + λ i j k or 1 (461.93 N m) (100.421 N m) (371.56 N m)= − ⋅ − ⋅ − ⋅M i j k Now 2 1 ( 3600 6300 1800 ) ( 461.93 100.421 371.56 ) (3138.1 N m) (6400.4 N m) (2171.6 N m) R O= − = − + + − − − − = − ⋅ + ⋅ + ⋅ M M M i j k i j k i j k For equivalence: www.elsolucionario.net www.elsolucionario.net
- 316. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 314 PROBLEM 3.140* (Continued) Thus, we require 2 ( )P y z= × = +M r R r j k Substituting: 3138.1 6400.4 2171.6 0 2300 500 1850 y z− + + = i j k i j k Equating coefficients: : 6400.4 2300 or 2.78 m : 2171.6 2300 or 0.944 m z z y y = = = − = − j k The axis of the wrench intersects the yz-plane at 0.944 m 2.78 my z= − = www.elsolucionario.net www.elsolucionario.net
- 317. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 315 PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane. SOLUTION First determine the resultant of the forces at D. We have 2 2 2 2 2 2 ( 12) (9) (8) 17 in. ( 6) (0) ( 8) 10 in. DA ED d d = − + + = = − + + − = Then 34 lb ( 12 9 8 ) 17 (24 lb) (18 lb) (16 lb) DA = = − + + = − + + F i j k i j k and 30 lb ( 6 8 ) 10 (18 lb) (24 lb) ED = = − − = − − F i k i k Then : DA EDΣ = +F R F F ( 24 18 16 ( 18 24 ) (42 lb) (18 lb) (8 lb) = − + + + − − = − + − i j k i k i j k For the applied couple 2 2 2 ( 6) ( 6) (18) 6 11 in.AKd = − + − + = Then 160 lb in. ( 6 6 18 ) 6 11 160 [ (1 lb in.) (1 lb in.) (3 lb in.) ] 11 ⋅ = − − + = − ⋅ − ⋅ + ⋅ M i j k i j k To be able to reduce the original forces and couple to a single equivalent force, R and M must be perpendicular. Thus ? 0⋅ =R M www.elsolucionario.net www.elsolucionario.net
- 318. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 316 PROBLEM 3.141* (Continued) Substituting ?160 ( 42 18 8 ) ( 3 ) 0 11 − + − ⋅ − − + =i j k i j k or ?160 [( 42)( 1) (18)( 1) ( 8)(3)] 0 11 − − + − + − = or 0 0= R and M are perpendicular so that the given system can be reduced to the single equivalent force. (42.0 lb) (18.00 lb) (8.00 lb)= − + −R i j k Then for equivalence, Thus, we require /P D= ×M r R where / (12 in.) [( 3)in.] ( in.)P D y z= − + − +r i j k Substituting: 160 ( 3 ) 12 ( 3) 11 42 18 8 [( 3)( 8) ( )(18)] [( )( 42) ( 12)( 8)] [( 12)(18) ( 3)( 42)] y z y z z y − − + = − − − − = − − − + − − − − + − − − − i j k i j k i j k Equating coefficients: 160 : 42 96 or 1.137 in. 11 480 : 216 42( 3) or 11.59 in. 11 z z y y − = − − = − = − + − = j k The line of action of R intersects the yz-plane at 0 11.59 in. 1.137 in.x y z= = = − www.elsolucionario.net www.elsolucionario.net
- 319. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 317 PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have : A GΣ + =F F F R (40 mm) (60 mm) (120 mm) (50 N) 70 N 140 mm (20 N) (30 N) (10 N) + − = + = + − i j k R k i j k and 37.417 NR = We have : ( ) R O O C OΣ Σ × + Σ =M r F M M 0 [(0.12 m) (50 N) ] {(0.16 m) [(20 N) (30 N) (60 N) ]} (160 mm) (120 mm) (10 N m) 200 mm (40 mm) (120 mm) (60 mm) (14 N m) 140 mm (18 N m) (8.4 N m) (10.8 N m) R O R = × + × + − − + ⋅ − + + ⋅ = ⋅ − ⋅ + ⋅ M j k i i j k i j i j k M i j k To be able to reduce the original forces and couples to a single equivalent force, R and M must be perpendicular. Thus, 0.⋅ =R M Substituting ? (20 30 10 ) (18 8.4 10.8 ) 0+ − ⋅ − + =i j k i j k or ? (20)(18) (30)( 8.4) ( 10)(10.8) 0+ − + − = or 0 0= R and M are perpendicular so that the given system can be reduced to the single equivalent force. (20.0 N) (30.0 N) (10.00 N)= + −R i j k www.elsolucionario.net www.elsolucionario.net
- 320. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 318 PROBLEM 3.142* (Continued) Then for equivalence, Thus, we require R O p p y z= × = +M r R r j k Substituting: 18 8.4 10.8 0 20 30 10 y z− + = − i j k i j k Equating coefficients: : 8.4 20 or 0.42 m : 10.8 20 or 0.54 m z z y y − = = − = − = − j k The line of action of R intersects the yz-plane at 0 0.540 m 0.420 mx y z= = − = − www.elsolucionario.net www.elsolucionario.net
- 321. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 319 PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y-axis and applied respectively at A and B. SOLUTION Express the forces at A and B as x z x z A A B B = + = + A i k B i k Then, for equivalence to the given force system, : 0x x xF A BΣ + = (1) :z z zF A B RΣ + = (2) : ( ) ( ) 0x z zM A a B a bΣ + + = (3) : ( ) ( )z x xM A a B a b MΣ − − + = (4) From Equation (1), x xB A= − Substitute into Equation (4): ( ) ( )x xA a A a b M− + + = andx x M M A B b b = = − From Equation (2), z zB R A= − and Equation (3), ( )( ) 0z zA a R A a b+ − + = 1z a A R b = + and 1z a B R R b = − + z a B R b = − Then 1 M a R b b = + + A i k M a R b b = − − B i k www.elsolucionario.net www.elsolucionario.net
- 322. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 320 PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane. SOLUTION First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as andx y z x y zR R R M M M M= + + = + +R i j k i j k while the unknown forces A and B can be expressed as andx y z x zA A A B B= + + = +A i j k B i k Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems, :x x x xF R A BΣ = + (1) :y y yF R AΣ = (2) :z z z zF R A BΣ = + (3) :x x z yM M yA zAΣ = − (4) :y y x z zM M zA xA bBΣ = − − (5) :z z y xM M xA yAΣ = − (6) Based on the above six independent equations for the six unknowns ( , , , , , ),x y z x zA A A B B b there exists a unique solution for A and B. From Equation (2), y yA R= www.elsolucionario.net www.elsolucionario.net
- 323. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 321 PROBLEM 3.144* (Continued) Equation (6): 1 ( )x y zA xR M y = − Equation (1): 1 ( )x x y zB R xR M y = − − Equation (4): 1 ( )z x yA M zR y = + Equation (3): 1 ( )z z x yB R M zR y = − + Equation (5): ( ) ( ) x y z x z y xM yM zM b M yR zR + + = − + www.elsolucionario.net www.elsolucionario.net
- 324. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 322 PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point. SOLUTION First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. We have andR M= =R j M j and are known. The unknown forces A and B can be expressed as andx y z x y zA A A B B B= + + = + +A i j k B i j k The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence, :xFΣ 0 x xA B= + (1) :yFΣ y yR A B= + (2) :zFΣ 0 z zA B= + (3) :xMΣ 0 yzB= − (4) :y z z xM M aA xB zBΣ = − − + (5) :zMΣ 0 y yaA xB= + (6) Since A and B are made perpendicular, 0 or 0x x y y z zA B A B A B⋅ = + + =A B (7) There are eight unknowns: , , , , , , ,x y z x y zA A A B B B x z But only seven independent equations. Therefore, there exists an infinite number of solutions. Next, consider Equation (4): 0 yzB= − If 0,yB = Equation (7) becomes 0x x z zA B A B+ = Using Equations (1) and (3), this equation becomes 2 2 0x zA A+ = www.elsolucionario.net www.elsolucionario.net
- 325. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 323 PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that 0,yB ≠ so that from Equation (4), 0.z = To obtain one possible solution, arbitrarily let 0.xA = (Note: Setting , , ory z zA A B equal to zero results in unacceptable solutions.) The defining equations then become 0 xB= (1)′ y yR A B= + (2) 0 z zA B= + (3) z zM aA xB= − − (5)′ 0 y yaA xB= + (6) 0y y z zA B A B+ = (7)′ Then Equation (2) can be written y yA R B= − Equation (3) can be written z zB A= − Equation (6) can be written y y aA x B = − Substituting into Equation (5)′, ( ) y z z y R B M aA a A B − = − − − − or z y M A B aR = − (8) Substituting into Equation (7)′, ( ) 0y y y y M M R B B B B aR aR − + − = or 2 3 2 2 2y a R B a R M = + Then from Equations (2), (8), and (3), 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 2 y z z a R RM A R a R M a R M M a R aR M A aR a R M a R M aR M B a R M = − = + + = − = − + + = + www.elsolucionario.net www.elsolucionario.net
- 326. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 324 PROBLEM 3.145* (Continued) In summary, 2 2 2 ( ) RM M aR a R M = − + A j k 2 2 2 2 ( ) aR aR M a R M = + + B j k Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if 0R > and 0,M > it follows from the equations found for A and B that 0yA > and 0.yB > From Equation (6), 0x < (assuming 0).a > Then, as a consequence of letting 0,xA = force A lies in a plane parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yz-plane but to the left to the origin, as shown in the figure below. www.elsolucionario.net www.elsolucionario.net
- 327. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 325 PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action. SOLUTION First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA′ by A x y zλ λ λ λ= + +i j k it follows that force A can be expressed as ( )A x y zA Aλ λ λ λ= = + +A i j k Force B can be expressed as x y zB B B= + +B i j k Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then for equivalence, : 0x x xF A BλΣ = + (1) :y y yF R A BλΣ = + (2) : 0z z zF A BλΣ = + (3) : 0x yM zBΣ = − (4) :y z x zM M aA zB xBλΣ = − + − (5) : 0x y yM aA xBλΣ = − + (6) Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution. www.elsolucionario.net www.elsolucionario.net
- 328. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 326 PROBLEM 3.146* (Continued) Case 1: Let 0z = to satisfy Equation (4). Now Equation (2): y yA R Bλ = − Equation (3): z zB Aλ= − Equation (6): ( ) y y y y aA a x R B B B λ = − = − − Substitution into Equation (5): ( )( ) 1 z y z y y z a M aA R B A B M A B aR λ λ λ = − − − − − = − Substitution into Equation (2): 2 1 y y y z z y z y M R B B aR aR B aR M λ λ λ λ λ = − + = − Then z y y z x x x z y z z z z y MR R A aRaR M M MR B A aR M MR B A aR M λ λ λ λ λ λ λ λ λ λ λ λ = − = − − = − = − = − = − In summary, A y z P aR M λ λ λ = − A ( )x z z z y R M aR M aR M λ λ λ λ λ = + + − B i j k and 2 1 1 y z y z R x a B aR M a R aR λ λ λ = − − = − or y z M x R λ λ = Note that for this case, the lines of action of both A and B intersect the x-axis. www.elsolucionario.net www.elsolucionario.net
- 329. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 327 PROBLEM 3.146* (Continued) Case 2: Let 0yB = to satisfy Equation (4). Now Equation (2): y R A λ = Equation (1): x x y B R λ λ = − Equation (3): z z y B R λ λ = − Equation (6): 0yaAλ = which requires 0a = Substitution into Equation (5): orx z z x y y y M M z R x R x z R λ λ λ λ λ λ λ = − − − − = This last expression is the equation for the line of action of force B. In summary, A y R λ λ = A ( )x x y R λ λ λ = − − B i k Assuming that , , 0,x y zλ λ λ Ͼ the equivalent force system is as shown below. Note that the component of A in the xz-plane is parallel to B. www.elsolucionario.net www.elsolucionario.net
- 330. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 328 PROBLEM 3.147 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. SOLUTION (a) (300 N)cos 25 271.89 N (300 N)sin 25 126.785 N (271.89 N) (126.785 N) x y F F = ° = = ° = = +F i j (0.1m) (0.2 m) [ (0.1m) (0.2 m) ] [(271.89 N) (126.785 N) ] (12.6785 N m) (54.378 N m) (41.700 N m) D D DA= = − − = × = − − × + = − ⋅ + ⋅ = ⋅ r i j M r F M i j i j k k k 41.7 N mD = ⋅M (b) The smallest force Q at B must be perpendicular to DB at 45° ( ) 41.700 N m (0.28284 m) D Q DB Q = ⋅ = M 147.4 NQ = 45.0° www.elsolucionario.net www.elsolucionario.net
- 331. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 329 PROBLEM 3.148 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A. SOLUTION First note 2 2 (12.0 in.) (2.33 in.) 12.2241in. CBd = + = Then 12.0 in. cos 12.2241in. 2.33 in. sin 12.2241in. θ θ = = and cos sin 125 lb [(12.0 in.) (2.33 in.) ] 12.2241in. CB CB CBF Fθ θ= − = − F i j i j Now /A B A CB= ×M r F where / (15.3 in.) (12.0 in. 2.33 in.) (15.3 in.) (14.33 in.) B A = − + = − r i j i j Then 125 lb [(15.3 in.) (14.33 in.) ] (12.0 2.33 ) 12.2241in. (1393.87 lb in.) A = − × − = ⋅ M i j i j k (116.156 lb ft)= ⋅ k or 116.2 lb ftA = ⋅M www.elsolucionario.net www.elsolucionario.net
- 332. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 330 PROBLEM 3.149 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION (a) We have /A E A DE= ×M r T where / 2 2 2 (2.3 m) (0.6 m) (3.3 m) (3 m) (810 N) (0.6) (3.3) (3) m (108 N) (594 N) (540 N) E A DE DE DET = = + − = + + = + − r j T λ i j k i j k 0 2.3 0 N m 108 594 540 (1242 N m) (248.4 N m) A = ⋅ − = − ⋅ − ⋅ i j k M i k or (1242 N m) (248 N m)A = − ⋅ − ⋅M i k (b) We have /A G A CG= ×M r T where / 2 2 2 (2.7 m) (2.3 m) (.6 m) (3.3 m) (3 m) (810 N) (.6) (3.3) (3) m (108 N) (594 N) (540 N) G A CG CG CGT = + = − + − = + + = − + − r i j T λ i j k i j k 2.7 2.3 0 N m 108 594 540 (1242 N m) (1458 N m) (1852 N m) A = ⋅ − − = − ⋅ + ⋅ + ⋅ i j k M i j k or (1242 N m) (1458 N m) (1852 N m)A = − ⋅ + ⋅ + ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 333. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 331 PROBLEM 3.150 Section AB of a pipeline lies in the yz-plane and forms an angle of 37° with the z-axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD. SOLUTION First note (sin 37 cos 37 ) ( cos 40 cos 55 sin 40 cos 40 sin 55 ) AB AB CD = ° − ° = − ° ° + ° − ° ° j k CD j j k Now ( )( )cosAB CD AB CD θ⋅ = or (sin 37 cos 37 ) ( cos 40 cos 55 sin 40 cos 40 sin 55 )AB CD° − ° ⋅ − ° ° + ° − ° °j k i j k ( )( )cosAB CD θ= or cos (sin 37 )(sin 40 ) ( cos 37 )( cos 40 sin 55 ) 0.88799 θ = ° ° + − ° − ° ° = or 27.4θ = ° www.elsolucionario.net www.elsolucionario.net
- 334. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 332 PROBLEM 3.151 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a. SOLUTION First note (2.2 m) (3.2 m) ( m)BA a= − −i j k Now /D A D BA= ×M r T where / (2.2 m) (1.6 m) (2.2 3.2 ) (N) A D BA BA BA T a d = + = − − r i j T i j k Then 2.2 1.6 0 2.2 3.2 { 1.6 2.2 [(2.2)( 3.2) (1.6)(2.2)] } BA D BA BA BA T d a T a a d = − − = − + + − − i j k M i j k Thus 2.2 (N m) 10.56 (N m) BA y BA BA z BA T M a d T M d = ⋅ = − ⋅ Then forming the ratio y z M M 2.2 (N m)120 N m 460 N m 10.56 (N m) BA BA BA BA T d T d ⋅⋅ = − ⋅ − ⋅ or 1.252 ma = www.elsolucionario.net www.elsolucionario.net
- 335. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 333 PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that 25 ,θ = ° Mx 61lb ft,= − ⋅ and 43 lb ft,zM = − ⋅ determine φ and d. SOLUTION We have /:O A O OΣ × =M r F M where / (4 in.) (11in.) ( ) (cos cos sin cos sin ) A O d F θ φ θ θ φ = − + − = − + r i j k F i j k For 70 lb, 25F θ= = ° (70 lb)[(0.90631cos ) 0.42262 (0.90631sin ) ] (70 lb) 4 11 in. 0.90631cos 0.42262 0.90631sin (70 lb)[(9.9694sin 0.42262 ) ( 0.90631 cos 3.6252sin ) (1.69048 9.9694cos ) ] in. O d d d φ φ φ φ φ φ φ φ = − + = − − − − = − + − + + − F i j k i j k M i j k and (70 lb)(9.9694sin 0.42262 ) in. (61lb ft)(12 in./ft)xM dφ= − = − ⋅ (1) (70 lb)( 0.90631 cos 3.6252sin ) in.yM d φ φ= − + (2) (70 lb)(1.69048 9.9694cos ) in. 43 lb ft(12 in./ft)zM φ= − = − ⋅ (3) From Equation (3): 1 634.33 cos 24.636 697.86 φ − = = ° or 24.6φ = ° From Equation (1): 1022.90 34.577 in. 29.583 d = = or 34.6 in.d = www.elsolucionario.net www.elsolucionario.net
- 336. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 334 PROBLEM 3.153 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent force- couple system (a) at A, (b) at B. SOLUTION (a) Based on : 560 lbAF F TΣ = = or 560 lbA =F 20.0° : ( sin50 )( )A A AM M T dΣ = ° (560 lb)sin50 (18 ft) 7721.7 lb ft = ° = ⋅ or 7720 lb ftA = ⋅M (b) Based on : 560 lbBF F TΣ = = or 560 lbB =F 20.0° : ( sin50 )( )B B BM M T dΣ = ° (560 lb)sin50°(10 ft) 4289.8 lb ft = = ⋅ or 4290 lb ftB = ⋅M www.elsolucionario.net www.elsolucionario.net
- 337. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 335 PROBLEM 3.154 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle. SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. We have 2.9 lb 2.65 lb 0.25 lb,BF = − = where the 2.65-lb force is part of the couple. Combining the two parallel forces, couple (2.65 lb)[(3.2 in. 2.8 in.)cos25 ] 14.4103 lb in. M = + ° = ⋅ and couple 14.4103 lb in.= ⋅M A single equivalent force will be located in the negative z direction. Based on : 14.4103 lb in. [(0.25 lb)cos 25 ]( )BM aΣ − ⋅ = ° 63.600 in.a = F′ (0.25 lb)(cos 25 sin 25 )= ° + °i k F′ (0.227 lb) (0.1057 lb)= +i k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B. www.elsolucionario.net www.elsolucionario.net
- 338. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 336 PROBLEM 3.155 Replace the 150-N force with an equivalent force-couple system at A. SOLUTION Equivalence requires : (150 N)( cos 35 sin 35 ) (122.873 N) (86.036 N) Σ = − ° − ° = − − F F j k j k /:A D AΣ = ×M M r F where / (0.18 m) (0.12 m) (0.1 m)D A = − +r i j k Then 0.18 0.12 0.1 N m 0 122.873 86.036 [( 0.12)( 86.036) (0.1)( 122.873)] [ (0.18)( 86.036)] [(0.18)( 122.873)] (22.6 N m) (15.49 N m) (22.1 N m) = − ⋅ − − = − − − − + − − + − = ⋅ + ⋅ − ⋅ i j k M i j k i j k The equivalent force-couple system at A is (122.9 N) (86.0 N)= − −F j k (22.6 N m) (15.49 N m) (22.1 N m)= ⋅ + ⋅ − ⋅M i j k www.elsolucionario.net www.elsolucionario.net
- 339. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 337 PROBLEM 3.156 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam. SOLUTION For equivalence, : 1300 400 400 600y a F R b Σ − + − − = − or 2300 400 N a R b = − (1) : 400 (400) ( )(600) 2 A a a M a a b LR b Σ − − + = − or 2 1000 600 200 2300 400 a a b bL a b + − = − Then with 24 10 9 31.5 m 8 23 3 a a b L a + − = = − (2) where a, L are in m. (a) Find value of a to maximize L. 2 2 8 8 4 8 10 23 10 9 3 3 3 3 8 23 3 a a a a dL da a − − − + − − = − www.elsolucionario.net www.elsolucionario.net
- 340. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 338 PROBLEM 3.156 (Continued) or 2 2184 80 64 80 32 230 24 0 3 3 9 3 9 a a a a a− − + + + − = or 2 16 276 1143 0a a− + = Then 2 276 ( 276) 4(16)(1143) 2(16) a ± − − = or 10.3435 m and 6.9065 ma a= = Since 9 m,AB = a must be less than 9 m 6.91 ma = (b) Using Eq. (1), 6.9065 2300 400 1.5 R = − or 458 NR = and using Eq. (2), 24 10(6.9065) 9 (6.9065) 3 3.16 m 8 23 (6.9065) 3 L + − = = − R is applied 3.16 m to the right of A. www.elsolucionario.net www.elsolucionario.net
- 341. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 339 PROBLEM 3.157 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force (8 lb) + (4 lb)=C i k and the couple (360 lb ·C =M in.)i, determine the forces applied at A and at B when 2 lb.zA = SOLUTION We have :ΣF + =A B C or : 8 lbx x xF A B+ = ( 8 lb)x xB A= − + (1) : 0y y yF A BΣ + = or y yA B= − (2) : 2 lb 4 lbz zF BΣ + = or 2 lbzB = (3) We have / /:C B C A C CΣ × + × =M r B r A M 8 0 2 8 0 8 lb in. (360 lb in.) 2 2x y x yB B A A + ⋅ = ⋅ i j k i j k i or (2 8 ) (2 16 8 16)y y x xB A B A− + − + −i j (8 8 ) (360 lb in.)y yB A+ + = ⋅k i From i-coefficient: 2 8 360 lb in.y yB A− = ⋅ (4) j-coefficient: 2 8 32 lb in.x xB A− + = ⋅ (5) k-coefficient: 8 8 0y yB A+ = (6) www.elsolucionario.net www.elsolucionario.net
- 342. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 340 PROBLEM 3.157 (Continued) From Equations (2) and (4): 2 8( ) 360y yB B− − = 36 lb 36 lby yB A= = From Equations (1) and (5): 2( 8) 8 32x xA A− − + = 1.6 lbxA = From Equation (1): (1.6 8) 9.6 lbxB = − + = − (1.600 lb) (36.0 lb) (2.00 lb)= − +A i j k (9.60 lb) (36.0 lb) (2.00 lb)= − + +B i j k www.elsolucionario.net www.elsolucionario.net
- 343. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 341 PROBLEM 3.158 A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat. SOLUTION From the statement of the problem, it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that 0 or 0 0O x zM MΣ = Σ = Σ =M For the applied loads: Then 0: (6 3 ft) (6 3 ft)(10 kips) (6 3 ft)(20 kips) (6 3 ft) 0 x B F F F Σ = + − − = M or 10B FF F− = (1) 0: (12 ft)(15 kips) (6 ft) (6 ft)(10 kips) (12 ft)(30 kips) (6 ft)(20 kips) (6 ft) 0 z B F F F Σ = + − − − + = M or 60B FF F+ = (2) Then Eqs.(1) (2)+ 35.0 kipsB =F and 25.0 kipsF =F www.elsolucionario.net www.elsolucionario.net
- 347. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 345 PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION Free-Body Diagram: 2 2 (350 kg)(9.81 m/s ) 3.4335 kN (1400 kg)(9.81m/s ) 13.7340 kNt W W = = = = (a) Rear wheels: 0: (1.7 m 2.05 m) (2.05 m) (1.2 m) 2 (3 m) 0B tM W W W AΣ = + + + − = (3.4335 kN)(3.75 m) (3.4335 kN)(2.05 m) (13.7340 kN)(1.2 m) 2 (3 m) 0A + + − = 6.0659 kNA = + 6.07 kN=A (b) Front wheels: 0: 2 2 0y tF W W W A BΣ = − − − + + = 3.4335 kN 3.4335 kN 13.7340 kN 2(6.0659 kN) 2 0B− − − + + = 4.2346 kNB = + 4.23 kN=B www.elsolucionario.net www.elsolucionario.net
- 348. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 346 PROBLEM 4.2 Solve Problem 4.1, assuming that crate D is removed and that the position of crate C is unchanged. PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION Free-Body Diagram: 2 2 (350 kg)(9.81 m/s ) 3.4335 kN (1400 kg)(9.81m/s ) 13.7340 kNt W W = = = = (a) Rear wheels: 0: (1.7 m 2.05 m) (1.2 m) 2 (3 m) 0B tM W W AΣ = + + − = (3.4335 kN)(3.75 m) (13.7340 kN)(1.2 m) 2 (3 m) 0A+ − = 4.8927 kNA = + 4.89 kN=A (b) Front wheels: 0: 2 2 0y tM W W A BΣ = − − + + = 3.4335 kN 13.7340 kN 2(4.8927 kN) 2 0B− − + + = 3.6911 kNB = + 3.69 kN=B www.elsolucionario.net www.elsolucionario.net
- 349. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 347 PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if 10 in.,a = (b) if 7 in.a = SOLUTION Free-Body Diagram: 0: 0x xF BΣ = = 0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0BM a a AΣ = − − + + = (40 160) 12 a A − = (1) 0: (40 lb)(6 in.) (50 lb)(12 in.) (30 lb)( 12 in.) (10 lb)( 20 in.) (12 in.) 0 A y M a a B Σ = − − − + − + + = (1400 40 ) 12 y a B + = Since (1400 40 ) 0, 12 x a B B + = = (2) (a) For 10 in.,a = Eq. (1): (40 10 160) 20.0 lb 12 A × − = = + 20.0 lb=A Eq. (2): (1400 40 10) 150.0 lb 12 B + × = = + 150.0 lb=B (b) For 7 in.,a = Eq. (1): (40 7 160) 10.00 lb 12 A × − = = + 10.00 lb=A Eq. (2): (1400 40 7) 140.0 lb 12 B + × = = + 140.0 lb=B www.elsolucionario.net www.elsolucionario.net
- 350. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 348 PROBLEM 4.4 For the bracket and loading of Problem 4.3, determine the smallest distance a if the bracket is not to move. PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if 10 in.,a = (b) if 7 in.a = SOLUTION Free-Body Diagram: For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to 0.=A 0: (40 lb)(6 in.) (30 lb) (10 lb)( 8 in.) (12 in.) 0BM a a AΣ = − − + + = (40 160) 12 a A − = 0: (40 160) 0A a= − = 4.00 in.a = www.elsolucionario.net www.elsolucionario.net
- 351. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 349 PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when 35 ,α = ° (b) the corresponding reaction at each of the two wheels. SOLUTION Free-Body Diagram: 2 1 2 (40 kg)(9.81 m/s ) 392.40 N (300 mm)sin (80 mm)cos (430 mm)cos (300 mm)sin (930 mm)cos W mg a a b α α α α α = = = = − = − = From free-body diagram of hand truck, Dimensions in mm 2 10: ( ) ( ) ( ) 0BM P b W a W aΣ = − + = (1) 0: 2 2 0yF P W BΣ = − + = (2) For 35α = ° 1 2 300sin35 80cos35 106.541 mm 430cos35 300sin35 180.162 mm 930cos35 761.81 mm a a b = ° − ° = = ° − ° = = ° = (a) From Equation (1): (761.81 mm) 392.40 N(180.162 mm) 392.40 N(106.54 mm) 0P − + = 37.921 NP = or 37.9 N=P (b) From Equation (2): 37.921 N 2(392.40 N) 2 0B− + = or 373 N=B www.elsolucionario.net www.elsolucionario.net
- 352. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 350 PROBLEM 4.6 Solve Problem 4.5 when α = 40°. PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels. SOLUTION Free-Body Diagram: 2 1 2 (40 kg)(9.81 m/s ) 392.40 N (300 mm)sin (80 mm)cos (430 mm)cos (300 mm)sin (930 mm)cos W mg W a a b α α α α α = = = = − = − = From F.B.D.: 2 10: ( ) ( ) ( ) 0BM P b W a W aΣ = − + = 2 1( )/P W a a b= − (1) 0: 2 0yF W W P BΣ = − − + + = 1 2 B W P= − (2) For 40 :α = ° 1 2 300sin 40 80cos40 131.553 mm 430cos40 300sin 40 136.563 mm 930cos40 712.42 mm a a b = ° − ° = = ° − ° = = ° = (a) From Equation (1): 392.40 N (0.136563 m 0.131553 m) 0.71242 m P − = 2.7595 NP = 2.76 N=P (b) From Equation (2): 1 392.40 N (2.7595 N) 2 B = − 391 N=B www.elsolucionario.net www.elsolucionario.net
- 353. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 351 PROBLEM 4.7 A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B. SOLUTION Free-Body Diagram: (a) Front wheels: 0: (1700 lb)(52 in.) (3200 lb)(12 in.) 2 (36 in.) 0BM AΣ = + − = 1761.11lbA = + 1761lb=A (b) Rear wheels: 0: 1700 lb 3200 lb 2(1761.11lb) 2 0yF BΣ = − − + + = 688.89 lbB = + 689 lb=B www.elsolucionario.net www.elsolucionario.net
- 354. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 352 PROBLEM 4.8 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC. SOLUTION Free-Body Diagram: (a) Reaction at A: 0: 0x xF AΣ = = 0: (15 lb)(28 in.) (20 lb)(22 in.) (35 lb)(14 in.) (20 lb)(6 in.) (6 in.) 0 B y M A Σ = + + + − = 245 lbyA = + 245 lb=A (b) Tension in BC: 0: (15 lb)(22 in.) (20 lb)(16 in.) (35 lb)(8 in.) (15 lb)(6 in.) (6 in.) 0 A BC M F Σ = + + − − = 140.0 lbBCF = + 140.0 lbBCF = Check: 0: 15 lb 20 lb 35 lb 20 lb 0 105 lb 245 lb 140.0 0 y BCF A FΣ = − − = − + − = − + − = 0 0 (Checks)= www.elsolucionario.net www.elsolucionario.net
- 355. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 353 PROBLEM 4.9 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward. SOLUTION Assume B is positive when directed . Sketch showing distance from D to forces. 0: (300 lb)(8 in. ) (300 lb)( 2 in.) (50 lb)(4 in.) 16 0DM a a BΣ = − − − − + = 600 2800 16 0a B− + + = (2800 16 ) 600 B a + = (1) For 100 lbB = 100 lb,= − Eq. (1) yields: [2800 16( 100)] 1200 2 in. 600 600 a + − ≥ = = 2.00 in.a ≥ For 200B = 200 lb,= + Eq. (1) yields: [2800 16(200)] 6000 10 in. 600 600 a + ≤ = = 10.00 in.a ≤ Required range: 2.00 in. 10.00 in.a≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 356. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 354 PROBLEM 4.10 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe. SOLUTION 0: 0x xF BΣ = = yB B= 0: (50 N) (100 N)(0.45 m ) (150 N)(0.9 m ) (0.9 m ) 0AM d d d B dΣ = − − − − + − = 50 45 100 135 150 0.9 0d d d B Bd− + − + + − = 180 N m (0.9 m) 300 B d A B ⋅ − = − (1) 0: (50 N)(0.9 m) (0.9 m ) (100 N)(0.45 m) 0BM A dΣ = − − + = 45 0.9 45 0A Ad− + + = (0.9 m) 90 N mA d A − ⋅ = (2) Since 180 N,B ≤ Eq. (1) yields 180 (0.9)180 18 0.15 m 300 180 120 d − ≥ = = − 150.0 mmd ≥ Since 180 N,A ≤ Eq. (2) yields (0.9)180 90 72 0.40 m 180 180 d − ≤ = = 400 mmd ≤ Range: 150.0 mm 400 mmd≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 357. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 355 PROBLEM 4.11 Three loads are applied as shown to a light beam supported by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P = 0. SOLUTION 0: (3.00 kN)(0.500 m) (2.25 m) (3.00 m) 0B DM T QΣ = + − = 0.500 kN (0.750) DQ T= + (1) 0: (3.00 kN)(2.75 m) (2.25 m) (0.750 m) 0D BM T QΣ = − − = 11.00 kN (3.00) BQ T= − (2) For cable B not to be slack, 0,BT ≥ and from Eq. (2), 11.00 kNQ ≤ For cable D not to be slack, 0,DT ≥ and from Eq. (1), 0.500 kNQ ≥ For neither cable to be slack, 0.500 kN 11.00 kNQ≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 358. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 356 PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0. SOLUTION 0: (3.00 kN)(0.500 m) (2.25 m) (3.00 m) 0B DM T QΣ = + − = 0.500 kN (0.750) DQ T= + (1) 0: (3.00 kN)(2.75 m) (2.25 m) (0.750 m) 0D BM T QΣ = − − = 11.00 kN (3.00) BQ T= − (2) For 4.00 kN,BT ≤ Eq. (2) yields 11.00 kN 3.00(4.00 kN)Q ≥ − 1.000 kNQ ≥ − For 4.00 kN,DT ≤ Eq. (1) yields 0.500 kN 0.750(4.00 kN)Q ≤ + 3.50 kNQ ≤ For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11, 0.500 kN 3.50 kNQ≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 359. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 357 PROBLEM 4.13 For the beam of Problem 4.12, determine the range of values of Q for which the loading is safe when P = 1 kN. PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0. SOLUTION 0: (3.00 kN)(0.500 m) (1.000 kN)(0.750 m) (2.25 m) (3.00 m) 0B DM T QΣ = − + − = 0.250 kN 0.75 DQ T= + (1) 0: (3.00 kN)(2.75 m) (1.000 kN)(1.50 m) (2.25 m) (0.750 m) 0 D B M T Q Σ = + − − = 13.00 kN 3.00 BQ T= − (2) For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN. Making 0 4.00 kNBT≤ ≤ in Eq. (2), we have 13.00 kN 3.00(4.00 kN) 13.00 kN 3.00(0)Q− ≤ ≤ − 1.000 kN 13.00 kNQ≤ ≤ (3) Making 0 4.00 kNDT≤ ≤ in Eq. (1), we have 0.250 kN 0.750(0) 0.250 kN 0.750(4.00 kN)Q+ ≤ ≤ + 0.250 kN 3.25 kNQ≤ ≤ (4) Combining Eqs. (3) and (4), 1.000 kN 3.25 kNQ≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 360. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 358 PROBLEM 4.14 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward. SOLUTION 0: 0x xF BΣ = = yB=B 0: (3 ft) (9 ft) (6 kips)(11ft) (6 kips)(13 ft) 0AM P BΣ = − + − − = 3 48 kipsP B= − (1) 0: (9 ft) (6 ft) (6 kips)(2 ft) (6 kips)(4 ft) 0BM A PΣ = − + − − = 1.5 6 kipsP A= + (2) Since 30 kips,B ≤ Eq. (1) yields (3)(30 kips) 48 kipsP ≤ − 42.0 kipsP ≤ Since 0 30 kips,A≤ ≤ Eq. (2) yields 0 6 kips (1.5)(30 kips)1.6 kipsP+ ≤ ≤ 6.00 kips 51.0 kipsP≤ ≤ Range of values of P for which beam will be safe: 6.00 kips 42.0 kipsP≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 361. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 359 PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION At B: 0.18 m 0.24 m y x T T = 3 4 y xT T= (1) (a) 0: (0.18 m) (240 N)(0.4 m) (240 N)(0.8 m) 0C xM TΣ = − − = 1600 NxT = + From Eq. (1): 3 (1600 N) 1200 N 4 yT = = 2 2 2 2 1600 1200 2000 Nx yT T T= + = + = 2.00 kNT = (b) 0: 0x x xF C TΣ = − = 1600 N 0 1600 Nx xC C− = = + 1600 Nx =C 0: 240 N 240 N 0y y yF C TΣ = − − − = 1200 N 480 N 0yC − − = 1680 NyC = + 1680 Ny =C 46.4 2320 NC α = ° = 2.32 kN=C 46.4° www.elsolucionario.net www.elsolucionario.net
- 362. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 360 PROBLEM 4.16 Solve Problem 4.15, assuming that 0.32 m.a = PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION At B: 0.32 m 0.24 m 4 3 y x y x T T T T = = 0: (0.32 m) (240 N)(0.4 m) (240 N)(0.8 m) 0C xM TΣ = − − = 900 NxT = From Eq. (1): 4 (900 N) 1200 N 3 yT = = 2 2 2 2 900 1200 1500 Nx yT T T= + = + = 1.500 kNT = 0: 0x x xF C TΣ = − = 900 N 0 900 Nx xC C− = = + 900 Nx =C 0: 240 N 240 N 0y y yF C TΣ = − − − = 1200 N 480 N 0yC − − = 1680 NyC = + 1680 Ny =C 61.8 1906 NC α = ° = 1.906 kN=C 61.8° www.elsolucionario.net www.elsolucionario.net
- 363. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 361 PROBLEM 4.17 The lever BCD is hinged at C and attached to a control rod at B. If P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C. SOLUTION Free-Body Diagram: (a) 0: (5 in.) (100 lb)(7.5 in.) 0CM TΣ = − = 150.0 lbT = (b) 3 0: 100 lb (150.0 lb) 0 5 x xF CΣ = + + = 190 lbxC = − 190 lbx =C 4 0: lb) 0 5 y yF CΣ = + (150.0 = 120 lbyC = − 120 lby =C 32.3α = ° 225 lbC = 225 lb=C 32.3° www.elsolucionario.net www.elsolucionario.net
- 364. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 362 PROBLEM 4.18 The lever BCD is hinged at C and attached to a control rod at B. Determine the maximum force P that can be safely applied at D if the maximum allowable value of the reaction at C is 250 lb. SOLUTION Free-Body Diagram: 0: (5 in.) (7.5 in.) 0CM T PΣ = − = 1.5T P= 3 0: (1.5 ) 0 5 x xF P C PΣ = + + = 1.9xC P= − 1.9x P=C 4 0: (1.5 ) 0 5 y yF C PΣ = + = 1.2= −yC P 1.2=y PC 2 2 2 2 (1.9 ) (1.2 ) x yC C C P P = + = + 2.2472C P= For 250 lb,C = 250 lb 2.2472P= 111.2 lbP = 111.2 lb=P www.elsolucionario.net www.elsolucionario.net
- 365. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 363 PROBLEM 4.19 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C. SOLUTION Free-Body Diagram: 0: (100 mm) (120 mm) 0C AB DEM F FΣ = − = 5 6 DE ABF F= (1) (a) For 720 NABF = 5 (720 N) 6 DEF = 600 NDEF = (b) 3 0: (720 N) 0 5 x xF CΣ = − + = 432 NxC = + 4 0: (720 N) 600 N 0 5 1176 N y y y F C C Σ = − + − = = + 1252.84 N 69.829 C α = = ° 1253 N=C 69.8° www.elsolucionario.net www.elsolucionario.net
- 366. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 364 PROBLEM 4.20 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N. SOLUTION See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1). 5 6 DE ABF F= (1) 3 3 0: 0 5 5 x AB x x ABF F C C FΣ = − + = = 4 0: 0 5 y AB y DEF F C FΣ = − + − = 4 5 0 5 6 49 30 AB y AB y AB F C F C F − + − = = 2 2 2 21 (49) (18) 30 1.74005 x y AB AB C C C F C F = + = + = For 1600 N, 1600 N 1.74005 ABC F= = 920 NABF = www.elsolucionario.net www.elsolucionario.net
- 367. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 365 PROBLEM 4.21 Determine the reactions at A and C when (a) 0,α = (b) 30 .α = ° SOLUTION (a) 0α = From F.B.D. of member ABC: 0: (300 N)(0.2 m) (300 N)(0.4 m) (0.8 m) 0CM AΣ = + − = 225 NA = or 225 N=A 0: 225 N 0y yF CΣ = + = 225 N or 225 Ny yC = − =C 0: 300 N 300 N 0x xF CΣ = + + = 600 N or 600 Nx xC = − =C Then 2 2 2 2 (600) (225) 640.80 Nx yC C C= + = + = and 1 1 225 tan tan 20.556 600 y x C C θ − − − = = = ° − or 641 N=C 20.6° (b) 30α = ° From F.B.D. of member ABC: 0: (300 N)(0.2 m) (300 N)(0.4 m) ( cos30 )(0.8 m) ( sin30 )(20 in.) 0 CM A A Σ = + − ° + ° = 365.24 NA = or 365 N=A 60.0° 0: 300 N 300 N (365.24 N)sin30 0x xF CΣ = + + ° + = 782.62xC = − www.elsolucionario.net www.elsolucionario.net
- 368. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 366 PROBLEM 4.21 (Continued) 0: (365.24 N)cos30 0y yF CΣ = + ° = 316.31 N or 316 Ny yC = − =C Then 2 2 2 2 (782.62) (316.31) 884.12 Nx yC C C= + = + = and 1 1 316.31 tan tan 22.007 782.62 y x C C θ − − − = = = ° − or 884 N=C 22.0° www.elsolucionario.net www.elsolucionario.net
- 369. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 367 PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION (a) 0α = 0: (20 in.) 75 lb(10 in.) 0 37.5 lb AM B B Σ = − = = 0: 0x xF AΣ = = + 0: 75 lb 37.5 lb 0 37.5 lb y y y F A A Σ = − + = = 37.5 lb= =A B (b) 90α = ° 0: (12 in.) 75 lb(10 in.) 0 62.5 lb AM B B Σ = − = = 0: 0 62.5 lb x x x F A B A Σ = − = = 0: 75 lb 0 75 lb y y y F A A Σ = − = = 2 2 2 2 (62.5 lb) (75 lb) 97.6 lb x yA A A= + = + = 75 tan 62.5 50.2 θ θ = = ° 97.6 lb=A 50.2 ; 62.51 lb° =B www.elsolucionario.net www.elsolucionario.net
- 370. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 368 PROBLEM 4.22 (Continued) (c) 30α = ° 0: ( cos 30°)(20 in.) ( sin30 )(12 in.) (75 lb)(10 in.) 0 AM B BΣ = + ° − = 32.161 lbB = 0: (32.161)sin30 0 16.0805 lb x x x F A A Σ = − ° = = 0: (32.161)cos30 75 0 47.148 lb y y y F A A Σ = + ° − = = 2 2 2 2 (16.0805) (47.148) 49.8 lb x yA A A= + = + = 47.148 tan 16.0805 71.2 θ θ = = ° 49.8 lb=A 71.2 ; 32.2 lb° =B 60.0° www.elsolucionario.net www.elsolucionario.net
- 371. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 369 PROBLEM 4.23 Determine the reactions at A and B when (a) 0,h = (b) 200 mm.h = SOLUTION Free-Body Diagram: 0: ( cos60 )(0.5 m) ( sin 60 ) (150 N)(0.25 m) 0AM B B hΣ = ° − ° − = 37.5 0.25 0.866 B h = − (1) (a) When 0,h = From Eq. (1): 37.5 150 N 0.25 B = = 150.0 N=B 30.0° 0: sin 60 0y xF A BΣ = − ° = (150)sin 60 129.9 NxA = ° = 129.9 Nx =A 0: 150 cos60 0y yF A BΣ = − + ° = 150 (150)cos60 75 NyA = − ° = 75 Ny =A 30 150.0 NA α = ° = 150.0 N=A 30.0° (b) When h = 200 mm = 0.2 m, From Eq. (1): 37.5 488.3 N 0.25 0.866(0.2) B = = − 488 N=B 30.0° 0: sin 60 0x xF A BΣ = − ° = (488.3)sin 60 422.88 NxA = ° = 422.88 Nx =A 0: 150 cos60 0y yF A BΣ = − + ° = 150 (488.3)cos60 94.15 NyA = − ° = − 94.15 Ny =A 12.55 433.2 NA α = ° = 433 N=A 12.55° www.elsolucionario.net www.elsolucionario.net
- 372. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 370 PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Free-Body Diagram: Geometry: (10 in.)cos20 9.3969 in. (10 in.)sin 20 3.4202 in. AC AC x y = ° = = ° = 12 in. 3.4202 in. 8.5798 in.DAy = − = 1 1 8.5798 tan tan 42.397 9.3969 DA AC y x α − − = = = ° 90 20 42.397 27.603β = ° − ° − ° = ° Equilibrium for lever: (a) 0: cos27.603 (10 in.) (75 lb)[(15 in.)cos 20°] 0C ADM TΣ = ° − = 119.293 lbADT = 119.3 lbADT = (b) 0: (119.293 lb)cos42.397 0x xF CΣ = + ° = 88.097 lbxC = − 0: 75 lb lb)sin 42.397 0y yF CΣ = − − (119.293 ° = 155.435yC = Thus, 2 2 2 2 ( 88.097) (155.435) 178.665 lbx yC C C= + = − + = and 1 1 155.435 tan tan 60.456 88.097 y x C C θ − − = = = ° 178.7 lb=C 60.5° www.elsolucionario.net www.elsolucionario.net
- 373. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 371 PROBLEM 4.25 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION (a) Free-Body Diagram: 0: (20 in.) (50 lb)(4 in.) (40 lb)(10 in.) 0AM BΣ = − − = 30 lbB = + 30.0 lb=B 0: 40 lb 0x xF AΣ = + = 40 lbxA = − 40.0 lbx =A 0: 50 lb 0y yF A BΣ = + − = 30 lb 50 lb 0yA + − = 20 lbyA = + 20.0 lby =A 26.56 44.72 lbA α = ° = 44.7 lb=A 26.6° www.elsolucionario.net www.elsolucionario.net
- 374. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 372 PROBLEM 4.25 (Continued) (b) Free-Body Diagram: 0: ( cos30 )(20 in.) (40 lb)(10 in.) (50 lb)(4 in.) 0AM BΣ = ° − − = 34.64 lbB = 34.6 lb=B 60.0° 0: sin30 40 lbx xF A BΣ = − ° + (34.64 lb)sin30 40 lb 0xA − ° + = 22.68 lbxA = − 22.68 lbx =A 0: cos30 50 lb 0y yF A BΣ = + ° − = (34.64 lb)cos30 50 lb 0yA + ° − = 20 lbyA = + 20.0 lby =A 41.4 30.24 lbA α = ° = 30.2 lb=A 41.4° www.elsolucionario.net www.elsolucionario.net
- 375. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 373 PROBLEM 4.26 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION (a) Free-Body Diagram: 0: (20 in.) (50 lb)(16 in.) (40 lb)(10 in.) 0BM AΣ = + − = 20 lbA = + 20.0 lb=A 0: 40 lb 0x xF BΣ = + = 40 lbxB = − 40 lbx =B 0: 50 lb 0y yF A BΣ = + − = 20 lb 50 lb 0yB+ − = 30 lbyB = + 30 lby =B 36.87α = ° 50 lbB = 50.0 lb=B 36.9° www.elsolucionario.net www.elsolucionario.net
- 376. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 374 PROBLEM 4.26 (Continued) (b) 0: ( cos30 )(20 in.) (40 lb)(10 in.) (50 lb)(16 in.) 0AM AΣ = − ° − + = 23.09 lbA = 23.1lb=A 60.0° 0: sin30 40 lb 0x xF A BΣ = ° + + = (23.09 lb)sin30 40 lb 8 0x° + + = 51.55 lbxB = − 51.55 lbx =B 0: cos30 50 lb 0y yF A BΣ = ° + − = (23.09 lb)cos30 50 lb 0yB° + − = 30 lbyB = + 30 lby =B 30.2 59.64 lbB α = ° = 59.6 lb=B 30.2° www.elsolucionario.net www.elsolucionario.net
- 377. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 375 PROBLEM 4.27 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that 200 mm,d = determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION Free-Body Diagram: (a) Move T along BD until it acts at Point D. 0: ( sin 45 )(0.2 m) (90 N)(0.1m) (90 N)(0.2 m) 0AM TΣ = ° + + = 190.919 NT = 190.9 NT = (b) 0: (190.919 N)cos45 0x xF AΣ = − ° = 135.0 NxA = + 135.0 Nx =A 0: 90 N 90 N (190.919 N)sin 45° 0y yF AΣ = − − + = 45.0 NyA = + 45.0 Ny =A 142.3 N=A 18.43° www.elsolucionario.net www.elsolucionario.net
- 378. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 376 PROBLEM 4.28 A rod AB, hinged at A and attached at B to cable BD, supports the loads shown. Knowing that 150 mm,d = determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION Free-Body Diagram: 10 tan ; 33.690° 15 α α= = (a) Move T along BD until it acts at Point D. 0: ( sin33.690 )(0.15 m) (90 N)(0.1 m) (90 N)(0.2 m) 0AM TΣ = ° − − = 324.50 NT = 324 NT = (b) 0: (324.50 N)cos33.690 0x xF AΣ = − ° = 270 NxA = + 270 Nx =A 0: 90 N 90 N (324.50 N)sin33.690 0y yF AΣ = − − + ° = 0yA = 270 N=A www.elsolucionario.net www.elsolucionario.net
- 379. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 377 PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D. SOLUTION Free-Body Diagram: (a) 5 12 0: (90 lb)(9 in.) (9 in.) (7 in.) (3 in.) 0 13 13 DM T T TΣ = − − + = 117 lbT = 117.0 lbT = (b) 5 0: 117 lb (117 lb) 90 0 13 x xF DΣ = − − + = 72 lbxD = + 12 0: lb) 0 13 y yF DΣ = + (117 = 108 lbyD = − 129.8 lb=D 56.3° www.elsolucionario.net www.elsolucionario.net
- 380. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 378 PROBLEM 4.30 Solve Problem 4.29 for a = 6 in. PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D. SOLUTION Free-Body Diagram: (a) 5 12 0: (90 lb)(6 in.) (6 in.) (7 in.) (6 in.) 0 13 13 DM T T TΣ = − − + = 195 lbT = 195.0 lbT = (b) 5 0: 195 lb (195 lb) 90 0 13 x xF DΣ = − − + = 180 lbxD = + 12 0: lb) 0 13 y yF DΣ = + (195 = 180 lbyD = − 255 lb=D 45.0° www.elsolucionario.net www.elsolucionario.net
- 381. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 379 PROBLEM 4.31 Neglecting friction, determine the tension in cable ABD and the reaction at support C. SOLUTION Free-Body Diagram: 0: (0.25 m) (0.1 m) (120 N)(0.1 m) 0CM T TΣ = − − = 80.0 NT = 0: 80 N 0 80 Nx x xF C CΣ = − = = + 80.0 Nx =C 0: 120 N 80 N 0 40 Ny y yF C CΣ = − + = = + 40.0 Ny =C 89.4 N=C 26.6° www.elsolucionario.net www.elsolucionario.net
- 382. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 380 PROBLEM 4.32 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C. SOLUTION Free-Body Diagram: Dimensions in mm Geometry: Distance: 2 2 (0.36) (0.150) 0.39 mAD = + = Distance: 2 2 (0.2) (0.15) 0.25 mBD = + = Equilibrium for beam: (a) 0.15 0.15 0: (120 N)(0.28 m) (0.36 m) (0.2 m) 0 0.39 0.25 CM T T Σ = − − = 130.000 NT = or 130.0 NT = (b) 0.36 0.2 0: (130.000 N) (130.000 N) 0 0.39 0.25 x xF C Σ = + + = 224.00 NxC = − 0.15 0.15 0: (130.00 N) (130.00 N) 120 N 0 0.39 0.25 y yF C Σ = + + − = 8.0000 NyC = − Thus, 2 2 2 2 ( 224) ( 8) 224.14 Nx yC C C= + = − + − = and 1 1 8 tan tan 2.0454 224 y x C C θ − − = = = ° 224 N=C 2.05° www.elsolucionario.net www.elsolucionario.net
- 383. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 381 PROBLEM 4.33 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Free-Body Diagram: 0: ( ) ( ) 0D xM C R P RΣ = − = xC P= + 0: sin 0x xF C B θΣ = − = sin 0 /sin P B B P θ θ − = = sin P θ =B θ 0: cos 0y yF C B PθΣ = + − = ( /sin )cos 0 1 1 tan y y C P P C P θ θ θ + − = = − For 30 ,θ = ° (a) /sin30 2B P P= ° = 2P=B 60.0° (b) x xC P C P= + = (1 1/tan30 ) 0.732/yC P P= − ° = − 0.7321y P=C 1.239P=C 36.2° www.elsolucionario.net www.elsolucionario.net
- 384. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 382 PROBLEM 4.34 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following expressions: 1 1 tan sin x y C P C P P B θ θ = + = − = For 60 ,θ = ° (a) /sin 60 1.1547B P P= ° = 1.155P=B 30.0° (b) x xC P C P= + = (1 1/tan 60 ) 0.4226yC P P= − ° = + 0.4226y P=C 1.086P=C 22.9° www.elsolucionario.net www.elsolucionario.net
- 385. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 383 PROBLEM 4.35 A movable bracket is held at rest by a cable attached at C and by frictionless rollers at A and B. For the loading shown, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION Free-Body Diagram: (a) 0: 600 N 0yF TΣ = − = 600 NT = (b) 0: 0xF B A B AΣ = − = ∴ = Note that the forces shown form two couples. 0: (600 N)(600 mm) (90 mm) 0M AΣ = − = 4000 NA = 4000 NB∴ = 4.00 kN=A ; 4.00 kN=B www.elsolucionario.net www.elsolucionario.net
- 386. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 384 PROBLEM 4.36 A light bar AB supports a 15-kg block at its midpoint C. Rollers at A and B rest against frictionless surfaces, and a horizontal cable AD is attached at A. Determine (a) the tension in cable AD, (b) the reactions at A and B. SOLUTION Free-Body Diagram: 2 (15 kg)(9.81 m/s ) 147.150 N W = = (a) 0: 105.107 N 0x ADF TΣ = − = 105.1 NADT = (b) 0: 0yF A WΣ = − = 147.150 N 0A − = 147.2 N=A 0: (350 mm) (147.150 N)(250 mm) 0 105.107 N AM B B Σ = − = = 105.1 N=B www.elsolucionario.net www.elsolucionario.net
- 387. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 385 PROBLEM 4.37 A light bar AD is suspended from a cable BE and supports a 50-lb block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D. SOLUTION Free-Body Diagram: 0:xF A DΣ = = 0:Σ =yF 50.0 lb=BET We note that the forces shown form two couples. 0: (8 in.) (50 lb)(3 in.) 0M AΣ = − = 18.75 lbA = 18.75 lb=A 18.75 lb=D www.elsolucionario.net www.elsolucionario.net
- 388. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 386 PROBLEM 4.38 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: 0: cos30 (120 lb)cos60 0xF AΣ = ° − ° = 69.28 lbA = 69.3 lb=A 0: (8 in.) (120 lb)(16 in.)cos30 (69.28 lb)(8 in.)sin30 0 BM CΣ = − ° + ° = 173.2 lbC = 173.2 lb=C 60.0° 0: (8 in.) (120 lb)(8 in.)cos30 (69.28 lb)(16 in.)sin30 0 CM BΣ = − ° + ° = 34.6 lbB = 34.6 lb=B 60.0° Check: 0: 173.2 34.6 (69.28)sin30 (120)sin60 0yFΣ = − − ° − ° = 0 0 (check)= www.elsolucionario.net www.elsolucionario.net
- 389. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 387 PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION Free-Body Diagram: 0: cos30 (80 N)cos30 0yF TΣ = − ° + ° = 80 NT = 80.0 NT = 0: ( sin30 )(0.4 m) (80 N)(0.2 m) (80 N)(0.2 m) 0CM AΣ = ° − − = 160 NA = + 160.0 N=A 30.0° 0: (80 N)(0.2 m) (80 N)(0.6 m) ( sin30 )(0.4 m) 0AM CΣ = − + ° = 160 NC = + 160.0 N=C 30.0° www.elsolucionario.net www.elsolucionario.net
- 390. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 388 PROBLEM 4.40 Solve Problem 4.39 if the cord BE is parallel to the rods (α = 30°). PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION Free-Body Diagram: 0: (80 N)cos30 0yF TΣ = − + ° = 69.282 NT = 69.3 NT = 0: (69.282 N)cos30 (0.2 m) (80 N)(0.2 m) ( sin30 )(0.4 m) 0 CM A Σ = − ° − + ° = 140.000 NA = + 140.0 N=A 30.0° 0: (69.282 N)cos30 (0.2 m) (80 N)(0.6 m) ( sin30 )(0.4 m) 0 AM C Σ = + ° − + ° = 180.000 NC = + 180.0 N=C 30.0° www.elsolucionario.net www.elsolucionario.net
- 391. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 389 PROBLEM 4.41 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when 30 .θ = ° SOLUTION Free-Body Diagram: 0: cos30 20 40 0yF EΣ = ° − − = 60 lb 69.282 lb cos 30° E = = 69.3 lb=E 60.0° 0: (20 lb)(4 in.) (40 lb)(4 in.) (3 in.) sin 30 (3 in.) 0 Σ = − − + ° = DM C E 80 3 69.282(0.5)(3) 0C− − + = 7.9743 lbC = 7.97 lb=C 0: sin30 0xF E C DΣ = ° + − = (69.282 lb)(0.5) 7.9743 lb 0D+ − = 42.615 lbD = 42.6 lb=D www.elsolucionario.net www.elsolucionario.net
- 392. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 390 PROBLEM 4.42 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E. SOLUTION Free-Body Diagram: 0: cos 20 40 0yF E θΣ = − − = 60 cos E θ = (1) 0: (20 lb)(4 in.) (40 lb)(4 in.) (3 in.) 60 + sin 3 in. 0 cos DM C θ θ Σ = − − = 1 (180tan 80) 3 C θ= − (a) For 0,C = 180tan 80θ = 4 tan 23.962 9 θ θ= = ° 24.0θ = ° From Eq. (1): 60 65.659 cos23.962 E = = ° 0: sin 0xF D C E θΣ = − + + = (65.659)sin 23.962 26.666 lb= =D (b) 0 26.7 lb= =C D 65.7 lb=E 66.0° www.elsolucionario.net www.elsolucionario.net
- 393. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 391 PROBLEM 4.43 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) 100W = lb, (b) 90 lb.W = SOLUTION (a) 100 lbW = From F.B.D. of beam AD: 0: 0x xF DΣ = = 0: 40 lb 40 lb 100 lb 0y yF DΣ = − − + = 20.0 lbyD = − or 20.0 lb=D 0: (100 lb)(5 ft) (40 lb)(8 ft) (40 lb)(4 ft) 0 D DM MΣ = − + + = 20.0 lb ftDM = ⋅ or 20.0 lb ftD = ⋅M (b) 90 lbW = From F.B.D. of beam AD: 0: 0x xF DΣ = = 0: 90 lb 40 lb 40 lb 0y yF DΣ = + − − = 10.00 lbyD = − or 10.00 lb=D 0: (90 lb)(5 ft) (40 lb)(8 ft) (40 lb)(4 ft) 0 D DM MΣ = − + + = 30.0 lb ftDM = − ⋅ or 30.0 lb ftD = ⋅M www.elsolucionario.net www.elsolucionario.net
- 394. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 392 PROBLEM 4.44 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft. SOLUTION For min ,W 40 lb ftDM = − ⋅ From F.B.D. of beam AD: min0: (40 lb)(8 ft) (5 ft) (40 lb)(4 ft) 40 lb ft 0 DM WΣ = − + − ⋅ = min 88.0 lbW = For max,W 40 lb ftDM = ⋅ From F.B.D. of beam AD: max0: (40 lb)(8 ft) (5 ft) (40 lb)(4 ft) 40 lb ft 0 DM WΣ = − + + ⋅ = max 104.0 lbW = or 88.0 lb 104.0 lbW≤ ≤ www.elsolucionario.net www.elsolucionario.net
- 395. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 393 PROBLEM 4.45 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case. SOLUTION 2 (8 kg)(9.81 m/s ) 78.480 NW mg= = = (a) 0: 0x xF AΣ = = 0: 0y yF A WΣ = − = 78.480 Ny =A 0: (1.6 m) 0A AM M WΣ = − = (78.480 N)(1.6 m)AM = + 125.568 N mA = ⋅M 78.5 N=A 125.6 N mA = ⋅M (b) 0: 0x xF A WΣ = − = 78.480x =A 0: 0y yF A WΣ = − = 78.480y =A (78.480 N) 2 110.987 N= =A 45° 0: (1.6 m) 0A AM M WΣ = − = (78.480 N)(1.6 m) 125.568 N mA AM = + = ⋅M 111.0 N=A 45° 125.6 N mA = ⋅M (c) 0: 0x xF AΣ = = 0: 2 0y yF A WΣ = − = 2 2(78.480 N) 156.960 NyA W= = = 0: 2 (1.6 m) 0A AM M WΣ = − = 2(78.480 N)(1.6 m)AM = + 251.14 N mA = ⋅M 157.0 N=A 251 N mA = ⋅M www.elsolucionario.net www.elsolucionario.net
- 396. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 394 PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C. SOLUTION Free-Body Diagram: 0: (20 N) 0x xF CΣ = + = 20 NxC = − 0: (20 N) 0y yF CΣ = − = 20 NyC = + 28.3 N=C 45.0° 0: (20 N)(0.160 m) (20 N)(0.055 m) 0C CM MΣ = + + = 4.30 N mCM = − ⋅ 4.30 N mC = ⋅M www.elsolucionario.net www.elsolucionario.net
- 397. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 395 PROBLEM 4.47 Solve Problem 4.46, assuming that 15-mm-radius pulleys are used. PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C. SOLUTION Free-Body Diagram: 0: (20 N) 0x xF CΣ = + = 20 NxC = − 0: (20 N) 0y yF CΣ = − = 20 NyC = + 28.3 N=C 45.0° 0: (20 N)(0.165 m) (20 N)(0.060 m) 0C CM MΣ = + + = 4.50 N mCM = − ⋅ 4.50 N mC = ⋅M www.elsolucionario.net www.elsolucionario.net
- 398. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 396 PROBLEM 4.48 The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance x = 12 ft from the vertical member DBE. If the tension in the cable is 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 ft above E. SOLUTION Free-Body Diagram: 0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0E EM M x T= + + − = 3.75 3600 7800EM T x= − − (1) (a) For 12 ft and 4000 lbs,x T= = 3.75(4000) 3600(12) 7800 36,000 lb ft EM = − − = ⋅ 0 0x xF EΣ = ∴ = 0: 3600 lb 1200 lb 4000 0y yF EΣ = − − − = 8800 lbyE = 8.80 kips=E ; 36.0 kip ftE = ⋅M www.elsolucionario.net www.elsolucionario.net
- 399. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 397 PROBLEM 4.48 (Continued) (b) 0: (3600 lb)(12 ft) (1200 lb)(6.5 ft) 0E EM MΣ = + + = 51,000 lb ftEM = − ⋅ 0 0x xF EΣ = ∴ = 0: 3600 lb 1200 lb 0y yF EΣ = − − = 4800 lbyE = 4.80 kips=E ; 51.0 kip ftE = ⋅M www.elsolucionario.net www.elsolucionario.net
- 400. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 398 PROBLEM 4.49 For the rig and crate of Prob. 4.48, and assuming that cable is anchored at F as shown, determine (a) the required tension in cable ADCF if the maximum value of the couple at E as x varies from 1.5 to 17.5 ft is to be as small as possible, (b) the corresponding maximum value of the couple. SOLUTION Free-Body Diagram: 0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0E EM M x T= + + − = 3.75 3600 7800EM T x= − − (1) For 1.5 ft, Eq. (1) becomesx = 1( ) 3.75 3600(1.5) 7800EM T= − − (2) For 17.5 ft, Eq. (1) becomesx = 2( ) 3.75 3600(17.5) 7800EM T= − − (a) For smallest max value of | |,EM we set 1 2( ) ( )E EM M− = 3.75 13,200 3.75 70,800T T− = − + 11.20 kips=T (b) From Equation (2), then 3.75(11.20) 13.20EM = − | | 28.8 kip ftEM = ⋅ www.elsolucionario.net www.elsolucionario.net
- 401. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 399 PROBLEM 4.50 A 6-m telephone pole weighing 1600 N is used to support the ends of two wires. The wires form the angles shown with the horizontal axis and the tensions in the wires are, respectively, 1 600 NT = and 2 375T = N. Determine the reaction at the fixed end A. SOLUTION Free-Body Diagram: 0: (375 N)cos20 (600 N)cos10 0x xF AΣ = + ° − ° = 238.50 NxA = + 0: 1600 N (600 N)sin10 (375 N)sin 20 0y yF AΣ = − − ° − ° = 1832.45 NyA = + 2 2 1 238.50 1832.45 1832.45 tan 238.50 A θ − = + = 1848 N=A 82.6° 0: (600 N)cos10 (6 m) (375 N)cos20 (6 m) 0A AM MΣ = + ° − ° = 1431.00 N mAM = − ⋅ 1431 N mA = ⋅M www.elsolucionario.net www.elsolucionario.net
- 402. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 400 PROBLEM 4.51 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W. SOLUTION Free-Body Diagram: (a) Triangle ABC is isosceles. We have ( )cos cos 2 2 CD BC l θ θ = = 0: ( cos ) cos 0 2 CM P l W l θ θ Σ = − = Setting 2 cos 2cos 1: 2 θ θ = − 2 2cos 1 cos 0 2 2 Pl Wl θ θ − − = 2 2 2 1 cos cos 0 2 2 2 2 1 cos 8 2 4 W P W W P P θ θ θ − − = = ± + 2 1 2 1 2cos 8 4 W W P P θ − = ± + www.elsolucionario.net www.elsolucionario.net
- 403. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 401 PROBLEM 4.51 (Continued) (b) For 2 ,P W= ( )1 1 1 1 cos 8 1 33 2 4 2 4 8 θ = ± + = ± cos 0.84307 and cos 0.59307 2 2 32.534 126.375 2 2 θ θ θ θ = = − = ° = ° 65.1θ = ° 252.75 (discard)θ = ° 65.1θ = ° www.elsolucionario.net www.elsolucionario.net
- 404. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 402 PROBLEM 4.52 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W. SOLUTION (a) Triangle ABC is isosceles. We have ( )cos cos 2 2 CD BC l θ θ = = 0: cos ( sin ) 0 2 CM W l P l θ θ Σ = − = Setting sin 2sin cos : cos 2 sin cos 0 2 2 2 2 2 Wl Pl θ θ θ θ θ θ = − = 2 sin 0 2 W P θ − = 1 2sin 2 W P θ − = (b) For 2 ,P W= sin 0.25 2 2 4 W W P W θ = = = 14.5 2 θ = ° 29.0θ = ° or 165.5 331 (discard) 2 θ θ= ° = ° www.elsolucionario.net www.elsolucionario.net
- 405. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 403 PROBLEM 4.53 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W. SOLUTION (a) From F.B.D. of rod AB: 1 0: ( sin ) cos ( cos ) 0 2 CM T l W T lθ θ θ Σ = + − = cos 2(cos sin ) W T θ θ θ = − Dividing both numerator and denominator by cos θ, 1 2 1 tan W T θ = − or ( )2 (1 tan ) W T θ = − (b) For 3 ,T W= ( )2 3 (1 tan ) 1 1 tan 6 W W θ θ = − − = or 1 5 tan 39.806 6 θ − = = ° or 39.8θ = ° www.elsolucionario.net www.elsolucionario.net
- 406. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 404 PROBLEM 4.54 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when 150 N · m,M = 200 N,P = and 600 mm.l = SOLUTION Free-Body Diagram: (a) From free-body diagram of rod AB: 0: ( cos ) ( sin ) 0CM P l P l Mθ θΣ = + − = or sin cos M Pl θ θ+ = (b) For 150 lb in., 20 lb, and 6 in.,M P l= ⋅ = = 150 lb in. 5 sin cos 1.25 (20 lb)(6 in.) 4 θ θ ⋅ + = = = Using identity 2 2 2 1/2 2 1/2 2 2 sin cos 1 sin (1 sin ) 1.25 (1 sin ) 1.25 sin 1 sin 1.5625 2.5sin sin θ θ θ θ θ θ θ θ θ + = + − = − = − − = − + 2 2sin 2.5sin 0.5625 0θ θ− + = Using quadratic formula ( 2.5) (625) 4(2)(0.5625) sin 2(2) 2.5 1.75 4 θ − − ± − = ± = or sin 0.95572 and sin 0.29428 72.886 and 17.1144 θ θ θ θ = = = ° = ° or 17.11 and 72.9θ θ= ° = ° www.elsolucionario.net www.elsolucionario.net
- 407. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 405 PROBLEM 4.55 Solve Sample Problem 4.5, assuming that the spring is unstretched when 90 .θ = ° SOLUTION First note: tension in springT ks= = where deformation of springs r F kr β β = = = From F.B.D. of assembly: 0 0: ( cos ) ( ) 0M W l F rβΣ = − = or 2 2 cos 0 cos Wl kr kr Wl β β β β − = = For 2 250 lb/in. 3 in. 8 in. 400 lb (250 lb/in.)(3 in.) cos (400 lb)(8 in.) k r l W β β = = = = = or cos 0.703125β β= Solving numerically, 0.89245 radβ = or 51.134β = ° Then 90 51.134 141.134θ = ° + ° = ° or 141.1θ = ° www.elsolucionario.net www.elsolucionario.net
- 408. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 406 PROBLEM 4.56 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when 0.θ = (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when 75 lb,W = 30 in.,l = and 3 lb/in.k = SOLUTION Free-Body Diagram: Spring force: ( cos ) (1 cos )sF ks k l l klθ θ= = − = − (a) 0: ( sin ) cos 0 2 D s l M F l Wθ θ Σ = − = (1 cos ) sin cos 0 2 W kl l lθ θ θ− − = (1 cos )tan 0 2 W kl θ θ− − = or (1 cos )tan 2 W kl θ θ− = (b) For given values of 75 lb 30 in. 3 lb/in. (1 cos )tan tan sin 75 lb 2(3 lb/in.)(30 in.) 0.41667 W l k θ θ θ θ = = = − = − = = Solving numerically, 49.710θ = ° or 49.7θ = ° www.elsolucionario.net www.elsolucionario.net
- 409. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 407 PROBLEM 4.57 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position terms of P, k, and l. (b) Determine the value of θ corresponding to equilibrium if P = 4 1 kl. SOLUTION Free-Body Diagram: (a) Triangle ABC is isosceles. We have 2( ) 2 sin ; cos 2 2 AB AD l CD l θ θ = = = Elongation of spring: ( ) ( ) 60 2 sin 2 sin30 2 x AB AB l l θ θ θ = − = ° = − ° 1 2 sin 2 2 T k x kl θ = = − 0: cos ( sin ) 0 2 CM T l P l θ θ Σ = − = www.elsolucionario.net www.elsolucionario.net
- 410. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 408 PROBLEM 4.57 (Continued) 1 2 1 2 sin cos 2sin cos 0 2 2 2 2 2 cos 0 or 2( )sin 0 2 2 180 (trivial) sin 2 kl l Pl kl P kl kl kl P θ θ θ θ θ θ θ θ − − = = − − = = ° = − 1 1 2sin /( ) 2 kl kl Pθ − = − (b) For 1 , 4 P kl= 1 2 3 4 2 sin 2 3 kl kl θ = = 41.8 2 θ = ° 83.6θ = ° www.elsolucionario.net www.elsolucionario.net
- 411. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 409 PROBLEM 4.58 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ 0.= (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that 300 N,W = l 500 mm,= and 800 N/m,k = determine the value of θ corresponding to equilibrium. SOLUTION First note: T ks= where spring constant elongation of spring cos (1 cos ) cos (1 cos ) cos k s l l l kl T θ θ θ θ θ = = = − = − = − (a) From F.B.D. of collar B: 0: sin 0yF T WθΣ = − = or (1 cos )sin 0 cos kl Wθ θ θ − − = or tan sin W kl θ θ− = (b) For 3 lb 6 in. 8 lb/ft 6 in. 0.5 ft 12 in./ft 3 lb tan sin 0.75 (8 lb/ft)(0.5 ft) W l k l θ θ = = = = = − = = Solving numerically, 57.957θ = ° or 58.0θ = ° www.elsolucionario.net www.elsolucionario.net
- 412. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 410 PROBLEM 4.59 Eight identical 500 750-mm× rectangular plates, each of mass 40 kg,m = are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions. SOLUTION 1. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained 196.2 N= =A C 2. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained 0, 196.2 N= = =B C D 3. Four non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained 294 Nx =A , 294 Nx =D ( 392 Ny y+ =A D ) 4. Three concurrent reactions (through D): (a) Plate: improperly constrained (b) Reactions: indeterminate (c) No equilibrium ( 0)DMΣ ≠ www.elsolucionario.net www.elsolucionario.net
- 413. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 411 PROBLEM 4.59 (Continued) 5. Two reactions: (a) Plate: partial constraint (b) Reactions: determinate (c) Equilibrium maintained 196.2 N= =C D 6. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained 294 N=B , 491 N=D 53.1° 7. Two reactions: (a) Plate: improperly constrained (b) Reactions determined by dynamics (c) No equilibrium ( 0)yFΣ ≠ 8. Four non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained 196.2 Ny= =B D ( 0)x+ =C D www.elsolucionario.net www.elsolucionario.net
- 414. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 412 PROBLEM 4.60 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb. SOLUTION 1. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained 120.2 lb=A 56.3°, 66.7 lb=B 2. Four concurrent, reactions (through A): (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium ( 0)AMΣ ≠ 3. Two reactions: (a) Bracket: partial constraint (b) Reactions: indeterminate (c) Equilibrium maintained 50 lb=A , 50 lb=C 4. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained 50 lb=A , 83.3 lb=B 36.9°, 66.7 lb=C www.elsolucionario.net www.elsolucionario.net
- 415. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 413 PROBLEM 4.60 (Continued) 5. Four non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained ( 0) 50 lbC yMΣ = =A 6. Four non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained 66.7 lbx =A 66.7 lbx =B ( 100 lby y+ =A B ) 7. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained 50 lb= =A C 8. Three concurrent, reactions (through A) (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium ( 0)AMΣ ≠ www.elsolucionario.net www.elsolucionario.net
- 416. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 414 PROBLEM 4.61 Determine the reactions at A and B when a = 150 mm. SOLUTION Free-Body Diagram: Force triangle 80 mm 80 mm tan 150 mm 28.072 a β β = = = ° 320 N sin 28.072 A = ° 680 N=A 28.1° 320 N tan 28.072 B = ° 600 N=B www.elsolucionario.net www.elsolucionario.net
- 417. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 415 PROBLEM 4.62 Determine the value of a for which the magnitude of the reaction at B is equal to 800 N. SOLUTION Free-Body Diagram: Force triangle 80 mm 80 mm tan tan a a β β = = (1) From force triangle: 320 N tan 0.4 800 N β = = From Eq. (1): 80 mm 0.4 a = 200 mma = www.elsolucionario.net www.elsolucionario.net
- 418. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 416 PROBLEM 4.63 Using the method of Sec. 4.7, solve Problem 4.22b. PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION Free-Body Diagram: (Three-force body) The line of action at A must pass through C, where B and the 75-lb load intersect. In triangle ACE: 10 in. tan 12 in. θ = 39.806θ = ° Force triangle (75 lb) tan 39.806° 62.5 lb 75 lb 97.6 cos39.806 B A = = = = ° ° 97.6 lb=A 50.2 ;° 62.5 lb=B www.elsolucionario.net www.elsolucionario.net
- 419. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 417 PROBLEM 4.64 A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft obstruction. A cable is wrapped around the tank and pulled horizontally as shown. Knowing that the corner of the obstruction at A is rough, find the required tension in the cable and the reaction at A. SOLUTION Free-Body Diagram: Force triangle 2 ft cos 0.5 60 4 ft GD AG α α= = = = ° 1 30 ( 60 ) 2 (500 lb) tan 30° 289 lbT T θ α β= = ° = ° = = 500 lb cos30 A = ° 577 lb=A 60.0° www.elsolucionario.net www.elsolucionario.net
- 420. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 418 PROBLEM 4.65 For the frame and loading shown, determine the reactions at A and C. SOLUTION Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions: 1 6 in. tan 30.964 10 in. β − = = ° Applying the law of sines to the force triangle for member BCD, 30 lb sin(45 ) sin sin135 B C β β = = ° − ° or 30 lb sin14.036 sin30.964 sin135 B C = = ° ° ° (30 lb)sin30.964 63.641 lb sin14.036 A B ° = = = ° or 63.6 lb=A 45.0° and (30 lb)sin135 87.466 lb sin14.036 C ° = = ° or 87.5 lb=C 59.0° www.elsolucionario.net www.elsolucionario.net
- 421. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 419 PROBLEM 4.66 For the frame and loading shown, determine the reactions at C and D. SOLUTION Since BD is a two-force member, the reaction at D must pass through Points B and D. Free-Body Diagram: (Three-force body) Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect. Triangle CEF: 4.5 ft tan 56.310 3 ft β β= = ° Triangle ABE: 1 tan 26.565 2 γ γ= = ° www.elsolucionario.net www.elsolucionario.net
- 422. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 420 PROBLEM 4.66 (Continued) Force Triangle Law of sines: 150 lb sin 29.745 sin116.565 sin33.690 C D = = ° ° ° 270.42 lb, 167.704 lb C D = = 270 lb=C 56.3 ;° 167.7 lb=D 26.6° www.elsolucionario.net www.elsolucionario.net
- 423. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 421 PROBLEM 4.67 Determine the reactions at B and D when 60 mm.b = SOLUTION Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D. Free-Body Diagram: (Three-force body) Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 220 mm tan 250 mm 41.348 β β = = ° Force triangle Law of sines: 80 N sin 3.652° sin 45 sin131.348 888.0 N 942.8 N B D B D = = ° ° = = 888 N=B 41.3° 943 N=D 45.0° www.elsolucionario.net www.elsolucionario.net
- 424. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 422 PROBLEM 4.68 Determine the reactions at B and D when 120 mm.b = SOLUTION Since CD is a two-force member, line of action of reaction at D must pass through C and D . Free-Body Diagram: (Three-force body) Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 280 mm tan 250 mm 48.24 β β = = ° Force triangle Law of sines: 80 N sin 3.24° sin135 sin 41.76 B D = = ° ° 1000.9 N 942.8 N B D = = 1001 N=B 48.2 943 N° =D 45.0° www.elsolucionario.net www.elsolucionario.net
- 425. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 423 PROBLEM 4.69 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 45°. SOLUTION Free-Body Diagram: (Three-force body) The line of action of C must pass through E, where A and the 300-N force intersect. Triangle ABE is isosceles: 400 mmEA AB= = In triangle CEF: 150 mm tan 700 mm CF CF EF EA AF θ = = = + 12.0948θ = ° www.elsolucionario.net www.elsolucionario.net
- 426. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 424 PROBLEM 4.69 (Continued) Force Triangle Law of sines: 300 N sin32.905 sin135 sin12.0948 A C = = ° ° ° 778 N=A ; 1012 N=C 77.9° www.elsolucionario.net www.elsolucionario.net
- 427. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 425 PROBLEM 4.70 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 60°. SOLUTION Free-Body Diagram: (400 mm) tan30° 230.94 mm EA = = In triangle CEF: tan CF CF EF EA AF θ = = + 150 tan 230.94 300 15.7759 θ θ = + = ° www.elsolucionario.net www.elsolucionario.net
- 428. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 426 PROBLEM 4.70 (Continued) Force Triangle Law of sines: 300 N sin 44.224 sin120 sin15.7759 770 N 956 N A C A C = = ° ° ° = = 770 N=A ; 956 N=C 74.2° www.elsolucionario.net www.elsolucionario.net
- 429. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 427 PROBLEM 4.71 A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right. SOLUTION Force Triangle Force Triangle Geometry: For each case as roller comes into contact with tile, 1 3.7 in. cos 4 in. 22.332 α α − = = ° (a) Roller pushed to left (three-force body): Forces must pass through O. Law of sines: 40 lb ; 24.87 lb sin37.668 sin 22.332 P P= = ° ° 24.9 lb=P 30.0° (b) Roller pulled to right (three-force body): Forces must pass through O. Law of sines: 40 lb ; 15.3361 lb sin97.668 sin 22.332 P P= = ° ° 15.34 lb=P 30.0° www.elsolucionario.net www.elsolucionario.net
- 430. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 428 PROBLEM 4.72 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction at A and the tension in the cord. SOLUTION Free-Body Diagram: (Three-force body) The line of action of reaction at A must pass through E, where T and the 40-lb load intersect. Force triangle 23 tan 12 62.447 5 tan 12 22.620 EF AF EH DH α α β β = = = ° = = = ° 40 lb sin67.380 sin 27.553 sin 85.067° A T = = ° ° 37.1 lb=A 62.4° 18.57 lbT = www.elsolucionario.net www.elsolucionario.net
- 431. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 429 PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that 1.5 m,a = determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION Three-force body: W and CDT intersect at E. 0.7497 m tan 1.5 m 26.556 β β = = ° Three forces intersect at E. 2 (50 kg) 9.81 m/s 490.50 N W = = Force triangle Law of sines: 490.50 N sin 61.556° sin 63.444 sin55 498.99 N 456.96 N CD CD T B T B = = ° ° = = (a) 499 NCDT = (b) 457 N=B 26.6° www.elsolucionario.net www.elsolucionario.net
- 432. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 430 PROBLEM 4.74 Solve Problem 4.73, assuming that 3 m.a = PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that 1.5 m,a = determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION W and CDT intersect at E. Free-Body Diagram: Three-Force Body 0.301 m tan 3 m 5.7295 AE AB β β = = = ° Three forces intersect at E. Force Triangle 2 (50 kg) 9.81 m/s 490.50 N W = = Law of sines: 490.50 N sin 29.271° sin95.730 sin55 998.18 N 821.76 N CD CD T B T B = = ° ° = = (a) 998 NCDT = (b) 822 N=B 5.73° www.elsolucionario.net www.elsolucionario.net
- 433. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 431 PROBLEM 4.75 Determine the reactions at A and B when 50°.β = SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through Point D where the 100-N force and B intersect. In right Δ BCD: 90 75 15 250tan75 933.01 mmBD α = ° − ° = ° = ° = In right Δ ABD: 150 mm tan 933.01mm 9.1333 AB BD γ γ = = = ° Force Triangle Law of sines: 100 N sin9.1333° sin15 sin155.867 163.1 N; 257.6 N A B A B = = ° ° = = 163.1 N=A 74.1° 258 N=B 65.0° Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 434. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 432 PROBLEM 4.76 Determine the reactions at A and B when β = 80°. SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through D where the 100-N force and B intersect. In right triangle BCD: 90 75 15α = ° − ° = ° tan 75 250 tan75 933.01 mm BD BC BD = ° = ° = In right triangle ABD: 150 mm tan 933.01 mm AB BD γ = = 9.1333γ = ° www.elsolucionario.net www.elsolucionario.net
- 435. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 433 PROBLEM 4.76 (Continued) Force Triangle Law of sines: 100 N sin 9.1333° sin15 sin155.867 A B = = ° ° 163.1 N=A 55.9° 258 N=B 65.0° www.elsolucionario.net www.elsolucionario.net
- 436. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 434 PROBLEM 4.77 Knowing that θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. In Δ CDE: ( 3 1) tan 3 1 36.2 R R β β − = = − = ° Force Triangle Law of sines: sin 23.8 sin126.2 sin30 2.00 1.239 P B C B P C P = = ° ° ° = = (a) 2P=B 60.0° (b) 1.239P=C 36.2° Free-Body Diagram: (Three-force body) www.elsolucionario.net www.elsolucionario.net
- 437. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 435 PROBLEM 4.78 Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. In ΔCDE: 3 tan 1 1 3 22.9 RR R β β − = = − = ° Force Triangle Law of sines: sin52.9 sin67.1 sin60 1.155 1.086 P B C B P C P = = ° ° ° = = (a) 1.155P=B 30.0° (b) 1.086P=C 22.9° Free-Body Diagram: (Three-force body) www.elsolucionario.net www.elsolucionario.net
- 438. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 436 PROBLEM 4.79 Using the method of Section 4.7, solve Problem 4.23. PROBLEM 4.23 Determine the reactions at A and B when (a) 0,h = (b) 200 mm.h = SOLUTION Force Triangle (a) h = 0 Reaction A must pass through C where the 150-N weight and B interect. Force triangle is equilateral. 150.0 N=A 30.0° 150.0 N=B 30.0° (b) h = 200 mm 55.662 tan 250 12.5521 β β = = ° Law of sines: 150 N sin17.4480° sin60 sin102.552 433.24 N 488.31 N A B A B = = ° ° = = 433 N=A 12.55° 488 N=B 30.0° Free-Body Diagram: www.elsolucionario.net www.elsolucionario.net
- 439. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 437 PROBLEM 4.80 Using the method of Section 4.7, solve Problem 4.24. PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Dimensions in in. Reaction at C must pass through E, where the 75-lb force and T intersect. 9.3969 in. tan 8.5798 in. 47.602 α α = = ° 14.0954 in. tan 24.870 in. 29.543 β β = = ° Force Triangle Law of sines: 75 lb sin18.0590° sin 29.543 sin132.398 T C = = ° ° (a) 119.3 lbT = (b) 178.7 lb=C 60.5° Free-Body Diagram: www.elsolucionario.net www.elsolucionario.net
- 440. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 438 PROBLEM 4.81 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Reaction at B must pass through D. 7 in. tan 12 in. 30.256 7 in. tan 24 in. 16.26 α α β β = = ° = = ° Force Triangle Law of sines: 72 lb sin59.744 sin13.996 sin106.26 (sin13.996 ) ( 72 lb)(sin 59.744°) (0.24185) ( 72)(0.86378) T T B T T T T − = = ° ° ° = − = − 100.00 lbT = 100.0 lbT = sin 106.26° (100 lb) sin59.744 111.14 lb B = ° = 111.1 lb=B 30.3° Free-Body Diagram: www.elsolucionario.net www.elsolucionario.net
- 441. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 439 PROBLEM 4.82 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Force Triangle Reaction at B must pass through D. 120 tan ; 36.9 160 α α= = ° 75 N 4 3 5 T T B− = = 3 4 300; 300 N 5 5 (300 N) 375 N 4 4 T T T B T = − = = = = 375 N=B 36.9° www.elsolucionario.net www.elsolucionario.net
- 442. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 440 PROBLEM 4.83 A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in the string, (c) the reaction at C. SOLUTION Free-Body Diagram: (Three-force body) The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point. Thus, points A, B, and G are in a straight line. (a) From triangle ACG: 2 2 2 2 ( ) ( ) (265 mm) (140 mm) 225.00 mm d AG CG= − = − = 225 mmd = Force Triangle 2 (2 kg)(9.81 m/s ) 19.6200 NW = = Law of sines: 19.6200 N 265 mm 140 mm 225.00 mm T C = = (b) 23.1 NT = (c) 12.21 N=C www.elsolucionario.net www.elsolucionario.net
- 443. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 441 PROBLEM 4.84 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium. SOLUTION Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angleθ of triangle DCA. 2α θ= The horizontal projections of , ( ),AEAE x and , ( ),AGAG x are equal. AE AG Ax x x= = or ( )cos2 ( )cosAE AGθ θ= and (2 )cos2 cosR Rθ θ= Now 2 cos2 2cos 1θ θ= − then 2 4cos 2 cosθ θ− = or 2 4cos cos 2 0θ θ− − = Applying the quadratic equation, cos 0.84307 and cos 0.59307θ θ= = − 32.534 and 126.375 (Discard)θ θ= ° = ° or 32.5θ = ° www.elsolucionario.net www.elsolucionario.net
- 444. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 442 PROBLEM 4.85 A slender rod BC of length L and weight W is held by two cables as shown. Knowing that cable AB is horizontal and that the rod forms an angle of 40° with the horizontal, determine (a) the angle θ that cable CD forms with the horizontal, (b) the tension in each cable. SOLUTION Free-Body Diagram: (Three-force body) (a) The line of action of TCD must pass through E, where TAB and W intersect. 1 2 tan sin 40 cos40 2tan 40 59.210 CF EF L L θ = ° = ° = ° = ° 59.2θ = ° (b) Force Triangle tan30.790 0.59588 ABT W W = ° = 0.596ABT W= cos30.790 1.16408 CD W T W = ° = 1.164CDT W= www.elsolucionario.net www.elsolucionario.net
- 445. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 443 PROBLEM 4.86 A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium. SOLUTION Free-Body Diagram (Three-force body) Reaction B must pass through D where B and W intersect. Note that ΔABC and ΔBGD are similar. cosAC AE L θ= = In Δ ABC: 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) (2 cos ) ( sin ) 4cos sin 4cos 1 cos 3cos 1 CE BE BC L L R R L R L R L θ θ θ θ θ θ θ + = + = = + = + − = + 2 2 1 cos 1 3 R L θ = − www.elsolucionario.net www.elsolucionario.net
- 446. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 444 PROBLEM 4.87 Knowing that for the rod of Problem 4.86, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B. SOLUTION See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following equation: 2 2 1 cos 1 3 R L θ = − For 15 in., 20 in., and 10 lb,L R W= = = (a) 2 2 1 20 in. cos 1 ; 59.39 3 15 in. θ θ = − = ° 59.4θ = ° In Δ ABC: sin 1 tan tan 2 cos 2 1 tan tan59.39 0.8452 2 40.2 BE L CE L θ α θ θ α α = = = = ° = = ° Force Triangle tan (10 lb)tan 40.2 8.45 lb (10 lb) 13.09 lb cos cos 40.2 A W W B α α = = ° = = = = ° (b) 8.45 lb=A 13.09 lb=B 49.8° www.elsolucionario.net www.elsolucionario.net
- 447. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 445 PROBLEM 4.88 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm. SOLUTION Free-Body Diagram: Since ,ED EDy x a= = slope of ED is 45 ;° slope of HC is 45 .° Also 2DE a= and 1 2 2 a DH HE DE = = = For triangles DHC and EHC, 2 sin 2 a a R R β = = Now sin(45 )c R β= ° − For 20 mm and 100 mm 20 mm sin 2(100 mm) 0.141421 8.1301 a R β β = = = = = ° and (100 mm)sin(45 8.1301 )c = ° − ° 60.00 mm= or 60.0 mmc = www.elsolucionario.net www.elsolucionario.net
- 448. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 446 PROBLEM 4.89 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β. SOLUTION As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry: Free-Body Diagram: tan GB AB x y β = where cosABy L θ= and 1 sin 2 GBx L θ= 1 2 sin tan cos 1 tan 2 L L θ β θ θ = = or tan 2tanθ β= www.elsolucionario.net www.elsolucionario.net
- 449. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 447 PROBLEM 4.90 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B. SOLUTION (a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces: Free-Body Diagram: tan CB BC x y β = where 1 sin 2 CBx L θ= and cos 1 tan tan 2 BCy L θ β θ = = or tan 2tanθ β= For 30β = ° tan 2tan30 1.15470 49.107 θ θ = ° = = ° or 49.1θ = ° (b) 2 (8 kg)(9.81 m/s ) 78.480 NW mg= = = From force triangle: tan (78.480 N)tan30 A W β= = ° 45.310 N= or 45.3 N=A and 78.480 N 90.621 N cos cos30 W B β = = = ° or 90.6 N=B 60.0° www.elsolucionario.net www.elsolucionario.net
- 450. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 448 PROBLEM 4.91 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. —OK 0: ( 120 ) ( ) (120 160 ) (80 200 ) ( 720 ) 0C x yD D TΣ = − × + + − × + − × − =M k i j j k i k i j 3 3 120 120 120 160 57.6 10 144 10 0x yD D T T− + − − + × + × =j i k j i k Equating to zero the coefficients of the unit vectors: :k 3 120 144 10 0T− + × = (a) 1200 NT = :i 3 120 57.6 10 0 480 Ny yD D+ × = = − : 120 160(1200 N) 0xD− − =j 1600 NxD = − 0:xFΣ = 0x xC D T+ + = 1600 1200 400 NxC = − = 0:yFΣ = 720 0y yC D+ − = 480 720 1200 NyC = + = 0:zFΣ = 0zC = (b) (400 N) (1200 N) ; (1600 N) (480 N)= + = − −C i j D i j www.elsolucionario.net www.elsolucionario.net
- 451. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 449 PROBLEM 4.92 Solve Problem 4.91, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical. PROBLEM 4.91 A 200-mm lever and a 240-mm- diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. 0: ( 120 ) ( ) (120 160 ) (80 173.21 ) ( 720 ) 0C x yD D TΣ = − × + + − × + − × − =M k i j j k i k i j 3 3 120 120 120 160 57.6 10 124.71 10 0x yD D T T− + − − + × + × =j i k j i k Equating to zero the coefficients of the unit vectors, 3 : 120 124.71 10 0 1039.2 NT T− + × = =k 1039 NT = 3 : 120 57.6 10 0 480 Ny yD D+ × = = −i : 120 160(1039.2)xD− −j 1385.6 NxD = − 0:xFΣ = 0x xC D T+ + = 1385.6 1039.2 346.4xC = − = 0:yFΣ = 720 0y yC D+ − = 480 720 1200 NyC = + = 0:zFΣ = 0zC = (b) (346 N) (1200 N) (1386 N) (480 N)= + = − −C i j D i j www.elsolucionario.net www.elsolucionario.net
- 452. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 450 PROBLEM 4.93 A 4 × 8-ft sheet of plywood weighing 40 lb has been temporarily propped against column CD. It rests at A and B on small wooden blocks and against protruding nails. Neglecting friction at all surfaces of contact, determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but equilibrium is maintained ( 0).xFΣ = / / /0: ( ) ( 40 lb) 0A B A y z C A G AM B B CΣ = × + + × + × − =r j k r k r j 5 0 0 4 4sin 60 4cos60 2 2sin 60 2cos60 0 0 0 0 0 40 0y zB B C + ° − ° + ° − ° = − i j k i j k i j k (4 sin 60 80cos60 ) ( 5 4 ) (5 80) 0z yC B C B° − ° + − − + − =i j k Equating the coefficients of the unit vectors to zero, :i 4 sin 60 80cos60 0C ° − ° = 11.5470 lbC = :j 5 4 0zB C− − = 9.2376 lbzB = :k 5 80 0yB − = 16.0000 lbyB = 0:yFΣ = 40 0y yA B+ − = 40 16.0000 24.000 lbyA = − = 0:zFΣ = 0z zA B C+ + = 9.2376 11.5470 2.3094 lbzA = − = − (24.0 lb) (2.31lb) ; (16.00 lb) (9.24 lb) ; (11.55 lb)= − = − =A j k B j k C k www.elsolucionario.net www.elsolucionario.net
- 453. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 451 PROBLEM 4.94 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 1.5 in. and the radius of spool C is 2 in. Knowing that TB = 20 lb and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle. SOLUTION Free-Body Diagram: We have six unknowns and six equations of equilibrium. 0: (4.5 1.5 ) ( 20 ) (10.5 2 ) ( ) (15 ) ( ) 0A C x y zM T D D DΣ = + × − + + × − + × + + =i k j i j k i i j k 90 30 10.5 2 15 15 0C C y zT T D D− + + − + − =k i j i k j Equate coefficients of unit vectors to zero: :i 30 2 0CT− = 15 lbCT = :j 10.5 15 0 10.5(15) 15 0C z zT D D− = − = 10.5 lbzD = :k 90 15 0yD− + = 6 lbyD = 0:xFΣ = 0xD = 0: 20 lb 0y y yF A DΣ = + − = 20 6 14 lbyA = − = 0: 15 lb 0z z zF A DΣ = + − = 15 10.5 4.5 lbzA = − = (14.00 lb) (4.50 lb) ; (6.00 lb) (10.50 lb)= + = +A j k D j k www.elsolucionario.net www.elsolucionario.net
- 454. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 452 PROBLEM 4.95 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION We replace BT and B′T by their resultant ( 180 N)− j and CT and C′T by their resultant ( 300 N) .− k Dimensions in mm We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but equilibrium is maintained ( 0).xMΣ = 0: (150 ) ( 180 ) (250 ) ( 300 ) (450 ) ( ) 0A y zD DΣ = × − + × − + × + =M i j i k i j k 3 3 27 10 75 10 450 450 0y zD D− × + × + − =k j k j Equating coefficients of j and k to zero, :j 3 75 10 450 0zD× − = 166.7 NzD = 3 : 27 10 450 0yD− × + =k 60.0 NyD = 0:xFΣ = 0xA = 0: 180 N 0y y yF A DΣ = + − = 180 60 120.0 NyA = − = 0: 300 N 0z z zF A DΣ = + − = 300 166.7 133.3 NzA = − = (120.0 N) (133.3 N) ; (60.0 N) (166.7 N)= + = +A j k D j k www.elsolucionario.net www.elsolucionario.net
- 455. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 453 PROBLEM 4.96 Solve Problem 4.95, assuming that the pulley rotates at a constant rate and that TB = 104 N, T′B = 84 N, TC = 175 N. PROBLEM 4.95 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. —OK 0: (150 250 ) ( 104 ) (150 250 ) ( 84 ) (250 125 ) ( 175 ) (250 125 ) ( ) 450 ( ) 0 A C y z T D D Σ = + × − + − × − + + × − + − × − + × + = M i k j i k j j k i ji i j k 150(104 84) 250(104 84) 250(175 ) 125(175 ) 450 450 0 C C y z T T D D ′ ′− + + − + + − − + − = k i j k j Equating the coefficients of the unit vectors to zero, : 250(104 84) 125(175 ) 0 175 40 135;C C CT T T′ ′ ′− − − = = = =i : 250(175 135) 450 0zD+ − =j 172.2 NzD = : 150(104 84) 450 0yD− + + =k 62.7 NyD = www.elsolucionario.net www.elsolucionario.net
- 456. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 454 PROBLEM 4.96 (Continued) 0:xFΣ = 0xA = 0:yFΣ = 104 84 62.7 0yA − − + = 125.3 NyA = 0:zFΣ = 175 135 172.2 0zA − − + = 137.8 NzA = (125.3 N) (137.8 N) ; (62.7 N) (172.2 N)= + = +A j k D j k www.elsolucionario.net www.elsolucionario.net
- 457. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 455 PROBLEM 4.97 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that 0.4 m,a = determine the tension in each wire. SOLUTION 1 2 0.6 1.2 W m g W m g ′= ′= / / 1 / 2 /0: ( ) ( ) 0D A D A E D F D C D CM T W W TΣ = × + × − + × − + × =r j r j r j r j 1 2( 0.4 0.6 ) ( 0.4 0.3 ) ( ) 0.2 ( ) 0.8 0A CT W W T− + × + − + × − + × − + × =i k j i k j i j i j 1 1 20.4 0.6 0.4 0.3 0.2 0.8 0A A CT T W W W T− − + + − + =k i k i k k Equate coefficients of unit vectors to zero: 1 1 1 1 : 0.6 0.3 0; 0.6 0.3 2 2 A AT W T W m g m g′ ′− + = = = =i 1 2: 0.4 0.4 0.2 0.8 0A CT W W T− + − + =k 0.4(0.3 ) 0.4(0.6 ) 0.2(1.2 ) 0.8 0Cm g m g m g T′ ′ ′− + − + = (0.12 0.24 0.24) 0.15 0.8 C m g T m g ′− − ′= = 1 20: 0y A C DF T T T W WΣ = + + − − = 0.3 0.15 0.6 1.2 0 1.35 D D m g m g T m g m g T m g ′ ′ ′ ′+ + − − = ′= 2 (8 kg/m)(9.81m/s ) 78.48 N/mm g′ = = 0.3 0.3 78.45AT m g′= = × 23.5 NAT = 0.15 0.15 78.45BT m g′= = × 11.77 NBT = 1.35 1.35 78.45CT m g′= = × 105.9 NCT = www.elsolucionario.net www.elsolucionario.net
- 458. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 456 PROBLEM 4.98 For the pipe assembly of Problem 4.97, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire. SOLUTION 1 2 0.6 1.2 W m g W m g ′= ′= / / 1 / 2 /0: ( ) ( ) 0D A D A E D F D C D CM T W W TΣ = × + × − + × − + × =r j r j r j r j 1 2( 0.6 ) ( 0.3 ) ( ) (0.6 ) ( ) (1.2 ) 0A Ca T a W a W a T− + × + − + × − + − × − + − × =i k j i k j i j i j 1 1 20.6 0.3 (0.6 ) (1.2 ) 0A A CT a T W a W W a T a− − + + − − + − =k i k i k k Equate coefficients of unit vectors to zero: 1 1 1 1 : 0.6 0.3 0; 0.6 0.3 2 2 A AT W T W m g m g′ ′− + = = = =i 1 2: (0.6 ) (1.2 ) 0A CT a W a W a T a− + − − + − =k 0.3 0.6 1.2 (0.6 ) (1.2 ) 0Cm ga m ga m g a T a′ ′ ′− + − − + − = 0.3 0.6 1.2(0.6 ) 1.2 C a a a T a − + − = − For maximum a and no tipping, 0.CT = (a) 0.3 1.2(0.6 ) 0 0.3 0.72 1.2 0 a a a a − + − = − + − = 1.5 0.72a = 0.480 ma = www.elsolucionario.net www.elsolucionario.net
- 459. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 457 PROBLEM 4.98 (Continued) (b) Reactions: 2 (8 kg/m) 9.81 m/s 78.48 N/mm g′ = = 0.3 0.3 78.48 23.544 NAT m g′= = × = 23.5 NAT = 1 20: 0y A C DF T T T W WΣ = + + − − = 0 0.6 1.2 0A DT T m g m g′ ′+ + − − = 1.8 1.8 78.48 23.544 117.72D AT m g T′= − = × − = 117.7 NDT = www.elsolucionario.net www.elsolucionario.net
- 460. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 458 PROBLEM 4.99 The 45-lb square plate shown is supported by three vertical wires. Determine the tension in each wire. SOLUTION Free-Body Diagram: / / /0: ( 45 lb) 0B C B C A B A G BM T TΣ = × + × + × − =r j r j r j [ (20 in.) (15 in.) ] (20 in.)C AT T− + × + ×i k j k j [ (10 in.) (10 in.) ] [ (45 lb) ] 0+ − + × − =i k j 20 15 20 450 450 0C C AT T T− − − + + =k i i k i Equating to zero the coefficients of the unit vectors, :k 20 450 0CT− + = 22.5 lbCT = :i 15(22.5) 20 450 0AT− − + = 5.625 lbAT = 0:yFΣ = 45 lb 0A B CT T T+ + − = 5.625 lb 22.5 lb 45 lb 0BT+ + − = 16.875 lbBT = 5.63 lb; 16.88 lb; 22.5 lbA B CT T T= = = www.elsolucionario.net www.elsolucionario.net
- 461. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 459 PROBLEM 4.100 The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table. SOLUTION 2 ft sin30 1ftr b r= = ° = We shall sum moments about AB. ( ) ( ) 0b r C a b P bW+ + − − = (1 2) ( 1)100 (1)30 0C a+ + − − = 1 [30 ( 1)100] 3 C a= − − If table is not to tip, 0.C ≥ [30 ( 1)100] 0 30 ( 1)100 a a − − ≥ ≥ − 1 0.3 1.3 ft 1.300 fta a a− ≤ ≤ = Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping. www.elsolucionario.net www.elsolucionario.net
- 462. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 460 PROBLEM 4.101 An opening in a floor is covered by a 1 1.2-m× sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. SOLUTION / / / 0.6 0.8 1.05 0.3 0.6 B A C A G A = = + = + r i r i k r i k (18 kg)9.81 176.58 N W mg W = = = / / /0: ( ) 0A B A C A G AM B C WΣ = × + × + × − =r j r j r j (0.6 ) (0.8 1.05 ) (0.3 0.6 ) ( ) 0B C W× + + × + + × − =i j i k j i k j 0.6 0.8 1.05 0.3 0.6 0B C C W W+ − − + =k k i k i Equate coefficients of unit vectors to zero: 0.6 : 1.05 0.6 0 176.58 N 100.90 N 1.05 C W C + = = = i : 0.6 0.8 0.3 0B C W+ − =k 0.6 0.8(100.90 N) 0.3(176.58 N) 0 46.24 NB B+ − = = − 0: 0yF A B C WΣ = + + − = 46.24 N 100.90 N 176.58 N 0 121.92 NA A− + + = = ( ) 121.9 N ( ) 46.2 N ( ) 100.9 Na A b B c C= = − = www.elsolucionario.net www.elsolucionario.net
- 463. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 461 PROBLEM 4.102 Solve Problem 4.101, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. PROBLEM 4.101 An opening in a floor is covered by a 1 1.2-m× sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. SOLUTION / / / 0.6 0.65 1.2 0.3 0.6 B A C A G A = = + = + r i r i k r i k 2 (18 kg) 9.81 m/s 176.58 N W mg W = = = / / /0: ( ) 0A B A C A G AM B C WΣ = × + × + × − =r j r j r j 0.6 (0.65 1.2 ) (0.3 0.6 ) ( ) 0B C W× + + × + + × − =i j i k j i k j 0.6 0.65 1.2 0.3 0.6 0B C C W W+ − − + =k k i k i Equate coefficients of unit vectors to zero: 0.6 : 1.2 0.6 0 176.58 N 88.29 N 1.2 C W C − + = = = i : 0.6 0.65 0.3 0B C W+ − =k 0.6 0.65(88.29 N) 0.3(176.58 N) 0 7.36 NB B+ − = = − 0: 0yF A B C WΣ = + + − = 7.36 N 88.29 N 176.58 N 0 95.648 NA A− + − = = ( ) 95.6 N ( ) 7.36 N ( ) 88.3 Na A b B c C= = − = www.elsolucionario.net www.elsolucionario.net
- 464. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 462 PROBLEM 4.103 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the tension in each wire. SOLUTION Free-Body Diagram: / / /0: ( 80 lb) 0B A B A C B C G BT TΣ = × + × + × − =M r j r j r j (60 in.) [(60 in.) (15 in.) ] [(30 in.) (30 in.) ] ( 80 lb) 0A CT T× + + × + + × − =k j i k j i k j 60 60 15 2400 2400 0A C CT T T− + − − + =i k i k i Equating to zero the coefficients of the unit vectors, :i 60 15(40) 2400 0AT − + = 30.0 lbAT = :k 60 2400 0CT − = 40.0 lbCT = 0:yFΣ = 80 lb 0A B CT T T+ + − = 30 lb 40 lb 80 lb 0BT+ + − = 10.00 lbBT = www.elsolucionario.net www.elsolucionario.net
- 465. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 463 PROBLEM 4.104 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal. SOLUTION Free-Body Diagram: Let bW− j be the weight of the block and x and z the block’s coordinates. Since tensions in wires are equal, let A B CT T T T= = = 0 0: ( ) ( ) ( ) ( ) ( ) ( ) 0A B C G bM T T T W x z WΣ = × + × + × + × − + + × − =r j r j r j r j i k j or (75 ) (15 ) (60 30 ) (30 45 ) ( ) ( ) ( ) 0bT T T W x z W× + × + + × + + × − + + × − =k j k j i k j i k j i k j or 75 15 60 30 30 45 0b bT T T T W W W W z− − + − − + − × + =i i k i k i k i Equate coefficients of unit vectors to zero: : 120 45 0bT W W z− + + =i (1) :k 60 30 0bT W W x− − = (2) Also, 0:yFΣ = 3 0bT W W− − = (3) Eq. (1) + 40 Eq. (3): 5 ( 40) 0bW z W+ − = (4) Eq. (2) – 20 Eq. (3): 10 ( 20) 0bW x W− − − = (5) www.elsolucionario.net www.elsolucionario.net
- 466. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 464 PROBLEM 4.104 (Continued) Solving Eqs. (4) and (5) for /bW W and recalling that 0 60 in.,x≤ ≤ 0 90 in.,z≤ ≤ Eq. (4): 5 5 0.125 40 40 0 bW W z = ≥ = − − Eq. (5): 10 10 0.5 20 20 0 bW W x = ≥ = − − Thus, min( ) 0.5 0.5(80) 40 lbbW W= = = min( ) 40.0 lbbW = Making 0.5bW W= in Eqs. (4) and (5): 5 ( 40)(0.5 ) 0W z W+ − = 30.0 in.z = 10 ( 20)(0.5 ) 0W x W− − − = 0 in.x = www.elsolucionario.net www.elsolucionario.net
- 467. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 465 PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).ACMΣ = 1.2 2.4 B A = = r k r k 0.8 0.6 2.4 2.6 m 0.8 1.2 2.4 2.8 m AD AD AE AE = − + − = = + − = i j k i j k ( 0.8 0.6 2.4 ) 2.6 (0.8 1.2 2.4 ) 2.8 AD AD AE AE TAD T AD TAE T AE = = − + − = = + − i j k i j k 0: ( 3 kN) 0C A AD A AE BMΣ = × + × + × − =r T r T r j 0 0 2.4 0 0 2.4 1.2 ( 3.6 kN) 0 2.6 2.8 0.8 0.6 2.4 0.8 1.2 2.4 AD AET T + + × − = − − − i j k i j k k j Equate coefficients of unit vectors to zero: : 0.55385 1.02857 4.32 0 : 0.73846 0.68671 0 AD AE AD AE T T T T − − + = − + = i j (1) 0.92857AD AET T= (2) From Eq. (1): 0.55385(0.92857) 1.02857 4.32 0AE AET T− − + = 1.54286 4.32 2.800 kN AE AE T T = = 2.80 kNAET = www.elsolucionario.net www.elsolucionario.net
- 468. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 466 PROBLEM 4.105 (Continued) From Eq. (2): 0.92857(2.80) 2.600 kNADT = = 2.60 kNADT = 0.8 0.8 0: (2.6 kN) (2.8 kN) 0 0 2.6 2.8 0.6 1.2 0: (2.6 kN) (2.8 kN) (3.6 kN) 0 1.800 kN 2.6 2.8 2.4 2.4 0: (2.6 kN) (2.8 kN) 0 4.80 kN 2.6 2.8 x x x y y y z z z F C C F C C F C C Σ = − + = = Σ = + + − = = Σ = − − = = (1.800 kN) (4.80 kN)= +C j k www.elsolucionario.net www.elsolucionario.net
- 469. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 467 PROBLEM 4.106 Solve Problem 4.105, assuming that the 3.6-kN load is applied at Point A. PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).ACMΣ = 0.8 0.6 2.4 2.6 m 0.8 1.2 2.4 2.8 m AD AD AE AE = − + − = = + − = i j k i j k ( 0.8 0.6 2.4 ) 2.6 (0.8 1.2 2.4 ) 2.8 AD AD AE AE TAD T AD TAE T AE = = − + − = = + − i j k i j k 0: ( 3.6 kN)C A AD A AE AMΣ = × + × + × −r T r T r j Factor :Ar ( (3.6 kN) )A AD AE× + −r T T j or (3 kN) 0AD AE+ − =T T j (Forces concurrent at A) Coefficient of i: (0.8) (0.8) 0 2.6 2.8 AD AET T − + = 2.6 2.8 AD AET T= (1) Coefficient of j: (0.6) (1.2) 3.6 kN 0 2.6 2.8 2.6 0.6 1.2 3.6 kN 0 2.8 2.6 2.8 0.6 1.2 3.6 kN 2.8 AD AE AE AE AE T T T T T + − = + − = + = 5.600 kNAET = 5.60 kNAET = www.elsolucionario.net www.elsolucionario.net
- 470. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 468 PROBLEM 4.106 (Continued) From Eq. (1): 2.6 (5.6) 5.200 kN 2.8 ADT = = 5.20 kNADT = 0.8 0.8 0: (5.2 kN) (5.6 kN) 0 0 2.6 2.8 0.6 1.2 0: (5.2 kN) (5.6 kN) 3.6 kN 0 0 2.6 2.8 2.4 2.4 0: (5.2 kN) (5.6 kN) 0 9.60 kN 2.6 2.8 x x x y y y z z z F C C F C C F C C Σ = − + = = Σ = + + − = = Σ = − − = = (9.60 kN)=C k Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2. www.elsolucionario.net www.elsolucionario.net
- 471. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 469 PROBLEM 4.107 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A. SOLUTION We have five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).xMΣ = Free-Body Diagram: ( 6 ft) (7 ft) (6 ft) 11ft ( 6 ft) (7 ft) (6 ft) 11ft ( 6 7 6 ) 11 ( 6 7 6 ) 11 BD BD BD BE BE BE BD BD BE BE TBD T T BD TBE T T BE = − + + = = − + − = = = − + + = = − + − i j k i j k i j k i j k 0: ( 840 ) 0A B BD B BE CM T TΣ = × + × + × − =r r r j 6 ( 6 7 6 ) 6 ( 6 7 6 ) 10 ( 840 ) 0 11 11 BD BET T × − + + + × − + − + × − =i i j k i i j k i j 42 36 42 36 8400 0 11 11 11 11 BD BD BE BET T T T− + + − =k j k j k Equate coefficients of unit vectors to zero: 36 36 : 0 11 11 BD BE BE BDT T T T− + = =i 42 42 : 8400 0 11 11 BD BET T+ − =k 42 2 8400 11 BDT = 1100 lbBDT = 1100 lbBET = www.elsolucionario.net www.elsolucionario.net
- 472. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 470 PROBLEM 4.107 (Continued) 6 6 0: (1100 lb) (1100 lb) 0 11 11 x xF AΣ = − − = 1200 lbxA = 7 7 0: (1100 lb) (1100 lb) 840 lb 0 11 11 y yF AΣ = + + − = 560 lbyA = − 6 6 0: (1100 lb) (1100 lb) 0 11 11 z zF AΣ = + − = 0zA = (1200 lb) (560 lb)= −A i j www.elsolucionario.net www.elsolucionario.net
- 473. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 471 PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A. SOLUTION Free-Body Diagram: There are five unknowns and six equations of equilibrium. The pole is free to rotate about the y-axis, but equilibrium is maintained under the given loading ( 0).yMΣ = Resolve BE and BF into components: (7.5 m) (8 m) (6 m)BE = − +i j k 12.5 mBE = (7.5 m) (8 m) (6 m)BF = − −i j k 12.5 mBF = Express BET and BFT in terms of components: (0.60 0.64 0.48 )BE BE BE BE T T BE = = − +T i j k (1) (0.60 0.64 0.48 )BF BF BF BF T T BF = = − −T i j k (2) www.elsolucionario.net www.elsolucionario.net
- 474. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 472 PROBLEM 4.108 (Continued) / / /0: ( 14 kN) 0A B A BE B A BF C AM T TΣ = × + × + × − =r r r i 8 (0.60 0.64 0.48 ) 8 (0.60 0.64 0.48 ) 12 ( 14 ) 0BE BFT T× − + + × − − + × − =j i j k j i j k j i 4.8 3.84 4.8 3.84 168 0BE BE BF BFT T T T− + − − + =k i k i k Equating the coefficients of the unit vectors to zero, :i 3.84 3.84 0BE BFT T− = BE BFT T= :k 4.8 4.8 168 0BE BFT T− − + = 17.50 kNBE BFT T= = 0:xFΣ = 2(0.60)(17.50 kN) 14 kN 0xA + − = 7.00 kNxA = 0:yFΣ = (0.64)(17.50 kN) 0yA z− = 22.4 kNyA = 0:zFΣ = 0 0zA + = 0zA = (7.00 kN) (22.4 kN)= − +A i j Because of the symmetry, we could have noted at the outset that BF BET T= and eliminated one unknown. www.elsolucionario.net www.elsolucionario.net
- 475. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 473 PROBLEM 4.109 Solve Problem 4.108, assuming that cable CD forms an angle φ = 25° with the vertical xy plane. PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A. SOLUTION Free-Body Diagram: (7.5 m) (8 m) (6 m) 12.5 m (7.5 m) (8 m) (6 m) 12.5 m (0.60 0.64 0.48 ) (0.60 0.64 0.48 ) BE BE BE BF BF BF BE BE BF BF BE T T BE BF T T BF = − + = = − − = = = − + = = − − i j k i j k T i j k T i j k / / /0: 0A B A BE B A BF C A CDMΣ = × + × + × =r T r T r T 8 (0.60 0.64 0.48 ) 8 (0.60 0.64 0.48 ) 12 (19 kN)( cos25 sin 25 ) 0 BE BFT T× − + + × − − + × − ° + ° = j i j k j i j k j i k 4.8 3.84 4.8 3.84 152.6 71.00 0BE BE BF BFT T T T− + − − + − =k i k i k i Equating the coefficients of the unit vectors to zero, 3.84 3.84 71.00 0; 18.4896BE BF BF BET T T T− + = − =i: 4.8 4.8 152.26 0; 31.721BE BF BF BET T T T− − + = + =k: Solving simultaneously, 6.6157 kN;BET = 25.105 kNBFT = 6.62 kN; 25.1 kNBE BFT T= = 0: (0.60)( ) 14cos25 0 12.6883 0.60(31.7207) 6.34 kN x x BF BE x x F A T T A A Σ = + + − ° = = − = − www.elsolucionario.net www.elsolucionario.net
- 476. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 474 PROBLEM 4.109 (Continued) 0: (0.64)( ) 0 0.64(31.721) 20.3 kN y y BF BE y y F A T T A A Σ = − + = = = 0: 0.48( ) 14sin 25 0 0.48(18.4893) 5.9167 2.96 kN z z BF BE z z F A T T A A Σ = − − + ° = = − = (6.34 kN) (20.3 kN) (2.96 kN)= − + +A i j k www.elsolucionario.net www.elsolucionario.net
- 477. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 475 PROBLEM 4.110 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).ACMΣ = T = Tension in both parts of cable DAE. 30 48 B A = = r k r k 20 48 52 in. 20 48 52 in. 16 30 34 in. AD AD AE AE BF BF = − − = = − = = − = i k j k i k ( 20 48 ) ( 5 12 ) 52 13 (20 48 ) (5 12 ) 52 13 (16 30 ) (8 15 ) 34 17 AD AE BF BF BF BF AD T T T AD AE T T T AE T TBF T BF = = − − = − − = = − = − = = − = − T i k i k T j k j k T i k i k 0: ( 320 lb) 0C A AD A AE B BF BΣ = × + × + × + × − =M r T r T r T r j 0 0 48 0 0 48 0 0 30 (30 ) ( 320 ) 0 13 13 17 5 0 12 0 5 12 8 0 15 BFTT T + + + × − = − − − − i j k i j k i j k k j Coefficient of i: 240 9600 0 520 lb 13 T T− + = = www.elsolucionario.net www.elsolucionario.net
- 478. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 476 PROBLEM 4.110 (Continued) Coefficient of j: 240 240 0 13 17 BDT T− + = 17 17 (520) 680 lb 13 13 BD BDT T T= = = 0: 320 0AD AE BF CΣ = + + − + =F T T T j Coefficient of i: 20 8 (520) (680) 0 52 17 xC− + + = 200 320 0 120 lbx xC C− + + = = − Coefficient of j: 20 (520) 320 0 52 yC− + = 200 320 0 120 lby yC C− + = = Coefficient of k: 48 48 30 (520) (520) (680) 0 52 52 34 zC− − − + = 480 480 600 0zC− − − + = 1560 lbzC = Answers: DAET T= 520 lbDAET = 680 lbBDT = (120.0 lb) (120.0 lb) (1560 lb)= − + +C i j k www.elsolucionario.net www.elsolucionario.net
- 479. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 477 PROBLEM 4.111 Solve Problem 4.110, assuming that the 320-lb load is applied at A. PROBLEM 4.110 A 48-in. boom is held by a ball-and- socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).ACMΣ = T = tension in both parts of cable DAE. 30 48 B A = = r k r k 20 48 52 in. 20 48 52 in. 16 30 34 in. AD AD AE AE BF BF = − − = = − = = − = i k j k i k ( 20 48 ) ( 5 12 ) 52 13 (20 48 ) (5 12 ) 52 13 (16 30 ) (8 15 ) 34 17 AD AE BF BF BF BF AD T T T AD AE T T T AE T TBF T BF = = − − = − − = = − = − = = − = − T i k i k T j k j k T i k i k 0: ( 320 lb) 0C A AD A AE B BF AMΣ = × + × + × + × − =r T r T r T r j 0 0 48 0 0 48 0 0 30 48 ( 320 ) 0 13 13 17 5 0 12 0 5 12 8 0 15 BFTT T + + + × − = − − − − i j k i j k i j k k j Coefficient of i: 240 15,360 0 832 lb 13 T T− + = = www.elsolucionario.net www.elsolucionario.net
- 480. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 478 PROBLEM 4.111 (Continued) Coefficient of j: 240 240 0 13 17 BDT T− + = 17 17 (832) 1088 lb 13 13 BD BDT T T= = = 0: 320 0AD AE BFΣ = + + − + =F T T T j C Coefficient of i: 20 8 (832) (1088) 0 52 17 xC− + + = 320 512 0 192 lbx xC C− + + = = − Coefficient of j: 20 (832) 320 0 52 yC− + = 320 320 0 0y yC C− + = = Coefficient of k: 48 48 30 (832) (852) (1088) 0 52 52 34 zC− − − + = 768 768 960 0 2496 lbz zC C− − − + = = Answers: DAET T= 832 lbDAET = 1088 lbBDT = (192.0 lb) (2496 lb)= − +C i k www.elsolucionario.net www.elsolucionario.net
- 481. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 479 PROBLEM 4.112 A 600-lb crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 200-lb boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A. SOLUTION Free-Body Diagram: 600 lb 200 lb C G W W = = We have five unknowns ( , , , , )DE DF x y zT T A A A and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since 0.ABMΣ = We have (30 ft) (22.5 ft) 37.5 ft 8.8 (13.8 ft) (22.5 ft) (6.6 ft) 12 (13.8 ft) (16.5 ft) (6.6 ft) 22.5 ft (13.8 ft) (16.5 ft) (6.6 ft) 22.5 ft BH BH DE DE DF DF = − = = − + = − + = = − − = i j i j k i j k i j k www.elsolucionario.net www.elsolucionario.net
- 482. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 480 PROBLEM 4.112 (Continued) Thus: 30 22.5 (600 lb) (480 lb) (360 lb) 37.5 (13.8 16.5 6.6 ) 22.5 (13.8 16.5 6.6 ) 22.5 BH BH DE DE DE DE DF DF BH BH TDE DE TDF DF − = = = − = = − + = = − − i j T T i j T T i j k T T i j k (a) 0: ( ) ( ) ( ) ( ) ( ) 0 (12 ) ( 600 ) (6 ) ( 200 ) (18 ) (480 360 ) A J C K G H BH E DE F DFΣ = × + × + × + × + × = − × − − × − + × − M r W r W r T r T r T i j i j i i j 5 0 6.6 5 0 6.6 0 22.5 22.5 13.8 16.5 6.6 13.8 16.5 6.6 DE DFT T + + − = − − − i j k i j k or 7200 1200 6480 4.84( )DE DFT T+ − + −k k k i 58.08 82.5 ( ) ( ) 0 22.5 22.5 DE DF DE DFT T T T+ − − + =j k Equating to zero the coefficients of the unit vectors, i or j: 0 *DE DF DE DFT T T T− = = 82.5 : 7200 1200 6480 (2 ) 0 22.5 DET+ − − =k 261.82 lbDET = 262 lbDE DFT T= = (b) 13.8 0: 480 2 (261.82) 0 801.17 lb 22.5 16.5 0: 600 200 360 2 (261.82) 0 1544.00 lb 22.5 x x x y y y F A A F A A Σ = + + = = − Σ = − − − − = = 0: 0z zF AΣ = = (801lb) (1544 lb)= − +A i j *Remark: The fact is that DE DFT T= could have been noted at the outset from the symmetry of structure with respect to xy plane. www.elsolucionario.net www.elsolucionario.net
- 483. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 481 PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION / / / (960 180) 780 960 450 90 2 2 390 225 600 450 B A G A C A − = = − + = + = + r i i r i k i k r i k Dimensions in mm T = Tension in cable DCE 690 675 450 1065 mm 270 675 450 855 mm CD CD CE CE = − + − = = + − = i j k i j k 2 ( 690 675 450 ) 1065 (270 675 450 ) 855 (100 kg)(9.81m/s ) (981 N) CD CE T T mg = − + − = + − = − = − = − T i j k T i j k W i j j / / / /0: ( ) 0A C A CD C A CE G A B AWΣ = × + × + × − + × =M r T r T r j r B 600 0 450 600 0 450 1065 855 690 675 450 270 675 450 390 0 225 780 0 0 0 0 981 0 0 y z T T B B + − − − + + = − i j k i j k i j k i j k www.elsolucionario.net www.elsolucionario.net
- 484. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 482 PROBLEM 4.113 (Continued) Coefficient of i: 3 (450)(675) (450)(675) 220.73 10 0 1065 855 T T − − + × = 344.64 NT = 345 NT = Coefficient of j: 344.64 344.64 ( 690 450 600 450) (270 450 600 450) 780 0 1065 855 zB− × + × + × + × − = 185.516 NzB = Coefficient of k: 3344.64 344.64 (600)(675) (600)(675) 382.59 10 780 0 113.178 N 1065 855 y yB B+ − × + = = (113.2 N) (185.5 N)= +B j k 0: 0CD CEΣ = + + + + =F A B T T W Coefficient of i: 690 270 (344.64) (344.64) 0 1065 855 xA − + = 114.5 NxA = Coefficient of j: 675 675 113.178 (344.64) (344.64) 981 0 377 N 1065 855 y yA A+ + + − = = Coefficient of k: 450 450 185.516 (344.64) (344.64) 0 1065 855 zA + − − = 141.5 NzA = (114.5 N) (377 N) (144.5 N)= + +A i j k www.elsolucionario.net www.elsolucionario.net
- 485. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 483 PROBLEM 4.114 Solve Problem 4.113, assuming that cable DCE is replaced by a cable attached to Point E and hook C. PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION See solution to Problem 4.113 for free-body diagram and analysis leading to the following: 1065 mm 855 mm CD CE = = Now, 2 ( 690 675 450 ) 1065 (270 675 450 ) 855 (100 kg)(9.81 m/s ) (981 N) CD CE T T mg = − + − = + − = − = − = − T i j k T i j k W i j j / / /0: ( ) 0A C A CE G A B AT W BΣ = × + × − + × =M r r j r 600 0 450 390 0 225 780 0 0 0 855 270 675 450 0 981 0 0 y z T B B + + = − − i j k i j k i j k Coefficient of i: 3 (450)(675) 220.73 10 0 855 T − + × = 621.31 NT = 621 NT = Coefficient of j: 621.31 (270 450 600 450) 780 0 364.74 N 855 z zB B× + × − = = Coefficient of k: 3621.31 (600)(675) 382.59 10 780 0 113.186 N 855 y yB B− × + = = (113.2 N) (365 N)= +B j k www.elsolucionario.net www.elsolucionario.net
- 486. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 484 PROBLEM 4.114 (Continued) 0: 0CEΣ = + + + =F A B T W Coefficient of i: 270 (621.31) 0 855 xA + = 196.2 NxA = − Coefficient of j: 675 113.186 (621.31) 981 0 855 yA + + − = 377.3 NyA = Coefficient of k: 450 364.74 (621.31) 0 855 zA + − = 37.7 N= −zA (196.2 N) (377 N) (37.7 N)= − + −A i j k www.elsolucionario.net www.elsolucionario.net
- 487. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 485 PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION / / / / / / (38 8) 30 (30 4) 20 26 20 38 10 2 19 10 8 25 20 33 in. (8 25 20 ) 33 0: ( 75 ) 0 = − = = − + = + = + = + = + − = = = + − Σ = × + × − + × = B A E A G A A E A G A B A EF EF AE T T AE T B r i i r i k i k r i k i k i j k T i j k M r r j r 26 0 20 19 0 10 30 0 0 0 33 8 25 20 0 75 0 0 y z T B B + + = − − i j k i j k i j k Coefficient of i: (25)(20) 750 0: 33 T − + = 49.5 lbT = Coefficient of j: 49.5 (160 520) 30 0: 34 lb 33 z zB B+ − = = Coefficient of k: 49.5 (26)(25) 1425 30 0: 15 lb 33 y yB B− + = = (15.00 lb) (34.0 lb)= +B j k www.elsolucionario.net www.elsolucionario.net
- 488. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 486 PROBLEM 4.115 (Continued) 0: (75 lb) 0Σ = + + − =F A B T j Coefficient of i: 8 (49.5) 0 12.00 lb 33 x xA A+ = = − Coefficient of j: 25 15 (49.5) 75 0 22.5 lb 33 y yA A+ + − = = Coefficient of k: 20 34 (49.5) 0 4.00 lb 33 z zA A+ − = = − (12.00 lb) (22.5 lb) (4.00 lb)= − + −A i j k www.elsolucionario.net www.elsolucionario.net
- 489. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 487 PROBLEM 4.116 Solve Problem 4.115, assuming that cable EF is replaced by a cable attached at points E and H. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION / / (38 8) 30 (30 4) 20 26 20 B A E A = − = = − + = + r i i r i k i k / 38 10 2 19 10 G A = + = + r i k i k 30 12 20 38 in. EH EH = − + − = i j k ( 30 12 20 ) 38 EH T T EH = = − + −T i j k / / /0: ( 75 ) 0Σ = × + × − + × =A E A G A B AM r T r j r B 26 0 20 19 0 10 30 0 0 0 38 30 12 20 0 75 0 0 y z T B B + + = − − − i j k i j k i j k Coefficient of i: (12)(20) 750 0 38 − + = T 118.75T = 118.8lbT = Coefficient of j: 118.75 ( 600 520) 30 0 8.33lb 38 − + − = = −z zB B Coefficient of k: 118.75 (26)(12) 1425 30 0 38 − + =yB 15.00 lb=yB (15.00 lb) (8.33 lb)= −B j k www.elsolucionario.net www.elsolucionario.net
- 490. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 488 PROBLEM 4.116 (Continued) 0:Σ =F (75 lb) 0+ + − =A B T j Coefficient of i: 30 (118.75) 0 93.75 lb 38 x xA A− = = Coefficient of j: 12 15 (118.75) 75 0 22.5 lb 38 y yA A+ + − = = Coefficient of k: 20 8.33 (118.75) 0 70.83 lb 38 z zA A− − = = (93.8 lb) (22.5 lb) (70.8 lb)= + +A i j k www.elsolucionario.net www.elsolucionario.net
- 491. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 489 PROBLEM 4.117 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust. SOLUTION Force exerted by CE: 2 / / / (cos75 ) (sin 75 ) (0.25882 0.96593 ) 20 kg(9.81m/s ) 196.2 N 0.6 0.9 0.6 0.45 0.3 (0.25882 0.96593 ) A B C B G B F F F W mg F = ° + ° = + = = = = = + = + = + F i j F i j r k r i k r i k F i j / / /0: ( 196.2 ) 0B G B C B A BΣ = × − + × + × =M r j r F r A (a) 0.45 0 0.3 0.9 0 0.6 0 0 0.6 0 0 196.2 0 0.25882 0.96593 0 0x y F A A + + = − + i j k i j k i j k Coefficient of i : 58.86 0.57956 0.6 0yF A+ − − = (1) Coefficient of j: 0.155292 0.6 0xF A+ + = (2) Coefficient of k: 88.29 0.86934 0: 101.56 NF F− + = = From Eq. (2): 58.86 0.57956(101.56) 0.6 0 0y yA A+ − − = = From Eq. (3): 0.155292(101.56) 0.6 0 26.286 Nx xA A+ + = = − (101.6 N)F = (b) : 0WΣ + + − =jF A B F Coefficient of i: 26.286 0.25882(101.56) 0 0x xB B+ + = = Coefficient of j: 0.96593(101.56) 196.2 0 98.1 Ny yB B+ − = = Coefficient of k: 0zB = (26.3 N) ; (98.1 N)= − =A i B j www.elsolucionario.net www.elsolucionario.net
- 492. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 490 PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Free-Body Diagram: ABHΔ is equilateral. Dimensions in mm / / / 50 250 300 350 250 (sin30 ) (cos30 ) (0.5 0.866 ) H C D C F C T T T = − + = = + = ° − ° = − r i j r i r i k T j k j k / /0: ( 400 ) 0C H C D F CΣ = × + × + × − =M r T r D r j 50 250 0 300 0 0 350 0 250 0 0 0.5 0.866 0 0 400 0y z T D D − + + = − − i j k i j k i j k Coefficient i: 3 216.5 100 10 0T− + × = 461.9 NT = 462 NT = Coefficient of j: 43.3 300 0zT D− − = 43.3(461.9) 300 0 66.67 Nz zD D− − = = − www.elsolucionario.net www.elsolucionario.net
- 493. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 491 PROBLEM 4.118 (Continued) Coefficient of k: 3 25 300 140 10 0yT D− + − × = 3 25(461.9) 300 140 10 0 505.1 Ny yD D− + − × = = (505 N) (66.7 N)= −D j k 0: 400 0Σ = + + − =F C D T j Coefficient i: 0xC = 0xC = Coefficient j: (461.9)0.5 505.1 400 0 336 Ny yC C+ + − = = − Coefficient k: (461.9)0.866 66.67 0zC − − = 467 NzC = (336 N) (467 N)= − +C j k www.elsolucionario.net www.elsolucionario.net
- 494. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 492 PROBLEM 4.119 Solve Problem 4.115, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION / / (30 4) 20 26 20 (0.5 38) 10 19 10 8 25 20 33 in. (8 25 20 ) 33 E A G A AE AE AE T T T AE = − + = + = × + = + = + − = = = + − r i k i k r i k i k i j k i j k / /0: ( 75 ) ( ) ( ) 0Σ = × + × − + + =A E A G A A y A zM MM r T r j j k 26 0 20 19 0 10 ( ) ( ) 0 33 8 25 20 0 75 0 A y A z T M M+ + + = − − i j k i j k j k Coefficient of i: (20)(25) 750 0 33 T − + = 49.5 lbT = Coefficient of j: 49.5 (160 520) ( ) 0 ( ) 1020 lb in. 33 + + = = − ⋅A y A yM M Coefficient of k: 49.5 (26)(25) 1425 ( ) 0 ( ) 450 lb in. 33 − + = = ⋅A z A zM M 0: 75 0Σ = + − =F A T j (1020 lb in.) (450 lb in.)A = − ⋅ + ⋅M j k Coefficient of i: 8 (49.5) 0 12.00 lb 33 x xA A+ = = Coefficient of j: 25 (49.5) 75 0 37.5 lb 33 y yA A+ − = = Coefficient of k: 20 (49.5) 0 33 zA − = 30.0 lb=zA (12.00 lb) (37.5 lb) (30.0 lb)= − + +A i j k www.elsolucionario.net www.elsolucionario.net
- 495. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 493 PROBLEM 4.120 Solve Problem 4.118, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes. PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Free-Body Diagram: ABHΔ is equilateral. Dimensions in mm / / 50 250 350 250 (sin30 ) (cos30 ) (0.5 0.866 ) H C F C T T T = − + = + = ° − ° = − r i j r i k T j k j k / /0: ( 400 ) ( ) ( ) 0Σ = × − + × + + =C F C H C C y C zT M MM r j r j k 350 0 250 50 250 0 ( ) ( ) 0 0 400 0 0 0.5 0.866 C y C zT M M+ − + + = − − i j k i j k j k Coefficient of i: 3 100 10 216.5 0 461.9 N+ × − = =T T 462 NT = Coefficient of j: 43.3(461.9) ( ) 0C yM− + = 3 ( ) 20 10 N mm ( ) 20.0 N m = × ⋅ = ⋅ C y C y M M www.elsolucionario.net www.elsolucionario.net
- 496. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 494 PROBLEM 4.120 (Continued) Coefficient of k: 3 140 10 25(461.9) ( ) 0C zM− × − + = 3 ( ) 151.54 10 N mm ( ) 151.5 N m = × ⋅ = ⋅ C z C z M M 0: 400 0F C TΣ = + − =j (20.0 N m) (151.5 N m)C = ⋅ + ⋅M j k Coefficient of i: 0=xC Coefficient of j: 0.5(461.9) 400 0 169.1 Ny yC C+ − = = Coefficient of k: 0.866(461.9) 0 400 Nz zC C− = = (169.1 N) (400 N)= +C j k www.elsolucionario.net www.elsolucionario.net
- 497. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 495 PROBLEM 4.121 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y-axis. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: First note: 2 2 2 2 (0.08 m) (0.06 m) (0.08) (0.06) m ( 0.8 0.6 ) (0.12 m) (0.09 m) (0.12) (0.09) m (0.8 0.6 ) − + = = + = − + − = = + = − CF CF CF CF CF DE DE DE DE DE T T T T T T i j T λ i j j k T λ j k (a) From F.B.D. of assembly: 0: 0.6 0.8 480 N 0Σ = + − =y CF DEF T T or 0.6 0.8 480 NCF DET T+ = (1) 0: (0.8 )(0.135 m) (0.6 )(0.08 m) 0Σ = − + =y CF DEM T T or 2.25DE CFT T= (2) Substituting Equation (2) into Equation (1), 0.6 0.8[(2.25) ] 480 NCF CFT T+ = 200.00 NCFT = or 200 NCFT = and from Equation (2): 2.25(200.00 N) 450.00DET = = or 450 NDET = www.elsolucionario.net www.elsolucionario.net
- 498. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 496 PROBLEM 4.121 (Continued) (b) From F.B.D. of assembly: 0: (0.6)(450.00 N) 0 270.00 NΣ = − = =z z zF A A 0: (0.8)(200.00 N) 0 160.000 NΣ = − = =x x xF A A or (160.0 N) (270 N)= +A i k 0: (480 N)(0.135 m) [(200.00 N)(0.6)](0.135 m) [(450 N)(0.8)](0.09 m) 0 Σ = + − − = xx AM M 16.2000 N mxAM = − ⋅ 0: (480 N)(0.08 m) [(200.00 N)(0.6)](0.08 m) [(450 N)(0.8)](0.08 m) 0 zz AM MΣ = − + + = 0zAM = or (16.20 N m)A = − ⋅M i www.elsolucionario.net www.elsolucionario.net
- 499. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 497 PROBLEM 4.122 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C. SOLUTION Free-Body Diagram: / / 4.2 2 1.6 2.4 A C E C = + = − r j k r i j / /0: ( 6 ) ( ) ( ) ( ) 0Σ = × − + × + + + =C A C E C C y C zM T M Mr j r i k j k (4.2 2 ) ( 6 ) (1.6 2.4 ) ( ) ( ) ( ) 0C y C zT M M+ × − + − × + + + =j k j i j i k j k Coefficient of i: 12 2.4 0− =T 5.00 lbT = Coefficient of j: 1.6(5 lb) ( ) 0 ( ) 8 lb in.C y C yM M− + = = ⋅ Coefficient of k: 2.4(5 lb) ( ) 0 ( ) 12 lb in.C z C zM M+ = = − ⋅ (8.00 lb in.) (12.00 lb in.)C = ⋅ − ⋅M j k 0: (6 lb) (5 lb) (5 lb) 0Σ = + + − + + =x y zF C C Ci j k j i k Equate coefficients of unit vectors to zero. 5 lb 6 lb 5 lbx y zC C C= − = = − (5.00 lb) (6.00 lb) (5.00 lb)= − + −C i j k www.elsolucionario.net www.elsolucionario.net
- 500. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 498 PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: / / / / / 12 12 8 12 16 12 24 12 32 = = − = − = − = − r i r j k r i k r i k r i k B A F A D A E A F A 12 9 15 in. 0.8 0.6BG BG BG = − + = = − + i k λ i k 12 16 ; 20 in.; 0.6 0.8 12 16 ; 20 in.; 0.6 0.8 DH FJ DH DH FJ FJ λ λ = − + = = − + = − + = = − + i j i j i j i j / / / / 0: ( 24 ) ( 24 ) 0 12 0 0 12 0 16 12 0 32 0.8 0 0.6 0.6 0.8 0 0.6 0.8 0 12 0 8 12 0 24 0 0 24 0 0 24 0 A B A BG BG DH DH DH F A FJ FJ F A E A BG DH FJT T T λ λ λΣ = × + × + × + × − + × − = + − + − − − − + − + − = − − M r T r T r T r j r j i j k i j k i j k i j k i j k Coefficient of i: 12.8 25.6 192 576 0DH FJT T+ + − − = (1) Coefficient of k: 9.6 9.6 288 288 0DH FJT T+ + − − = (2) 3 4 Eq. (1) − Eq. (2): 9.6 0FJT = 0FJT = www.elsolucionario.net www.elsolucionario.net
- 501. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 499 PROBLEM 4.123 (Continued) From Eq. (1): 12.8 268 0DHT − = 60 lbDHT = Coefficient of j: 7.2 (16 0.6)(60.0 lb) 0− + × =BGT 80.0 lbBGT = 0: 24 24 0BG BG DH DH FJT T TΣ = + + + − − =F A j jλ λ Coefficient of i: (80)( 0.8) (60.0)( 0.6) 0 100.0 lb+ − + − = =x xA A Coefficient of j: (60.0)(0.8) 24 24 0yA + − − = 0yA = Coefficient of k: (80.0)( 0.6) 0zA + + = 48.0 lbzA = − (100.0 lb) (48.0 lb)= −A i j Note: The value 0yA = can be confirmed by considering 0.BFMΣ = www.elsolucionario.net www.elsolucionario.net
- 502. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 500 PROBLEM 4.124 Solve Problem 4.123, assuming that the load at C has been removed. PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: / / / / 12 12 16 12 24 12 32 B A B A E A F A = = − = − = − r i r i k r i k r i k 12 9 ; 15 in.; 0.8 0.6 12 16 ; 20 in.; 0.6 0.8 12 16 ; 20 in.; 0.6 0.8 BG DH FJ BG BG DH DH FJ FJ = − + = = − + = − + = = − + = − + = = − + i k i k i j i j i j i j λ λ λ / /0:A B A BG BG D A DHT TΣ = × + ×M r rλ / / ( 24 ) 0DH F A FJ FJ E AT+ × + × − =r r jλ λ 12 0 0 12 0 16 12 0 32 12 0 24 0 0.8 0 0.6 0.6 0.8 0 0.6 0.8 0 0 24 0 BG DH FJT T T+ − + − + − = − − − − i j k i j k i j k i j k : 12.8 25.6 576 0DH FJT T+ + − =i (1) :k 9.6 9.6 288 0DH FJT T+ + − = (2) www.elsolucionario.net www.elsolucionario.net
- 503. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 501 PROBLEM 4.124 (Continued) Multiply Eq. (1) by 3 4 and subtract Eq. (2): 9.6 144 0FJT − = 15.00 lbFJT = From Eq. (1): 12.8 25.6(15.00) 576 0DHT + − = 15.00 lbDHT = :j 7.2 (16)(0.6)(15) (32)(0.6)(15) 0BGT− + + = 7.2 432 0BGT− + = 60.0 lbBGT = 0: 24 0BG BG DA DH FJ FJF T T TΣ = + + + − =A jλ λ λ : (60)( 0.8) (15)( 0.6) (15)( 0.6) 0xA + − + − + − =i 66.0 lbxA = :j (15)(0.8) (15)(0.8) 24 0yA + + − = 0yA = :k (60)(0.6) 0zA + = 36.0 lbzA = − (66.0 lb) (36.0 lb)= −A i k www.elsolucionario.net www.elsolucionario.net
- 504. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 502 PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable. SOLUTION Free-Body Diagram: Dimensions in mm In this problem: 210 mma = We have (240 mm) (320 mm) 400 mm (420 mm) (240 mm) (320 mm) 580 mm (420 mm) (320 mm) 528.02 mm CD CD BD BD BE BE = − = = − + − = = − = j k i j k i k Thus, (0.6 0.8 ) ( 0.72414 0.41379 0.55172 ) (0.79542 0.60604 ) CD CD CD BD BD BD BE BE BE CD T T T CD BD T T T BD BE T T T BE = = − = = − + − = = − j k i j k i k 0: ( ) ( ) ( ) ( ) 0Σ = × + × + × + × =A C CD B BD B BE WM r T r T r T r W www.elsolucionario.net www.elsolucionario.net
- 505. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 503 PROBLEM 4.125 (Continued) Noting that (420 mm) (320 mm) (320 mm) (320 mm) C B W a = − + = = − + r i k r k r i k and using determinants, we write 420 0 320 0 0 320 0 0.6 0.8 0.72414 0.41379 0.55172 0 0 320 0 320 0 0.79542 0 0.60604 0 1.8 0 CD BD BE T T T a − + − − − + + − = − − i j k i j k i j k i j k Equating to zero the coefficients of the unit vectors, i: 192 132.413 576 0CD BDT T− − + = (1) :j 336 231.72 254.53 0CD BD BET T T− − + = (2) k: 252 1.8 0CDT a− + = (3) Recalling that 210 mm,a = Eq. (3) yields 1.8(210) 1.500 kN 252 CDT = = 1.500 kNCDT = From Eq. (1): 192(1.5) 132.413 576 0BDT− − + = 2.1751 kNBDT = 2.18 kNBDT = From Eq. (2): 336(1.5) 231.72(2.1751) 254.53 0BET− − + = 3.9603 kNBET = 3.96 kNBET = www.elsolucionario.net www.elsolucionario.net
- 506. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 504 PROBLEM 4.126 Solve Problem 4.125, assuming that the 1.8-kN load is applied at C. PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable. SOLUTION See solution of Problem 4.125 for free-body diagram and derivation of Eqs. (1), (2), and (3): 192 132.413 576 0CD BDT T− − + = (1) 336 231.72 254.53 0CD BD BET T T− − + = (2) 252 1.8 0CDT a− + = (3) In this problem, the 1.8-kN load is applied at C and we have 420 mm.a = Carrying into Eq. (3) and solving for ,CDT 3.00CDT = 3.00 kNCDT = From Eq. (1): (192)(3) 132.413 576 0BDT− − + = 0BDT = From Eq. (2): 336(3) 0 254.53 0BET− − + = 3.96 kNBET = www.elsolucionario.net www.elsolucionario.net
- 507. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 505 PROBLEM 4.127 The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F. SOLUTION / / / / 200 80 ( ) / 2 80 200 ( ) / 2 200 80 ( ) / 2 80 200 = − + = − = − = − + = + = − + = + r i j T i j r j k T j k r i j T i j r j k E F B B B F C C C F D D D E T T T / / / /0: ( ) 0F B F B C F C D F D E FM T PΣ = × + × + × + × − =r T r T r r j 0 80 200 200 80 0 0 80 200 200 80 0 0 2 2 2 1 1 0 0 1 1 1 1 0 0 0 CB DTT T P − + + + − = − − − − − i j k i j k i j k i j k Equate coefficients of unit vectors to zero and multiply each equation by 2. :i 200 80 200 0B C DT T T− + + = (1) :j 200 200 200 0B C DT T T− − − = (2) :k 80 200 80 200 2 0B C DT T T P− − + + = (3) 80 (2): 200 80 80 80 0B C DT T T− − − = (4) Eqs. (3) (4):+ 160 280 200 2 0B CT T P− − + = (5) Eqs. (1) (2):+ 400 120 0B CT T− − = 120 0.3 400 B C CT T T= − − (6) www.elsolucionario.net www.elsolucionario.net
- 508. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 506 PROBLEM 4.127 (Continued) Eqs. (6) (5): 160( 0.3 ) 280 200 2 0C CT T P− − − + = 232 200 2 0CT P− + = 1.2191CT P= 1.219CT P= From Eq. (6): 0.3(1.2191 ) 0.36574BT P P= − = − = 0.366BT P= − From Eq. (2): 200( 0.3657 ) 200(1.2191 ) 200 0DP P Tθ− − − − = 0.8534DT P= − 0.853DT P= − 0:Σ =F 0B C D P+ + + − =F T T T j ( 0.36574 ) ( 0.8534 ) : 0 2 2 x P P F − − + − =i 0.3448 0.345x xF P F P= − = − ( 0.36574 ) (1.2191 ) ( 0.8534 ) : 200 0 2 2 2 y P P P F − − − − − − =j y yF P F P= = (1.2191 ) : 0 2 z P F + =k 0.8620 0.862z zF P F P= − = − 0.345 0.862P P P= − + −F i j k www.elsolucionario.net www.elsolucionario.net
- 509. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 507 PROBLEM 4.128 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained ( 0).ABMΣ = 2 (10 kg)9.81m/s 98.1 N = = = W mg W / / 300 200 225 425 mm ( 300 200 225 ) 425 600 400 150 mm 300 200 75 mm B A G A GC GC GC T T GC = − + − = = = − + − = − + + = − + + i j k T i j k r i j r i j / / /0: ( ) 0A B A G A G AM WΣ = × + × + × − =r B r T r j 600 400 150 300 200 75 300 200 75 425 0 0 300 200 225 0 98.1 0 T B − + − + − − − − i j k i j k i j k Coefficient of : ( 105.88 35.29) 7357.5 0T− − + =i 52.12 NT = 52.1 NT = www.elsolucionario.net www.elsolucionario.net
- 510. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 508 PROBLEM 4.128 (Continued) Coefficient of 52.12 : 150 (300 75 300 225) 0 425 B − × + × =j 73.58 NB = (73.6 N)=B i 0: 0WΣ = + + − =F A B T j Coefficient of :i 300 73.58 52.15 0 425 xA + − = 36.8 NxA = − Coefficient of :j 200 52.15 98.1 0 425 yA + − = 73.6 NyA = Coefficient of :k 225 52.15 0 425 zA − = 27.6 NzA = www.elsolucionario.net www.elsolucionario.net
- 511. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 509 PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when 240P = lb, 12 in., 8 in.,a b= = and 10 in.c = SOLUTION From F.B.D. of weldment: / / /0: 0O A O B O C OΣ = × + × + × =M r A r B r C 12 0 0 0 8 0 0 0 10 0 0 0 0y z x z x yA A B B C C + + = i j k i j k i j k ( 12 12 ) (8 8 ) ( 10 10 ) 0z y z x y xA A B B C C− + + − + − + =j k i k i j From i-coefficient: 8 10 0z yB C− = or 1.25z yB C= (1) j-coefficient: 12 10 0z xA C− + = or 1.2x zC A= (2) k-coefficient: 12 8 0y xA B− = or 1.5x yB A= (3) 0: 0Σ = + + − =F A B C P or ( ) ( 240 lb) ( ) 0x x y y z zB C A C A B+ + + − + + =i j k From i-coefficient: 0x xB C+ = or x xC B= − (4) j-coefficient: 240 lb 0y yA C+ − = or 240 lby yA C+ = (5) www.elsolucionario.net www.elsolucionario.net
- 512. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 510 PROBLEM 4.129 (Continued) k-coefficient: 0z zA B+ = or z zA B= − (6) Substituting xC from Equation (4) into Equation (2), 1.2z zB A− = (7) Using Equations (1), (6), and (7), 1 1.25 1.25 1.25 1.2 1.5 x xz z y B BB A C − = = = = (8) From Equations (3) and (8): 1.5 or 1.5 y y y y A C C A= = and substituting into Equation (5), 2 240 lbyA = 120 lby yA C= = (9) Using Equation (1) and Equation (9), 1.25(120 lb) 150.0 lbzB = = Using Equation (3) and Equation (9), 1.5(120 lb) 180.0 lbxB = = From Equation (4): 180.0 lbxC = − From Equation (6): 150.0 lbzA = − Therefore, (120.0 lb) (150.0 lb)= −A j k (180.0 lb) (150.0 lb)= +B i k (180.0 lb) (120.0 lb)= − +C i j www.elsolucionario.net www.elsolucionario.net
- 513. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 511 PROBLEM 4.130 Solve Problem 4.129, assuming that the force P is removed and is replaced by a couple (600 lb in.)= + ⋅M j acting at B. PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when 240P = lb, 12 in., 8 in.,a b= = and c 10 in.= SOLUTION From F.B.D. of weldment: / / /0: 0O A O B O C OΣ = × + × + × + =M r A r B r C M 12 0 0 0 8 0 0 0 10 (600 lb in.) 0 0 0 0y z x z x yA A B B C C + + + ⋅ = i j k i j k i j k j ( 12 12 ) (8 8 ) ( 10 10 ) (600 lb in.) 0z y z x y xA A B B C C− + + − + − + + ⋅ =j k j k i j j From i-coefficient: 8 10 0z yB C− = or 0.8y zC B= (1) j-coefficient: 12 10 600 0z xA C− + + = or 1.2 60x zC A= − (2) k-coefficient: 12 8 0y xA B− = or 1.5x yB A= (3) 0: 0Σ = + + =F A B C ( ) ( ) ( ) 0x x y y z zB C A C A B+ + + + + =i j k From i-coefficient: x xC B= − (4) j-coefficient: y yC A= − (5) k-coefficient: z zA B= − (6) www.elsolucionario.net www.elsolucionario.net
- 514. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 512 PROBLEM 4.130 (Continued) Substituting xC from Equation (4) into Equation (2), 50 1.2 x z B A = − (7) Using Equations (1), (6), and (7), 2 0.8 0.8 40 3 y z z xC B A B = = − = − (8) From Equations (3) and (8): 40y yC A= − Substituting into Equation (5), 2 40yA = 20.0 lbyA = From Equation (5): 20.0 lbyC = − Equation (1): 25.0 lbzB = − Equation (3): 30.0 lbxB = Equation (4): 30.0 lbxC = − Equation (6): 25.0 lbzA = Therefore, (20.0 lb) (25.0 lb)= +A j k (30.0 lb) (25.0 lb)= −B i k (30.0 lb) (20.0 lb)= − −C i j www.elsolucionario.net www.elsolucionario.net
- 515. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 513 PROBLEM 4.131 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench (48 N) , (90 N m) .= − = − ⋅F k M k Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. SOLUTION From F.B.D. of pipe assembly ABCD: 0: 0x xF BΣ = = ( -axis) 0: (48 N)(2.5 m) (2 m) 0D x zM BΣ = − = 60.0 NzB = and (60.0 N)=B k ( -axis) 0: (3 m) 90 N m 0D z yM CΣ = − ⋅ = 30.0 NyC = ( -axis) 0: (3 m) (60.0 N)(4 m) (48 N)(4 m) 0D y zM CΣ = − − + = 16.00 NzC = − and (30.0 N) (16.00 N)= −C j k 0: 30.0 0y yF DΣ = + = 30.0 NyD = − 0: 16.00 N 60.0 N 48 N 0z zF DΣ = − + − = 4.00 NzD = and (30.0 N) (4.00 N)= − +D j k www.elsolucionario.net www.elsolucionario.net
- 516. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514 PROBLEM 4.132 Solve Problem 4.131, assuming that the plumber exerts a force F (48 N)= − k and that the motor is turned off ( 0).=M PROBLEM 4.131 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench (48 N) , (90 N= − = − ⋅F k M m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. SOLUTION From F.B.D. of pipe assembly ABCD: 0: 0x xF BΣ = = ( -axis) 0: (48 N)(2.5 m) (2 m) 0D x zM BΣ = − = 60.0 NzB = and (60.0 N)=B k ( -axis) 0: (3 m) (2 m) 0D z y xM C BΣ = − = 0yC = ( -axis) 0: (3 m) (60.0 N)(4 m) (48 N)(4 m) 0D y zM CΣ = − + = 16.00 NzC = − and (16.00 N)= −C k 0: 0y y yF D CΣ = + = 0yD = 0: 0z z z zF D B C FΣ = + + − = 60.0 N 16.00 N 48 N 0zD + − − = 4.00 NzD = and (4.00 N)=D k www.elsolucionario.net www.elsolucionario.net
- 517. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 515 PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. SOLUTION Free-Body Diagram: 2 (50 kg)(9.81 m/s ) 490.50 N 240 600 400 760 mm ( 240 600 400 ) 760 480 200 1 (12 5 ) 520 13 AB W mg W CE CE CE T T CE AB AB = = = = − + − = = = − + − − = = = − i j k T i j k i j λ i j / /0: ( ) ( ) 0AB AB E A AB G AT WΣ = ⋅ × + ⋅ × − =M λ r λ r j / /240 400 ; 240 100 200E A G A= + = − +r i j r i j k 12 5 0 12 5 0 1 240 400 0 240 100 200 0 13 20 13 240 600 400 0 0 ( 12 400 400 5 240 400) 12 200 0 760 T W T W − − + − = × − − − − × × − × × + × = 0.76 0.76(490.50 N)T W= = 373 NT = www.elsolucionario.net www.elsolucionario.net
- 518. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 516 PROBLEM 4.134 Solve Problem 4.133, assuming that wire CE is replaced by a wire connecting E and D. PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. SOLUTION Free-Body Diagram: Dimensions in mm 2 (50 kg)(9.81 m/s ) 490.50 N 240 400 400 614.5 mm (240 400 400 ) 614.5 480 200 1 (12 5 ) 520 13 AB W mg W DE DE DE T T DE AB AB = = = = − + − = = = + − − = = = − i j k T i j k i j λ i j / /240 400 ; 240 100 200E A G A= + = − +r i j r i j k 12 5 0 12 5 0 1 240 400 0 240 100 200 0 13 614.5 13 240 400 400 0 0 ( 12 400 400 5 240 400) 12 200 0 614.5 T W T W − + − = × − − − × × − × × + × × = 0.6145 0.6145(490.50 N)T W= = 301 NT = www.elsolucionario.net www.elsolucionario.net
- 519. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 517 PROBLEM 4.135 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C. SOLUTION First note: 2 2 2 / / (6 in.) (9 in.) (12 in.) (6) (9) (12) in. 1 ( 6 9 12 ) 16.1555 (6 in.) (80 lb) (8 in.) ( ) − − + = + + = − − + = − = = = i j k i j k r i P k r i C j BD A B C D C λ From the F.B.D. of the plates: / /0: ( P C 0BD BD A B BD C DMΣ = ⋅ ) + ⋅( ) =r rλ × λ × 6 9 12 6 9 12 6(80) (8) 1 0 0 1 0 0 0 16.1555 16.1555 0 0 1 0 1 0 ( 9)(6)(80) (12)(8) 0 C C − − − − − + = − + = 45.0 lbC = or (45.0 lb)=C j www.elsolucionario.net www.elsolucionario.net
- 520. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 518 PROBLEM 4.136 Two 2 4-ft× plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram: 1 2 / / / 4 2 4 6ft 1 (2 2 ) 3 2 4 2 4 = − − = = − − = − = − − = i j k λ i j k r i j r i j k r i AF G A G A B A AF AF 1 2/ / /0: ( ( 12 ) ( ( 12 )) ( ) 0AF AF G A AF G A AF B AM TΣ = ⋅ × − + ⋅ × − + ⋅ × =λ r j λ r j λ r / 2 1 2 2 1 2 1 1 2 1 0 4 1 2 ( ) 0 3 3 0 12 0 0 12 0 − − − − − + − − + ⋅ × = − − λ r TAF B A / 1 1 (2 2 12) ( 2 2 12 2 4 12) ( ) 0 3 3 AF B A× × + − × × + × × + ⋅ × =λ r T / / /( ) 32 or ( ) 32⋅ × = − ⋅ × = −λ r T T λ rAF B A A F B A (1) www.elsolucionario.net www.elsolucionario.net
- 521. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 519 PROBLEM 4.136 (Continued) Projection of T on /( )AF B A×λ r is constant. Thus, minT is parallel to / 1 1 (2 2 ) 4 ( 8 4 ) 3 3 AF B A× = − − × = − +λ r i j k i j k Corresponding unit vector is 1 5 ( 2 ).− +j k min 1 ( 2 ) 5 T T= − +j k (2) From Eq. (1): 1 ( 2 ) (2 2 ) 4 32 35 1 ( 2 ) ( 8 4 ) 32 35 T T − + ⋅ − − × = − − + ⋅ − + = − j k i j k i j k j k 3 5(32) (16 4) 32 4.8 5 203 5 T T+ = − = − = 10.7331 lbT = From Eq. (2): min min 1 ( 2 ) 5 1 4.8 5( 2 ) 5 (9.6 lb) (4.8 lb ) T T= − + = − + = − + j k j k T j k Since minT has no i component, wire BH is parallel to the yz plane, and 4 ft.x = (a) 4.00 ft; 8.00 ftx y= = (b) min 10.73 lbT = www.elsolucionario.net www.elsolucionario.net
- 522. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 520 PROBLEM 4.137 Solve Problem 4.136, subject to the restriction that H must lie on the y-axis. PROBLEM 4.136 Two 2 4-ft× plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and- socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram: 1 2 / / / 4 2 4 1 (2 2 ) 3 2 4 2 4 AF G A G A B A AF = − − = − − = − = − − = i j k λ i j k r i j r i j k r i 2/ / /0: ( ( 12 ) ( ( 12 )) ( ) 0AF AF G A AF G A AF B AM TΣ = ⋅ × − + ⋅ × − + ⋅ × =λ r j λ r j λ r / 2 1 2 2 1 2 1 1 2 1 0 4 1 2 ( ) 0 3 3 0 12 0 0 12 0 − − − − + − − + ⋅ × = − − λ r TAF B A / 1 1 (2 2 12) ( 2 2 12 2 4 12) ( ) 0 3 3 AF B A× × + − × × + × × + ⋅ × =λ r T /( ) 32AF B A⋅ × = −λ r T (1) 2 1/2 4 4 (32 )BH y BH y= − + − = +i j k 2 1/2 4 4 (32 ) BH y T T BH y − + − = = + i j k T www.elsolucionario.net www.elsolucionario.net
- 523. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 521 PROBLEM 4.137 (Continued) From Eq. (1): / 2 1/2 2 1 2 ( ) 4 0 0 32 3(32 ) 4 4 AF B A T T y y − − ⋅ × = = − + − − λ r 2 1/2 2 1/2 (32 ) ( 16 8 ) 3 32(32 ) 96 8 16 + − − = − × + = + y y T y T y (2) 2 1/2 2 1/21 2 2 (8 +16) (32 ) (2 ) (32 ) (8) 0: 96 (8 16) y y y ydT dy y − + + + = + Numerator = 0: 2 (8 16) (32 )8y y y+ = + 2 2 8 16 32 8 8y y y+ = × + 0 ft; 16.00 ftx y= = From Eq. (2): 2 1/2 (32 16 ) 96 11.3137 lb 8 16 16 T + = = × + min 11.31 lbT = www.elsolucionario.net www.elsolucionario.net
- 524. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 522 PROBLEM 4.138 The frame ACD is supported by ball-and-socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. SOLUTION Free-Body Diagram: (1 m) (0.75 m) 1.25 m 0.8 0.6 0.5 0.925 0.4 1.125 0.375 0.75 0.75 1.125 AD AD BG BG BG BH BH BH AD AD BG T BG T BH T T BH T − = = = − = − + − = = + − = i k i k T i j k i j k λ λ www.elsolucionario.net www.elsolucionario.net
- 525. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 523 PROBLEM 4.138 (Continued) / /(0.5 m) ; (1 m) ; (268 N)B A C A= = = −r i r i P j To eliminate the reactions at A and D, we shall write 0:ADΣ =M / / /( ) ( ) ( ) 0AD B A BG AD B A BH AD C A⋅ × + ⋅ × + ⋅ × =r T r T r Pλ λ λ (1) Substituting for terms in Eq. (1) and using determinants, 0.8 0 0.6 0.8 0 0.6 0.8 0 0.6 0.5 0 0 0.5 0 0 1 0 0 0 1.125 1.125 0.5 0.925 0.4 0.375 0.75 0.75 0 268 0 BG BHT T − − − + + = − − − − Multiplying all terms by (–1.125), 0.27750 0.22500 180.900BG BHT T+ = (2) For this problem, BG BHT T T= = (0.27750 0.22500) 180.900T+ = 360 NT = www.elsolucionario.net www.elsolucionario.net
- 526. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 524 PROBLEM 4.139 Solve Prob. 4.138, assuming that cable GBH is replaced by a cable GB attached at G and B. PROBLEM 4.138 The frame ACD is supported by ball-and- socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. SOLUTION Free-Body Diagram: (1 m) (0.75 m) 1.25 m 0.8 0.6 0.5 0.925 0.4 1.125 0.375 0.75 0.75 1.125 AD AD BG BG BG BH BH BH AD AD BG T T BG T BH T T BH T − = = = − = − + − = = + − = i k i k i j k i j k λ λ www.elsolucionario.net www.elsolucionario.net
- 527. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 525 PROBLEM 4.139 (Continued) / /(0.5 m) ; (1 m) ; (268 N)B A C A= = = −r i r i P j To eliminate the reactions at A and D, we shall write / / /0: ( ) ( ) ( ) 0AD AD B A BG AD B A BH AD C AΣ = ⋅ × + ⋅ × + ⋅ × =M r T r T r Pλ λ λ (1) Substituting for terms in Eq. (1) and using determinants, 0.8 0 0.6 0.8 0 0.6 0.8 0 0.6 0.5 0 0 0.5 0 0 1 0 0 0 1.125 1.125 0.5 0.925 0.4 0.375 0.75 0.75 0 268 0 BG BHT T − − − + + = − − − − Multiplying all terms by (–1.125), 0.27750 0.22500 180.900BG BHT T+ = (2) For this problem, 0.BHT = Thus, Eq. (2) reduces to 0.27750 180.900BGT = 652 NBGT = www.elsolucionario.net www.elsolucionario.net
- 528. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 526 PROBLEM 4.140 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable. SOLUTION Free-Body Diagram: / / 16 11 8 21in. ( 16 11 8 ) 21 16 16 14 7 24 25 D E C E EA DF DF DE T T DF EA EA = − + − = = = − + − = = − − = = i j k T i j k r i r i k i k λ / /0: ( ) ( ( 60 )) 0EA EA B E EA C EMΣ = ⋅ × + ⋅ ⋅ − =λ r T λ r j 7 0 24 7 0 24 1 16 0 0 16 0 14 0 21 25 25 16 11 8 0 60 0 24 16 11 7 14 60 24 16 60 0 21 25 25 201.14 17,160 0 T T T − − + − = × − − − × × − × × + × × − + = × + = 85.314 lbT = 85.3 lbT = www.elsolucionario.net www.elsolucionario.net
- 529. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 527 PROBLEM 4.141 Solve Problem 4.140, assuming that cable DF is replaced by a cable connecting B and F. SOLUTION Free-Body Diagram: / / 9 9 10 B A C A = = + r i r i k 16 11 16 25.16 in. ( 16 11 16 ) 25.16 7 24 AE BF BF BF T T BF AE AE = − + + = = = − + + − = = i j k T i j k i k λ 25 / /0: ( ) ( ( 60 )) 0AE AF B A AE C AMΣ = ⋅ × + ⋅ ⋅ − =λ r T λ r j 7 0 24 7 0 24 1 9 0 0 9 0 10 0 25 25.16 25 16 11 16 0 60 0 T − − + = × − − 24 9 11 24 9 60 7 10 60 0 25 25.16 25 T × × × × + × × − + = × 94.436 17,160 0− =T 181.7 lbT = www.elsolucionario.net www.elsolucionario.net
- 530. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 528 PROBLEM 4.142 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle? SOLUTION Free-Body Diagram: 0: (2 )(1 m) (60 N)(0.15 m) (250 N)(0.3 m) 0AM FΣ = − − = 42.0 N=F www.elsolucionario.net www.elsolucionario.net
- 531. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 529 PROBLEM 4.143 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C. SOLUTION Free-Body Diagram: 7 in.BC = (a) 0: (15 in.) (200 lb)(6.062 in.) 0Σ = − =CM P 80.83 lbP = 80.8 lb=P (b) 0: 200 lb 0y xF CΣ = − = 200 lbx =C 0: 0 80.83 lb 0y y yF C P CΣ = − = − = 80.83 lby =C 22.0 215.7 lbC α = ° = 216 lb=C 22.0° www.elsolucionario.net www.elsolucionario.net
- 532. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 530 PROBLEM 4.144 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Triangle ACD is isosceles with 90 30 120C = ° + ° = ° 1 (180 120 ) 30 . 2 A D= = ° − ° = ° Thus, DA forms angle of 60° with the horizontal axis. (a) We resolve ADF into components along AB and perpendicular to AB. 0: ( sin30 )(250 mm) (500 N)(100 mm) 0C ADM FΣ = ° − = 400 NADF = (b) 0: (400 N)cos60 500 N 0x xF CΣ = − ° + − = 300 NxC = + 0: (400 N)sin 60° 0y yF CΣ = − + = 346.4 NyC = + 458 N=C 49.1° www.elsolucionario.net www.elsolucionario.net
- 533. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 531 PROBLEM 4.145 A force P of magnitude 280 lb is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at A. SOLUTION Free-Body Diagram: (a) 0: (280 lb)(8 in.) 7 (12 in.) (12 in.) 25 24 (8 in.) 0 25 AM T T T Σ = − − − = (12 11.04) 840T− = 875 lbT = (b) 7 0: (875 lb) 875 lb 0 25 x xF AΣ = + + = 1120xA = − 1120 lbx =A 24 0: 280 lb (875 lb) 0 25 y yF AΣ = − − = 1120yA = + 1120 lby =A 1584 lb=A 45.0° www.elsolucionario.net www.elsolucionario.net
- 534. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 532 PROBLEM 4.146 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F. SOLUTION Free-Body Diagram: (a) 0: 15 lb sin30 0xF BΣ = − ° = 30.0 lb=B 60.0° (b) 0: (30 lb)(4 in.) sin30 (3 in.) cos30 (11in.) (13 in.) 0AM B B FΣ = − + ° + ° − = 120 lb in. (30 lb)sin30 (3 in.) (30 lb)cos30 (11in.) (13 in.) 0F− ⋅ + ° + ° − = 16.2145 lbF = + 16.21lb=F (a) 0: 30 lb cos30 0yF A B FΣ = − + ° − = 30 lb (30 lb)cos30 16.2145 lb 0A − + ° − = 20.23 lbA = + 20.2 lb=A www.elsolucionario.net www.elsolucionario.net
- 535. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 533 PROBLEM 4.147 Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the frame shown. SOLUTION 1300 N 5 13 500 N 12 13 1200 N x y T T T T T = = = = = 0: 450 N 500 N 0 50 Nx x xM C CΣ = − + = = − 50 Nx =C 0: 750 N 1200 N 0 1950 Ny y yF C CΣ = − − = = + 1950 Ny =C 1951 N=C 88.5° 0: (750 N)(0.5 m) (4.50 N)(0.4 m) (1200 N)(0.4 m) 0 C CM MΣ = + + − = 75.0 N mCM = − ⋅ 75.0 N mC = ⋅M www.elsolucionario.net www.elsolucionario.net
- 536. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 534 PROBLEM 4.148 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B. SOLUTION Free-Body Diagram: (Three-force body) The line of action of A must pass through D, where B and P intersect. 3sin50 tan 3cos50 15 0.135756 7.7310 α α ° = ° + = = ° 60 lb sin 7.7310° 446.02 lb 60 lb tan 7.7310° 441.97 lb A B = = = = Force triangle 446 lb=A 7.73° 442 lb=B www.elsolucionario.net www.elsolucionario.net
- 537. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 535 PROBLEM 4.149 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B. SOLUTION The reaction at A must pass through D where C and the 170-N force intersect. 160 mm tan 300 mm 28.07 α α = = ° We note that triangle ABD is isosceles (since AC = BC) and, therefore, 28.07CAD α= = ° Also, since ,CD CB⊥ reaction C forms angle 28.07α = ° with the horizontal axis. Force triangle We note that A forms angle 2α with the vertical axis. Thus, A and C form angle 180 (90 ) 2 90α α α° − ° − − = ° − Force triangle is isosceles, and we have 170 N 2(170 N)sin 160.0 N A C α = = = 170.0 N=A 33.9°; 160.0 N=C 28.1° Free-Body Diagram: (Three-force body) www.elsolucionario.net www.elsolucionario.net
- 538. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 536 PROBLEM 4.150 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when 10a = in., (b) the value of a for which the tension in each wire is 8 lb. SOLUTION / / / 30 30 15 15 B A C A G A a a = + = + = + r i k r i k r i k By symmetry, .B C= / /0: ( ) 0A B A C G AM B C WΣ = × + × + × − =r j r j r j ( 30 ) (30 ) (15 15 ) ( ) 0a B a B W+ × + + × + + × − =i k j i k j i k j 30 30 15 15 0Ba B B Ba W W− + − − + =k i k i k i Equate coefficient of unit vector i to zero: : 30 15 0B Ba W− − + =i 15 15 30 30 W W B C B a a = = = + + (1) 0: 0yF A B C WΣ = + + − = 15 2 0; 30 30 W aW A W A a a + − = = + + (2) (a) For 10 in.a = From Eq. (1): 15(24 lb) 9.00 lb 30 10 C B= = = + From Eq. (2): 10(24 lb) 6.00 lb 30 10 A = = + 6.00 lb; 9.00 lbA B C= = = www.elsolucionario.net www.elsolucionario.net
- 539. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 537 PROBLEM 4.150 (Continued) (b) For tension in each wire = 8 lb, From Eq. (1): 15(24 lb) 8 lb 30 a = + 30 in. 45a+ = 15.00 in.a = www.elsolucionario.net www.elsolucionario.net
- 540. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 538 PROBLEM 4.151 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For 150 mm,a = determine the tension in each cable and the reaction at A. SOLUTION First note: 2 2 (0.48 m) (0.14 m) (0.48) (0.14) m DG DG DG DGT T − + = = + i j T λ 0.48 0.14 0.50 (24 7 ) 25 DG DG T T − + = = + i j i j 2 2 (0.48 m) (0.2 m) (0.48) (0.2) m BE BE BE BET T − + = = + i k T λ 0.48 0.2 0.52 ( 12 5 ) 13 BE BE T T − + = = − + i k j k From F.B.D. of frame ABCD: 7 0: (0.3 m) (350 N)(0.15 m) 0 25 x DGM T Σ = − = or 625 NDGT = 24 5 0: 625 N (0.3 m) (0.48 m) 0 25 13 y BEM T Σ = × − = or 975 NBET = 7 0: (0.14 m) 625 N (0.48 m) (350 N)(0.48 m) 0 25 Σ = + × − = z CFM T or 600 NCFT = www.elsolucionario.net www.elsolucionario.net
- 541. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 539 PROBLEM 4.151 (Continued) 0: ( ) ( ) 0x x CF BE x DG xF A T T TΣ = + + + = 12 24 600 N 975 N 625 N 0 13 25 xA − − × − × = 2100 NxA = 0: ( ) 350 N 0y y DG yF A TΣ = + − = 7 625 N 350 N 0 25 yA + × − = 175.0 NyA = 0: ( ) 0z z BE zF A TΣ = + = 5 975 N 0 13 zA + × = 375 NzA = − Therefore, (2100 N) (175.0 N) (375 N)= + −A i j k www.elsolucionario.net www.elsolucionario.net
- 542. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 540 PROBLEM 4.152 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown. SOLUTION Free-Body Diagram: Dimensions in mm / / 480 160 240 560 mm 480 160 240 560 6 2 3 7 200 480 160 AE AE B A D A AE AE AE AE = + − = + − = = + − = = = + i j k i j k λ i j k λ r i r i j 480 330 240 ; 630 mm 480 330 240 16 11 8 630 21 DF DF DF DF DF DF DF T T T DF = − + − = − + − − + − = = = i j k i j k i j k T / /( ) ( ( 600 )) 0AE AE D A DF AE B AMΣ = ⋅ × + ⋅ × − =λ r T λ r j 6 2 3 6 2 3 1 480 160 0 200 0 0 0 21 7 7 16 11 8 0 640 0 DFT − − + = × − − − 6 160 8 2 480 8 3 480 11 3 160 16 3 200 640 0 21 7 7 DFT − × × + × × − × × − × × × × + = × 3 1120 384 10 0DFT− + × = 342.86 NDFT = 343 NDFT = www.elsolucionario.net www.elsolucionario.net
- 543. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 541 PROBLEM 4.153 A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports. SOLUTION (a) 0: ( sin 45 )2 (cos45 ) 0A aM P C a aΣ = − + ° + ° = 3 2 C P= 2 3 C P= 0.471P=C 45° 2 1 0: 3 2 x xF A P Σ = − 3 x P A = 2 1 2 0: 3 32 y y y P F A P P A Σ = − + = 0.745P=A 63.4° (b) 0: ( cos30 )2 ( sin 30 ) 0Σ = + − ° + ° =CM Pa A a A a (1.732 0.5) 0.812A P A P− = = 0.812P=A 60.0° 0: (0.812 )sin30 0x xF P CΣ = ° + = 0.406xC P= − 0: (0.812 )cos30 0y yF P P CΣ = ° − + = 0.297yC P= − 0.503P=C 36.2° www.elsolucionario.net www.elsolucionario.net
- 544. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 542 PROBLEM 4.153 (Continued) (c) 0: ( cos30 )2 ( sin30 ) 0CM Pa A a A aΣ = + − ° + ° = (1.732 0.5) 0.448A P A P+ = = 0.448A P= 60.0° 0: (0.448 )sin30 0 0.224x x xF P C C PΣ = − ° + = = 0: (0.448 )cos30 0 0.612y y yF P P C C PΣ = ° − + = = 0.652P=C 69.9° (d) Force T exerted by wire and reactions A and C all intersect at Point D. 0: 0D aM PΣ = = Equilibrium is not maintained. Rod is improperly constrained. www.elsolucionario.net www.elsolucionario.net
- 545. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 543 PROBLEM 4.F1 For the frame and loading shown, draw the free-body diagram needed to determine the reactions at A and E when α = 30°. SOLUTION Free-Body Diagram of Frame: www.elsolucionario.net www.elsolucionario.net
- 546. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 544 PROBLEM 4.F2 Neglecting friction, draw the free-body diagram needed to determine the tension in cable ABD and the reaction at C when θ = 60°. SOLUTION Free-Body Diagram of Member ACD: www.elsolucionario.net www.elsolucionario.net
- 547. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 545 PROBLEM 4.F3 Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Draw the free-body diagram needed to determine the tension in cable BD and the reactions at A and C. SOLUTION Free-Body Diagram of Bar AC: Note: By similar triangles 0.15 m 0.075 m 0.25 m 0.5 m B B y y= = www.elsolucionario.net www.elsolucionario.net
- 548. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 546 PROBLEM 4.F4 Draw the free-body diagram needed to determine the tension in each cable and the reaction at D. SOLUTION Free-Body Diagram of Member ABCD: www.elsolucionario.net www.elsolucionario.net
- 549. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 547 PROBLEM 4.F5 A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction on all surfaces, draw the free-body diagram needed to determine the reactions at A, B, and C. SOLUTION Free-Body Diagram of Plywood sheet: www.elsolucionario.net www.elsolucionario.net
- 550. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 548 PROBLEM 4.F6 Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 2.5 in. and the sheave at C has a radius of 2 in. Knowing that the system rotates at a constant rate, draw the free-body diagram needed to determine the tension T and the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and axle. SOLUTION Free-Body Diagram of axle-sheave system: www.elsolucionario.net www.elsolucionario.net
- 551. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 549 PROBLEM 4.F7 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. Draw the free-body diagram needed to determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram of Pole: www.elsolucionario.net www.elsolucionario.net
- 555. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 553 PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION 2 , inA , inx , iny 3 ,inxA 3 ,inyA 1 8 0.5 4 4 32 2 3 2.5 2.5 7.5 7.5 Σ 11 11.5 39.5 X A x AΣ = 2 3 (11in ) 11.5 inX = 1.045 in.X = Y A y AΣ = Σ (11) 39.5Y = 3.59 in.Y = www.elsolucionario.net www.elsolucionario.net
- 556. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 554 PROBLEM 5.2 Locate the centroid of the plane area shown. SOLUTION For the area as a whole, it can be concluded by observation that 2 (72 mm) 3 Y = or 48.0 mmY = Dimensions in mm 2 , mmA , mmx 3 , mmxA 1 1 30 72 1080 2 × × = 20 21,600 2 1 48 72 1728 2 × × = 46 79,488 Σ 2808 101,088 Then X A x A= Σ (2808) 101,088X = or 36.0 mmX = www.elsolucionario.net www.elsolucionario.net
- 557. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 555 PROBLEM 5.3 Locate the centroid of the plane area shown. SOLUTION 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 126 54 6804× = 9 27 61,236 183,708 2 1 126 30 1890 2 × × = 30 64 56,700 120,960 3 1 72 48 1728 2 × × = 48 16− 82,944 27,648− Σ 10,422 200,880 277,020 Then X A xAΣ = Σ 2 2 (10,422 m ) 200,880 mmX = or 19.27 mmX = and Y A yAΣ = Σ 2 3 (10,422 m ) 270,020 mmY = or 26.6 mmY = www.elsolucionario.net www.elsolucionario.net
- 558. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 556 PROBLEM 5.4 Locate the centroid of the plane area shown. SOLUTION 2 , inA , inx , iny 3 , inxA 3 , inyA 1 1 (12)(6) 36 2 = 4 4 144 144 2 (6)(3) 18= 9 7.5 162 135 Σ 54 306 279 Then XA xA= Σ (54) 306X = 5.67 in.X = (54) 279 YA yA Y = Σ = 5.17 in.Y = www.elsolucionario.net www.elsolucionario.net
- 559. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 557 PROBLEM 5.5 Locate the centroid of the plane area shown. SOLUTION By symmetry, X Y= Component 2 , inA , in.x 3 , inxA I Quarter circle 2 (10) 78.54 4 π = 4.2441 333.33 II Square 2 (5) 25− = − 2.5 62.5− Σ 53.54 270.83 2 3 : (53.54 in ) 270.83 inX A x A XΣ = Σ = 5.0585 in.X = 5.06 in.X Y= = www.elsolucionario.net www.elsolucionario.net
- 560. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 558 PROBLEM 5.6 Locate the centroid of the plane area shown. SOLUTION 2 , inA , in.x , in.y 3 , inxA 3 , inyA 1 14 20 280× = 7 10 1960 2800 2 2 (4) 16π π− = − 6 12 –301.59 –603.19 Σ 229.73 1658.41 2196.8 Then 1658.41 229.73 xA X A Σ = = Σ 7.22 in.X = 2196.8 229.73 yA Y A Σ = = Σ 9.56 in.Y = www.elsolucionario.net www.elsolucionario.net
- 561. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 559 PROBLEM 5.7 Locate the centroid of the plane area shown. SOLUTION By symmetry, 0X = Component 2 , inA , in.y 3 , inyA I Rectangle (3)(6) 18= 1.5 27.0 II Semicircle 2 (2) 6.28 2 π − = − 2.151 13.51− Σ 11.72 13.49 2 3 (11.72 in. ) 13.49 in 1.151in. Y A y A Y Y Σ = Σ = = 0X = 1.151in.Y = www.elsolucionario.net www.elsolucionario.net
- 562. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 560 PROBLEM 5.8 Locate the centroid of the plane area shown. SOLUTION 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 (60)(120) 7200= –30 60 3 216 10− × 3 432 10× 2 2 (60) 2827.4 4 π = 25.465 95.435 3 72.000 10× 3 269.83 10× 3 2 (60) 2827.4 4 π − = − –25.465 25.465 3 72.000 10× 3 72.000 10− × Σ 7200 3 72.000 10− × 3 629.83 10× Then 3 (7200) 72.000 10XA x A X= Σ = − × 10.00 mmX = − 3 (7200) 629.83 10YA y A Y= Σ = × 87.5 mmY = www.elsolucionario.net www.elsolucionario.net
- 563. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 561 PROBLEM 5.9 Locate the centroid of the plane area shown. SOLUTION 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 1 (120)(75) 4500 2 = 80 25 3 360 10× 3 112.5 10× 2 (75)(75) 5625= 157.5 37.5 3 885.94 10× 3 210.94 10× 3 2 (75) 4417.9 4 π − = − 163.169 43.169 3 720.86 10− × 3 190.716 10− × Σ 5707.1 3 525.08 10× 3 132.724 10× Then 3 (5707.1) 525.08 10XA x A X= Σ = × 92.0 mmX = 3 (5707.1) 132.724 10YA y A Y= Σ = × 23.3 mmY = www.elsolucionario.net www.elsolucionario.net
- 564. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 562 PROBLEM 5.10 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies 0X = 2 , inA , in.y 3 , inyA 1 2 (8) 100.531 2 π − = − 3.3953 –341.33 2 2 (12) 226.19 2 π = 5.0930 1151.99 Σ 125.659 810.66 Then 3 2 810.66 in 125.66 in y A Y A Σ = = Σ or 6.45 in.Y = www.elsolucionario.net www.elsolucionario.net
- 565. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 563 PROBLEM 5.11 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies 0X = 2 , mA , my 3 , myA 1 4 4.5 3 18 3 × × = 1.2 21.6 2 2 (1.8) 5.0894 2 π − = − 0.76394 −3.8880 Σ 12.9106 17.7120 Then 3 2 17.7120 m 12.9106 m y A Y A Σ = = Σ or 1.372 mY = www.elsolucionario.net www.elsolucionario.net
- 566. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 564 PROBLEM 5.12 Locate the centroid of the plane area shown. SOLUTION Area mm2 , mmx , mmy 3 , mmxA 3 , mmyA 1 31 (200)(480) 32 10 3 = × 360 60 6 11.52 10× 6 1.92 10× 2 31 (50)(240) 4 10 3 − = × 180 15 6 0.72 10− × 6 0.06 10− × Σ 3 28 10× 6 10.80 10× 6 1.86 10× :X A x AΣ = Σ 3 2 6 3 (28 10 mm ) 10.80 10 mmX × = × 385.7 mmX = 386 mmX = :Y A y AΣ = Σ 3 2 6 3 (28 10 mm ) 1.86 10 mmY × = × 66.43 mmY = 66.4 mmY = www.elsolucionario.net www.elsolucionario.net
- 567. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 565 PROBLEM 5.13 Locate the centroid of the plane area shown. SOLUTION 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 (15)(80) 1200= 40 7.5 3 48 10× 3 9 10× 2 1 (50)(80) 1333.33 3 = 60 30 3 80 10× 3 40 10× Σ 2533.3 3 128 10× 3 49 10× Then XA xA= Σ 3 (2533.3) 128 10X = × 50.5 mmX = 3 (2533.3) 49 10 YA yA Y = Σ = × 19.34 mmY = www.elsolucionario.net www.elsolucionario.net
- 568. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 566 PROBLEM 5.14 Locate the centroid of the plane area shown. SOLUTION Dimensions in in. 2 , inA , in.x , in.y 3 , inxA 3 , inyA 1 2 (4)(8) 21.333 3 = 4.8 1.5 102.398 32.000 2 1 (4)(8) 16.0000 2 − = − 5.3333 1.33333 85.333 −21.333 Σ 5.3333 17.0650 10.6670 Then X A xAΣ = Σ 2 3 (5.3333 in ) 17.0650 inX = or 3.20 in.X = and Y A yAΣ = Σ 2 3 (5.3333 in ) 10.6670 inY = or 2.00 in.Y = www.elsolucionario.net www.elsolucionario.net
- 569. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 567 PROBLEM 5.15 Locate the centroid of the plane area shown. SOLUTION Dimensions in mm 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 47 26 1919.51 2 π × × = 0 11.0347 0 21,181 2 1 94 70 3290 2 × × = −15.6667 −23.333 −51,543 −76,766 Σ 5209.5 −51,543 −55,584 Then 51,543 5209.5 x A X A Σ − = = Σ 9.89 mmX = − 55,584 5209.5 y A Y A Σ − = = Σ 10.67 mmY = − www.elsolucionario.net www.elsolucionario.net
- 570. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 568 PROBLEM 5.16 Determine the x coordinate of the centroid of the trapezoid shown in terms of h1, h2, and a. SOLUTION A x xA 1 1 1 2 h a 1 3 a 2 1 1 6 h a 2 2 1 2 h a 2 3 a 2 2 2 6 h a Σ 1 2 1 ( ) 2 a h h+ 2 1 2 1 ( 2 ) 6 a h h+ 21 1 26 1 1 22 ( 2 ) ( ) a h hx A X A a h h +Σ = = Σ + 1 2 1 2 21 3 h h X a h h + = + www.elsolucionario.net www.elsolucionario.net
- 571. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 569 PROBLEM 5.17 For the plane area of Problem 5.5, determine the ratio a/r so that the centroid of the area is located at point B. SOLUTION By symmetry, .X Y= For centroid to be at B, .X a= Area x x A I Quarter circle 21 4 rπ 4 3 r π 31 3 r II Square 2 a− 1 2 a 31 2 a− Σ 2 2 4 r a π − 3 31 1 3 2 r a− :X A x AΣ = Σ 2 2 3 31 1 4 3 2 X r a r a π − = − Set :X a= 2 2 3 31 1 4 3 2 a r a r a π − = − 3 2 31 1 0 2 4 3 a r a r π − + = Divide by 31 : 2 r 3 2 0 2 3 a a r r π − + = 0.508 a r = www.elsolucionario.net www.elsolucionario.net
- 572. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 570 PROBLEM 5.18 Determine the y coordinate of the centroid of the shaded area in terms of r1, r2, and α. SOLUTION First, determine the location of the centroid. From Figure 5.8A: ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 sin2 3 2 2 cos 3 y r A r r π π π α π α α α α − = = − − = − Similarly, ( ) 2 1 1 1 1 2 2 cos 3 2 y r A rπ α π α α = = − − Then ( ) ( ) ( ) 2 2 2 2 1 1 2 2 3 3 2 1 2 cos 2 cos 3 2 3 2 2 cos 3 yA r r r r r r π π α π α π α α α α α Σ = − − − − − = − and ( ) 2 2 2 1 2 2 2 1 2 2 2 A r r r r π π α α π α Σ = − − − = − − Now Y A yAΣ = Σ ( ) ( )2 2 3 3 2 1 2 1 2 cos 2 3 Y r r r r π α α − − = − 3 3 2 1 2 2 2 1 2 2cos 3 2 r r Y r r α π α − = −− www.elsolucionario.net www.elsolucionario.net
- 573. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 571 PROBLEM 5.19 Show that as r1 approaches r2, the location of the centroid approaches that for an arc of circle of radius 1 2( )/2.r r+ SOLUTION First, determine the location of the centroid. From Figure 5.8A: ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 sin2 3 2 2 cos 3 y r A r r π π π α π α α α α − = = − − = − Similarly, ( ) 2 1 1 1 1 2 2 cos 3 2 y r A rπ α π α α = = − − Then ( ) ( ) ( ) 2 2 2 2 1 1 2 2 3 3 2 1 2 cos 2 cos 3 2 3 2 2 cos 3 yA r r r r r r π π α π α π α α α α α Σ = − − − − − = − and ( ) 2 2 2 1 2 2 2 1 2 2 2 A r r r r π π α α π α Σ = − − − = − − Now Y A yAΣ = Σ ( ) ( )2 2 3 3 2 1 2 1 3 3 2 1 2 2 2 1 2 cos 2 3 2 2cos 3 2 Y r r r r r r Y r r π α α α π α − − = − − = −− www.elsolucionario.net www.elsolucionario.net
- 574. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 572 PROBLEM 5.19 (Continued) Using Figure 5.8B, Y of an arc of radius 1 2 1 ( ) 2 r r+ is 2 1 2 2 1 2 2 sin( )1 ( ) 2 ( ) 1 cos ( ) 2 ( ) Y r r r r π π π α α α α − = + − = + − (1) Now ( )2 23 3 2 1 2 1 2 12 1 2 2 2 1 2 12 1 2 2 2 1 2 1 2 1 ( ) ( )( ) r r r r r rr r r r r rr r r r r r r r − + +− = − +− + + = + Let 2 1 r r r r = + Δ = − Δ Then 1 2 1 ( ) 2 r r r= + and 3 3 2 2 2 1 2 2 2 1 2 2 ( ) ( )( )( ) ( ) ( ) 3 2 r r r r r r r rr r r r − + Δ + + Δ − Δ − Δ = + Δ + − Δ− + Δ = In the limit as Δ 0 (i.e., 1 2 ),r r= then 3 3 2 1 2 2 2 1 1 2 3 2 3 1 ( ) 2 2 r r r r r r r − = − = × + So that 1 2 2 2 3 cos ( ) 3 4 Y r r π α α = × + − or 1 2 cos ( ) 2 Y r r α π α = + − which agrees with Equation (1). www.elsolucionario.net www.elsolucionario.net
- 575. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 573 PROBLEM 5.20 The horizontal x-axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x-axis, and explain the results obtained. SOLUTION Length of BD: 0.84in. 0.48 in. (1.44 in. 0.48 in.) 0.48 0.56 1.04 in. 0.84 in. 0.60 in. BD = + − = + = × Area above x-axis (consider two triangular areas): 1 3 3 1 1 (0.28 in.) (0.84 in.)(1.04 in.) (0.56 in.) (0.84 in.)(0.48 in.) 2 2 0.122304 in 0.112896 in AQ y = Σ = + = + 3 1 0.2352 inQ = Area below x-axis: 2 3 3 1 1 (0.40 in.) (0.60 in.)(1.44 in.) (0.20 in.) (0.60 in.) 2 2 0.1728 in 0.0624 in Q yA = Σ = − − = − − 3 2 0.2352 inQ = − 2| | | |,Q Q= since C is centroid and thus, 0Q y A= Σ = www.elsolucionario.net www.elsolucionario.net
- 576. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 574 PROBLEM 5.21 The horizontal x-axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x-axis, and explain the results obtained. SOLUTION Area above x-axis (Area A1): 1 3 3 (25)(20 80) (7.5)(15 20) 40 10 2.25 10 Q y A= Σ = × + × = × + × 3 3 1 42.3 10 mmQ = × Area below x-axis (Area A2): 2 ( 32.5)(65 20)Q y A= Σ = − × 3 3 2 42.3 10 mmQ = − × 1 2| | | |,Q Q= since C is centroid and thus, 0Q y A= Σ = Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 577. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 575 PROBLEM 5.22 A composite beam is constructed by bolting four plates to four 60 × 60 × 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x-axis of the red-shaded areas shown, respectively, in parts a and b of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B. SOLUTION From the problem statement, F is proportional to .xQ Therefore, ( ) , or ( ) ( ) ( ) x BA B B A x A x B x A QF F F F Q Q Q = = For the first moments, 3 3 12 ( ) 225 (300 12) 2 831,600 mm 12 ( ) ( ) 2 225 (48 12) 2(225 30)(12 60) 2 1,364,688 mm x A x B x Q Q Q A = + × = = + − × + − × = Then 1,364,688 (280 N) 831,600 BF = or 459 NBF = www.elsolucionario.net www.elsolucionario.net
- 578. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 576 PROBLEM 5.23 The first moment of the shaded area with respect to the x-axis is denoted by Qx. (a) Express Qx in terms of b, c, and the distance y from the base of the shaded area to the x-axis. (b) For what value of y is xQ maximum, and what is that maximum value? SOLUTION Shaded area: ( ) 1 ( )[ ( )] 2 x A b c y Q yA c y b c y = − = = + − (a) 2 21 ( ) 2 xQ b c y= − (b) For max,Q 1 0 or ( 2 ) 0 2 dQ b y dy = − = 0y = For 0,y = 21 ( ) 2 xQ bc= 21 ( ) 2 xQ bc= www.elsolucionario.net www.elsolucionario.net
- 579. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 577 PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. L, mm , mmx , mmy 2 , mmyL 2 , mmyL 1 2 2 30 72 78+ = 15 36 1170.0 2808.0 2 2 2 48 72 86.533+ = 54 36 4672.8 3115.2 3 78 39 72 3042.0 5616.0 Σ 242.53 8884.8 11,539.2 Then (242.53) 8884.8 X L x L X Σ = Σ = or 36.6 mmX = and (242.53) 11,539.2 Y L yL Y Σ = Σ = or 47.6 mm=Y www.elsolucionario.net www.elsolucionario.net
- 580. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 578 PROBLEM 5.25 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION L, mm , mmx , mmy 2 , mmxL 2 , mmyL 1 2 2 72 48 86.533+ = 36 −24 3115.2 −2076.8 2 132 72 18 9504.0 2376.0 3 2 2 126 30 129.522+ = 9 69 1165.70 8937.0 4 54 −54 27 −2916.0 1458.0 5 54 −27 0 −1458.0 0 Σ 456.06 9410.9 10,694.2 Then X L x LΣ = Σ (456.06) 9410.9X = or 20.6 mm=X Y L y LΣ = Σ (456.06) 10,694.2Y = or 23.4 mm=Y www.elsolucionario.net www.elsolucionario.net
- 581. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 579 PROBLEM 5.26 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION L, in. , in.x , in.y 2 , inxL 2 , inyL 1 2 2 12 6 13.4164+ = 6 3 80.498 40.249 2 3 12 7.5 36 22.5 3 6 9 9 54 54.0 4 3 6 7.5 18 22.5 5 6 3 6 18 36.0 6 6 0 3 0 18.0 Σ 37.416 206.50 193.249 Then (37.416) 206.50X L x L XΣ = Σ = 5.52 in.=X (37.416) 193.249Y L y L YΣ = Σ = 5.16 in.=Y www.elsolucionario.net www.elsolucionario.net
- 582. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 580 PROBLEM 5.27 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION By symmetry, .X Y= L, in. , in.x 2 , inyL 1 1 (10) 15.7080 2 π = 2(10) 6.3662 π = 100 2 5 0 0 3 5 2.5 12.5 4 5 5 25 5 5 7.5 37.5 Σ 35.708 175 Then (35.708) 175X L x L XΣ = Σ = 4.90 in.= =X Y www.elsolucionario.net www.elsolucionario.net
- 583. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 581 PROBLEM 5.28 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion BCD of the wire is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C. Further, because the wire is homogeneous, the center of gravity of the wire will coincide with the centroid of the corresponding line. Thus, 0=X so that 0X LΣ = Then ( 40 mm)(80 mm) ( 40 mm)(100 mm) 0 2 + − + − = L 2 2 14,400 mmL = 120.0 mm=L www.elsolucionario.net www.elsolucionario.net
- 584. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 582 PROBLEM 5.29 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion AB of the wire is horizontal. SOLUTION I II III80 100W w W w W Lw= = = 0: (80 )(32) (100 )(14) ( )(0.4 ) 0CM w w Lw LΣ = + − = 2 9900L = 99.5 mmL = www.elsolucionario.net www.elsolucionario.net
- 585. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 583 PROBLEM 5.30 The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of θ for which the wire is in equilibrium for the indicated position. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus, 0X = so that 0x LΣ = Then 1 2 cos ( ) cos ( ) 0 2 r r r r rθ θ π π − + − = or 4 cos 1 2 0.54921 θ π = + = or 56.7θ = ° www.elsolucionario.net www.elsolucionario.net
- 586. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 584 PROBLEM 5.31 A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C. SOLUTION For quarter circle, 2r r π = (a) 2 0: 0C r M W Tr π Σ = − = 2 2 (8 lb)T W π π = = 5.09 lbT = (b) 0: 0 5.09 lb 0x x xF T C CΣ = − = − = 5.09 lbx =C 0: 0 8 lb 0y y yF C W CΣ = − = − = 8 lby =C 9.48 lb=C 57.5° www.elsolucionario.net www.elsolucionario.net
- 587. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 585 PROBLEM 5.32 Determine the distance h for which the centroid of the shaded area is as far above line BB′ as possible when (a) k = 0.10, (b) k = 0.80. SOLUTION A y yA 1 1 2 ba 1 3 a 21 6 a b 2 1 ( ) 2 kb h− 1 3 h 21 6 kbh− Σ ( ) 2 b a kh− 2 2 ( ) 6 b a kh− Then 2 2 ( ) ( ) 2 6 Y A yA b b Y a kh a kh Σ = Σ − = − or 2 2 3( ) a kh Y a kh − = − (1) and 2 2 2 1 2 ( ) ( )( ) 0 3 ( ) dY kh a kh a kh k dh a kh − − − − − = = − or 2 2 2 ( ) 0h a kh a kh− − + = (2) Simplifying Eq. (2) yields 2 2 2 0kh ah a− + = www.elsolucionario.net www.elsolucionario.net
- 588. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 586 PROBLEM 5.32 (Continued) Then 2 2 2 ( 2 ) 4( )( ) 2 1 1 a a k a h k a k k ± − − = = ± − Note that only the negative root is acceptable since .h a< Then (a) 0.10k = 1 1 0.10 0.10 a h = − − or 0.513h a= (b) 0.80k = 1 1 0.80 0.80 a h = − − or 0.691h a= www.elsolucionario.net www.elsolucionario.net
- 589. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 587 PROBLEM 5.33 Knowing that the distance h has been selected to maximize the distance y from line BB′ to the centroid of the shaded area, show that 2 /3.y h= SOLUTION See solution to Problem 5.32 for analysis leading to the following equations: 2 2 3( ) a kh Y a kh − = − (1) 2 2 2 ( ) 0h a kh a kh− − + = (2) Rearranging Eq. (2) (which defines the value of h which maximizes )Y yields 2 2 2 ( )a kh h a kh− = − Then substituting into Eq. (1) (which defines ),Y 1 2 ( ) 3( ) Y h a kh a kh = × − − or 2 3 Y h= www.elsolucionario.net www.elsolucionario.net
- 590. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 588 PROBLEM 5.34 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION We have h y x a = and ( ) 1 dA h y dx x h dx a = − = − 1 ( ) 2 1 2 EL EL x x y h y h x a = = + = + Then 2 0 0 1 1 2 2 a a x x A dA h dx h x ah a a = = − = − = and 2 3 2 0 0 1 1 2 3 6 a a EL x x x x dA x h dx h a h a a = − = − = 0 2 2 2 3 2 2 20 0 1 1 2 1 1 2 2 33 a EL a a h x x y dA h dx a a h x h x dx x ah a a = + − = − = − = 21 1 : 2 6 ELxA x dA x ah a h = = 2 3 x a= 21 1 : 2 3 ELy A y dA y ah ah = = 2 3 y h= www.elsolucionario.net www.elsolucionario.net
- 591. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 589 PROBLEM 5.35 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION 3 (1 )y h kx= − For , 0.x a y= = 3 0 (1 )= −h k a 3 1 k a ∴ = 3 3 1 x y h a = − 1 , 2 EL ELx x y y dA ydx= = = 3 4 3 30 0 0 3 1 44 a a a x x A dA ydx h dx h x ah a a = = = − = − = 4 2 5 2 3 30 0 0 3 2 105 a a a EL x x x x dA xydx h x dx h a h a a = = − = − = 3 2 3 6 2 3 3 60 0 0 1 1 2 1 1 2 2 2 a a a EL x h x x y dA y ydx h dx dx a a a = = − = − + 2 4 7 2 3 6 0 9 2 282 7 a h x x x ah a a = − + = 23 3 : 4 10 ELxA x dA x ah a h = = 2 5 =x a 23 9 : 4 28 ELyA y dA y ah ah = = 3 7 =y h www.elsolucionario.net www.elsolucionario.net
- 592. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 590 PROBLEM 5.36 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION At ( , ),a h 2 1 :y h ka= or 2 h k a = 2 :y h ma= or h m a = Now 1 2 1 ( ) 2 EL EL x x y y y = = + and 2 2 1 2 2 2 ( ) ( ) h h dA y y dx x x dx a a h ax x dx a = − = − = − Then 2 2 3 2 20 0 1 1 ( ) 2 3 6 a a h h a A dA ax x dx x x ah a a = = − = − = and ( ) 2 3 4 2 2 20 0 2 2 1 2 2 1 2 1 1 1 ( ) 3 4 12 1 1 ( )[( ) ] 2 2 a a EL EL h h a x dA x ax x dx x x a h a a y dA y y y y dx y y dx = − = − = = + − = − 2 2 2 4 2 40 2 2 3 5 4 0 2 1 2 1 1 2 3 5 1 15 a a h h x x dx a a h a x x a ah = − = − = www.elsolucionario.net www.elsolucionario.net
- 593. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 591 PROBLEM 5.36 (Continued) 21 1 : 6 12 ELxA x dA x ah a h = = 1 2 x a= 21 1 : 6 15 ELyA y dA y ah ah = = 2 5 y h= www.elsolucionario.net www.elsolucionario.net
- 594. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 592 PROBLEM 5.37 Determine by direct integration the centroid of the area shown. SOLUTION 2 2 1, x y y k x k = = But 2 ,a ka= thus, 2 1 k a = 2 2 1, x y ax y a = = 2 2 1 2 0 3 3/2 2 0 ( ) 2 1 3 3 3 EL a a x x x dA y y dx ax dx a x A dA ax dx a x ax a a = = − = − = = − = − = 2 3 4 3/2 5/2 3 0 0 0 2 3 5 4 20 a a a EL x x x x dA x ax dx ax dx ax a a a a = − = − = − = 2 31 3 9 : 3 20 20 EL a xA x dA x a a x = = = By symmetry, 9 20 = = a y x www.elsolucionario.net www.elsolucionario.net
- 595. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 593 PROBLEM 5.38 Determine by direct integration the centroid of the area shown. SOLUTION For the element (EL) shown, 2 2b y a x a = − and ( ) ( ) 2 2 2 2 ( ) 1 ( ) 2 2 EL EL dA b y dx b a a x dx a x x y y b b a a x a = − = − − = = + = + − Then ( )2 2 0 a b A dA a a x dx a = = − − To integrate, let 2 2 sin : cos , cosx a a x a dx a dθ θ θ θ= − = = Then /2 0 /2 2 2 0 ( cos )( cos ) 2 sin sin 2 4 1 4 b A a a a d a b a a a ab π π θ θ θ θ θ θ π = − = − + = − and ( )2 2 0 /2 2 2 2 3/2 0 3 1 ( ) 2 3 1 6 a EL b x dA x a a x dx a b a x a x a a b π = − − = + − = www.elsolucionario.net www.elsolucionario.net
- 596. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 594 PROBLEM 5.38 (Continued) ( ) ( )2 2 2 2 0 2 2 3 2 2 20 0 2 2 ( ) 32 2 1 6 a EL a a b b y dA a a x a a x dx a a b b x x dx a a ab = + − − − = = = 21 : 1 4 6 ELxA x dA x ab a b π = − = 2 or 3(4 ) a x π = − 21 : 1 4 6 ELyA y dA y ab ab π = − = 2 or 3(4 ) b y π = − www.elsolucionario.net www.elsolucionario.net
- 597. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 595 PROBLEM 5.39 Determine by direct integration the centroid of the area shown. SOLUTION First note that symmetry implies 0x = For the element (EL) shown, 2 (Figure 5.8B)EL r y dA rdr π π = = Then ( ) 2 2 1 1 2 2 2 2 1 2 2 r r r r r A dA rdr r r π π π = = = = − and ( ) 2 2 1 1 3 3 3 2 1 2 1 2 ( ) 2 3 3 r r EL r r r y dA rdr r r rπ π = = = − So ( ) ( )2 2 3 3 2 1 2 1 2 : 2 3 ELyA y dA y r r r r π = − = − or 3 3 2 1 2 2 2 1 4 3 r r y r rπ − = − www.elsolucionario.net www.elsolucionario.net
- 598. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 596 PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION 2 2 2 1 1 1 1 2 4 4 4 2 2 2 2 4 4 2 2 1 2 2 1 2 4 2 2 2 but but ( ) 1 ( ) 2 2 EL EL b y k x b k a y x a b y k x b k a y x a b x dA y y dx x dx a a x x y y y b x x a a = = = = = = = − = − = = + = + 4 2 2 20 3 5 2 2 0 4 2 2 20 5 3 2 20 4 6 2 2 0 2 3 5 2 15 4 6 1 12 a a a EL a a b x A dA x dx a a b x x a a ba b x x dA x x dx a a b x x dx a a b x x a a a b = = − = − = = − = − = − = www.elsolucionario.net www.elsolucionario.net
- 599. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 597 PROBLEM 5.40 (Continued) 4 4 2 2 2 2 2 20 2 8 4 4 40 2 5 9 2 4 4 0 2 2 2 5 452 9 a EL a a b x b x y dA x x dx a a a a b x x dx a a b x x ab a a = + − = − = − = 22 1 : 15 12 ELxA x dA x ba a b = = 5 8 x a= 22 2 : 15 45 ELyA y dA y ba ab = = 1 3 y b= www.elsolucionario.net www.elsolucionario.net
- 600. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 598 PROBLEM 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION First note that symmetry implies y b= At ,x a y b= = 2 1:y b ka= or 2 b k a = Then 2 1 2 b y x a = 2 2: 2y b b ca= − or 2 b c a = Then 2 2 2 2 x y b a = − Now 2 2 2 1 2 2 2 2 2 ( ) 2 2 1 x b dA y y dx b x dx a a x b dx a = − = − − = − and ELx x= Then 2 3 2 20 0 4 2 1 2 33 a a x x A dA b dx b x ab a a = − = − = and 2 2 4 2 2 20 0 1 2 1 2 2 24 a a EL x x x x dA x b dx b a b a a = − = − = 24 1 : 3 2 ELxA x dA x ab a b = = 3 8 x a= www.elsolucionario.net www.elsolucionario.net
- 601. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 599 PROBLEM 5.42 Determine by direct integration the centroid of the area shown. SOLUTION 1 2 EL ELx x y y dA y dx= = = 2 2 3 2 20 0 2 2 3 2 20 0 2 3 4 2 2 0 2 5 1 2 2 3 6 1 2 2 1 2 1 2 3 4 3 L L L L EL L x x x x A dA h dx h x hL L LL L x x x x x dA xh dx h x dx L LL L x x x h hL L L = = + − = + − = = + − = + − = + − = 25 1 : 6 3 = = ELxA x dA x hL hL 2 5 =x L 2 2 5 1 1 2 6 2 = = = + − EL x x A hL y y y h L L 2 2 2 2 20 2 2 4 2 3 2 4 2 30 2 3 5 2 3 4 2 2 4 2 3 0 1 1 2 2 2 1 4 2 4 4 2 4 4 4 2 103 5 3 L EL L L h x x y dA y dx dx L L h x x x x x dx LL L L L h x x x x x x h L LL L L L = = + − = + + + − − = + + + − − = 25 4 : 6 10 ELyA y dA y hL h L = = 12 25 y h= www.elsolucionario.net www.elsolucionario.net
- 602. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 600 PROBLEM 5.43 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For y1 at ,x a= 2 2 2 2 , 2 , or b y b b ka k a = = = Then 2 1 2 2b y x a = By observation, 2 ( 2 ) 2 b x y x b b a a = − + = − Now ELx x= and for 0 ,x a≤ ≤ 2 2 1 12 2 1 2 and 2 EL b b y y x dA y dx x dx a a = = = = For 2 ,a x a≤ ≤ 2 2 1 2 and 2 2 2 EL b x x y y dA y dx b dx a a = = − = = − Then 2 2 20 2 23 2 0 0 2 2 2 7 2 3 2 6 a a a aa b x A dA x dx b dx aa b x a x b ab aa = = + − = + − − = and 2 2 20 2 4 3 2 2 0 0 2 2 2 2 3 2 2 2 2 4 3 1 1 (2 ) ( ) (2 ) ( ) 2 3 7 6 a a EL a a a b x x dA x x dx x b dx aa b x x b x aa a b b a a a a a a b = + − = + − = + − + − = www.elsolucionario.net www.elsolucionario.net
- 603. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 601 PROBLEM 5.43 (Continued) 2 2 2 2 20 0 2 32 5 2 4 0 2 2 2 2 2 2 2 5 2 3 17 30 a a EL aa a b b b x x y dA x x dx b dx a aa a b x b a x aa ab = + − − = + − − = Hence, 27 7 : 6 6 ELxA x dA x ab a b = = x a= 27 17 : 6 30 ELyA y dA y ab ab = = 17 35 y b= www.elsolucionario.net www.elsolucionario.net
- 604. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 602 PROBLEM 5.44 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For y2 at ,x a= 2 2 , , or a y b a kb k b = = = Then 1/2 2 b y x a = Now ELx x= and for 0 , 2 a x≤ ≤ 1/2 2 1/2 2 2 2 EL y b x y a x dA y dx b dx a = = = = For , 2 a x a≤ ≤ 1/2 1 2 1 1 ( ) 2 2 2 EL b x x y y y a a = + = − + 1/2 2 1 1 ( ) 2 x x dA y y dx b dx aa = − = − + Then 1/2 1/2/2 0 /2 1 2 a a a x x x A dA b dx b dx aa a = = + − + /2 3/2 2 3/2 0 /2 3/2 3/2 3/2 2 2 2 2 1 3 3 2 2 2 ( ) 3 2 2 1 1 ( ) ( ) 2 2 2 2 13 24 aa a b x x x b x aa a b a a a a a a b a a a ab = + − + = + − + − − + − = www.elsolucionario.net www.elsolucionario.net
- 605. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 603 PROBLEM 5.44 (Continued) and 1/2 1/2/2 0 /2 1 2 a a EL a x x x x dA x b dx x b dx aa a = + − + /2 5/2 3 4 5/2 0 /2 5/2 5/2 5/2 3 2 3 2 2 2 2 5 5 3 4 2 ( ) 5 2 2 1 1 ( ) ( ) 3 2 4 2 71 240 aa a b x x x x b aa a b a a a a a a b a a a a b = + − + = + − + − − + − = 1/2 1/2/2 0 1/2 1/2 /2 /2 32 2 2 2 0 /2 2 2 2 2 2 1 1 2 2 2 1 1 1 2 2 2 2 3 2 1 ( ) 4 2 2 6 2 2 a EL a a a a a b x x y dA b dx a a b x x x x b dx a aa a b b x x x a a a a b a a b a a a a = + − + − + = + − − = + − − − 3 211 48 ab = Hence, 213 71 : 24 240 ELxA x dA x ab a b = = 17 0.546 130 x a a= = 213 11 : 24 48 ELyA y dA y ab ab = = 11 0.423 26 y b b= = www.elsolucionario.net www.elsolucionario.net
- 606. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 604 PROBLEM 5.45 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line. Now cos andELx r dL rdθ θ= = Then 7 /4 7 /4 /4 /4 3 [ ] 2 L dL r d r r π π π π θ θ π= = = = and 7 /4 /4 7 /42 /4 2 2 cos ( ) [sin ] 1 1 2 2 2 ELx dL r rd r r r π π π π θ θ θ = = = − − = − Thus 23 : 2 2 xL xdL x r rπ = = − 2 2 3 x r π = − www.elsolucionario.net www.elsolucionario.net
- 607. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 605 PROBLEM 5.46 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line. Now 3 2 2 cos andELx a dL dx dyθ= = + where 3 2 3 2 cos : 3 cos sin sin : 3 sin cos x a dx a d y a dy a d θ θ θ θ θ θ θ θ = = − = = Then 2 2 2 2 1/2 2 2 1/2 /2 /2 2 0 0 [( 3 cos sin ) (3 sin cos ) ] 3 cos sin (cos sin ) 3 cos sin 1 3 cos sin 3 sin 2 3 2 dL a d a d a d a d L dL a d a a π π θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = − + = + = = = = = and /2 3 0 /2 2 5 2 0 cos (3 cos sin ) 1 3 3 cos 5 5 ELx dL a a d a a π π θ θ θ θ θ = = − = Hence, 23 3 : 2 5 ELxL x dL x a a = = 2 5 x a= Alternative Solution: 2/3 3 2 2/3 3 2 cos cos sin sin x x a a y y a a θ θ θ θ = = = = 2/3 2/3 2/3 2/3 3/2 1 or ( ) x y y a x a a + = = − www.elsolucionario.net www.elsolucionario.net
- 608. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 606 PROBLEM 5.46 (Continued) Then 2/3 2/3 1/2 1/3 ( ) ( ) dy a x x dx − = − − Now ELx x= and { } 2 1/2 2 2/3 2/3 1/2 1/3 1 1 ( ) ( ) dy dL dx dx a x x dx− = + = + − − Then 1/3 1/3 2/3 1/30 0 3 3 2 2 a a a L dL dx a x a x = = = = and 1/3 1/3 5/3 2 1/30 0 3 3 5 5 a a EL a x dL x dx a x a x = = = Hence 23 3 : 2 5 ELxL x dL x a a = = 2 5 x a= www.elsolucionario.net www.elsolucionario.net
- 609. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 607 PROBLEM 5.47* A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. We have at ,x a= 3/2 1 , , ory a a ka k a = = = Then 3/21 y x a = and 1/23 2 dy x dx a = Now ELx x= and 2 1/22 1/2 1 3 1 2 1 4 9 2 dy dL dx dx x dx a a x dx a = + = + = + Then 0 3/2 0 3/2 1 4 9 2 1 2 1 (4 9 ) 3 92 [(13) 8] 27 1.43971 a a L dL a x dx a a x a a a = = + = × + = − = and 0 1 4 9 2 a ELx dL x a x dx a = + www.elsolucionario.net www.elsolucionario.net
- 610. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 608 PROBLEM 5.47* (Continued) Use integration by parts with 3/2 4 9 2 (4 9 ) 27 u x dv a x dx du dx v a x = = + = = + Then 3/2 3/2 0 0 3/2 2 5/2 0 2 3/2 5/2 2 1 2 2 (4 9 ) (4 9 ) 27 272 (13) 1 2 (4 9 ) 27 4527 2 (13) [(13) 32] 27 45 0.78566 a a EL a x dL x a x a x dx a a a x a a a = × + − + = − + = − − = 2 : (1.43971 ) 0.78566ELxL x dL x a a= = or 0.546x a= www.elsolucionario.net www.elsolucionario.net
- 611. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 609 PROBLEM 5.48 Determine by direct integration the centroid of the area shown. SOLUTION /2 /2 0 0 /2 0 sin 1 , , 2 sin cos EL EL L L L x y a L x x y y dA y dx x A y dx a dx L L x aL A a L π π π π π = = = = = = = − = /2 /2 0 0 sin L L EL x x dA xydx xa dx L π = = Setting , x u L π = we have , L x u π = , L dx du π = 2 /2 /2 0 0 sin sinEL L L L x dA u a u du a u xdu π π π π π = = Integrating by parts, 2 2/2 /2 0 20 2/2 /2 /2 2 2 2 2 0 0 0 /22 2/2 2 2 0 0 [ cos ] cos 1 1 sin sin 2 2 2 1 1 1 (1 cos2 ) sin 2 2 4 2 82 EL L L EL L aL x dA a u u u du x a L y dA y dx a dx u du L a L a L u du u u a L π π π π π π π π π ππ = − + = = = = = − = − = 2 2 : π π = = EL aL aL xA x dA x π = L x 21 : 8π = = EL aL yA y dA y a L 8 π =y a www.elsolucionario.net www.elsolucionario.net
- 612. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 610 PROBLEM 5.49* Determine by direct integration the centroid of the area shown. SOLUTION We have 2 2 cos cos 3 3 2 2 sin sin 3 3 EL EL x r ae y r ae θ θ θ θ θ θ = = = = and 2 21 1 ( )( ) 2 2 dA r rd a e dθ θ θ= = Then 2 2 2 2 0 0 2 2 2 1 1 1 2 2 2 1 ( 1) 4 133.623 A dA a e d a e a e a π π θ θ π θ = = = = − = and 2 2 0 3 3 0 2 1 cos 3 2 1 cos 3 ELx dA ae a e d a e d π θ θ π θ θ θ θ θ = = To proceed, use integration by parts, with 3 u e θ = and 3 3du e dθ θ= cosdv dθ θ= and sinv θ= Then 3 3 3 cos sin sin (3 )e d e e dθ θ θ θ θ θ θ θ= − Now let 3 u e θ = then 3 3du e dθ θ= sin ,dv dθ θ= then cosv θ= − Then 3 3 3 3 cos sin 3 cos ( cos )(3 )e d e e e dθ θ θ θ θ θ θ θ θ θ = − − − − www.elsolucionario.net www.elsolucionario.net
- 613. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 611 PROBLEM 5.49* (Continued) so that 3 3 3 3 0 3 3 3 cos (sin 3cos ) 10 1 (sin 3cos ) 3 10 ( 3 3) 1239.26 30 EL e e d e x dA a a e a θ θ πθ π θ θ θ θ θ θ = + = + = − − = − Also, 2 2 0 3 3 0 2 1 sin 3 2 1 sin 3 ELy dA ae a e d a e d π θ θ π θ θ θ θ θ = = Use integration by parts, as above, with 3 u e θ = and 3 3du e dθ θ= sindv dθ θ= and cosv θ= − Then 3 3 3 sin cos ( cos )(3 )e d e e dθ θ θ θ θ θ θ θ= − − − so that 3 3 3 3 0 3 3 3 sin ( cos 3sin ) 10 1 ( cos 3sin ) 3 10 ( 1) 413.09 30 EL e e d e y dA a a e a θ θ πθ π θ θ θ θ θ θ = − + = − + = + = Hence, 2 3 : (133.623 ) 1239.26ELxA x dA x a a= = − or 9.27x a= − 2 3 : (133.623 ) 413.09ELyA y dA y a a= = or 3.09y a= www.elsolucionario.net www.elsolucionario.net
- 614. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 612 PROBLEM 5.50 Determine the centroid of the area shown when 2a = in. SOLUTION We have 1 1 1 1 2 2 EL EL x x y y x = = = − and 1 1dA ydx dx x = = − Then 2 1 1 1 1 [ ln ] ( ln 1) in 2 a adx A dA x x a a x = = − = − = − − and 2 2 3 1 1 1 1 1 in 2 2 2 a a EL x a x dA x dx x a x = − = − = − + 21 1 3 1 1 1 1 1 2 1 1 1 1 2 2 1 1 1 1 2ln 2ln in 2 2 a a EL a y dA dx dx x x x x x x a a x a = − − = − + = − − = − − 2 1 2 2 1 : in. ln 1 2ln : in. 2( ln 1) a EL a EL a xA x dA x a a a a yA y dA y a a − + = = − − − − = = − − Find x and y when 2 in.=a We have 21 1 2 2 (2) 2 2 ln 2 1 x − + = − − or 1.629 in.x = and 1 2 2 2ln 2 2(2 ln 2 1) y − − = − − or 0.1853 in.y = www.elsolucionario.net www.elsolucionario.net
- 615. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 613 PROBLEM 5.51 Determine the value of a for which the ratio /x y is 9. SOLUTION We have 1 1 1 1 2 2 EL EL x x y y x = = = − and 1 1dA y dx dx x = = − Then 1 1 2 1 1 [ ln ] 2 ( ln 1) in a adx A dA x x x a a = = − = − = − − and 2 1 1 2 3 1 1 2 1 in 2 2 a a EL x x dA x dx x x a a = − = − = − + 21 1 1 3 1 1 1 1 2 1 1 1 1 2 2 1 1 2ln 2 1 1 2ln in 2 a a EL a y dA dx dx x x x x x x x a a a = − − = − + = − − = − − 2 1 2 2 1 : in. ln 1 2ln : in. 2( ln 1) a EL a EL a xA x dA x a a a a yA y dA y a a − + = = − − − − = = − − www.elsolucionario.net www.elsolucionario.net
- 616. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 614 PROBLEM 5.51 (Continued) Find a so that 9. x y = We have EL EL x dAx xA y yA y dA = = Then ( ) 21 1 2 2 11 2 9 2 ln a a a a a − + = − − or 3 2 11 18 ln 9 0a a a a a− + + + = Using trial and error or numerical methods, and ignoring the trivial solution a = 1 in., we find 1.901in. and 3.74 in.a a= = www.elsolucionario.net www.elsolucionario.net
- 617. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 615 PROBLEM 5.52 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the x-axis, (b) the y-axis. SOLUTION From the solution of Problem 5.1, we have 2 3 3 11in 11.5 in 39.5 in A xA yA = Σ = Σ = Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x-axis: area 3 Volume 2 2 2 (39.5 in ) y A yAπ π π = = Σ = or 3 Volume 248 in= line line 2 2 3 3 4 4 5 5 6 6 7 7 8 8 Area 2 2 ( ) 2 ( ) 2 [(1)(2) (2)(3) (2.5)(1) (3)(3) (5.5)(5) (8)(1) (4)(8)] π π π π = = Σ = + + + + + + = + + + + + + y L y L y L y L y L y L y L y L y L or 2 Area 547 in= (b) Rotation about the y-axis: area 3 Volume 2 2 2 (11.5 in ) x A xAπ π π = = Σ = or 3 Volume 72.3 in= line line 1 1 2 2 3 3 4 4 5 5 6 6 7 7 Area 2 2 ( ) 2 ( ) 2 [(0.5)(1) (1)(2) (2.5)(3) (4)(1) (2.5)(3) (1)(5) (0.5)(1)] x L x L x L x L x L x L x L x L x L π π π π = = Σ = + + + + + + = + + + + + + or 2 Area 169.6 in= www.elsolucionario.net www.elsolucionario.net
- 618. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 616 PROBLEM 5.53 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.2 about (a) the line y = 72 mm, (b) the x-axis. SOLUTION From the solution of Problem 5.2, we have 2 2808 mm 36 mm 48 mm = = = A x y Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the line 72 mm:y = Volume 2 (72 ) 2 (72 48)(2808) π π = − = − y A 3 3 Volume 423 10 mm= × line line 1 1 3 3 Area 2 2 ( ) 2 ( ) y L y L y L y L π π π = = Σ = + where 1y and 3y are measured with respect to line 72 mm.=y ( ) ( )2 2 2 2 Area 2 (36) 48 72 (36) 30 72π = + + + 3 2 Area 37.2 10 mm= × (b) Rotation about the x-axis: areaVolume 2 2 (48)(2808) y Aπ π = = 3 3 Volume 847 10 mm= × ( ) ( ) line line 1 1 2 2 3 3 2 2 2 2 Area 2 2 ( ) 2 ( ) 2 (36) 48 72 (72)(78) (36) 30 72 y L y L y L y L y L π π π π = = Σ = + + = + + + + 3 2 Area 72.5 10 mm= × www.elsolucionario.net www.elsolucionario.net
- 619. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 617 PROBLEM 5.54 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.8 about (a) the line x = −60 mm, (b) the line y = 120 mm. SOLUTION From the solution of Problem 5.8, we have 2 3 3 3 3 7200 mm 72 10 mm 629.83 10 mm A x A y A = Σ = − × Σ = × Applying the theorems of Pappus-Guldinus, we have (a) Rotation about line 60 mm:x = − 3 Volume 2 ( 60) 2 ( 60 ) 2 [ 72 10 60(7200)] x A xA Aπ π π = + = Σ + = − × + 6 3 Volume 2.26 10 mm= × line line 1 1 2 2 3 3 Area 2 2 ( ) 2 ( ) 2(60) (60) 2(60) (60) 2 60 60 (60)(120) 2 2 x L x L x L x L x L π π π π π π π π = = Σ = + + = − + + + where 1 2 3, ,x x x are measured with respect to line 60 mm.x = − 3 2 Area 116.3 10 mm= × (b) Rotation about line 120 mm:y = 3 Volume 2 (120 ) 2 (120 ) 2 [120(7200) 629.83 10 ] y A A yAπ π π = − = −Σ = − × 6 3 Volume 1.471 10 mm= × line line 1 1 2 2 4 4 Area 2 2 ( ) 2 ( ) y L y L y L y L y L π π π = = Σ = + + where 1 2 4, ,y y y are measured with respect to line 120 mm.y = 2(60) (60) 2(60) (60) Area 2 120 (60)(120) 2 2 π π π π π = − + + 3 2 Area 116.3 10 mm= × www.elsolucionario.net www.elsolucionario.net
- 620. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 618 PROBLEM 5.55 Determine the volume of the solid generated by rotating the parabolic area shown about (a) the x-axis, (b) the axis AA′. SOLUTION First, from Figure 5.8a, we have 4 3 2 5 A ah y h = = Applying the second theorem of Pappus-Guldinus, we have (a) Rotation about the x-axis: Volume 2 2 4 2 5 3 yA h ah π π = = or 216 Volume 15 ahπ= (b) Rotation about the line :AA′ Volume 2 (2 ) 4 2 (2 ) 3 a A a ah π π = = or 216 Volume 3 a hπ= www.elsolucionario.net www.elsolucionario.net
- 621. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 619 PROBLEM 5.56 Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm. SOLUTION The area A and circumference C of the cross section of the bar are 2 and . 4 A d C d π π= = Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V, side end2( ) 2( ) 2( ) 2( ) 2( ) V V V AL RA L R A π π = + = + = + or 2 2[30 mm (10 mm)] (6 mm) 4 V π π = + 3 3470 mm= 3 or 3470 mmV = For the area A, side end2( ) 2( ) 2( ) 2( ) 2( ) A A A CL RC L R C π π = + = + = + or 2[30 mm (10 mm)][ (6 mm)]A π π= + 2 2320 mm= 2 or 2320 mmA = www.elsolucionario.net www.elsolucionario.net
- 622. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 620 PROBLEM 5.57 Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 264 are correct. SOLUTION Following the second theorem of Pappus-Guldinus, in each case, a specific generating area A will be rotated about the x-axis to produce the given shape. Values of y are from Figure 5.8a. (1) Hemisphere: the generating area is a quarter circle. We have 24 2 2 3 4 a V y A a π π π π = = or 32 3 V aπ= (2) Semiellipsoid of revolution: the generating area is a quarter ellipse. We have 4 2 2 3 4 a V y A ha π π π π = = or 22 3 V a hπ= (3) Paraboloid of revolution: the generating area is a quarter parabola. We have 3 2 2 2 8 3 V y A a ahπ π = = or 21 2 V a hπ= (4) Cone: the generating area is a triangle. We have 1 2 2 3 2 a V y A haπ π = = or 21 3 V a hπ= www.elsolucionario.net www.elsolucionario.net
- 623. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 621 PROBLEM 5.58 Determine the volume and weight of the solid brass knob shown, knowing that the specific weight of brass is 0.306 lb/in3 . SOLUTION Area, in2 , in.y 3 , inyA 1 2 (0.75) 0.4418 4 π = 0.8183 0.3615 2 (0.5)(0.75) 0.375= 0.25 0.0938 3 (1.25)(0.75) 0.9375= 0.625 0.5859 4 2 (0.75) 0.4418 4 π− = − 0.9317 −0.4116 Σ 0.6296 3 3 2 2 (0.6296 in ) 3.9559 inV y Aπ π= Σ = = 3 3.96 inV = 3 3 (0.306 lb/in )(3.9559 in )W Vγ= = 1.211lb=W Volume of knob is obtained by rotating area at left about the x-axis. Consider area as made of components shown below. www.elsolucionario.net www.elsolucionario.net
- 624. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 622 PROBLEM 5.59 Determine the total surface area of the solid brass knob shown. SOLUTION Area is obtained by rotating lines shown about the x-axis. L, in. , in.y 2 , inyL 1 0.5 0.25 0.1250 2 (0.75) 1.1781 2 π = 0.9775 1.1516 3 (0.75) 1.1781 2 π = 0.7725 0.9101 4 0.5 0.25 0.1250 Σ 2.3117 2 2 2 (2.3117 in )A y Lπ π= Σ = 2 14.52 inA = www.elsolucionario.net www.elsolucionario.net
- 625. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 623 PROBLEM 5.60 The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that the density of aluminum is 2800 kg/m3 , determine the mass of the shade. SOLUTION The mass of the lamp shade is given by m V Atρ ρ= = where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have 1 1 2 2 3 3 4 4 2 2 2 ( ) A yL yL y L y L y L y L π π π = = Σ = + + + or 2 2 2 2 2 2 2 13 mm 13 16 2 (13 mm) mm (32 mm) (3 mm) 2 2 16 28 mm (8 mm) (12 mm) 2 28 33 mm (28 mm) (5 mm) 2 2 (84.5 466.03 317.29 867.51) 10,903.4 mm A π π + = + × + + + × + + + × + = + + + = Then 3 3 2 (2800 kg/m )(10.9034 10 m )(0.001 m) m Atρ − = = × or 0.0305 kgm = www.elsolucionario.net www.elsolucionario.net
- 626. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 624 PROBLEM 5.61 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from brass. Knowing that the density of brass is 8470 kg/m3 , determine the mass of the escutcheon. SOLUTION The mass of the escutcheon is given by (density) ,m V= where V is the volume. V can be generated by rotating the area A about the x-axis. From the figure: 2 2 1 2 75 12.5 73.9510 m 37.5 76.8864 mm tan 26 L L = − = = = ° 2 1 1 2.9324 mm 12.5 sin 9.5941 75 26 9.5941 8.2030 0.143168 rad 2 a L L φ α − = − = = = ° ° − ° = = ° = Area A can be obtained by combining the following four areas: Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have 2 2V yA yAπ π= = Σ Seg. 2 , mmA , mmy 3 , mmyA 1 1 (76.886)(37.5) 1441.61 2 = 1 (37.5) 12.5 3 = 18,020.1 2 2 (75) 805.32α− = − 2(75)sin sin ( ) 15.2303 3 α α φ α + = −12,265.3 3 1 (73.951)(12.5) 462.19 2 − = − 1 (12.5) 4.1667 3 = −1925.81 4 (2.9354)(12.5) 36.693− = − 1 (12.5) 6.25 2 = −229.33 Σ 3599.7 www.elsolucionario.net www.elsolucionario.net
- 627. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 625 PROBLEM 5.61 (Continued) Then 3 3 3 6 3 2 2 (3599.7 mm ) 22,618 mm (density) (8470 kg/m )(22.618 10 m ) V yA m V π π − = Σ = = = = × 0.191574 kg= or 0.1916 kgm = www.elsolucionario.net www.elsolucionario.net
- 628. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 626 PROBLEM 5.62 A 3 4 - in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process. SOLUTION The required volume can be generated by rotating the area shown about the y-axis. Applying the second theorem of Pappus-Guldinus, we have 2 3 1 1 1 1 1 2 in. in. in. 8 3 4 2 4 4 V x Aπ π = = + × × × 3 0.0900 inV = www.elsolucionario.net www.elsolucionario.net
- 629. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 627 PROBLEM 5.63 Knowing that two equal caps have been removed from a 10-in.-diameter wooden sphere, determine the total surface area of the remaining portion. SOLUTION The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of Pappus-Guldinus, we have 1 1 2 2 2 2 2 (2 ) A XL x L x L x L π π π = = Σ = + Now 4 tan 3 α = or 53.130α = ° Then 2 180 5 in. sin53.130 53.130 4.3136 in. x π ° × ° = °× = and 2 2 53.130 (5 in.) 180 9.2729 in. 3 2 2 in. (3 in.) (4.3136 in.)(9.2729 in.) 2 L A π π = °× ° = = + or 2 308 inA = www.elsolucionario.net www.elsolucionario.net
- 630. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 628 PROBLEM 5.64 Determine the capacity, in liters, of the punch bowl shown if R = 250 mm. SOLUTION The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have 1 1 2 2 2 6 3 3 3 3 3 2 2 2 ( ) 1 1 1 1 3 2 sin30 2 3 2 2 2 2 3 6 2 16 3 2 3 3 3 8 3 3 (0.25 m) 8 0.031883 m V xA xA x A x A R R R R R R R R π π π π π π π π π = = Σ = + ° = × × × + × = + = = = Since 3 3 10 l 1m= 3 3 3 10 l 0.031883 m 1 m V = × 31.9 lV = www.elsolucionario.net www.elsolucionario.net
- 631. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 629 PROBLEM 5.65* The shade for a wall-mounted light is formed from a thin sheet of translucent plastic. Determine the surface area of the outside of the shade, knowing that it has the parabolic cross section shown. SOLUTION First note that the required surface area A can be generated by rotating the parabolic cross section through π radians about the y-axis. Applying the first theorem of Pappus-Guldinus, we have A xLπ= Now at 2 1 100 mm, 250 mm 250 (100) or 0.025 mm x y k k − = = = = and 2 1 ELx x dy dL dx dx = = + where 2 dy kx dx = Then 2 2 1 4dL k x dx= + We have ( )100 2 2 0 1 4ELxL x dL x k x dx= = + { } 100 2 2 3/2 2 0 2 2 3/2 3/2 2 2 1 1 (1 4 ) 3 4 1 1 [1 4(0.025) (100) ] (1) 12 (0.025) 17,543.3 mm xL k x k = + = + − = Finally, 2 (17,543.3 mm )A π= or 3 2 55.1 10 mmA = × www.elsolucionario.net www.elsolucionario.net
- 632. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 630 PROBLEM 5.66 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION (a) I II I II 1 (1100 N/m)(6 m) 2200 N 3 (900N/m)(6m) 5400 N 2200 5400 7600 N R R R R R = = = = = + = + = : (7600) (2200)(1.5) (5400)(3)XR xR X= Σ = + 2.5658 mX = 7.60 kN=R , 2.57 m=X (b) 0: (6 m) (7600 N)(2.5658 m) 0 3250.0 N AM B B Σ = − = = 3.25 kN=B 0: 3250.0 N 7600 N 0 4350.0 N Σ = + − = = yF A A 4.35 kN=A www.elsolucionario.net www.elsolucionario.net
- 633. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 631 PROBLEM 5.67 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION I II I II 1 (150lb/ft)(9 ft) 675 lb 2 1 (120 lb/ft)(9ft) 540lb 2 675 540 1215 lb : (1215) (3)(675) (6)(540) 4.3333 ft R R R R R XR x R X X = = = = = + = + = = Σ = + = (a) 1215 lb=R 4.33 ftX = (b) Reactions: 0: (9 ft) (1215 lb)(4.3333 ft) 0AM BΣ = − = 585.00 lbB = 585 lb=B 0: 585.00 lb 1215 lb 0yF AΣ = + − = 630.00 lbA = 630 lb=A www.elsolucionario.net www.elsolucionario.net
- 634. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 632 PROBLEM 5.68 Determine the reactions at the beam supports for the given loading. SOLUTION First replace the given loading by the loadings shown below. Both loading are equivalent since they are both defined by a linear relation between load and distance and have the same values at the end points. 1 2 1 (900 N/m)(1.5 m) 675 N 2 1 (400 N/m)(1.5 m) 300 N 2 R R = = = = 0: (675 N)(1.4 m) (300 N)(0.9 m) (2.5 m)AM B CΣ = − + + = 270 N=B 270 N=B 0: 675 N 300 N 270 N 0yF AΣ = − + + = 105.0 N=A 105.0 N=A www.elsolucionario.net www.elsolucionario.net
- 635. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 633 PROBLEM 5.69 Determine the reactions at the beam supports for the given loading. SOLUTION I I II II (200 lb/ft)(15 ft) 3000 lb 1 (200 lb/ft)(6 ft) 2 600 lb R R R R = = = = 0: (3000 lb)(1.5 ft) (600 lb)(9 ft 2ft) (15 ft) 0AM BΣ = − − + + = 740 lbB = 740 lb=B 0: 740 lb 3000 lb 600 lb 0yF AΣ = + − − = 2860 lbA = 2860 lb=A www.elsolucionario.net www.elsolucionario.net
- 636. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 634 PROBLEM 5.70 Determine the reactions at the beam supports for the given loading. SOLUTION I II (200 lb/ft)(4 ft) 800 lb 1 (150 lb/ft)(3 ft) 225 lb 2 R R = = = = 0: 800 lb 225 lb 0yF AΣ = − + = 575 lb=A 0: (800 lb)(2 ft) (225 lb)(5 ft) 0A AM MΣ = − + = 475 lb ftA = ⋅M www.elsolucionario.net www.elsolucionario.net
- 637. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 635 PROBLEM 5.71 Determine the reactions at the beam supports for the given loading. SOLUTION I II 1 (4 kN/m)(6 m) 2 12 kN (2 kN/m)(10m) 20 kN R R = = = = 0: 12 kN 20 kN 0yF AΣ = − − = 32.0 kN=A 0: (12 kN)(2 m) (20 kN)(5 m) 0A AM MΣ = − − = 124.0 kN mA = ⋅M www.elsolucionario.net www.elsolucionario.net
- 638. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 636 PROBLEM 5.72 Determine the reactions at the beam supports for the given loading. SOLUTION First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the same. We have I II (6 m)(300 N/m) 1800 N 2 (6 m)(1200 N/m) 4800 N 3 R R = = = = Then 0: 0x xF AΣ = = 0: 1800 N 4800 N 0y yF AΣ = + − = or 3000 NyA = 3.00 kN=A 15 0: (3 m)(1800 N) m (4800 N) 0 4 A AM M Σ = + − = or 12.6 kN mAM = ⋅ 12.60 kN m= ⋅AM www.elsolucionario.net www.elsolucionario.net
- 639. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 637 PROBLEM 5.73 Determine the reactions at the beam supports for the given loading. SOLUTION We have I II 1 (12 ft)(200 lb/ft) 800 lb 3 1 (6 ft)(100 lb/ft) 200 lb 3 R R = = = = Then 0: 0x xF AΣ = = 0: 800 lb 200 lb 0y yF AΣ = − − = or 1000 lbyA = 1000 lb=A 0: (3 ft)(800 lb) (16.5 ft)(200 lb) 0A AM MΣ = − − = or 5700 lb ftAM = ⋅ 5700 lb ftA = ⋅M www.elsolucionario.net www.elsolucionario.net
- 640. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 638 PROBLEM 5.74 Determine the reactions at the beam supports for the given loading when wO = 400 lb/ft. SOLUTION I II 1 1 (12 ft) (400 lb/ft)(12 ft) 2400 lb 2 2 1 (300 lb/ft)(12 ft) 1800 lb 2 OR w R = = = = = 0: (2400 lb)(1ft) (1800 lb)(3 ft) (7 ft) 0Σ = − + =BM C 428.57 lb=C 429 lb=C 0: 428.57 lb 2400 lb 1800 lb 0Σ = + − − =yF B 3771lb=B 3770 lb=B www.elsolucionario.net www.elsolucionario.net
- 641. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 639 PROBLEM 5.75 Determine (a) the distributed load wO at the end A of the beam ABC for which the reaction at C is zero, (b) the corresponding reaction at B. SOLUTION For ,Ow I II 1 (12 ft) 6 2 1 (300 lb/ft)(12 ft) 1800 lb 2 O OR w w R = = = = (a) For 0,C = 0: (6 )(1ft) (1800 lb)(3 ft) 0B OM wΣ = − = 900 lb/ftOw = (b) Corresponding value of I:R I 6(900) 5400 lb= =R 0: 5400 lb 1800 lb 0Σ = − − =yF B 7200 lb=B www.elsolucionario.net www.elsolucionario.net
- 642. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 640 PROBLEM 5.76 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports. SOLUTION (a) We have I II 1 ( m)(1800 N/m) 900 N 2 1 [(4 ) m](600 N/m) 300(4 ) N 2 R a a R a a = = = − = − Then 0: 900 300(4 ) 0y y yF A a a BΣ = − − − + = or 1200 600y yA B a+ = + Now 600 300 (N)y y y yA B A B a= = = + (1) Also, 0: (4 m) 4 m [(900 ) N] 3 B y a M A a Σ = − + − 1 (4 ) m [300(4 ) N] 0 3 a a + − − = or 2 400 700 50yA a a= + − (2) Equating Eqs. (1) and (2), 2 600 300 400 700 50a a a+ = + − or 2 8 4 0a a− + = Then 2 8 ( 8) 4(1)(4) 2 a ± − − = or 0.53590 ma = 7.4641ma = Now 4 ma ≤ 0.536 ma = www.elsolucionario.net www.elsolucionario.net
- 643. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 641 PROBLEM 5.76 (Continued) (b) We have 0: 0x xF AΣ = = From Eq. (1): 600 300(0.53590) y yA B= = + 761 N= 761 N= =A B www.elsolucionario.net www.elsolucionario.net
- 644. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 642 PROBLEM 5.77 Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports. SOLUTION (a) We have I II 1 ( m)(1800 N/m) 900 N 2 1 [(4 )m](600 N/m) 300(4 ) N 2 R a a R a a = = = − = − Then 8 0: m (900 N) m [300(4 )N] (4 m) 0 3 3 A y a a M a a B + Σ = − − − + = or 2 50 100 800yB a a= − + (1) Then 100 100 0 ydB a da = − = or 1.000 ma = (b) From Eq. (1): 2 50(1) 100(1) 800 750 NyB = − + = 750 N=B and 0: 0x xF AΣ = = 0: 900(1) N 300(4 1) N 750 N 0y yF AΣ = − − − + = or 1050 NyA = 1050 N=A www.elsolucionario.net www.elsolucionario.net
- 645. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 643 PROBLEM 5.78 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine the values of wBC and wDE corresponding to equilibrium when 600Aw = N/m. SOLUTION We have I II 1 (6 m)(600 N/m) 1800 N 2 1 (6 m)(1200 N/m) 3600 N 2 (0.8 m)( N/m) (0.8 ) N (1.0 m)( N/m) ( ) N BC BC BC DE DE DE R R R w w R w w = = = = = = = = Then 0: (1 m)(1800 N) (3 m)(3600 N) (4 m)( N) 0G DEM wΣ = − − + = or 3150 N/mDEw = and 0: (0.8 ) N 1800 N 3600 N 3150 N 0y BCF wΣ = − − + = or 2812.5 N/mBCw = 2810 N/mBCw = www.elsolucionario.net www.elsolucionario.net
- 646. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 644 PROBLEM 5.79 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine (a) the value of wA so that wBC = wDE, (b) the corresponding values of wBC and wDE. SOLUTION (a) We have I II 1 (6 m)( N/m) (3 ) N 2 1 (6 m)(1200 N/m) 3600 N 2 (0.8 m)( N/m) (0.8 ) N (1m)( N/m) ( ) N A A BC BC BC DE DE DE R w w R R w w R w w = ⋅ = = = = = = Then 0: (0.8 ) N (3 ) N 3600 N ( ) N 0y BC A DEF w w wΣ = − − + = or 0.8 3600 3BC DE Aw w w+ = + Now 5 2000 3 BC DE BC DE Aw w w w w= = = + (1) Also, 0: (1 m)(3 N) (3 m)(3600 N) (4 m)( N) 0G A DEM w wΣ = − − + = or 3 2700 4 DE Aw w= + (2) Equating Eqs. (1) and (2), 5 3 2000 2700 3 4 A Aw w+ = + or 8400 N/m 11 Aw = 764 N/mAw = (b) Eq. (1) 5 8400 2000 3 11 BC DEw w = = + or 3270 N/mBC DEw w= = www.elsolucionario.net www.elsolucionario.net
- 647. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 645 PROBLEM 5.80 The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam. SOLUTION (a) Consider free body made of dam and section BDE of water. (Thickness = 1 m) 3 2 (3 m)(10 kg/m )(9.81m/s )p = 3 3 2 1 3 3 2 2 3 3 2 3 3 3 2 (1.5 m)(4 m)(1 m)(2.4 10 kg/m )(9.81m/s ) 144.26 kN 1 (2 m)(3 m)(1m)(2.4 10 kg/m )(9.81 m/s ) 47.09 kN 3 2 (2 m)(3 m)(1 m)(10 kg/m )(9.81m/s ) 39.24 kN 3 1 1 (3 m)(1m)(3 m)(10 kg/m )(9.81 m/s ) 44.145 kN 2 2 = × = = × = = = = = = W W W P Ap 0: 44.145 kN 0Σ = − =xF H 44.145 kN=H 44.1 kN=H 0: 141.26 47.09 39.24 0Σ = − − − =yF V 227.6 kN=V 228 kN=V www.elsolucionario.net www.elsolucionario.net
- 648. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 646 PROBLEM 5.80 (Continued) 1 2 3 1 (1.5 m) 0.75 m 2 1 1.5 m (2 m) 2 m 4 5 1.5 m (2 m) 2.75 m 8 x x x = = = + = = + = 0: (1 m) 0Σ = − Σ + =AM xV xW P (227.6 kN) (141.26 kN)(0.75 m) (47.09 kN)(2 m) (39.24 kN)(2.75 m) (44.145 kN)(1 m) 0 (227.6 kN) 105.9 94.2 107.9 44.145 0 (227.6) 263.9 0 x x x − − − + = − − − + = − = 1.159 mx = (to right of A) (b) Resultant of face BC: Consider free body of section BDE of water. 59.1kN− =R 41.6° 59.1 kN=R 41.6° www.elsolucionario.net www.elsolucionario.net
- 649. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 647 PROBLEM 5.81 The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam. SOLUTION (a) Consider free body made of dam and triangular section of water shown. (Thickness = 1 m.) 3 3 2 (7.2m)(10 kg/m )(9.81m/s )=p 3 3 2 1 3 3 2 2 3 3 2 3 3 3 2 2 (4.8 m)(7.2 m)(1m)(2.4 10 kg/m )(9.81m/s ) 3 542.5 kN 1 (2.4 m)(7.2 m)(1 m)(2.4 10 kg/m )(9.81m/s ) 2 203.4 kN 1 (2.4 m)(7.2 m)(1 m)(10 kg/m )(9.81m/s ) 2 84.8 kN 1 1 (7.2 m)(1m)(7.2 m)(10 kg/m )(9.81 m/s ) 2 2 = × = = × = = = = = = W W W P Ap 254.3 kN 0: 254.3 kN 0xF HΣ = − = 254 kN=H 0: 542.5 203.4 84.8 0yF VΣ = − − − = 830.7 kNV = 831kN=V www.elsolucionario.net www.elsolucionario.net
- 650. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 648 PROBLEM 5.81 (Continued) (b) 1 2 3 5 (4.8 m) 3 m 8 1 4.8 (2.4) 5.6 m 3 2 4.8 (2.4) 6.4 m 3 x x x = = = + = = + = 0: (2.4 m) 0AM xV xW PΣ = − Σ + = (830.7 kN) (3 m)(542.5 kN) (5.6 m)(203.4 kN) (6.4 m)(84.8 kN) (2.4 m)(254.3 kN) 0 (830.7) 1627.5 1139.0 542.7 610.3 0 (830.7) 2698.9 0 x x x − − − + = − − − + = − = 3.25 mx = (to right of A) (c) Resultant on face BC: Direct computation: 3 3 2 2 2 2 2 (10 kg/m )(9.81m/s )(7.2 m) 70.63 kN/m (2.4) (7.2) 7.589 m 18.43 1 2 1 (70.63 kN/m )(7.589 m)(1m) 2 P gh P BC R PA ρ θ = = = = + = = ° = = 268 kN=R 18.43° Alternate computation: Use free body of water section BCD. 268 kN− =R 18.43° 268 kN=R 18.43° www.elsolucionario.net www.elsolucionario.net
- 651. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 649 PROBLEM 5.82 An automatic valve consists of a 9 × 9-in. square plate that is pivoted about a horizontal axis through A located at a distance h = 3.6 in. above the lower edge. Determine the depth of water d for which the valve will open. SOLUTION Since valve is 9 in. wide, 9 9 ,w p hγ= = where all dimensions are in inches. 1 2 I 1 II 2 9 ( 9), 9 1 1 (9 in.) (9)(9 )( 9) 2 2 1 1 (9 in.) (9)(9 ) 2 2 w d w d P w d P w d γ γ γ γ = − = = = − = = Valve opens when 0.=B I II0: (6 in. 3.6 in.) (3.6 in. 3 in.) 0Σ = − − − =AM P P 1 1 (9)(9 )( 9) (2.4) (9)(9 ) (0.6) 0 2 2 d dγ γ − − = ( 9)(2.4) (0.6) 0 1.8 21.6 0 − − = − = d d d 12.00 in.=d www.elsolucionario.net www.elsolucionario.net
- 652. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 650 PROBLEM 5.83 An automatic valve consists of a 9 × 9-in. square plate that is pivoted about a horizontal axis through A. If the valve is to open when the depth of water is d = 18 in., determine the distance h from the bottom of the valve to the pivot A. SOLUTION Since valve is 9 in. wide, 9 9 ,w p hγ= = where all dimensions are in inches. 1 2 9 ( 9) 9 w d w d γ γ = − = For 18 in.,d = 1 2 I II 9 (18 9) 81 9 (18) 162 1 1 (9)(9 )(18 9) (729 ) 2 2 1 (9)(9 )(18) 729 2 w w P P γ γ γ γ γ γ γ γ = − = = = = − = = = Valve opens when 0.B = 1 II0: (6 ) ( 3) 0Σ = − − − =AM P h P h 1 729 (6 ) 729( 3) 0 2 1 3 3 0 2 6 1.5 0 h h h h h γ − − − = − − + = − = 4.00 in.=h www.elsolucionario.net www.elsolucionario.net
- 653. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 651 PROBLEM 5.84 The 3 × 4-m side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 200 kN, and the design specifications require the force in the rod not to exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank. SOLUTION Consider the free-body diagram of the side. We have 1 1 ( ) 2 2 P Ap A gdρ= = Now 0: 0 3 A d M hT PΣ = − = where 3 mh = Then for ,maxd 3 3 3 21 (3 m)(0.2 200 10 N) (4 m ) (10 kg/m 9.81 m/s ) 0 3 2 max max max d d d × × − × × × × = or 3 2 120 N m 6.54 N/m 0maxd⋅ − = or 2.64 mmaxd = www.elsolucionario.net www.elsolucionario.net
- 654. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 652 PROBLEM 5.85 The 3 × 4-m side of an open tank is hinged at its bottom A and is held in place by a thin rod BC. The tank is to be filled with glycerine, whose density is 1263 kg/m3 . Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 2.9 m. SOLUTION Consider the free-body diagram of the side. We have 3 2 1 1 ( ) 2 2 1 [(2.9 m)(4 m)] [(1263 kg/m )(9.81 m/s )(2.9 m)] 2 = 208.40 kN P Ap A gdρ= = = Then 0: 0y yF AΣ = = 2.9 0: (3 m) m (208.4 kN) 0 3 AM T Σ = − = or 67.151 kNT = 67.2 kN=T 0: 208.40 kN 67.151 kN 0x xF AΣ = + − = or 141.249 kNxA = − 141.2 kN=A www.elsolucionario.net www.elsolucionario.net
- 655. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 653 PROBLEM 5.86 The friction force between a 6 × 6-ft square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb. SOLUTION Consider the free-body diagram of the gate. Now 2 3 I I 1 1 [(6 6) ft ][(62.4 lb/ft )(9 ft)] 2 2 10,108.8 lb P Ap= = × = 2 3 II II 1 1 [(6 6) ft ][(62.4 lb/ft )(15 ft)] 2 2 16,848 lb P Ap= = × = Then I II0.1 0.1( ) 0.1(10,108.8 16,848) lb 2695.7 lb F P P P= = + = + = Finally 0: 2695.7 lb 1000 lb 0yF TΣ = − − = or 3.70 kips=T www.elsolucionario.net www.elsolucionario.net
- 656. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 654 PROBLEM 5.87 A tank is divided into two sections by a 1 × 1-m square gate that is hinged at A. A couple of magnitude 490 N · m is required for the gate to rotate. If one side of the tank is filled with water at the rate of 0.1 m3 /min and the other side is filled simultaneously with methyl alcohol (density ρma = 789 kg/m3 ) at the rate of 0.2 m3 /min, determine at what time and in which direction the gate will rotate. SOLUTION Consider the free-body diagram of the gate. First note baseV A d= and .V rt= Then 3 3 0.1 m / min (min) 0.25 (m) (0.4 m)(1m) 0.2 m / min (min) (m) (0.2 m)(1m) W MA t d t t d t × = = × = = Now 1 1 ( ) 2 2 P Ap A ghρ= = so that 3 3 2 2 3 2 2 1 [(0.25 ) m (1 m)][(10 kg/m )(9.81 m/s )(0.25 ) m] 2 306.56 N 1 [( ) m (1 m)][(789 kg/m )(9.81m/s )( ) m] 2 3870 N W MA P t t t P t t t = × = = × = Now assume that the gate will rotate clockwise and when 0.6 m.≤MAd When rotation of the gate is impending, we require 1 1 : 0.6 m 0.6 m 3 3 Σ = − − − A R MA MA W WM M d P d P Substituting 2 21 1 490 N m 0.6 m (3870 ) N 0.6 0.25 m (306.56 ) N 3 3 t t t t ⋅ = − × − − × × www.elsolucionario.net www.elsolucionario.net
- 657. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 655 PROBLEM 5.87 (Continued) Simplifying 3 2 1264.45 2138.1 490 0− + =t t Solving (positive roots only) 0.59451 mint = and 1.52411 mint = Now check assumption using the smaller root. We have ( ) m 0.59451m 0.6 mMAd t= = < 0.59451min 35.7 st∴ = = and the gate rotates clockwise. www.elsolucionario.net www.elsolucionario.net
- 658. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 656 PROBLEM 5.88 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance 0.10 mh = below the center of gravity C of the gate. Determine the depth of water d for which the gate will open. SOLUTION First note that when the gate is about to open (clockwise rotation is impending), yB 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then (0.25 ) 3 d a h= − − and 2 8 (0.4) 3 15 3 d b = − Now 8 15 a b = so that ( ) 3 82 3 15 3 (0.25 ) 8 15(0.4) d d h− − = − Simplifying yields 289 70.6 15 45 12 d h+ = (1) Alternative solution: Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now 2 2 1 1 ( 1m)( ) 2 2 1 (N) 2 1 8 1 m 2 15 4 (N) 15 P Ap d gd gd W gV g d d gd ρ ρ ρ ρ ρ ′ ′= = × = ′ = = × × × = www.elsolucionario.net www.elsolucionario.net
- 659. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 657 PROBLEM 5.88 (Continued) Then with 0yB = (as explained above), we have 2 22 1 8 4 1 0: (0.4) (0.25 ) 0 3 3 15 15 3 2 A d M d gd h gdρ ρ Σ = − − − − = Simplifying yields 289 70.6 15 45 12 d h+ = as above. Find d: 0.10 mh = Substituting into Eq. (1), 289 70.6 15(0.10) 45 12 d + = or 0.683 m=d www.elsolucionario.net www.elsolucionario.net
- 660. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 658 PROBLEM 5.89 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. Determine the distance h if the gate is to open when 0.75 m.d = SOLUTION First note that when the gate is about to open (clockwise rotation is impending), yB 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then (0.25 ) 3 d a h= − − and 2 8 (0.4) 3 15 3 d b = − Now 8 15 a b = so that ( ) 3 82 3 15 3 (0.25 ) 8 15(0.4) d d h− − = − Simplifying yields 289 70.6 15 45 12 d h+ = (1) Alternative solution: Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now 2 2 1 1 ( 1m)( ) 2 2 1 (N) 2 1 8 1 m 2 15 4 (N) 15 P Ap d gd gd W gV g d d gd ρ ρ ρ ρ ρ ′ ′= = × = ′ = = × × × = www.elsolucionario.net www.elsolucionario.net
- 661. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 659 PROBLEM 5.89 (Continued) Then with 0yB = (as explained above), we have 2 22 1 8 4 1 0: (0.4) (0.25 ) 0 3 3 15 15 3 2 A d M d gd h gdρ ρ Σ = − − − − = Simplifying yields 289 70.6 15 45 12 d h+ = as above. Find h: 0.75 md = Substituting into Eq. (1), 289 70.6 (0.75) 15 45 12 h+ = or 0.0711 mh = www.elsolucionario.net www.elsolucionario.net
- 662. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 660 PROBLEM 5.90 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water 3.5d = ft, determine the force exerted on the gate by the shear pin. SOLUTION First consider the force of the water on the gate. We have 1 2 1 ( ) 2 P Ap A hγ = = Then 2 3 I 2 3 II 1 (1.8 ft) (62.4 lb/ft )(1.7 ft) 2 171.850 lb 1 (1.8 ft) (62.4 lb/ft ) (1.7 1.8cos30 ) ft 2 329.43 lb P P = = = × + ° = Now I II 1 2 0: 0 3 3 A AB AB AB BM L P L P L F Σ = + − = or 1 2 (171.850 lb) (329.43 lb) 0 3 3 BF+ − = or 276.90 lbBF = 277 lbB =F 30.0° www.elsolucionario.net www.elsolucionario.net
- 663. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 661 PROBLEM 5.91 A long trough is supported by a continuous hinge along its lower edge and by a series of horizontal cables attached to its upper edge. Determine the tension in each of the cables, at a time when the trough is completely full of water. SOLUTION Consider free body consisting of 20-in. length of the trough and water. 20-in.l = length of free body 2 2 4 1 1 1 ( ) 2 2 2 A A W v r l P r P P rl r rl r l π γ γ γ γ γ = = = = = = 1 0: 0 3 AM Tr Wr P r Σ = − − = 2 24 1 1 0 4 3 2 3 r Tr r l r l r π γ γ π − − = 2 2 21 1 1 3 6 2 T r l r l r lγ γ γ= + = Data: 3 24 20 62.4 lb/ft ft 2 ft ft 12 12 r lγ = = = = Then 3 21 20 (62.4 lb/ft )(2 ft) ft 2 12 T = 208.00 lb= 208 lbT = www.elsolucionario.net www.elsolucionario.net
- 664. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 662 PROBLEM 5.92 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack. SOLUTION First consider the force of the water on the gate. We have 1 1 ( ) 2 2 ρ= =P Ap A gh so that 3 3 2 I 1 [(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.45 m)] 2 882.9 N P = × = 3 3 2 II 1 [(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.93 m)] 2 1824.66 N P = × = Reactions at A and B when T = 0: We have 1 2 0: (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) (0.8 m) 0 3 3 AM BΣ = − = or 1510.74 NB = or 1511 N=B 53.1° 0: 1510.74 N 882.9 N 1824.66 N 0F AΣ = + − − = or 1197 N=A 53.1° www.elsolucionario.net www.elsolucionario.net
- 665. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 663 PROBLEM 5.93 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the minimum tension required in cable BCD to open the gate. SOLUTION First consider the force of the water on the gate. We have 1 1 ( ) 2 2 ρ= =P Ap A gh so that 3 3 2 I 1 [(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.45 m)] 2 882.9 N P = × = 3 3 2 II 1 [(0.5 m)(0.8 m)] [(10 kg/m )(9.81 m/s )(0.93 m)] 2 1824.66 N P = × = T to open gate: First note that when the gate begins to open, the reaction at B 0. Then 1 2 0: (0.8 m)(882.9 N)+ (0.8 m)(1824.66 N) 3 3 8 (0.45 0.27)m 0 17 AM T Σ = − + × = or 235.44 973.152 0.33882 0T+ − = or 3570 N=T www.elsolucionario.net www.elsolucionario.net
- 666. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 664 PROBLEM 5.94 A 4 2-ft× gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor. SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. We have 2 sin 30 4 θ θ= = ° Then (3 ft)tan30SPx = ° and 828 lb/ft 3 ft tan30° 1434.14 lb SP SPF kx= = × × = Assume 4 ftd ≥ We have 1 1 ( ) 2 2 γ= =P Ap A h Then 3 I 1 [(4 ft)(2 ft)] [(62.4 lb/ft )( 4) ft] 2 249.6( 4) lb P d d = × − = − 3 II 1 [(4 ft)(2 ft)] [(62.4 lb/ft )( 4 4cos30 )] 2 249.6( 0.53590 ) lb P d d = × − + ° = − ° For mind so that the gate opens, 0W = Using the above free-body diagrams of the gate, we have 4 8 0: ft [249.6( 4) lb] ft [249.6( 0.53590) lb] 3 3 AM d d Σ = − + − (3 ft)(1434.14 lb) 0− = or (332.8 1331.2) (665.6 356.70) 4302.4 0d d− + − − = or 6.00 ftd = 4 ftd ≥ assumption correct 6.00 ftd = www.elsolucionario.net www.elsolucionario.net
- 667. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 665 PROBLEM 5.95 Solve Problem 5.94 if the gate weighs 1000 lb. PROBLEM 5.94 A 4 2-ft× gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor. SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. We have 2 sin 30 4 θ θ= = ° Then (3 ft)tan30SPx = ° and 828 lb/ft 3 ft tan30° 1434.14 lb SP SPF kx= = × × = Assume 4 ftd ≥ We have 1 1 ( ) 2 2 γ= =P Ap A h Then 3 I 1 [(4 ft)(2 ft)] [(62.4 lb/ft )( 4) ft] 2 249.6( 4) lb P d d = × − = − 3 II 1 [(4 ft)(2 ft)] [(62.4 lb/ft )( 4 4cos30 )] 2 249.6( 0.53590 ) lb P d d = × − + ° = − ° For mind so that the gate opens, 1000 lb=W Using the above free-body diagrams of the gate, we have 4 8 0: ft [249.6( 4) lb] ft [249.6( 0.53590) lb] 3 3 (3 ft)(1434.14 lb) (1ft)(1000 lb) 0 AM d d Σ = − + − − − = or (332.8 1331.2) (665.6 356.70) 4302.4 1000 0− + − − − =d d or 7.00 ftd = 4 ftd ≥ assumption correct 7.00 ftd = www.elsolucionario.net www.elsolucionario.net
- 668. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 666 PROBLEM 5.96 A hemisphere and a cone are attached as shown. Determine the location of the centroid of the composite body when (a) 1.5 ,=h a (b) 2 .=h a SOLUTION V y yV Cone I 21 3 πa h 4 h 2 21 12 πa h Hemisphere II 32 3 πa 3 8 − a 41 4 π− a 2 2 2 2 1 ( 2 ) 3 1 ( 3 ) 12 π π = + Σ = − V a h a yV a h a (a) For 1.5 ,h a= 2 21 7 (1.5 2 ) 3 6 π π= + =V a a a a 2 2 2 41 1 [(1.5 ) 3 ] 12 16 π πΣ = − = −yV a a a a 3 47 1 3 : 6 16 56 π π =Σ = − = − Y V yV Y a a Y a Centroid is 0.0536a below base of cone. www.elsolucionario.net www.elsolucionario.net
- 669. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 667 PROBLEM 5.96 (Continued) (b) For 2 ,h a= 2 31 4 (2 2 ) 3 3 π π= + =V a a a a 2 2 2 41 1 [(2 ) 3 ] 12 12 yV a a a aπ πΣ = − = 3 44 1 1 : 3 12 16 π π = Σ = = Y V yV Y a a Y a Centroid is 0.0625a above base of cone. www.elsolucionario.net www.elsolucionario.net
- 670. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 668 PROBLEM 5.97 Consider the composite body shown. Determine (a) the value of x when /2,h L= (b) the ratio h/L for which .x L= SOLUTION V x xV Rectangular prism Lab 1 2 L 21 2 L ab Pyramid 1 3 2 b a h 1 4 L h+ 1 1 6 4 abh L h + Then 2 1 6 1 1 3 6 4 V ab L h xV ab L h L h Σ = + Σ = + + Now X V xVΣ = Σ so that 2 21 1 1 3 6 6 4 X ab L h ab L hL h + = + + or 2 2 1 1 1 1 3 6 6 4 h h h X L L L L + = + + (1) (a) ?X = when 1 . 2 h L= Substituting 1 into Eq. (1), 2 h L = 2 1 1 1 1 1 1 1 3 6 2 6 2 4 2 X L + = + + or 57 104 X L= 0.548X L= www.elsolucionario.net www.elsolucionario.net
- 671. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 669 PROBLEM 5.97 (Continued) (b) ? h L = when .X L= Substituting into Eq. (1), 2 2 1 1 1 1 3 6 6 4 h h h L L L L L + = + + or 2 2 1 1 1 1 1 6 2 6 24 h h h L L L + = + + or 2 2 12 h L = 2 3 h L = www.elsolucionario.net www.elsolucionario.net
- 672. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 670 PROBLEM 5.98 Determine the y coordinate of the centroid of the body shown. SOLUTION First note that the values of Y will be the same for the given body and the body shown below. Then V y yV Cone 21 3 a hπ 1 4 h− 2 21 12 a hπ− Cylinder 2 21 2 4 a b a bπ π − = − 1 2 b− 2 21 8 a bπ Σ 2 (4 3 ) 12 a h b π − 2 2 2 (2 3 ) 24 a h b π − − We have Σ = ΣY V yV Then 2 2 2 2 (4 3 ) (2 3 ) 12 24 Y a h b a h b π π − = − − or 2 2 2 3 2(4 3 ) h b Y h b − = − − www.elsolucionario.net www.elsolucionario.net
- 673. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 671 PROBLEM 5.99 Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.) SOLUTION First note that the body can be formed by removing a half cylinder from a half cone, as shown. V z zV Half cone 21 6 a hπ a π − 31 6 a h− Half cylinder 2 2 2 2 8 a b a b π π − = − 4 2 3 2 3 a a π π − = − 31 12 a b Σ 2 (4 3 ) 24 a h b π − 31 (2 ) 12 a h b− − From Sample Problem 5.13: We have Σ = ΣZ V zV Then 2 31 (4 3 ) (2 ) 24 12 Z a h b a h b π − = − − or 4 2 4 3 a h b Z h bπ − = − − www.elsolucionario.net www.elsolucionario.net
- 674. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 672 PROBLEM 5.100 For the machine element shown, locate the y coordinate of the center of gravity. SOLUTION For half-cylindrical hole, III 1.25 in. 4(1.25) 2 3 1.470 in. r y π = = − = For half-cylindrical plate, IV 2 in. 4(2) 7 7.85 in. 3π = = + = r z 3 , inV , in.y , in.z 4 , inyV 4 , inzV I Rectangular plate (7)(4)(0.75) 21.0= –0.375 3.5 –7.875 73.50 II Rectangular plate (4)(2)(1) 8.0= 1.0 2 8.000 16.00 III –(Half cylinder) 2 (1.25) (1) 2.454 2 π − = 1.470 2 –3.607 –4.908 IV Half cylinder 2 (2) (0.75) 4.712 2 π = –0.375 –7.85 –1.767 36.99 V –(Cylinder) 2 (1.25) (0.75) 3.682π− = − –0.375 7 1.381 –25.77 Σ 27.58 –3.868 95.81 3 4 (27.58 in ) 3.868 in Y V yV Y Σ = Σ = − 0.1403 in.Y = − www.elsolucionario.net www.elsolucionario.net
- 675. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 673 PROBLEM 5.101 For the machine element shown, locate the y coordinate of the center of gravity. SOLUTION First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. 3 , mmV , mmx , mmy 4 , mmxV 4 , mmyV I (100)(18)(90) 162,000= 50 9 8,100,000 1,458,000 II (16)(60)(50) 48,000= 92 48 4,416,000 2,304,000 III 2 (12) (10) 4523.9π = 105 54 475,010 244,290 IV 2 (13) (18) 9556.7π− = − 28 9 –267,590 –86,010 Σ 204,967.2 12,723,420 3,920,280 We have Y V yVΣ = Σ 3 4 (204,967.2 mm ) 3,920,280 mmY = or 19.13 mmY = www.elsolucionario.net www.elsolucionario.net
- 676. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 674 PROBLEM 5.102 For the machine element shown, locate the x coordinate of the center of gravity. SOLUTION First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. 3 , mmV , mmx , mmy 4 , mmxV 4 , mmyV I (100)(18)(90) 162,000= 50 9 8,100,000 1,458,000 II (16)(60)(50) 48,000= 92 48 4,416,000 2,304,000 III 2 (12) (10) 4523.9π = 105 54 475,010 244,290 IV 2 (13) (18) 9556.7π− = − 28 9 –267,590 –86,010 Σ 204,967.2 12,723,420 3,920,280 We have X V xVΣ = Σ 3 4 (204,967.2 mm ) 12,723,420 mmX = 62.1mmX = www.elsolucionario.net www.elsolucionario.net
- 677. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 675 PROBLEM 5.103 For the machine element shown, locate the z coordinate of the center of gravity. SOLUTION For half-cylindrical hole, III 1.25 in. 4(1.25) 2 3 1.470 in. r y π = = − = For half-cylindrical plate, IV 2 in. 4(2) 7 7.85 in. 3π = = + = r z 3 , inV , in.y , in.z 4 , inyV 4 , inzV I Rectangular plate (7)(4)(0.75) 21.0= –0.375 3.5 –7.875 73.50 II Rectangular plate (4)(2)(1) 8.0= 1.0 2 8.000 16.00 III –(Half cylinder) 2 (1.25) (1) 2.454 2 π − = 1.470 2 –3.607 –4.908 IV Half cylinder 2 (2) (0.75) 4.712 2 π = –0.375 –7.85 –1.767 36.99 V –(Cylinder) 2 (1.25) (0.75) 3.682π− = − –0.375 7 1.381 –25.77 Σ 27.58 –3.868 95.81 Now Z V zVΣ = 3 4 (27.58 in ) 95.81inZ = 3.47 in.Z = www.elsolucionario.net www.elsolucionario.net
- 678. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 676 PROBLEM 5.104 For the machine element shown, locate the x coordinate of the center of gravity. SOLUTION 3 , mmV , mmx , mmz 4 , mmxV 4 , mmzV I Rectangular plate 3 (10)(90)(38) 34.2 10= × 19 45 649.8 × 103 1539 × 103 II Half cylinder 2 3 (20) (10) 6.2832 10 2 π = × 46.5 20 292.17 × 103 125.664 × 103 III –(Cylinder) 2 3 (12) (10) 4.5239 10π− = − × 38 20 −171.908 × 103 −90.478 × 103 IV Rectangular prism 3 (30)(10)(24) 7.2 10= × 5 78 36 × 103 561.6 × 103 V Triangular prism 31 (30)(9)(24) 3.24 10 2 = × 13 78 42.12 × 103 252.72 × 103 Σ 46.399 × 103 848.18 × 103 2388.5 × 103 3 4 3 3 848.18 10 mm 46.399 10 mm Σ = Σ Σ × = = Σ × X V xV xV X V 18.28 mm=X Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 679. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 677 PROBLEM 5.105 For the machine element shown, locate the z coordinate of the center of gravity. SOLUTION 3 , mmV , mmx , mmz 4 , mmxV 4 , mmzV I Rectangular plate 3 (10)(90)(38) 34.2 10= × 19 45 649.8 × 103 1539 × 103 II Half cylinder 2 3 (20) (10) 6.2832 10 2 π = × 46.5 20 292.17 × 103 125.664 × 103 III –(Cylinder) 2 3 (12) (10) 4.5239 10π− = − × 38 20 −171.908 × 103 −90.478 × 103 IV Rectangular prism 3 (30)(10)(24) 7.2 10= × 5 78 36 × 103 561.6 × 103 V Triangular prism 31 (30)(9)(24) 3.24 10 2 = × 13 78 42.12 × 103 252.72 × 103 Σ 46.399 × 103 848.18 × 103 2388.5 × 103 3 4 3 3 2388.5 10 mm 46.399 10 mm Σ = Σ Σ × = = Σ × Z V zV zV Z V 51.5 mm=Z Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 680. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 678 PROBLEM 5.106 Locate the center of gravity of the sheet-metal form shown. SOLUTION By symmetry, 80.0 mm=Y I II 4(80) 33.953 mm 3 2(60) 38.197 mm π π = = = − = − z z 2 , mmA , mmx , mmz 3 , mmxA 3 , mmzA I 2 (80) 10,053 2 π = 0 33.953 0 3 341.33 10× II (60)(160) 30,159π = 60 −38.197 3 1809.54 10× 3 1151.98 10− × Σ 40,212 3 1809.54 10× 3 810.65 10− × 3 : (40,212) 1809.54 10X A xA XΣ = Σ = × 45.0 mm=X 3 : (40,212) 810.65 10Z A zA ZΣ = Σ = − × 20.2 mm= −Z www.elsolucionario.net www.elsolucionario.net
- 681. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 679 PROBLEM 5.107 Locate the center of gravity of the sheet-metal form shown. SOLUTION First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. I I II II IV 1 0.18 (0.12) 0.22 m 3 1 (0.2 m) 3 2 0.18 0.36 m 4 0.05 0.34 3 0.31878 m y z x y x π π π = + = = × = = = × = − = 2 , mA , mx , my , mz 3 , mxA 3 , myA 3 , mzA I 1 (0.2)(0.12) 0.012 2 = 0 0.22 0.2 3 0 0.00264 0.0008 II (0.18)(0.2) 0.018 2 π π= 0.36 π 0.36 π 0.1 0.00648 0.00648 0.005655 III (0.16)(0.2) 0.032= 0.26 0 0.1 0.00832 0 0.0032 IV 2 (0.05) 0.00125 2 π π− = − 0.31878 0 0.1 –0.001258 0 –0.000393 Σ 0.096622 0.013542 0.00912 0.009262 www.elsolucionario.net www.elsolucionario.net
- 682. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 680 PROBLEM 5.107 (Continued) We have 2 3 : (0.096622 m ) 0.013542 mX V xV XΣ = Σ = or 0.1402 mX = 2 3 : (0.096622 m ) 0.00912 mY V yV YΣ = Σ = or 0.0944 mY = 2 3 : (0.096622 m ) 0.009262 mZ V zV ZΣ = Σ = or 0.0959 mZ = www.elsolucionario.net www.elsolucionario.net
- 683. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 681 PROBLEM 5.108 A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area. II VI II VI IV IV 2 2 II VI 2 IV (4)(25) 4 14.6103 in. 3 (4)(25) 100 in. 3 3 (2)(25) 4 19.9155 in. (2)(25) 50 in. (25) 490.87 in 4 (25)(34) 1335.18 in 2 y y z z y z A A A π π π π π π π π = = + = = = = = + = = = = = = = = 2 , inA , in.y , in.z 3 , inyA 3 , inzA I (4)(25) 100= 2 12.5 200 1250 II 490.87 14.6103 100 3π 7171.8 5208.3 III (4)(34) 136= 2 25 272 3400 IV 1335.18 19.9155 50 π 26,591 21,250 V (4)(25) 100= 2 12.5 200 1250 VI 490.87 14.6103 100 3π 7171.8 5208.3 Σ 2652.9 41,607 37,567 www.elsolucionario.net www.elsolucionario.net
- 684. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 682 PROBLEM 5.108 (Continued) Now, symmetry implies 17.00 in.X = and 2 3 : (2652.9 in ) 41,607 inY A yA YΣ = Σ = or 15.68 in.Y = 2 3 : (2652.9 in ) 37,567 inZ A zA ZΣ = Σ = or 14.16 in.Z = www.elsolucionario.net www.elsolucionario.net
- 685. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 683 PROBLEM 5.109 A thin sheet of plastic of uniform thickness is bent to form a desk organizer. Locate the center of gravity of the organizer. SOLUTION First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with the centroid of the corresponding area. Now note that symmetry implies 30.0 mmZ = 2 4 8 10 2 4 8 10 6 2 2 4 8 10 2 6 2 6 6 2.1803 mm 2 6 36 39.820 mm 2 6 58 54.180 mm 2 6 133 136.820 mm 2 6 6 2.1803 mm 2 5 75 78.183 mm 6 60 565.49 mm 2 5 60 942.48 mm x x x x y y y y y A A A A A π π π π π π π π × = − = × = + = × = − = × = + = × = = = = − = × = + = = = = = × × = = × × = www.elsolucionario.net www.elsolucionario.net
- 686. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 684 PROBLEM 5.109 (Continued) 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 (74)(60) 4440= 0 43 0 190,920 2 565.49 2.1803 2.1803 1233 1233 3 (30)(60) 1800= 21 0 37,800 0 4 565.49 39.820 2.1803 22,518 1233 5 (69)(60) 4140= 42 40.5 173,880 167,670 6 942.48 47 78.183 44,297 73,686 7 (69)(60) 4140= 52 40.5 215,280 167,670 8 565.49 54.180 2.1803 30,638 1233 9 (75)(60) 4500= 95.5 0 429,750 0 10 565.49 136.820 2.1803 77,370 1233 Σ 22,224.44 1,032,766 604,878 We have 2 3 : (22,224.44 mm ) 1,032,766 mmX A xA XΣ = Σ = or 46.5 mmX = 2 3 : (22,224.44 mm ) 604,878 mmY A yA YΣ = Σ = or 27.2 mmY = www.elsolucionario.net www.elsolucionario.net
- 687. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 685 PROBLEM 5.110 A wastebasket, designed to fit in the corner of a room, is 16 in. high and has a base in the shape of a quarter circle of radius 10 in. Locate the center of gravity of the wastebasket, knowing that it is made of sheet metal of uniform thickness. SOLUTION By symmetry, X Z= For III (Cylindrical surface), 2 2 2(10) 6.3662 in. (10)(16) 251.33 in 2 2 r x A rh π π π π = = = = = = For IV (Quarter-circle bottom), 2 2 2 4 4(10) 4.2441in. 3 3 (10) 78.540 in 4 4 r x A r π π π π = = = = = = 2 , inA , in.x , in.x 3 , inxA 3 , inyA I (10)(16) 160= 5 8 800 1280 II (10)(16) 160= 0 8 0 1280 III 251.33 6.3662 8 1600.0 2010.6 IV 78.540 4.2441 0 333.33 0 Σ 649.87 2733.3 4570.6 :X A xAΣ = Σ 2 3 (649.87 in ) 2733.3 inX = 4.2059 in.X = 4.21in.X Z= = :Y A yAΣ = Σ 2 3 (649.87 in ) 4570.6 inY = 7.0331in.Y = 7.03 in.Y = www.elsolucionario.net www.elsolucionario.net
- 688. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 686 PROBLEM 5.111 A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram) V 2 V 2 4(0.625) 2.25 3 1.98474 in. (0.625) 2 0.61359 in z A π π = − = = − = − 2 , inA , in.x , in.y , in.z 3 , inxA 3 , inyA 3 , inzA I (2.5)(6) 15= 1.25 0 3 18.75 0 45 II (1.25)(6) 7.5= 2.5 –0.625 3 18.75 –4.6875 22.5 III (0.75)(6) 4.5= 2.875 –1.25 3 12.9375 –5.625 13.5 IV 5 (3) 3.75 4 − = − 1.0 0 3.75 3.75 0 –14.0625 V 0.61359− 1.0 0 1.98474 0.61359 0 –1.21782 Σ 22.6364 46.0739 10.3125 65.7197 We have X A xAΣ = Σ 2 3 (22.6364 in ) 46.0739 inX = or 2.04 in.X = 2 3 (22.6364 in ) 10.3125 in Y A yA Y Σ = Σ = − or 0.456 in.Y = − 2 3 (22.6364 in ) 65.7197 in Z A zA Z Σ = Σ = or 2.90 in.Z = www.elsolucionario.net www.elsolucionario.net
- 689. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 687 PROBLEM 5.112 An 8-in.-diameter cylindrical duct and a 4 × 8-in. rectangular duct are to be joined as indicated. Knowing that the ducts were fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly. SOLUTION Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area. By symmetry, 0.z = 2 , inA , in.x , in.y 3 , inxA 3 , inyA 1 (8)(12) 96π π= 0 6 0 576π 2 (8)(4) 16 2 π π− = − 2(4) 8 π π = 10 128− 160π− 3 2 (4) 8 2 π π= 4(4) 16 3 3π π − = − 12 42.667− 96π 4 (8)(12) 96= 6 12 576 1152 5 (8)(12) 96= 6 8 576 768 6 2 (4) 8 2 π π− = − 4(4) 16 3 3π π = 8 42.667− 64π− 7 (4)(12) 48= 6 10 288 480 8 (4)(12) 48= 6 10 288 480 Σ 539.33 1514.6 4287.4 Then 1514.67 in. 539.33 xA X A Σ = = Σ or 2.81in.X = 4287.4 in. 539.33 yA Y A Σ = = Σ or 7.95 in.Y = www.elsolucionario.net www.elsolucionario.net
- 690. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 688 PROBLEM 5.113 An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies 38.0 mmY = I I II II IV IV V V 2 Note that 400 (400) 145.352 mm 2 400 (200) 272.68 mm 2 300 (200) 172.676 mm 4 400 (400) 230.23 mm 3 4 400 (200) 315.12 mm 3 4 300 (200) 215.12 mm 3 x z x z x z x z π π π π π π = = − = = − = = − = = = − = = − = = − = Also note that the corresponding top and bottom areas will contribute equally when determining and .x z Thus, 2 , mmA , mmx , mmz 3 , mmxA 3 , mmzA I (400)(76) 47,752 2 π = 145.352 145.352 6,940,850 6,940,850 II (200)(76) 23,876 2 π = 272.68 172.676 6,510,510 4,122,810 III 100(76) 7600= 200 350 1,520,000 2,660,000 IV 2 2 (400) 251,327 4 π = 230.23 230.23 57,863,020 57,863,020 V 2 2 (200) 62,832 4 π − = − 315.12 215.12 –19,799,620 –13,516,420 VI 2(100)(200) 40,000− = − 300 350 –12,000,000 –14,000,000 Σ 227,723 41,034,760 44,070,260 www.elsolucionario.net www.elsolucionario.net
- 691. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 689 PROBLEM 5.113 (Continued) We have 2 3 : (227,723 mm ) 41,034,760 mmX A xA XΣ = Σ = or 180.2 mmX = 2 3 : (227,723 mm ) 44,070,260 mmZ A zA ZΣ = Σ = or 193.5 mmZ = www.elsolucionario.net www.elsolucionario.net
- 692. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 690 PROBLEM 5.114 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter. SOLUTION By symmetry, 0X = L, in. , in.y , in.z 2 , inyL 2 , inzL AB 2 2 30 16 34+ = 15 0 510 0 AD 2 2 30 16 34+ = 15 8 510 272 AE 2 2 30 16 34+ = 15 0 510 0 BDE (16) 50.265π = 0 2(16) 10.186 π = 0 512 Σ 152.265 1530 784 2 : (152.265 in.) 1530 inY L y L YΣ = Σ = 10.048 in.Y = 10.05 in.Y = 2 : (152.265 in.) 784 inZ L z L ZΣ = Σ = 5.149 in.Z = 5.15 in.Z = www.elsolucionario.net www.elsolucionario.net
- 693. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 691 PROBLEM 5.115 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter. SOLUTION Uniform rod: 2 2 2 2 (1m) (0.6 m) (1.5 m)AB = + + 1.9 mAB = , mL , mx , my , mz 2 , mxL 2 , myL , mLΣ AB 1.9 0.5 0.75 0.3 0.95 1.425 0.57 BD 0.6 1.0 0 0.3 0.60 0 0.18 DO 1.0 0.5 0 0 0.50 0 0 OA 1.5 0 0.75 0 0 1.125 0 Σ 5.0 2.05 2.550 0.75 2 : (5.0 m) 2.05 mX L x L XΣ = Σ = 0.410 mX = 2 : (5.0 m) 2.55 mY L y L YΣ = Σ = 0.510 mY = 2 : (5.0 m) 0.75 mZ L z L ZΣ = Σ = 0.1500 mZ = www.elsolucionario.net www.elsolucionario.net
- 694. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 692 PROBLEM 5.116 A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity. SOLUTION First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line. 2 2 2 2.4 4.8 mx z π π × = = = , mL , mx , my , mz 2 , mxL 2 , myL 2 , mzL 1 2.6 1.2 0.5 0 3.12 1.3 0 2 2.4 1.2 2 π π× = 4.8 π 0 4.8 π 5.76 0 5.76 3 2.4 0 0 1.2 0 0 2.88 4 1.0 0 0.5 0 0 0.5 0 Σ 9.7699 8.88 1.8 8.64 We have 2 : (9.7699 m) 8.88 mΣ = Σ =X L x L X or 0.909 m=X 2 : (9.7699 m) 1.8 mΣ = Σ =Y L y L Y or 0.1842 m=Y 2 : (9.7699 m) 8.64 mΣ = Σ =Z L z L Z or 0.884 m=Z www.elsolucionario.net www.elsolucionario.net
- 695. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 693 PROBLEM 5.117 The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown. SOLUTION First assume that the channels are homogeneous so that the center of gravity of the frame will coincide with the centroid of the corresponding line. 8 9 8 9 2 3 6 ft 2 3 5 6.9099 ft x x y y π π π × = = = × = = + = , ftL , ftx , fty , ftz 2 , ftxL 2 , ftyL 2 , ftzL 1 2 3 0 1 6 0 2 2 3 1.5 0 2 4.5 0 6 3 5 3 2.5 0 15 12.5 0 4 5 3 2.5 2 15 12.5 10 5 8 0 4 2 0 32 16 6 2 3 5 1 6 10 2 7 3 1.5 5 2 4.5 15 6 8 3 4.7124 2 π × = 6 π 6.9099 0 9 32.562 0 9 3 4.7124 2 π × = 6 π 6.9099 2 9 32.562 9.4248 10 2 0 8 1 0 16 2 Σ 39.4248 69 163.124 53.4248 We have 2 : (39.4248 ft) 69 ftX L x L XΣ = Σ = or 1.750 ftX = 2 : (39.4248 ft) 163.124 ftY L y L YΣ = Σ = or 4.14 ftY = 2 : (39.4248 ft) 53.4248 ftZ L z L ZΣ = Σ = or 1.355 ftZ = www.elsolucionario.net www.elsolucionario.net
- 696. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 694 PROBLEM 5.118 Three brass plates are brazed to a steel pipe to form the flagpole base shown. Knowing that the pipe has a wall thickness of 8 mm and that each plate is 6 mm thick, determine the location of the center of gravity of the base. (Densities: brass = 8470 kg/m3 ; steel = 7860 kg/m3 .) SOLUTION Since brass plates are equally spaced, we note that the center of gravity lies on the y-axis. Thus, 0x z= = Steel pipe: 2 2 6 3 3 6 3 [(0.064 m) (0.048 m) ](0.192 m) 4 270.22 10 m (7860 kg/m )(270.22 10 m ) 2.1239 kg V m V π ρ − − = − = × = = × = Each brass plate: 6 3 3 6 3 1 (0.096 m)(0.192 m)(0.006 m) 55.296 10 m 2 (8470 kg/m )(55.296 10 m ) 0.46836 kgρ − − = = × = = × = V m V Flagpole base: 2.1239 kg 3(0.46836 kg) 3.5290 kg (0.096 m)(2.1239 kg) 3[(0.064 m)(0.46836 kg)] 0.29382 kg m : (3.5290 kg) 0.29382 kg m m ym Y m ym Y Σ = + = Σ = + = ⋅ Σ = Σ = ⋅ 0.083259 m=Y 83.3 mm=Y above the base www.elsolucionario.net www.elsolucionario.net
- 697. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 695 PROBLEM 5.119 A brass collar, of length 2.5 in., is mounted on an aluminum rod of length 4 in. Locate the center of gravity of the composite body. (Specific weights: brass = 0.306 lb/in3 , aluminum = 0.101 lb/in3 ) SOLUTION Aluminum rod: 3 2 (0.101lb/in ) (1.6 in.) (4 in.) 4 0.81229 lb W Vγ π = = = Brass collar: 3 2 2 (0.306 lb/in. ) [(3 in.) (1.6 in.) ](2.5 in.) 4 3.8693 lb W Vγ π = = − = Component W(lb) (in.)y (lb in.)yW ⋅ Rod 0.81229 2 1.62458 Collar 3.8693 1.25 4.8366 Σ 4.6816 6.4612 : (4.6816 lb) 6.4612 lb in.Y W yW YΣ = Σ = ⋅ 1.38013 in.Y = 1.380 in.Y = www.elsolucionario.net www.elsolucionario.net
- 698. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 696 PROBLEM 5.120 A bronze bushing is mounted inside a steel sleeve. Knowing that the specific weight of bronze is 0.318 lb/in3 and of steel is 0.284 lb/in3 , determine the location of the center of gravity of the assembly. SOLUTION First, note that symmetry implies 0X Z= = Now ( )W g Vρ= 3 2 2 2 I I 3 2 2 2 II II 3 2 III III 0.20 in. (0.284 lb/in ) [(1.8 0.75 ) in ](0.4 in.) 0.23889 lb 4 0.90 in. (0.284 lb/in ) [(1.125 0.75 ) in ](1in.) 0.156834 lb 4 0.70 in. (0.318 lb/in ) [(0.75 0 4 y W y W y W π π π = = − = = = − = = = − 2 2 .5 ) in ](1.4 in.) 0.109269 lb = We have (0.20 in.)(0.23889 lb) (0.90 in.)(0.156834 lb) (0.70 in.)(0.109269 lb) 0.23889 lb 0.156834 lb 0.109269 lb Y W yW Y Σ = Σ + + = + + or 0.526 in.Y = (above base) www.elsolucionario.net www.elsolucionario.net
- 699. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 697 PROBLEM 5.121 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3 and of steel is 3 7860 kg/m , locate the center of gravity of the awl. SOLUTION First, note that symmetry implies 0Y Z= = I 3 3 I 3 II 3 2 II 3 III 3 2 III 5 (12.5 mm) 7.8125 mm 8 2 (1030 kg/m ) (0.0125 m) 3 4.2133 10 kg 52.5 mm (1030 kg/m ) (0.025 m) (0.08 m) 4 40.448 10 kg 92.5 mm 25 mm 67.5 mm (1030 kg/m ) (0.0035 m) (0. 4 x W x W x W π π π − − = = = = × = = = × = − = = − 3 05 m) 0.49549 10 kg− = − × IV 3 2 2 3 IV V 3 2 3 V 182.5 mm 70 mm 112.5 mm (7860 kg/m ) (0.0035 m) (0.14 m) 10.5871 10 kg 4 1 182.5 mm (10 mm) 185 mm 4 (7860 kg/m ) (0.00175 m) (0.01 m) 0.25207 10 kg 3 x W x W π π − − = − = = = × = + = = = × www.elsolucionario.net www.elsolucionario.net
- 700. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 698 PROBLEM 5.121 (Continued) , kgW , mmx , kg mmxW ⋅ I 3 4.123 10− × 7.8125 3 32.916 10− × II 3 40.948 10− × 52.5 3 2123.5 10− × III 3 0.49549 10− − × 67.5 3 33.447 10− − × IV 3 10.5871 10− × 112.5 3 1191.05 10− × V 3 0.25207 10− × 185 3 46.633 10− × Σ 3 55.005 10− × 3 3360.7 10− × We have 3 3 : (55.005 10 kg) 3360.7 10 kg mmX W xW X − − Σ = Σ × = × ⋅ or 61.1mmX = (from the end of the handle) www.elsolucionario.net www.elsolucionario.net
- 701. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 699 PROBLEM 5.122 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A hemisphere SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx x xπ= = The equation of the generating curve is 2 2 2 x y a+ = so that 2 2 2 r a x= − and then 2 2 ( )dV a x dxπ= − Component 1: /2 3/2 2 2 2 1 0 0 3 ( ) 3 11 24 a a x V a x dx a x a π π π = − = − = and /2 2 2 1 0 /2 2 4 2 0 4 ( ) 2 4 7 64 a EL a x dV x a x dx x x a a π π π = − = − = Now 3 4 1 1 1 1 11 7 : 24 64 ELx V x dV x a aπ π = = 1 21 or 88 x a= Component 2: ( ) 3 2 2 2 2 /2 /2 3 3 22 2 3 ( ) 3 ( ) 3 2 3 5 24 a a a a a x V a x dx a x a a a a a a π π π π = − = − = − − − = www.elsolucionario.net www.elsolucionario.net
- 702. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 700 PROBLEM 5.122 (Continued) and ( ) ( ) 2 4 2 2 2 2 /2 /2 2 4 2 4 2 22 2 4 ( ) 2 4 ( ) ( ) 2 4 2 4 9 64 a a EL a a a a x x x dV x a x dx a a a a a a π π π π = − = − = − − − = Now 3 4 2 2 2 2 5 9 : 24 64 ELx V x dV x a aπ π = = 2 27 or 40 x a= www.elsolucionario.net www.elsolucionario.net
- 703. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 701 PROBLEM 5.123 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A semiellipsoid of revolution SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx x xπ= = The equation of the generating curve is 2 2 2 2 1 x y h a + = so that 2 2 2 2 2 ( ) a r h x h = − and then 2 2 2 2 ( ) a dV h x dx h π= − Component 1: /2 2 2 3/2 2 2 2 1 2 20 0 2 ( ) 3 11 24 h h a a x V h x dx h x h h a h π π π = − = − = and 2/2 2 2 21 0 /2 2 2 4 2 2 0 2 2 ( ) 2 4 7 64 h EL h a x dV x h x dx h a x x h h a h π π π = − = − = Now 2 2 2 1 1 1 1 11 7 : 24 64 ELx V x dV x a h a hπ π = = 1 21 or 88 x h= Component 2: ( ) ( ) 2 2 3 2 2 2 2 2 2/2 /2 3 2 3 22 2 2 2 ( ) 3 ( ) 3 2 3 5 24 h h h h h a a x V h x dx h x h h a h h h h h h a h π π π π = − = − = − − − = www.elsolucionario.net www.elsolucionario.net
- 704. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 702 PROBLEM 5.123 (Continued) and 2 2 2 22 /2 2 2 4 2 2 /2 ( ) 2 4 h EL h h h a x dV x h x dx h a x x h h π π = − = − ( ) ( ) 2 4 2 2 4 2 22 2 2 2 2 ( ) ( ) 2 4 2 4 9 64 h h a h h h h h a h π π = − − − = Now 2 2 2 2 2 2 2 5 9 : 24 64 ELx V x dV x a h a hπ π = = 2 27 or 40 x h= www.elsolucionario.net www.elsolucionario.net
- 705. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 703 PROBLEM 5.124 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A paraboloid of revolution SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx x xπ= = The equation of the generating curve is 2 2 h x h y a = − so that 2 2 ( ). a r h x h = − and then 2 ( ) a dV h x dx h π= − Component 1: 2/2 1 0 /2 2 2 0 2 ( ) 2 3 8 h h a V h x dx h a x hx h a h π π π = − = − = and 2/2 1 0 /2 2 2 3 2 2 0 ( ) 1 2 3 12 h EL h a x dV x h x dx h a x x h a h h π π π = − = − = Now 2 2 2 1 1 1 1 3 1 : 8 12 ELx V x dV x a h a hπ π = = 1 2 or 9 x h= Component 2: ( ) 2 2 2 2 /2 /2 2 2 2 2 2 ( ) 2 ( ) ( ) 2 2 2 1 8 h h h h h a a x V h x dx hx h h a h h h h h h a h π π π π = − = − = − − − = www.elsolucionario.net www.elsolucionario.net
- 706. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 704 PROBLEM 5.124 (Continued) and ( ) ( ) 2 2 2 3 2 /2 /2 2 3 2 2 3 2 2 2 2 ( ) 2 3 ( ) ( ) 2 3 2 3 1 12 h h EL h h h h a a x x x dV x h x dx h h h a h h h h h a h π π π π = − = − = − − − = Now 2 2 2 2 2 2 2 1 1 : 8 12 ELx V x dV x a h a hπ π = = 2 2 or 3 x h= www.elsolucionario.net www.elsolucionario.net
- 707. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 705 PROBLEM 5.125 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis. SOLUTION First note that symmetry implies 0y = 0z = Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx x xπ= = Now 1/3 r kx= so that 2 2/3 dV k x dxπ= at , ,x h y a= = 1/3 a kh= or 1/3 a k h = Then 2 2/3 2/3 a dV x dx h π= and 2 2/3 2/30 2 5/3 2/3 0 2 3 5 3 5 h h a V x dx h a x h a h π π π = = = Also 2 2 2/3 8/3 2/3 2/30 0 2 2 3 8 3 8 h h EL a a x dV x x dx x h h a h π π π = = = Now :xV xdV= 2 2 23 3 5 8 x a h a hπ π = or 5 8 x h= www.elsolucionario.net www.elsolucionario.net
- 708. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 706 PROBLEM 5.126 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis. SOLUTION First, note that symmetry implies 0y = 0z = Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx x xπ= = Now 1 1r x = − so that 2 2 1 1 2 1 1 dV dx x dx x x π π = − = − + Then 3 3 21 1 3 2 1 1 1 2ln 1 1 3 2ln3 1 2ln1 3 1 (0.46944 ) m V dx x x x xx π π π π = − + = − − = − − − − − = and 3 23 21 1 2 3 2 1 1 2 ln 2 3 1 2(3) ln3 2(1) ln1 2 2 (1.09861 ) m EL x x dV x dx x x x x π π π π = − + = − + = − + − − + = Now 3 4 : (0.46944 m ) 1.09861 mELxV x dV x π π= = or 2.34 mx = www.elsolucionario.net www.elsolucionario.net
- 709. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 707 PROBLEM 5.127 Locate the centroid of the volume obtained by rotating the shaded area about the line .x h= SOLUTION First, note that symmetry implies x h= 0z = Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dy y yπ= = Now 2 2 2 2 2 ( ) h x a y a = − so that 2 2 . h r h a y a = − − Then ( ) 2 2 2 2 2 h dV a a y dy a π= − − and ( ) 2 2 2 2 20 a h V a a y dy a π= − − Let sin cosy a dy a dθ θ θ= = Then ( ) 2 2/2 2 2 2 2 0 2 /2 2 2 2 2 2 0 /2 2 2 2 0 sin cos 2 ( cos ) ( sin ) cos (2cos 2cos sin cos ) h V a a a a d a h a a a a a a d a ah d π π π π θ θ θ π θ θ θ θ π θ θ θ θ θ = − − = − + − = − − /2 2 3 0 2 2 2 sin 2 1 2sin 2 sin 2 4 3 1 2 2 2 3 0.095870 ah ah ah π π θ θ π θ θ π π = − + − = − − = www.elsolucionario.net www.elsolucionario.net
- 710. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 708 PROBLEM 5.127 (Continued) and ( ) ( ) 2 2 2 2 20 2 2 2 2 3 2 0 2 2 a EL a h y dV y a a y dy a h a y ay a y y dy a π π = − − = − − − 2 2 2 2 2 3/2 4 2 0 2 2 2 4 2 3/2 2 2 2 2 1 ( ) 3 4 1 2 ( ) ( ) 4 3 1 12 a h a y a a y y a h a a a a a a a h π π π = + − − = − − = Now :ELyV y dV= 2 2 21 (0.095870 ) 12 y ah a hπ π= or 0.869y a= www.elsolucionario.net www.elsolucionario.net
- 711. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 709 PROBLEM 5.128* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x-axis. SOLUTION First, note that symmetry implies 0y = 0z = Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx x xπ= = Now sin 2 x r b a π = so that 2 2 sin 2 x dV b dx a π π= Then ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 sin 2 sin 2 2 1 2 a a a x a a a a a x V b dx a x b b ab π π π π π π π = = − = − = and 2 2 2 sin 2 a EL a x x dV x b dx a π π = Use integration by parts with 2 2 sin 2 sin 2 x a a x u x dV a x du dx V π π π = = = = − www.elsolucionario.net www.elsolucionario.net
- 712. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 710 PROBLEM 5.128* (Continued) Then 2 2 2 2 2 2 2 2 2 2 sin sin 2 2 2 1 2 cos 2 2 4 2 a x xa a a EL a a aa a a x x x dV b x dx a a a x b a a x a π π π π π π π π = − − − = − − + 2 2 2 2 2 2 2 2 2 2 2 2 2 3 1 1 (2 ) ( ) 2 4 42 2 3 1 4 0.64868 a a b a a a a b a b π π π π π π = − + − + = − = Now 2 2 21 : 0.64868 2 ELxV x dV x ab a bπ π = = or 1.297x a= www.elsolucionario.net www.elsolucionario.net
- 713. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 711 PROBLEM 5.129* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y-axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.) SOLUTION First note that symmetry implies 0x = 0z = Choose as the element of volume a cylindrical shell of radius r and thickness dr. Then 1 (2 )( )( ), 2 ELdV r y dr y yπ= = Now sin 2 r y b a π = so that 2 sin 2 r dV br dr a π π= Then 2 2 sin 2 a a r V br dr a π π= Use integration by parts with sin 2 2 cos 2 r u rd dv dr a a r du dr v a π π π = = = = − Then [ ] 2 2 2 2 2 2 2 2 ( ) cos cos 2 2 2 4 2 ( )( ) sin2 1 2 a a a a a a a r a r V b r dr a a a a r b a a π π π π π π π π π = − − = − + − 2 2 2 2 2 4 4 2 1 8 1 5.4535 a a V b a b a b π π π π = − = − = www.elsolucionario.net www.elsolucionario.net
- 714. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 712 PROBLEM 5.129* (Continued) Also 2 1 2 2 2 2 sin 2 sin 2 2 sin 2 a EL a a a r r y dV b br dr a a r b r dr a π π π π π = = Use integration by parts with 2 2 sin 2 sin 2 r a a r u r dv dr a r du dr v π π π = = = = − Then 2 2 2 2 2 2 2 2 2 2 sin sin ( ) 2 2 2 (2 ) ( ) cos 2 2 4 2 a r ra a a EL a a aa a a r r y dV b r dr a a r a r b a a a π π π π π π π π = − − − = − − + 2 2 2 2 2 2 2 2 2 2 2 2 2 3 (2 ) ( ) 2 4 42 2 3 1 4 2.0379 a a a a b a a b a b π π π π π = − + − + = − = Now 2 2 2 : (5.4535 ) 2.0379ELyV y dV y a b a b= = or 0.374y b= www.elsolucionario.net www.elsolucionario.net
- 715. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 713 PROBLEM 5.130* Show that for a regular pyramid of height h and n sides ( 3, 4, )n = the centroid of the volume of the pyramid is located at a distance h/4 above the base. SOLUTION Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by 2 baseA kb= where ( );k k N= see note below. Using similar triangles, we have s h y b h − = or ( ) b s h y h = − Then 2 2 2 slice 2 ( ) b dV A dy ks dy k h y dy h = = = − and 2 2 2 3 2 20 0 2 1 ( ) ( ) 3 1 3 h h b b V k h y dy k h y h h kb h = − = − − = Also, ELy y= so that 2 2 2 2 2 3 2 20 0 2 2 2 3 4 2 2 2 0 ( ) ( 2 ) 1 2 1 1 2 3 4 12 h h EL h b b y dV y k h y dy k h y hy y dy h h b k h y hy y kb h h = − = − + = − + = Now 2 2 21 1 : 3 12 ELyV y dV y kb h kb h = = or 1 Q.E.D. 4 y h= Note: 2 base tan 2 2 1 2 4 tan ( ) N b N A N b N b k N b π π = × × = = www.elsolucionario.net www.elsolucionario.net
- 716. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 714 PROBLEM 5.131 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R. SOLUTION First note that symmetry implies 0x = The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now ( )( ) 2 EL dA r Rd r y π θ π = = − where sinr R θ= so that 2 sin 2 sinEL dA R d R y π θ θ θ π = = − Then /2 2 2 /2 0 0 2 sin [ cos ]A R d R R π π π θ θ π θ π = = − = and /2 2 0 /2 3 0 3 2 sin ( sin ) sin 2 2 2 4 2 EL R y dA R d R R π π θ π θ θ π θ θ π = − = − − = − Now 2 3 : ( ) 2 ELyA y dA y R R π π= = − or 1 2 y R= − Symmetry implies z y= 1 2 z R= − www.elsolucionario.net www.elsolucionario.net
- 717. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 715 PROBLEM 5.132 The sides and the base of a punch bowl are of uniform thickness t. If t R<< and R = 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch. SOLUTION (a) Bowl: First note that symmetry implies 0x = 0z = for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y-axis. Then wall (2 sin )( )dA R Rdπ θ θ= and wall( ) cosELy R θ= − Then /2 2 wall /6 /22 /6 2 2 sin 2 [ cos ] 3 A R d R R π π π π π θ θ π θ π = = − = and wall wall wall /2 2 /6 /23 2 /6 3 ( ) ( cos )(2 sin ) [cos ] 3 4 ELy A y dA R R d R R π π π π θ π θ θ π θ π = = − = = − By observation, 2 base base 3 , 4 2 A R y R π = = − Now y A yAΣ = Σ or 2 2 3 23 3 3 4 4 4 2 y R R R R R π π π π + = − + − or 0.48763 250 mmy R R= − = 121.9 mmy = − www.elsolucionario.net www.elsolucionario.net
- 718. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 716 PROBLEM 5.132 (Continued) (b) Punch: First note that symmetry implies 0x = 0z = and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then 2 , ELdV x dy y yπ= = Now 2 2 2 x y R+ = so that 2 2 ( )dV R y dyπ= − Then 0 2 2 3/2 0 2 3 3/2 3 2 3 ( ) 1 3 3 1 3 3 3 2 3 2 8 R R V R y dy R y y R R R R π π π π − − = − = − = − − − − = and ( ) 0 2 2 3/2 0 2 2 4 3/2 2 4 2 4 ( ) 1 1 2 4 1 3 1 3 15 2 2 4 2 64 EL R R y dV y R y dy R y y R R R R π π π π − − = − = − = − − − − = − Now 3 43 15 : 3 8 64 ELyV y dV y R Rπ π = = − or 5 250 mm 8 3 y R R= − = 90.2 mmy = − www.elsolucionario.net www.elsolucionario.net
- 719. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 717 PROBLEM 5.133 Locate the centroid of the section shown, which was cut from a thin circular pipe by two oblique planes. SOLUTION First note that symmetry implies 0x = Assume that the pipe has a uniform wall thickness t and choose as the element of volume a vertical strip of width adθ and height 2 1( ).y y− Then 2 1 1 2 1 ( ) , ( ) 2 EL ELdV y y ta d y y y z zθ= − = + = Now 3 1 2 6 = + h h y z a 2 3 2 2 2 3 = − + h y z h a ( ) 6 = + h z a a ( 2 ) 3 = − + h z a a and cosz a θ= Then 2 1( ) ( cos 2 ) ( cos ) 3 6 (1 cos ) 2 h h y y a a a a a a h θ θ θ − = − + − + = − and 1 2( ) ( cos ) ( cos 2 ) 6 3 (5 cos ) 6 (1 cos ) (5 cos ), cos 2 12 EL EL h h y y a a a a a a h aht h dV d y z a θ θ θ θ θ θ θ + = + + − + = − = − = − = www.elsolucionario.net www.elsolucionario.net
- 720. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 718 PROBLEM 5.133 (Continued) Then 0 0 2 (1 cos ) [ sin ] 2 aht V d aht aht π π θ θ θ θ π = − = − = and 0 2 2 0 2 0 2 2 (5 cos ) (1 cos ) 12 2 (5 6cos cos ) 12 sin 2 5 6sin 12 2 4 11 24 EL h aht y dV d ah t d ah t ah t π π π θ θ θ θ θ θ θ θ θ θ π = − − = − + = − + + = 0 2 0 2 2 cos (1 cos ) 2 sin 2 sin 2 4 1 2 EL aht z dV a d a ht a ht π π θ θ θ θ θ θ π = − = − − = − Now 211 : ( ) 24 ELyV y dV y aht ah tπ π= = or 11 24 y h= and 21 : ( ) 2 π π= = − ELzV z dV z aht a ht or 1 2 z a= − www.elsolucionario.net www.elsolucionario.net
- 721. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 719 PROBLEM 5.134* Locate the centroid of the section shown, which was cut from an elliptical cylinder by an oblique plane. SOLUTION First note that symmetry implies 0x = Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then 1 2 , , 24 EL ELdV xy dz y z z= = = Now 2 2a x b z b = − and /2 ( ) 2 2 = − + = − h h h y z b z b b Then 2 2 2 ( ) 2− = − − b b a h V b z b z dz b b Let sin cosz b dz b dθ θ θ= = Then /2 2 /2 /2 2 2 /2 /2 3 /2 ( cos )[ (1 sin )] cos (cos sin cos ) sin 2 1 cos 2 4 3 1 2 ah V b b b d b abh d abh V abh π π π π π π θ θ θ θ θ θ θ θ θ θ θ π − − = − = − = + + = www.elsolucionario.net www.elsolucionario.net
- 722. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 720 PROBLEM 5.134* (Continued) and 2 2 2 2 2 2 3 1 ( ) 2 ( ) 2 2 2 1 ( ) 4 − − = × − − − = − − b EL b b b h a h y dV b z b z b z dz b b b ah b z b z dz b Let sin cosz b dz b dθ θ θ= = Then 2 /2 2 3 /2 /2 2 2 2 2 2 /2 1 [ (1 sin )] ( cos ) ( cos ) 4 1 (cos 2sin cos sin cos ) 4 π π π π θ θ θ θ θ θ θ θ θ θ − − = − × = − + EL ah y dV b b b d b abh d Now 2 21 1 sin (1 cos2 ) cos (1 cos2 ) 2 2 θ θ θ θ= − = + so that 2 2 21 sin cos (1 cos 2 ) 4 θ θ θ= − Then /2 2 2 2 2 /2 /2 2 3 /2 2 1 1 cos 2sin cos (1 cos 2 ) 4 4 1 sin 2 1 1 1 sin 4 cos 4 2 4 3 4 4 2 8 5 32 ELy dV abh d abh abh π π π π θ θ θ θ θ θ θ θ θ θ θ π − − = − + − = + + + − + = Also, 2 2 2 2 2 2 ( ) 2 ( ) − − = − − = − − b EL b b b a h z dV z a z b z dz b b ah z b z b z dz b Let sin cosz b dz b dθ θ θ= = Then /2 2 /2 /2 2 2 2 2 /2 ( sin )[ (1 sin )]( cos ) ( cos ) (sin cos sin cos ) EL ah z dV b b b b d b ab h d π π π π θ θ θ θ θ θ θ θ θ θ − − = − × = − Using 2 2 21 sin cos (1 cos 2 ) 4 θ θ θ= − from above, /2 2 2 2 /2 1 sin cos (1 cos 2 ) 4 ELz dV ab h d π π θ θ θ θ − = − − /2 2 3 2 /2 1 1 1 sin 4 1 cos 3 4 4 2 8 8 ab h ab h π π θ θ θ θ π − = − − + + = − www.elsolucionario.net www.elsolucionario.net
- 723. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 721 PROBLEM 5.134* (Continued) Now 21 5 : 2 32 ELyV y dV y abh abhπ π = = or 5 16 y h= and 21 1 : 2 8 ELz V z dV z abh ab hπ π = = − or 1 4 z b= − www.elsolucionario.net www.elsolucionario.net
- 724. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 722 PROBLEM 5.135 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 3 in. of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom surface of the gravel is an oblique plane, which can be represented by the equation y = a + bx + cz.) SOLUTION The centroid can be found by integration. The equation for the bottom of the gravel is ,y a bx cz= + + where the constants a, b, and c can be determined as follows: For 0x = and 0,z = 3 in., and therefore,y = − 3 1 ft , or ft 12 4 a a− = = − For 30 ft and 0, 5 in.,x z y= = = − and therefore, 5 1 1 ft ft (30 ft), or 12 4 180 b b− = − + = − For 0 and 50 ft, 6 in.,x z y= = = − and therefore, 6 1 1 ft ft (50 ft), or 12 4 200 c c− = − + = − Therefore, 1 1 1 ft 4 180 200 y x z= − − − Now ELx dV x V = A volume element can be chosen as | |dV y dxdz= or 1 1 1 1 4 45 50 dV x z dxdz = + + and ELx x= www.elsolucionario.net www.elsolucionario.net
- 725. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 723 PROBLEM 5.135 (Continued) Then 50 30 0 0 1 1 1 4 45 50 EL x x dV x z dxdz = + + 30 250 3 2 0 0 50 0 50 2 0 4 1 1 4 2 135 100 1 (650 9 ) 4 1 9 650 4 2 10937.5 ft x z x x dz z dz z z = + + = + = + = The volume is 50 30 0 0 1 1 1 1 4 45 50 V dV x z dxdz = + + 30 50 2 0 0 50 0 50 2 0 3 1 1 4 90 50 1 3 40 4 5 1 3 40 4 10 687.50 ft z x x x dz z dz z z = + + = + = + = Then 4 3 10937.5ft 15.9091 ft 687.5 ft ELx dV x V = = = Therefore, 3 688 ftV = 15.91ftx = www.elsolucionario.net www.elsolucionario.net
- 726. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 724 PROBLEM 5.136 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface y = 16h(ax − x2 )(bz − z2 )/a2 b2 . SOLUTION First note that symmetry implies 2 a x = 2 b z = Choose as the element of volume a filament of base dx dz× and height y. Then 1 , 2 ELdV y dxdz y y= = or 2 2 2 2 16 ( )( ) h dV ax x bz z dxdz a b = − − Then 2 2 2 20 0 16 ( )( ) b a h V ax x bz z dxdz a b = − − 2 2 3 2 2 0 0 2 3 2 3 2 2 0 2 3 2 16 1 ( ) 3 16 1 1 ( ) ( ) 2 3 2 3 8 1 ( ) ( ) 2 33 4 9 a b b h a V bz z x x dz za b h a b a a z z a b ah b b b b abh = − − = − − = − = www.elsolucionario.net www.elsolucionario.net
- 727. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 725 PROBLEM 5.136 (Continued) and 2 2 2 2 2 2 2 20 0 2 2 2 3 4 2 2 3 4 4 4 0 0 2 2 2 2 3 4 3 4 5 2 4 0 0 1 16 16 ( )( ) ( )( ) 2 128 ( 2 )( 2 ) 128 1 ( 2 ) 3 2 5 b a EL b a a b h h y dV ax x bz z ax x bz z dxdz a b a b h a x ax x b z bz z dxdz a b h a a b z bz z x x x dz a b = − − − − = − + − + = − + − + 2 2 2 3 4 5 3 4 5 4 4 0 2 3 3 4 5 2 4 128 1 1 ( ) ( ) ( ) 3 2 5 3 5 64 1 32 ( ) ( ) ( ) 3 2 5 22515 b h a a b b a a a z z z za b ah b b b b b abh b = − + − + = − + = Now 24 32 : 9 225 ELyV y dV y abh abh = = or 8 25 y h= www.elsolucionario.net www.elsolucionario.net
- 728. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 726 PROBLEM 5.137 Locate the centroid of the plane area shown. SOLUTION 2 , inA , in.x , in.y 3 , inxA 3 , inyA 1 2 (38) 2268.2 2 π = 0 16.1277 0 36,581 2 20 16 320− × = −10 8 3200 −2560 Σ 1948.23 3200 34,021 Then 3200 1948.23 xA X A Σ = = Σ 1.643 in.X = 34,021 1948.23 yA Y A Σ = = Σ 17.46 in.Y = www.elsolucionario.net www.elsolucionario.net
- 729. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 727 PROBLEM 5.138 Locate the centroid of the plane area shown. SOLUTION 2 , mmA , mmx , mmy 3 , mmxA 3 , mmyA 1 2 (75)(120) 6000 3 = 28.125 48 168,750 288,000 2 1 (75)(60) 2250 2 − = − 25 20 –56,250 –45,000 Σ 3750 112,500 243,000 Then X A xAΣ = Σ 2 3 (3750 mm ) 112,500 mmX = or 30.0 mmX = and Y A yAΣ = Σ 2 3 (3750 mm ) 243,000 mmY = or 64.8 mmY = www.elsolucionario.net www.elsolucionario.net
- 730. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 728 PROBLEM 5.139 The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C. SOLUTION First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide with the centroid of the corresponding line. L, m , mx 2 , mxL 1 1.35 0.675 0.91125 2 0.6 0.3 0.18 3 0.75 0 0 4 0.75 0.2 0.15 5 (0.75) 1.17810 2 π = 1.07746 1.26936 Σ 4.62810 2.5106 Then (4.62810) 2.5106 X L x L X Σ = Σ = or 0.54247 mX = The free-body diagram of the frame is then where 2 ( ) 4.73 kg/m 4.62810 m 9.81 m/s 214.75 N W m L g′= Σ = × × = www.elsolucionario.net www.elsolucionario.net
- 731. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 729 PROBLEM 5.139 (Continued) Equilibrium then requires (a) 3 0: (1.55 m) (0.54247 m)(214.75 N) 0 5 C BAM T Σ = − = or 125.264 NBAT = or 125.3 NBAT = (b) 3 0: (125.264 N) 0 5 x xF CΣ = − = or 75.158 Nx =C 4 0: (125.264 N) (214.75 N) 0 5 y yF CΣ = + − = or 114.539 Ny =C Then 137.0 N=C 56.7° www.elsolucionario.net www.elsolucionario.net
- 732. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 730 PROBLEM 5.140 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION For the element (EL) shown, at 3 3 , , or h x a y h h ka k a = = = = Then 1/3 1/3 a x y h = Now 1/3 1/3 1/3 1/3 1 1 2 2 EL EL a dA xdy y dy h a x x y h y y = = = = = Then ( )1/3 4/3 1/3 1/30 0 3 3 4 4 h h a a A dA y dy y ah h h = = = = and 1/3 1/3 5/3 2 1/3 1/3 2/30 0 1/3 7/3 2 1/3 1/30 0 1 1 3 3 2 2 5 10 3 3 7 7 h h EL h h EL a a a x dA y y dy y a h h h h a a y dA y y dy y ah h h = = = = = = Hence 23 3 : 4 10 ELxA x dA x ah a h = = 2 5 x a= 23 3 : 4 7 ELyA y dA y ah ah = = 4 7 y h= www.elsolucionario.net www.elsolucionario.net
- 733. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 731 PROBLEM 5.141 Determine by direct integration the centroid of the area shown. SOLUTION We have 2 2 2 2 1 1 2 2 1 EL EL x x a x x y y L L x x dA y dx a dx L L = = = − + = = − + Then 2 2 2 32 2 20 0 1 2 3 8 3 L L x x x x A dA a dx a x L LL L aL = = − + = − + = and 2 2 2 3 42 2 20 0 2 1 2 3 4 10 3 L L EL x x x x x x dA x a dx a L LL L aL = − + = − + = 2 22 2 20 2 2 3 4 2 3 40 2 2 2 3 4 5 2 3 4 0 2 1 1 2 1 2 3 2 2 2 2 5 11 5 L EL EL L a x x x x y dA a dx L LL L a x x x x dx L L L L a x x x x x L L L L a L = − + − + = − + − + = − + − + = Hence, 28 10 : 3 3 ELxA x dA x aL aL = = 5 4 x L= 21 11 : 8 5 ELyA y dA y a a = = 33 40 y a= www.elsolucionario.net www.elsolucionario.net
- 734. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 732 PROBLEM 5.142 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design. SOLUTION SOLUTION Applying the first theorem of Pappus-Guldinus, the contact area CA of a belt is given by CA yL yLπ π= = Σ where the individual lengths are the lengths of the belt cross section that are in contact with the pulley. (a) 1 1 2 2[2( ) ] 0.125 0.125 in. 2 3 in. [(3 0.125) in.](0.625 in.) 2 cos20 CA y L y Lπ π = + = − + − ° or 2 8.10 inCA = (b) 1 1[2( )] 0.375 0.375 in. 2 3 0.08 in. 2 cos20 CA y Lπ π = = − − ° or 2 6.85 inCA = (c) 1 1[2( )] 2(0.25) 3 in. [ (0.25 in.)] CA y Lπ π π π = = − or 2 7.01inCA = www.elsolucionario.net www.elsolucionario.net
- 735. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 733 PROBLEM 5.143 Determine the reactions at the beam supports for the given loading. SOLUTION We have I II III 1 (3ft)(480 lb/ft) 720 lb 2 1 (6 ft)(600 lb/ft) 1800 lb 2 (2ft)(600 lb/ft) 1200 lb R R R = = = = = = Then 0: 0x xF BΣ = = 0: (2 ft)(720 lb) (4 ft)(1800 lb) (6 ft) (7 ft)(1200 lb) 0B yM CΣ = − + − = or 2360 lbyC = 2360 lb=C 0: 720 lb 1800 lb 2360 lb 1200 lb 0y yF BΣ = − + − + − = or 1360 lbyB = 1360 lb=B www.elsolucionario.net www.elsolucionario.net
- 736. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 734 PROBLEM 5.144 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of ωA and ωB corresponding to equilibrium. SOLUTION I II 1 (1.8 m) 0.9 2 1 (1.8 m) 0.9 2 A A B B R R ω ω ω ω = = = = 0: (24 kN)(1.2 ) (30 kN)(0.3 m) (0.9 )(0.6 m) 0D AM a ωΣ = − − − = (1) For 0.6 m,a = 24(1.2 0.6) (30)(0.3) 0.54 0aω− − − = 14.4 9 0.54 0Aω− − = 10.00 kN/mAω = 0: 24 kN 30 kN 0.9(10 kN/m) 0.9 0y BF ωΣ = − − + + = 50.0 kN/mBω = www.elsolucionario.net www.elsolucionario.net
- 737. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 735 PROBLEM 5.145 The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density 3 3 1.76 10 kg/m )sρ = × is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe. SOLUTION First determine force on dam without the silt, 3 3 2 allow 1 1 ( ) 2 2 1 [(6.6 m)(1m)][(10 kg/m )(9.81m/s )(6.6 m)] 2 213.66 kN 1.2 (1.5)(213.66 kN) 256.39 kN ww p w P A A gh P P ρ= = = = = = = Next determine the force P′on the dam face after a depth d of silt has settled. We have 3 3 2 2 3 3 2 I 2 3 3 2 II 2 1 [(6.6 ) m (1 m)][(10 kg/m )(9.81m/s )(6.6 ) m] 2 4.905(6.6 ) kN ( ) [ (1 m)][(10 kg/m )(9.81m/s )(6.6 ) m] 9.81(6.6 ) kN 1 ( ) [ (1m)][(1.76 10 kg/m )(9.81 m/s )( ) m] 2 8.6328 kN w s s P d d d P d d d d P d d d ′ = − × − = − = − = − = × = 2 I II 2 2 2 ( ) ( ) [4.905(43.560 13.2000 ) 9.81(6.6 ) 8.6328 ] kN [3.7278 213.66] kN w s sP P P P d d d d d d ′ ′= + + = − + + − + = + Now it’s required that allowP P′ = to determine the maximum value of d. 2 (3.7278 213.66) kN 256.39 kNd + = or 3.3856 md = Finally, 3 m 3.3856 m 12 10 N year − = × × or 282 yearsN = www.elsolucionario.net www.elsolucionario.net
- 738. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 736 PROBLEM 5.146 Determine the location of the centroid of the composite body shown when (a) 2 ,h b= (b) 2.5 .h b= SOLUTION V x xV Cylinder I 2 a bπ 1 2 b 2 21 2 a bπ Cone II 21 3 a hπ 1 4 b h+ 21 1 3 4 a h b hπ + 2 2 2 2 1 3 1 1 1 2 3 12 V a b h xV a b hb h π π = + Σ = + + (a) For 2 ,h b= 2 21 5 (2 ) 3 3 V a b b a bπ π = + = 2 2 2 2 2 2 2 1 1 1 (2 ) (2 ) 2 3 12 1 2 1 3 2 3 3 2 xV a b b b b a b a b π π π Σ = + + = + + = 2 2 25 3 9 : 3 2 10 XV xV X a b a b X bπ π =Σ = = Centroid is 1 10 b to left of base of cone. www.elsolucionario.net www.elsolucionario.net
- 739. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 737 PROBLEM 5.146 (Continued) (b) For 2.5 ,h b= 2 21 (2.5 ) 1.8333 3 V a b b a bπ π = + = 2 2 2 2 2 2 2 1 1 1 (2.5 ) (2.5 ) 2 3 12 [0.5 0.8333 0.52083] 1.85416 xV a b b b b a b a b π π π Σ = + + = + + = 2 2 2 : (1.8333 ) 1.85416 1.01136XV xV X a b a b X bπ π= Σ = = Centroid is 0.01136b to right of base of cone. Note: Centroid is at base of cone for 6 2.449 .h b b= = www.elsolucionario.net www.elsolucionario.net
- 740. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 738 PROBLEM 5.147 Locate the center of gravity of the sheet-metal form shown. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. I I III 1 (1.2) 0.4 m 3 1 (3.6) 1.2 m 3 4(1.8) 2.4 m 3 y z x π π = − = − = = = − = − 2 , mA , mx , my , mz 3 , mxA 3 , myA 3 , mzA I 1 (3.6)(1.2) 2.16 2 = 1.5 −0.4 1.2 3.24 0.864− 2.592 II (3.6)(1.7) 6.12= 0.75 0.4 1.8 4.59 2.448 11.016 III 2 (1.8) 5.0894 2 π = 2.4 π − 0.8 1.8 −3.888 4.0715 9.1609 Σ 13.3694 3.942 5.6555 22.769 We have 2 3 : (13.3694 m ) 3.942 mX V xV XΣ = Σ = or 0.295 mX = 2 3 : (13.3694 m ) 5.6555 mY V yV YΣ = Σ = or 0.423 mY = 2 3 : (13.3694 m ) 22.769 mZ V zV ZΣ = Σ = or 1.703 mZ = www.elsolucionario.net www.elsolucionario.net
- 741. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 739 PROBLEM 5.148 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis. SOLUTION First note that symmetry implies 0y = and 0z = We have 2 ( )y k X h= − At 0, ,x y a= = 2 ( )a k h= − or 2 a k h = Choose as the element of volume a disk of radius r and thickness dx. Then 2 , ELdV r dx X xπ= = Now 2 2 ( ) a r x h h = − so that 2 4 4 ( ) a dV x h dx h π= − Then 2 2 4 5 04 40 2 ( ) [( ) ] 5 1 5 h ha a V x h dx x h h h a h π π π = − = − = and 2 4 40 2 5 4 2 3 3 2 4 4 0 2 6 5 2 4 3 3 4 2 4 0 2 2 ( ) ( 4 6 4 ) 1 4 3 4 1 6 5 2 3 2 1 30 h EL h h a x dV x x h dx h a x hx h x h x h x dx h a x hx h x h x h x h a h π π π π = − = − + − + = − + − + = Now 2 2 2 : 5 30 ELxV x dV x a h a h π π = = 1 or 6 x h= www.elsolucionario.net www.elsolucionario.net
- 745. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 743 PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: 0: 0 0y y yF BΣ = = =B 0: (3.2 m) (48 kN)(7.2 m) 0C xM BΣ = − − = 108 kN 108 kNx xB = − =B 0: 108 kN 48 kN 0xF CΣ = − + = 60 kNC = 60 kN=C Free body: Joint B: 108 kN 5 4 3 BCAB FF = = 180.0 kNABF T= 144.0 kNBCF T= Free body: Joint C: 60 kN 13 12 5 AC BCF F = = 156.0 kNACF C= 144 kN (checks)BCF = www.elsolucionario.net www.elsolucionario.net
- 746. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 744 PROBLEM 6.2 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: 0: 1260 lbAMΣ = =C 0: 0x xFΣ = =A 0: 960 lby yFΣ = =A Joint B: 300 lb 12 13 5 BCAB FF = = 720 lbABF T= 780 lbBCF C= Joint A: 4 0: 960 lb 0 5 y ACF FΣ = − − = 1200 lbACF = 1200 lbACF C= 3 0: 720 lb (1200 lb) 0 (checks) 5 xFΣ = − = www.elsolucionario.net www.elsolucionario.net
- 747. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 745 PROBLEM 6.3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION 2 2 2 2 3 1.25 3.25 m 3 4 5 m AB BC = + = = + = Reactions: 0: (84 kN)(3 m) (5.25 m) 0AM CΣ = − = 48 kN=C 0: 0x xF A CΣ = − = 48 kNx =A 0: 84 kN 0y yF AΣ = = = 84 kNy =A Joint A: 12 0: 48 kN 0 13 x ABF FΣ = − = 52 kNABF = + 52.0 kNABF T= 5 0: 84 kN (52 kN) 0 13 y ACF FΣ = − − = 64.0 kNACF = + 64.0 kNACF T= Joint C: 48 kN 5 3 BCF = 80.0 kNBCF C= www.elsolucionario.net www.elsolucionario.net
- 748. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 746 PROBLEM 6.4 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: From the symmetry of the truss and loading, we find 600 lb= =C D Free body: Joint B: 300 lb 2 15 BCAB FF = = 671lbABF T= 600 lbBCF C= Free body: Joint C: 3 0: 600 lb 0 5 y ACF FΣ = + = 1000 lbACF = − 1000 lbACF C= 4 0: ( 1000 lb) 600 lb 0 5 x CDF FΣ = − + + = 200 lbCDF T= From symmetry: 1000 lb , 671lb ,AD AC AE ABF F C F F T= = = = 600 lbDE BCF F C= = www.elsolucionario.net www.elsolucionario.net
- 749. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 747 PROBLEM 6.5 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: 0: (24) (4 2.4)(12) (1)(24) 0D yM FΣ = − + − = 4.2 kipsy =F 0: 0x xFΣ = =F 0: (1 4 1 2.4) 4.2 0yF DΣ = − + + + + = 4.2 kips=D Joint A: 0: 0x ABF FΣ = = 0ABF = 0: 1 0y ADF FΣ = − − = 1 kipADF = − 1.000 kipADF C= Joint D: 8 0: 1 4.2 0 17 y BDF FΣ = − + + = 6.8 kipsBDF = − 6.80 kipsBDF C= 15 0: ( 6.8) 0 17 x DEF FΣ = − + = 6 kipsDEF = + 6.00 kipsDEF T= www.elsolucionario.net www.elsolucionario.net
- 750. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 748 PROBLEM 6.5 (Continued) Joint E: 0: 2.4 0y BEF FΣ = − = 2.4 kipsBEF = + 2.40 kipsBEF T= Truss and loading symmetrical about c .L www.elsolucionario.net www.elsolucionario.net
- 751. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 749 PROBLEM 6.6 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION 2 2 2 2 5 12 13 ft 12 16 20 ft AD BCD = + = = + = Reactions: 0: 0x xF DΣ = = 0: (21ft) (693 lb)(5 ft) 0E yM DΣ = − = 165 lby =D 0: 165 lb 693 lb 0yF EΣ = − + = 528 lb=E Joint D: 5 4 0: 0 13 5 x AD DCF F FΣ = + = (1) 12 3 0: 165 lb 0 13 5 y AD DCF F FΣ = + + = (2) Solving Eqs. (1) and (2) simultaneously, 260 lbADF = − 260 lbADF C= 125 lbDCF = + 125 lbDCF T= www.elsolucionario.net www.elsolucionario.net
- 752. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 750 PROBLEM 6.6 (Continued) Joint E: 5 4 0: 0 13 5 x BE CEF F FΣ = + = (3) 12 3 0: 528 lb 0 13 5 y BE CEF F FΣ = + + = (4) Solving Eqs. (3) and (4) simultaneously, 832 lbBEF = − 832 lbBEF C= 400 lbCEF = + 400 lbCEF T= Joint C: Force polygon is a parallelogram (see Fig. 6.11, p. 209). 400 lbACF T= 125.0 lbBCF T= Joint A: 5 4 0: (260 lb) (400 lb) 0 13 5 x ABF FΣ = + + = 420 lbABF = − 420 lbABF C= 12 3 0: (260 lb) (400 lb) 0 13 5 0 0 (Checks) yFΣ = − = = www.elsolucionario.net www.elsolucionario.net
- 753. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 751 PROBLEM 6.7 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: 0: 2(5 kN) 0x xF CΣ = + = 10 kN 10 kNxxC = − =C 0: (2 m) (5 kN)(8 m) (5 kN)(4 m) 0CM DΣ = − − = 30 kN 30 kND = + =D 0: 30 kN 0 30 kN 30 kNyy y yF C CΣ = + = = − =C Free body: Joint A: 5 kN 4 117 AB ADF F = = 20.0 kNABF T= 20.6 kNADF C= Free body: Joint B: 1 0: 5 kN 0 5 x BDF FΣ = + = 5 5 kNBDF = − 11.18 kNBDF C= 2 0: 20 kN ( 5 5 kN) 0 5 y BCF FΣ = − − − = 30 kNBCF = + 30.0 kNBCF T= www.elsolucionario.net www.elsolucionario.net
- 754. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 752 PROBLEM 6.7 (Continued) Free body: Joint C: 0: 10 kN 0x CDF FΣ = − = 10 kNCDF = + 10.00 kNCDF T= 0: 30 kN 30 kN 0 (checks)yFΣ = − = www.elsolucionario.net www.elsolucionario.net
- 755. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 753 PROBLEM 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: 0: 16 kNC xMΣ = =A 0: 9 kNy yFΣ = =A 0: 16 kNxFΣ = =C Joint E: 3 kN 5 4 3 BE DEF F = = 5.00 kNBEF T= 4.00 kNDEF C= Joint B: 4 0: (5 kN) 0 5 x ABF FΣ = − = 4 kNABF = + 4.00 kNABF T= 3 0: 6 kN (5 kN) 0 5 y BDF FΣ = − − − = 9 kNBDF = − 9.00 kNBDF C= Joint D: 3 0: 9 kN 0 5 y ADF FΣ = − + = 15 kNADF = + 15.00 kNADF T= 4 0: 4 kN (15 kN) 0 5 x CDF FΣ = − − − = 16 kNCDF = − 16.00 kNCDF C= www.elsolucionario.net www.elsolucionario.net
- 756. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 754 PROBLEM 6.9 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xFΣ = =H Because of the symmetry of the truss and loading, 1 total load 2 y= =A H 1200 lby= =A H Free body: Joint A: 900 lb 5 4 3 ACAB FF = = 1500 lbAB C=F 1200 lbAC T=F Free body: Joint C: BC is a zero-force member. 0BC =F 1200 lbCEF T= Free body: Joint B: 24 4 4 0: (1500 lb) 0 25 5 5 x BD BEF F FΣ = + + = or 24 20 30,000 lbBD BEF F+ = − (1) 7 3 3 0: (1500) 600 0 25 5 5 y BD BEF F FΣ = − + − = or 7 15 7,500 lbBD BEF F− = − (2) www.elsolucionario.net www.elsolucionario.net
- 757. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 755 PROBLEM 6.9 (Continued) Multiply Eq. (1) by 3, Eq. (2) by 4, and add: 100 120,000 lbBDF = − 1200 lbBDF C= Multiply Eq. (1) by 7, Eq. (2) by –24, and add: 500 30,000 lbBEF = − 60.0 lbBEF C= Free body: Joint D: 24 24 0: (1200 lb) 0 25 25 x DFF FΣ = + = 1200 lbDFF = − 1200 lbDFF C= 7 7 0: (1200 lb) ( 1200 lb) 600 lb 0 25 25 y DEF FΣ = − − − − = 72.0 lbDEF = 72.0 lbDEF T= Because of the symmetry of the truss and loading, we deduce that =EF BEF F 60.0 lbEFF C= EG CEF F= 1200 lbEGF T= FG BCF F= 0FGF = FH ABF F= 1500 lbFHF C= GH ACF F= 1200 lbGHF T= Note: Compare results with those of Problem 6.11. www.elsolucionario.net www.elsolucionario.net
- 758. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 756 PROBLEM 6.10 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xFΣ = =H Because of the symmetry of the truss and loading, 1 total load 2 yA H= = 1200 lby= =A H Free body: Joint A: 900 lb 5 4 3 ACAB FF = = 1500 lbAB C=F 1200 lbAC T=F Free body: Joint C: BC is a zero-force member. 0BC =F 1200 lbCE T=F www.elsolucionario.net www.elsolucionario.net
- 759. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 757 PROBLEM 6.10 (Continued) Free body: Joint B: 4 4 4 0: (1500 lb) 0 5 5 5 x BD BCF F FΣ = + + = or 1500 lbBD BEF F+ = − (1) 3 3 3 0: (1500 lb) 600 lb 0 5 5 5 y BD BEF F FΣ = − + − = or 500 lbBD BEF F− = − (2) Add Eqs. (1) and (2): 2 2000 lbBDF = − 1000 lbBDF C= Subtract Eq. (2) from Eq. (1): 2 1000 lbBEF = − 500 lbBEF C= Free Body: Joint D: 4 4 0: (1000 lb) 0 5 5 x DFF FΣ = + = 1000 lbDFF = − 1000 lbDFF C= 3 3 0: (1000 lb) ( 1000 lb) 600 lb 0 5 5 y DEF FΣ = − − − − = 600 lbDEF = + 600 lbDEF T= Because of the symmetry of the truss and loading, we deduce that =EF BEF F 500 lbEFF C= EG CEF F= 1200 lbEGF T= FG BCF F= 0FGF = FH ABF F= 1500 lbFHF C= GH ACF F= 1200 lbGHF T= www.elsolucionario.net www.elsolucionario.net
- 760. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 758 PROBLEM 6.11 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xF AΣ = = Due to symmetry of truss and load, 1 total load 21 kN 2 yA H= = = Free body: Joint A: 15.3 kN 37 35 12 ACAB FF = = 47.175 kN 44.625 kNAB ACF F= = 47.2 kNABF C= 44.6 kNACF T= Free body: Joint B: From force polygon: 47.175 kN, 10.5 kNBD BCF F= = 10.50 kNBCF C= 47.2 kNBDF C= www.elsolucionario.net www.elsolucionario.net
- 761. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 759 PROBLEM 6.11 (Continued) Free body: Joint C: 3 0: 10.5 0 5 y CDF FΣ = − = 17.50 kNCDF T= 4 0: (17.50) 44.625 0 5 x CEF FΣ = + − = 30.625 kNCEF = 30.6 kNCEF T= Free body: Joint E: DE is a zero-force member. 0DEF = Truss and loading symmetrical about .cL www.elsolucionario.net www.elsolucionario.net
- 762. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 760 PROBLEM 6.12 Determine the force in each member of the Fink roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xFΣ = =A Because of the symmetry of the truss and loading, 1 total load 2 y = =A G 6.00 kNy = =A G Free body: Joint A: 4.50 kN 2.462 2.25 1 ACAB FF = = 11.08 kNABF C= 10.125 kNACF = 10.13 kNACF T= Free body: Joint B: 3 2.25 2.25 0: (11.08 kN) 0 5 2.462 2.462 x BC BDF F FΣ = + + = (1) 4 11.08 kN 0: 3 kN 0 5 2.462 2.462 BD y BC F F FΣ = − + + − = (2) Multiply Eq. (2) by –2.25 and add to Eq. (1): 12 6.75 kN 0 2.8125 5 BC BCF F+ = = − 2.81kNBCF C= Multiply Eq. (1) by 4, Eq. (2) by 3, and add: 12 12 (11.08 kN) 9 kN 0 2.462 2.462 BDF + − = 9.2335 kNBDF = − 9.23 kNBDF C= www.elsolucionario.net www.elsolucionario.net
- 763. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 761 PROBLEM 6.12 (Continued) Free body: Joint C: 4 4 0: (2.8125 kN) 0 5 5 y CDF FΣ = − = 2.8125 kN,CDF = 2.81kNCDF T= 3 3 0: 10.125 kN (2.8125 kN) (2.8125 kN) 0 5 5 x CEF FΣ = − + + = 6.7500 kNCEF = + 6.75 kNCEF T= Because of the symmetry of the truss and loading, we deduce that DE CDF F= 2.81kNCDF T= DF BDF F= 9.23 kNDFF C= EF BCF F= 2.81kNEFF C= EG ACF F= 10.13 kNEGF T= FG ABF F= 11.08 kNFGF C= www.elsolucionario.net www.elsolucionario.net
- 764. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 762 PROBLEM 6.13 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: (18 m) (2 kN)(4 m) (2 kN)(8 m) (1.75 kN)(12 m) (1.5 kN)(15 m) (0.75 kN)(18 m) 0 AM HΣ = − − − − − = 4.50 kN=H 0: 0 0: 9 0 9 4.50 x x y y y F A F A H A Σ = = Σ = + − = = − 4.50 kNy =A Free body: Joint A: 3.50 kN 2 15 7.8262 kN ACAB AB FF F C = = = 7.83 kNABF C= 7.00 kNACF T= www.elsolucionario.net www.elsolucionario.net
- 765. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 763 PROBLEM 6.13 (Continued) Free body: Joint B: 2 2 1 0: (7.8262 kN) 0 5 5 2 x BD BCF F FΣ = + + = or 0.79057 7.8262 kNBD BCF F+ = − (1) 1 1 1 0: (7.8262 kN) 2 kN 0 5 5 2 y BD BCF F FΣ = + − − = or 1.58114 3.3541BD BCF F− = − (2) Multiply Eq. (1) by 2 and add Eq. (2): 3 19.0065 6.3355 kN BD BD F F = − = − 6.34 kNBDF C= Subtract Eq. (2) from Eq. (1): 2.37111 4.4721 1.8861kN BC BC F F = − = − 1.886 kNBCF C= Free body: Joint C: 2 1 0: (1.8861kN) 0 5 2 y CDF FΣ = − = 1.4911kNCDF = + 1.491kNCDF T= 1 1 0: 7.00 kN (1.8861kN) (1.4911 kN) 0 2 5 x CEF FΣ = − + + = 5.000 kNCEF = + 5.00 kNCEF T= Free body: Joint D: 2 1 2 1 0: (6.3355 kN) (1.4911kN) 0 5 2 5 5 x DF DEF F FΣ = + + − = or 0.79057 5.5900 kNDF DEF F+ = − (1) 1 1 1 2 0: (6.3355 kN) (1.4911kN) 2 kN 0 5 2 5 5 y DF DEF F FΣ = − + − − = or 0.79057 1.1188 kNDF DEF F− = − (2) Add Eqs. (1) and (2): 2 6.7088 kNDFF = − 3.3544 kNDFF = − 3.35 kNDFF C= Subtract Eq. (2) from Eq. (1): 1.58114 4.4712 kNDEF = − 2.8278 kNDEF = − 2.83 kNDEF C= www.elsolucionario.net www.elsolucionario.net
- 766. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 764 PROBLEM 6.13 (Continued) Free body: Joint F: 1 2 0: (3.3544 kN) 0 2 5 x FGF FΣ = + = 4.243 kNFGF = − 4.24 kNFGF C= 1 1 0: 1.75 kN (3.3544 kN) ( 4.243 kN) 0 5 2 y EFF FΣ = − − + − − = 2.750 kNEFF = 2.75 kNEFF T= Free body: Joint G: 1 1 1 0: (4.243 kN) 0 2 2 2 x GH EGF F FΣ = − + = or 4.243 kNGH EGF F− = − (1) 1 1 1 0: (4.243 kN) 1.5 kN 0 2 2 2 y GH EGF F FΣ = − − − − = or 6.364 kNGH EGF F+ = − (2) Add Eqs. (1) and (2): 2 10.607GHF = − 5.303GHF = − 5.30 kNGHF C= Subtract Eq. (1) from Eq. (2): 2 2.121kNEGF = − 1.0605 kNEGF = − 1.061 kNEGF C= Free body: Joint H: 3.75 kN 1 1 EHF = 3.75 kNEHF T= We can also write 3.75 kN 12 GHF = 5.30 kN (Checks)GHF C= www.elsolucionario.net www.elsolucionario.net
- 767. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 765 PROBLEM 6.14 The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: 0: (800 N)(7.5 m) (800 N)(3.75 m) (2 m) 0FM AΣ = + − = 2250 NA = + 2250 N=A 0: 2250 N 0y yF FΣ = + = 2250 N 2250 Ny yF = − =F 0: 800 N 800 N 0x xF FΣ = − − + = 1600 N 1600 Nx xF = + =F Joint D: 800 N 8 15 17 DE BDF F = = 1700 NBDF C= 1500 NDEF T= Joint A: 2250 N 15 17 8 ACAB FF = = 2250 NABF C= 1200 NACF T= Joint F: 0: 1600 N 0x CFF FΣ = − = 1600 NCFF = + 1600 NCFF T= 0: 2250 N 0y EFF FΣ = − = 2250 NEFF = + 2250 NEFF T= www.elsolucionario.net www.elsolucionario.net
- 768. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 766 PROBLEM 6.14 (Continued) Joint C: 8 0: 1200 N 1600 N 0 17 x CEF FΣ = − + = 850 NCEF = − 850 NCEF C= 15 0: 0 17 y BC CEF F FΣ = + = 15 15 ( 850 N) 17 17 BC CEF F= − = − − 750 NBCF = + 750 NBCF T= Joint E: 8 0: 800 N (850 N) 0 17 x BEF FΣ = − − + = 400 NBEF = − 400 NBEF C= 15 0: 1500 N 2250 N (850 N) 0 17 0 0 (checks) yFΣ = − + = = www.elsolucionario.net www.elsolucionario.net
- 769. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 767 PROBLEM 6.15 Determine the force in each of the members located to the left of line FGH for the studio roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xFΣ = =A Because of symmetry of loading, 1 2 yA L= = total load 1200 lby = =A L Zero-Force Members: Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus, 0BC EH= =F F Also, CE ACF F= (1) Free body: Joint A: 1000 lb 2 15 2236 lb ACAB AB FF F C = = = 2240 lbABF C= 2000 lbACF T= From Eq. (1): 2000 lbCEF T= Free body: Joint B: 2 2 2 0: (2236 lb) 0 5 5 5 x BD BEF F FΣ = + + = or 2236 lbBD BEF F+ = − (2) 1 1 1 0: (2236 lb) 400 lb 0 5 5 5 Σ = − + − =y BD BEF F F or 1342 lbBD BEF F− = − (3) www.elsolucionario.net www.elsolucionario.net
- 770. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 768 PROBLEM 6.15 (Continued) Add Eqs. (2) and (3): 2 3578 lbBDF = − 1789 lbBDF C= Subtract Eq. (3) from Eq. (1): 2 894 lbBEF = − 447 lbBEF C= Free body: Joint E: 2 2 0: (447 lb) 2000 lb 0 5 5 x EGF FΣ = + − = 1789 lbEGF T= 1 1 0: (1789 lb) (447 lb) 0 5 5 y DEF FΣ = + − = 600 lbDEF = − 600 lbDEF C= Free body: Joint D: 2 2 2 0: (1789 lb) 0 5 5 5 x DF DGF F FΣ = + + = or 1789 lbDF DGF F+ = − (4) 1 1 1 0: (1789 lb) 5 5 5 600 lb 400 lb 0 y DF DGF F FΣ = − + + − = or 2236 lb− = −DF DGF F (5) Add Eqs. (4) and (5): 2 4025 lbDFF = − 2010 lbDFF C= Subtract Eq. (5) from Eq. (4): 2 447 lbDGF = 224 lbDGF T= www.elsolucionario.net www.elsolucionario.net
- 771. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 769 PROBLEM 6.16 Determine the force in member FG and in each of the members located to the right of FG for the studio roof truss shown. State whether each member is in tension or compression. SOLUTION Reaction at L: Because of the symmetry of the loading, 1 2 L = total load, 1200 lb=L (See F.B. diagram to the left for more details.) Free body: Joint L: 1 1 9 tan 26.57 18 3 tan 9.46 18 1000 lb sin63.43 sin99.46 sin17.11° JL KLF F α β − − = = ° = = ° = = ° ° 3040 lbJLF T= 3352.7 lbKLF C= 3350 lbKLF C= Free body: Joint K: 2 2 2 0: (3352.7 lb) 0 5 5 5 Σ = − − − =x IK JKF F F or 3352.7 lb+ = −IK JKF F (1) 1 1 1 0: (3352.7) 400 0 5 5 5 y IK JKF F FΣ = − + − = or 2458.3 lbIK JKF F− = − (2) Add Eqs. (1) and (2): 2 5811.0IKF = − 2905.5 lbIKF = − 2910 lbIKF C= Subtract Eq. (2) from Eq. (1): 2 894.4JKF = − 447.2 lbJKF = − 447 lbJKF C= www.elsolucionario.net www.elsolucionario.net
- 772. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 770 PROBLEM 6.16 (Continued) Free body: Joint J: 2 6 6 2 0: (3040 lb) (447.2) 0 13 37 37 5 x IJ GJF F FΣ = − − + − = (3) 3 1 1 1 0: (3040 lb) (447.2) 0 13 37 37 5 y IJ GJF F FΣ = + − − = (4) Multiply Eq. (4) by 6 and add to Eq. (3): 16 8 (447.2) 0 13 5 360.54 lb IJ IJ F F − = = 361lbIJF T= Multiply Eq. (3) by 3, Eq. (4) by 2, and add: 16 8 ( 3040) (447.2) 0 37 5 2431.7 lb GJ GJ F F − − − = = 2430 lbGJF T= Free body: Joint I: 2 2 2 2 0: (2905.5) (360.54) 0 5 5 5 13 Σ = − − − + =x FI GIF F F or 2681.9 lbFI GIF F+ = − (5) 1 1 1 3 0: (2905.5) (360.54) 400 0 5 5 5 13 Σ = − + − − =y FI GIF F F or 1340.3 lbFI GIF F− = − (6) Add Eqs. (5) and (6): 2 4022.2FIF = − 2011.1lbFIF = − 2010 lbFIF C= Subtract Eq. (6) from Eq. (5): 2 1341.6 lbGIF = − 671lbGIF C= Free body: Joint F: From 0: 2011.1lbx DF FIF F F CΣ = = = 1 0: 400 lb 2 2011.1lb 0 5 y FGF F Σ = − + = 1400 lbFGF = + 1400 lbFGF T= www.elsolucionario.net www.elsolucionario.net
- 773. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 771 PROBLEM 6.17 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression. SOLUTION Free Body: Truss: 0: 0x xFΣ = =A 0: (1 kN)(12 m) (2 kN)(10 m) (2 kN)(8 m) (1kN)(6 m) (12 m) 0L yM AΣ = + + + − = 4.50 kNy =A We note that BC is a zero-force member: 0BCF = Also, CE ACF F= (1) Free body: Joint A: 1 2 0: 0 2 5 x AB ACF F FΣ = + = (2) 1 1 0: 3.50 kN 0 2 5 y AB ACF F FΣ = + + = (3) Multiply Eq. (3) by –2 and add Eq. (2): 1 7 kN 0 2 ABF− − = 9.90 kNABF C= www.elsolucionario.net www.elsolucionario.net
- 774. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 772 PROBLEM 6.17 (Continued) Subtract Eq. (3) from Eq. (2): 1 3.50 kN 0 7.826 kN 5 AC ACF F− = = 7.83 kNACF T= From Eq. (1): 7.826 kNCE ACF F= = 7.83 kNCEF T= Free body: Joint B: 1 1 0: (9.90 kN) 2 kN 0 2 2 y BDF FΣ = + − = 7.071kNBDF = − 7.07 kNBDF C= 1 0: (9.90 7.071) kN 0 2 x BEF FΣ = + − = 2.000 kNBEF = − 2.00 kNBEF C= Free body: Joint E: 2 0: ( 7.826 kN) 2.00 kN 0 5 x EGF FΣ = − + = 5.590 kNEGF = 5.59 kNEGF T= 1 0: (7.826 5.590) kN 0 5 y DEF FΣ = − − = 1.000 kNDEF = 1.000 kNDEF T= Free body: Joint D: 2 1 0: ( ) (7.071kN) 5 2 x DF DGF F FΣ = + + or 5.590 kNDF DGF F+ = − (4) 1 1 0: ( ) (7.071kN) 2 kN 1kN 0 5 2 y DF DGF F FΣ = − + = − = or 4.472DE DGF F− = − (5) Add Eqs. (4) and (5): 2 10.062 kNDFF = − 5.03 kNDFF C= Subtract Eq. (5) from Eq. (4): 2 1.1180 kNDGF = − 0.559 kNDGF C= www.elsolucionario.net www.elsolucionario.net
- 775. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 773 PROBLEM 6.18 Determine the force in member FG and in each of the members located to the right of FG for the scissors roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: (12 m) (2 kN)(2 m) (2kN)(4 m) (1 kN)(6 m) 0AM LΣ = − − − = 1.500 kN=L Angles: tan 1 45 1 tan 26.57 2 α α β β = = ° = = ° Zero-force members: Examining successively joints K, J, and I, we note that the following members to the right of FG are zero- force members: JK, IJ, and HI. Thus, 0HI IJ JKF F F= = = We also note that GI IK KLF F F= = (1) and HJ JLF F= (2) www.elsolucionario.net www.elsolucionario.net
- 776. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 774 PROBLEM 6.18 (Continued) Free body: Joint L: 1.500 kN sin116.57 sin 45 sin18.43° JL KLF F = = ° ° 4.2436 kNJLF = 4.24 kNJLF C= 3.35 kNKLF T= From Eq. (1): GI IK KLF F F= = 3.35 kNGI IKF F T= = From Eq. (2): 4.2436 kNHJ JLF F= = 4.24 kNHJF C= Free body: Joint H: 4.2436 sin108.43 sin18.43 sin53.14 GHFH FF = = ° ° ° 5.03 kNFHF C= 1.677 kNGHF T= Free body: Joint F: 0: cos26.57 (5.03 kN)cos26.57 0x DFF FΣ = − ° − ° = 5.03 kNDFF = − 0: 1kN (5.03 kN)sin 26.57 ( 5.03 kN)sin 26.57 0y FGF FΣ = − − + ° − − ° = 3.500 kN=FGF 3.50 kNFGF T= www.elsolucionario.net www.elsolucionario.net
- 777. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 775 PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xF AΣ = = Due to symmetry of truss and loading, 1 total load 6 kips 2 yA G= = = Free body: Joint A: 6 kips 5 3 4 ACAB FF = = 7.50 kipsABF C= 4.50 kipsACF T= Free body: Joint B: 7.5 kips 5 6 5 BC BDF F = = 7.50 kipsBCF T= 9.00 kipsBDF C= Free body: Joint C: 4 4 0: (7.5) 6 0 5 5 y CDF FΣ = + − = 0CDF = 3 0: 4.5 (7.5) 0 5 x CEF FΣ = − − = 9 kipsCEF = + 9.00 kipsCEF T= Truss and loading is symmetrical about c .L www.elsolucionario.net www.elsolucionario.net
- 778. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 776 PROBLEM 6.20 Solve Problem 6.19 assuming that the load applied at E has been removed. PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xF AΣ = = 0: 6(36) (54) 0 4 kipsG y yM AΣ = − = =A 0: 4 6 0 2 kipsyF GΣ = − + = =G Free body: Joint A: 4 kips 5 3 4 ACAB FF = = 5.00 kipsABF C= 3.00 kipsACF T= Free body Joint B: 5 kips 5 6 5 BC BDF F = = 5.00 kipsBCF T= 6.00 kipsBDF C= Free body Joint C: 4 4 0: (5) 6 0 5 5 y CDM FΣ = + − = 2.50 kipsCDF T= 3 3 0: (2.5) (5) 3 0 5 5 x CEF FΣ = + − − = 4.50 kipsCEF T= www.elsolucionario.net www.elsolucionario.net
- 779. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 777 PROBLEM 6.20 (Continued) Free body: Joint D: 4 4 0: (2.5) 0 5 5 y DEF FΣ = − − = 2.5 kipsDEF = − 2.50 kipsDEF C= 3 3 0: 6 (2.5) (2.5) 0 5 5 x DFF FΣ = + − − = 3 kipsDFF = − 3.00 kipsDFF C= Free body: Joint F: 3 kips 5 5 6 FGEF FF = = 2.50 kipsEFF T= 2.50 kipsFGF C= Free body: Joint G: 2 kips 3 4 EGF = 1.500 kipsEGF T= Also, 2 kips 5 4 FGF = 2.50 kips (Checks)FGF C= www.elsolucionario.net www.elsolucionario.net
- 780. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 778 PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0z xFΣ = =A 0: (36 ft) (4 kips)(9 ft) (4 kips)(18 ft) (4 kips)(27 ft) 0 AM HΣ = − − − = 6 kips=H 0: 6 kips 12 kips 0 6 kipsy y yF AΣ = + − = =A Free body: Joint A: 6 kips 5 3 4 ACAB FF = = 7.50 kipsABF C= 4.50 kipsACF T= Free body: Joint C: 0:xFΣ = 4.50 kipsCEF T= 0:yFΣ = 4.00 kipsBCF T= Free body: Joint B: 4 4 0: (7.50 kips) 4.00 kips 0 5 5 y BEF FΣ = − + − = 2.50 kipsBEF T= 8 3 0: (7.50 kips) (2.50 kips) 0 5 5 x BDF FΣ = + + = 6.00 kipsBDF = − 6.00 kipsBDF C= www.elsolucionario.net www.elsolucionario.net
- 781. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 779 PROBLEM 6.21 (Continued) Free body: Joint D: We note that DE is a zero-force member: 0DEF = Also, 6.00 kipsDFF C= From symmetry: FE BEF F= 2.50 kipsEFF T= EG CEF F= 4.50 kipsEGF T= FG BCF F= 4.00 kipsFGF T= FH ABF F= 7.50 kipsFHF C= GH ACF F= 4.50 kipsGHF T= www.elsolucionario.net www.elsolucionario.net
- 782. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 780 PROBLEM 6.22 Solve Problem 6.21 assuming that the load applied at G has been removed. PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xFΣ = =A 0: (36 ft) (4 kips)(9 ft) (4 kips)(18 ft) 0A HΣ = − − =M 3.00 kips=H 0: 5.00 kipsy yFΣ = +A We note that DE and FG are zero-force members. Therefore, 0,DEF = 0FGF = Also, BD DFF F= (1) and EG GHF F= (2) Free body: Joint A: 5 kips 5 3 4 ACAB FF = = 6.25 kipsABF C= 3.75 kipsACF T= Free body: Joint C: 0:xFΣ = 3.75 kipsCEF T= 0:yFΣ = 4.00 kipsBCF T= Free body: Joint B: 4 4 0: (6.25 kips) 4.00 kips 0 5 5 x BEF FΣ = − − = 1.250 kipsBEF T= 3 3 0: (6.25 kips) (1.250 kips) 0 5 5 x BDF FΣ = + + = 4.50 kipsBDF = − 4.50 kipsBDF C= www.elsolucionario.net www.elsolucionario.net
- 783. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 781 PROBLEM 6.22 (Continued) Free body: Joint F: We recall that 0,FGF = and from Eq. (1) that DF BDF F= 4.50 kipsDFF C= 4.50 kips 5 5 6 EF FHF F = = 3.75 kipsEFF T= 3.75 kipsFHF C= Free body: Joint H: 3.00 kips 3 4 GHF = 2.25 kipsGHF T= Also, 3.00 kips 5 4 FHF = 3.75 kips (checks)FHF C= From Eq. (2): EG GHF F= 2.25 kipsEGF T= www.elsolucionario.net www.elsolucionario.net
- 784. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 782 PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Free body: Joint A: 1.2 kN 2.29 2.29 1.2 ACAB FF = = 2.29 kNABF T= 2.29 kNACF C= Free body: Joint F: 1.2 kN 2.29 2.29 2.1 = =DF EFF F 2.29 kNDFF T= 2.29 kNEFF C= Free body: Joint D: 2.29 kN 2.21 0.6 2.29 BD DEF F = = 2.21 kNBDF T= 0.600 kNDEF C= Free body: Joint B: 4 2.21 0: 2.21 kN (2.29 kN) 0 5 2.29 x BEF FΣ = + − = 0BEF = 3 0.6 0: (0) (2.29 kN) 0 5 2.29 y BCF FΣ = − − − = 0.600 kNBCF = − 0.600 kNBCF C= www.elsolucionario.net www.elsolucionario.net
- 785. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 783 PROBLEM 6.23 (Continued) Free body: Joint C: 2.21 0: (2.29 kN) 0 2.29 x CEF FΣ = + = 2.21kNCEF = − 2.21 kNCEF C= 0.6 0: 0.600 kN (2.29 kN) 0 2.29 y CHF FΣ = − − − = 1.200 kNCHF = − 1.200 kNCHF C= Free body: Joint E: 2.21 4 0: 2.21kN (2.29 kN) 0 2.29 5 Σ = − − =x EHF F 0EHF = 0.6 0: 0.600 kN (2.29 kN) 0 0 2.29 y EJF FΣ = − − − − = 1.200 kNEJF = − 1.200 kNEJF C= www.elsolucionario.net www.elsolucionario.net
- 786. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 784 PROBLEM 6.24 For the tower and loading of Problem 6.23 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Free body: Joint G: 1.2 kN 3.03 3.03 1.2 GH GIF F = = 3.03 kNGHF T= 3.03 kNGIF C= Free body: Joint L: 1.2 kN 3.03 3.03 1.2 = =JL KLF F 3.03 kNJLF T= 3.03 kNKLF C= Free body: Joint J: 2.97 0: (3.03 kN) 0 3.03 Σ = − + =x HJF F 2.97 kNHJF T= 0.6 0: 1.2 kN (3.03 kN) 0 3.03 y JKF F= − − − = 1.800 kNJKF = − 1.800 kNJKF C= Free body: Joint H: 4 2.97 0: 2.97 kN (3.03 kN) 0 5 3.03 x HKF FΣ = + − = 0HKF = 0.6 3 0: 1.2 kN (3.03) kN (0) 0 3.03 5 y HIF FΣ = − − − − = 1.800 kNHIF = − 1.800 kNHIF C= www.elsolucionario.net www.elsolucionario.net
- 787. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 785 PROBLEM 6.24 (Continued) Free body: Joint I: 2.97 0: (3.03 kN) 0 3.03 x IKF FΣ = + = 2.97 kNIKF = − 2.97 kNIKF C= 0.6 0: 1.800 kN (3.03 kN) 0 3.03 y INF FΣ = − − − = 2.40 kNINF = − 2.40 kNINF C= Free body: Joint K: 4 2.97 0: 2.97 kN (3.03 kN) 0 5 3.03 x KNF FΣ = − + − = 0KNF = 0.6 3 0: (3.03 kN) 1.800 kN (0) 0 3.03 5 y KOF FΣ = − − − − = 2.40 kNKOF = − 2.40 kNKOF C= www.elsolucionario.net www.elsolucionario.net
- 788. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 786 PROBLEM 6.25 Solve Problem 6.23 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Zero-Force Members: Considering joint F, we note that DF and EF are zero-force members: 0= =DF EFF F Considering next joint D, we note that BD and DE are zero-force members: 0BD DEF F= = Free body: Joint A: 1.2 kN 2.29 2.29 1.2 ACAB FF = = 2.29 kNABF T= 2.29 kNACF C= Free body: Joint B: 4 2.21 0: (2.29 kN) 0 5 2.29 Σ = − =x BEF F 2.7625 kNBEF = 2.76 kNBEF T= 0.6 3 0: (2.29 kN) (2.7625 kN) 0 2.29 5 y BCF FΣ = − − − = 2.2575 kNBCF = − 2.26 kNBCF C= www.elsolucionario.net www.elsolucionario.net
- 789. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 787 PROBLEM 6.25 (Continued) Free body: Joint C: 2.21 0: (2.29 kN) 0 2.29 Σ = + =x CEF F 2.21 kNCEF C= 0.6 0: 2.2575 kN (2.29 kN) 0 2.29 y CHF FΣ = − − − = 2.8575 kNCHF = − 2.86 kNCHF C= Free body: Joint E: 4 4 0: (2.7625 kN) 2.21 kN 0 5 5 x EHF FΣ = − − + = 0EHF = 3 3 0: (2.7625 kN) (0) 0 5 5 y EJF FΣ = − + − = 1.6575 kNEJF = + 1.658 kNEJF T= www.elsolucionario.net www.elsolucionario.net
- 790. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 788 PROBLEM 6.26 Determine the force in each of the members connecting joints A through F of the vaulted roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: 0x xFΣ = =A 0: (1.2 kN)6 (2.4 kN)5 (2.4 kN)4 (1.2 kN)3KM a a a aΣ = + + + (6 ) 0 5.40 kNy yA a− = =A Free body: Joint A: 4.20 kN 2 15 9.3915 kN ACAB AB FF F = = = 9.39 kNABF C= 8.40 kNACF T= Free body: Joint B: 2 1 2 0: (9.3915) 0 5 2 5 x BD BCF F FΣ = + + = (1) 1 1 1 0: (9.3915) 2.4 0 5 2 5 y BD BCF F FΣ = − + − = (2) Add Eqs. (1) and (2): 3 3 (9.3915 kN) 2.4 kN 0 5 5 BDF + − = 7.6026 kNBDF = − 7.60 kNBDF C= Multiply Eq. (2) by –2 and add Eq. (1): 3 4.8 kN 0 2 BF + = 2.2627 kNBCF = − 2.26 kNBCF C= www.elsolucionario.net www.elsolucionario.net
- 791. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 789 PROBLEM 6.26 (Continued) Free body: Joint C: 1 4 1 0: (2.2627) 8.40 0 5 17 2 x CD CEF F FΣ = + + − = (3) 2 1 1 0: (2.2627) 0 5 17 2 y CD CEF F FΣ = + − = (4) Multiply Eq. (4) by −4 and add Eq. (1): 7 5 (2.2627) 8.40 0 5 2 CDF− + − = 0.1278 kNCDF = − 0.128 kNCDF C= Multiply Eq. (1) by 2 and subtract Eq. (2): 7 3 (2.2627) 2(8.40) 0 17 2 + − =CEF 7.068 kNCEF = 7.07 kNCEF T= Free body: Joint D: 2 1 2 0: (7.6026) 1.5245 5 x DF DEF F FΣ = + + 1 (0.1278) 0 5 + = (5) 1 1.15 1 0: (7.6026) 1.5245 5 y DF DEF F FΣ = − + 2 (0.1278) 2.4 0 5 + − = (6) Multiply Eq. (5) by 1.15 and add Eq. (6): 3.30 3.30 3.15 (7.6026) (0.1278) 2.4 0 5 5 5 DFF + + − = 6.098 kNDFF = − 6.10 kNDFF C= Multiply Eq. (6) by –2 and add Eq. (5): 3.30 3 (0.1278) 4.8 0 1.524 5 DEF − + = 2.138 kNDEF = − 2.14 kNDEF C= www.elsolucionario.net www.elsolucionario.net
- 792. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 790 PROBLEM 6.26 (Continued) Free body: Joint E: 0.6 4 1 0: ( ) (2.138) 0 2.04 1.52417 x EF EH CEF F F FΣ = + − + = (7) 1.95 1 1.15 0: ( ) (2.138) 0 2.04 1.52417 y EF EH CEF F F FΣ = + − − = (8) Multiply Eq. (8) by 4 and subtract Eq. (7): 7.2 7.856 kN 0 2.04 EFF − = 2.23 kNEFF T= www.elsolucionario.net www.elsolucionario.net
- 793. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 791 PROBLEM 6.27 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: 0: (20 ft) (15 kips)(16 ft) (40 kips)(15 ft) 0FM GΣ = − − = 42 kips=G 0: 15 kips 0x xF FΣ = + = 15 kipsx =F 0: 40 kips 42 kips 0y yF FΣ = − + = 2 kipsy =F Free body: Joint F: 1 0: 15 kips 0 5 x DFF FΣ = − = 33.54 kipsDFF = 33.5 kipsDFF T= 2 0: 2 kips (33.54 kips) 0 5 y BFF FΣ = − + = 28.00 kipsBFF = − 28.0 kipsBFF C= Free body: Joint B: 5 5 0: 15 kips 0 29 61 x AB BDF F FΣ = + + = (1) 2 6 0: 28 kips 0 29 61 y AB BDF F FΣ = − + = (2) www.elsolucionario.net www.elsolucionario.net
- 794. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 792 PROBLEM 6.27 (Continued) Multiply Eq. (1) by 6, Eq. (2) by 5, and add: 40 230 kips 0 29 30.96 kips AB AB F F + = = − 31.0 kipsABF C= Multiply Eq. (1) by 2, Eq. (2) by –5, and add: 40 110 kips 0 61 21.48 kips BD BD F F − = = 21.5 kipsBDF T= Free body: Joint D: 2 2 6 0: (33.54) (21.48) 0 5 5 61 y ADF FΣ = − + = 15.09 kipsADF T= 1 5 0: (15.09 33.54) (21.48) 0 5 61 x DEF FΣ = + − − = 22.0 kipsDEF T= Free body: Joint A: 5 1 5 1 0: (30.36) (15.09) 0 29 5 29 5 x AC AEF F FΣ = + + − = (3) 2 2 2 2 0: (30.96) (15.09) 0 29 5 29 5 y AC AEF F FΣ = − − + − = (4) Multiply Eq. (3) by 2 and add Eq. (4): 8 12 4 (30.96) (15.09) 0 29 29 5 ACF + − = 28.27 kips,ACF = − 28.3 kipsACF C= Multiply Eq. (3) by 2, Eq. (4) by 5, and add: 8 20 12 (30.96) (15.09) 0 5 29 5 AEF− + − = 9.50 kipsAEF T= www.elsolucionario.net www.elsolucionario.net
- 795. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 793 PROBLEM 6.27 (Continued) Free body: Joint C: From force triangle: 28.27 kips 861 29 CE CGF F = = 41.0 kipsCEF T= 42.0 kipsCGF C= Free body: Joint G: 0:xFΣ = 0EGF = 0: 42 kips 42 kips 0 (Checks)yFΣ = − = www.elsolucionario.net www.elsolucionario.net
- 796. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 794 PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: 0:xFΣ = 0x =E 0:FMΣ = 45 kipsy =E 0:yFΣ = 60 kips=F Joint D: 15 kips 12 13 5 CD DHF F = = 36.0 kipsCDF T= 39.0 kipsDHF C= Joint H: 0:FΣ = 0CHF = 0:FΣ = 39.0 kipsGHF C= Joint C: 0:FΣ = 0CGF = 0:FΣ = 36.0 kipsBCF T= Joint G: 0:FΣ = 0BGF = 0:FΣ = 39.0 kipsFGF C= Joint B: 0:FΣ = 0BFF = 0:FΣ = 36.0 kipsABF T= www.elsolucionario.net www.elsolucionario.net
- 797. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 795 PROBLEM 6.28 (Continued) Joint A: 2 2 12 15 19.21 ftAE = + = 36 kips tan38.7 AFF ° = 45.0 kipsAFF C= 36 kips sin38.7 AEF ° = 57.6 kipsAEF T= Joint E: 0: (57.6 kips)sin38.7 0x EFF FΣ = + ° + = 36.0 kipsEFF = − 36.0 kipsEFF C= 0: (57.6 kips)cos38.7 45 kips 0yFΣ = ° − = (Checks) www.elsolucionario.net www.elsolucionario.net
- 798. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 796 PROBLEM 6.29 Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a are simple trusses. SOLUTION Truss of Problem 6.31a: Starting with triangle HDI and adding two members at a time, we obtain successively joints A, E, J, and B, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.32a: Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33a: Starting with triangle ABD and adding two members at a time, we obtain successively joints H, G, F, E, I, C, and J, thus completing the truss. Therefore, this is a simple truss. www.elsolucionario.net www.elsolucionario.net
- 799. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 797 PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses. SOLUTION Truss of Problem 6.31b: Starting with triangle CGM and adding two members at a time, we obtain successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss. Truss of Problem 6.32b: Starting with triangle ABC and adding two members at a time, we obtain successively joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33b: Starting with triangle GFH and adding two members at a time, we obtain successively joints D, E, C, A, and B, thus completing the truss. Therefore, this is a simple truss. www.elsolucionario.net www.elsolucionario.net
- 800. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 798 PROBLEM 6.31 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): : Joint : 0=BJFB B F : Joint : 0=DIFB D F : Joint : 0=EIFB E F : Joint : 0=AIFB I F : Joint : 0=FKFB F F : Joint : 0=GKFB G F : Joint : 0=CKFB K F The zero-force members, therefore, are , , , , , ,AI BJ CK DI EI FK GK Truss (b): : Joint : 0=FKFB K F : Joint :FB O 0IOF = The zero-force members, therefore, are andFK IO All other members are either in tension or compression. (b) (a) www.elsolucionario.net www.elsolucionario.net
- 801. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 799 PROBLEM 6.32 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): : Joint : 0=BCFB B F : Joint : 0=CDFB C F : Joint : 0=IJFB J F : Joint : 0=ILFB I F : Joint : 0=MNFB N F : Joint : 0=LMFB M F The zero-force members, therefore, are , , , , ,BC CD IJ IL LM MN Truss (b): : Joint : 0=BCFB C F : Joint : 0=BEFB B F : Joint : 0=FGFB G F : Joint : 0=EFFB F F : Joint : 0=DEFB E F : Joint : 0=IJFB I F : Joint : 0=MNFB M F : Joint : 0=KNFB N F The zero-force members, therefore, are , , , , , , ,BC BE DE EF FG IJ KN MN (a) (b) www.elsolucionario.net www.elsolucionario.net
- 802. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 800 PROBLEM 6.33 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): Note: Reaction at F is vertical ( 0).xF = Joint :G 0,FΣ = 0DGF = Joint :D 0,FΣ = 0DBF = Joint :F 0,FΣ = 0FGF = Joint :G 0,FΣ = 0GHF = Joint :J 0,FΣ = 0IJF = Joint :I 0,FΣ = 0HIF = www.elsolucionario.net www.elsolucionario.net
- 803. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 801 PROBLEM 6.33 (Continued) Truss (b): Joint :A 0,FΣ = 0ACF = Joint :C 0,FΣ = 0CEF = Joint :E 0,FΣ = 0EFF = Joint :F 0,FΣ = 0FGF = Joint :G 0,FΣ = 0GHF = www.elsolucionario.net www.elsolucionario.net
- 804. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 802 PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.26, (b) Problem 6.28. SOLUTION (a) Truss of Problem 6.26: : Joint : 0IJFB I F = : Joint : 0GJFB J F = : Joint : 0GHFB G F = The zero-force members, therefore, are , ,GH GJ IJ (b) Truss of Problem 6.28: : Joint : 0BFFB B F = : Joint : 0BGFB B F = : Joint : 0CGFB C F = : Joint : 0CHFB C F = The zero-force members, therefore, are , , ,BF BG CG CH www.elsolucionario.net www.elsolucionario.net
- 805. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 803 PROBLEM 6.35* The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Determine the force in each of the members for the given loading. SOLUTION Free body: Truss: From symmetry: and= =x x y yD B D B 0: (10 ft) (400 lb)(24 ft) 0zM AΣ = − − = 960 lbA = − 0: 0x x xF B D AΣ = + + = 2 960 lb 0, 480 lbx xB B− = = 0: 400 lb 0y y yF B DΣ = + − = 2 400 lb 200 lb y y B B = = + Thus, (480 lb) (200 lb)= +B i j Free body: C: ( 24 10 ) 26 ( 24 7 ) 25 ( 24 7 ) 25 AC CA AC BC CB BC CD CD CD FCA F F CA FCB F F CB FCD F F CD = = − + = = − + = = − − i j i k i k 0: (400 lb) 0CA CB CDΣ = + + − =F F F F j www.elsolucionario.net www.elsolucionario.net
- 806. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 804 PROBLEM 6.35* (Continued) Substituting for , , ,CA CB CDF F F and equating to zero the coefficients of , , :i j k i: 24 24 ( ) 0 26 25 − − + =AC BC CDF F F (1) j: 10 400 lb 0 26 ACF − = 1040 lbACF T= k: 7 ( ) 0 25 BC CD CD BCF F F F− = = Substitute for ACF and CDF in Eq. (1): 24 24 (10.40 lb) (2 ) 0 500lb 26 25 BC BCF F− − = = − 500 lbBC CDF F C= = Free body: B: (500 lb) (480 lb) (140 lb) (10 7 ) 12.21 BC AB BA AB BD BD CB F CB FBA F F BA F F = = − + = = − = − i k j k k 0: (480 lb) (200 lb) 0BA BD BCΣ = + + + + =F F F F i j Substituting for , ,BA BD BCF F F and equating to zero the coefficients of j and k: j: 10 200 lb 0 244.2 lb 12.21 + = = −AB ABF F 244 lbABF C= k: 7 140 lb 0 12.21 AB BDF F− − + = 7 ( 244.2 lb) 140 lb 280 lb 12.21 BDF = − − + = + 280 lbBDF T= From symmetry: AD ABF F= 244 lbADF C= www.elsolucionario.net www.elsolucionario.net
- 807. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 805 PROBLEM 6.36* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = (−2184 N)j and Q = 0. SOLUTION Free body: Truss: From symmetry: andx x y yD B D B= = 0: 2 0x xF BΣ = = 0x xB D= = 0: 0z zF BΣ = = 0: 2 (2.8 m) (2184 N)(2 m) 0c z yM BΣ = − + = 780 NyB = Thus, (780 N)=B j Free body: A: ( 0.8 4.8 2.1 ) 5.30 (2 4.8 ) 5.20 ( 0.8 4.8 2.1 ) 5.30 AB AB AB AC AC AC AD AD AD FAB F F AB FAC F F AC FAD F F AD = = − − + = = − = = + − − i j k i j i j k 0: (2184 N) 0AB AC ADΣ = + + − =F F F F j www.elsolucionario.net www.elsolucionario.net
- 808. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 806 PROBLEM 6.36* (Continued) Substituting for , , ,AB AC ADF F F and equating to zero the coefficients of , , :i j k i: 0.8 2 ( ) 0 5.30 5.20 AB AD ACF F F− + + = (1) j: 4.8 4.8 ( ) 2184 N 0 5.30 5.20 AB AD ACF F F− + − − = (2) k: 2.1 ( ) 0 5.30 AB ADF F− = AD ABF F= Multiply Eq. (1) by –6 and add Eq. (2): 16.8 2184 N 0, 676 N 5.20 AC ACF F − − = = − 676 NACF C= Substitute for ACF and ADF in Eq. (1): 0.8 2 2 ( 676 N) 0, 861.25 N 5.30 5.20 − + − = = − AB ABF F 861 NAB ADF F C= = Free body: B: (861.25 N) (130 N) (780 N) (341.25 N) 2.8 2.1 (0.8 0.6 ) 3.5 AB BC BC BC BD BD AB AB F F F = = − − + − = = − = − F i j k i k F i k F k 0: (780 N) 0AB BC BDΣ = + + + =F F F F j Substituting for , ,AB BC BDF F F and equating to zero the coefficients of i and ,k i: 130 N 0.8 0 162.5 NBC BCF F− + = = + 162.5 NBCF T= k: 341.25 N 0.6 0BC BDF F− − = 341.25 0.6(162.5) 243.75 N= − = +BDF 244 NBDF T= From symmetry: CD BCF F= 162.5 NCDF T= www.elsolucionario.net www.elsolucionario.net
- 809. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 807 PROBLEM 6.37* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = 0 and Q = (2968 N)i. SOLUTION Free body: Truss: From symmetry: andx x y yD B D B= = 0: 2 2968 N 0x xF BΣ = + = 1484 Nx xB D= = − 0: 2 (2.8 m) (2968 N)(4.8 m) 0cz yM B′Σ = − − = 2544 NyB = − Thus, (1484 N) (2544 N)= − −B i j Free body: A: ( 0.8 4.8 2.1 ) 5.30 (2 4.8 ) 5.20 ( 0.8 4.8 2.1 ) 5.30 AB AB AB AC AC AC AD AD AD AB F F AB F FAC F F AC AD F F AD F = = − − + = = − = = − − − i j k i j i j k 0: (2968 N) 0AB AC ADΣ = + + + =F F F F i www.elsolucionario.net www.elsolucionario.net
- 810. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 808 PROBLEM 6.37* (Continued) Substituting for , , ,AB AC ADF F F and equating to zero the coefficients of , , ,i j k i: 0.8 2 ( ) 2968 N 0 5.30 5.20 AB AD ACF F F− + + + = (1) j: 4.8 4.8 ( ) 0 5.30 5.20 AB AD ACF F F− + − = (2) k: 2.1 ( ) 0 5.30 AB ADF F− = AD ABF F= Multiply Eq. (1) by –6 and add Eq. (2): 16.8 6(2968 N) 0, 5512 N 5.20 AC ACF F − − = = − 5510 NACF C= Substitute for ACF and ADF in Eq. (2): 4.8 4.8 2 ( 5512 N) 0, 2809 N 5.30 5.20 AB ABF F − − − = = + 2810 NAB ADF F T= = Free body: B: (2809 N) (424 N) (2544 N) (1113 N) 2.8 2.1 (0.8 0.6 ) 3.5 AB BC BC BC BD BD BA BA F F F = = + − − = = − = − F i j k i k F i k F k 0: (1484 N) (2544 N) 0AB BC BDΣ = + + − − =F F F F i j Substituting for , ,AB BC BDF F F and equating to zero the coefficients of i and ,k i: 24 N 0.8 1484 N 0, 1325 NBC BCF F+ + − = = + 1325 NBCF T= k: 1113 N 0.6 0BC BDF F− − − = 1113 N 0.6(1325 N) 1908 N,BDF = − − = − 1908 NBDF C= From symmetry: CD BCF F= 1325 NCDF T= www.elsolucionario.net www.elsolucionario.net
- 811. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 809 PROBLEM 6.38* The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading. SOLUTION Free body: Truss: From symmetry: 0z zA B= = 0: 0x xF AΣ = = 0: (6 ft) (1600 lb)(7.5 ft) 0BC yM AΣ = + = 2000 lbyA = − (2000 lb)= −A j From symmetry: yB C= 0: 2 2000 lb 1600 lb 0y yF BΣ = − − = 1800 lbyB = (1800 lb)=B j Free body: A: 0: (2000 lb) 0AB AC ADΣ = + + − =F F F F j (0.6 0.8 ) (2000 lb) 0 2 2 + − + + + − = i k i k i j jAB AC ADF F F Factoring i, j, k and equating their coefficient to zero, 1 1 0.6 0 2 2 AB AC ADF F F+ + = (1) 0.8 2000 lb 0ADF − = 2500 lbADF T= 1 1 0 2 2 AB ACF F− = AC ABF F= www.elsolucionario.net www.elsolucionario.net
- 812. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 810 PROBLEM 6.38* (Continued) Substitute for ADF and ACF into Eq. (1): 2 0.6(2500 lb) 0, 1060.7 lb, 2 AB ABF + = = −F 1061lbAB ACF F C= = Free body: B: (1060.7 lb) (750 lb)( ) 2 (0.8 0.6 ) (7.5 8 6 ) 12.5 BA AB BC BC BD BD BE BE BE BA F BA F F FBE F BE + = = + = + = − = − = = + − i k F i k F k F j k F i j k 0: (1800 lb) 0BA BC BD BEΣ = + + + + =F F F F F j Substituting for , , , andBA BC BD BEF F F F and equating to zero the coefficients of , , ,i j k i: 7.5 750 lb 0, 1250 lb 12.5 BE BEF F + = = − 1250 lbBEF C= j: 8 0.8 ( 1250 lb) 1800 lb 0 12.5 BDF + − + = 1250 lbBDF C= k: 6 750 lb 0.6( 1250 lb) ( 1250 lb) 0 12.5 BCF− − − − − = 2100 lbBCF T= From symmetry: 1250 lbBD CDF F C= = Free body: D: 0: 0DA DB DC DEΣ = + + + =F F F F F i We now substitute for , ,DA DB DCF F F and equate to zero the coefficient of i. Only DAF contains i and its coefficient is 0.6 0.6(2500 lb) 1500 lbADF− = − = − i: 1500 lb 0DEF− + = 1500 lbDEF T= www.elsolucionario.net www.elsolucionario.net
- 813. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 811 PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0. SOLUTION Free body: Truss: 0: 1.8 (1.8 3 ) ( )B yC D DΣ = × + − × +M i j i k j k (0.6 0.75 ) ( 1200 ) 0+ − × − =i k j 1.8 1.8 1.8y zC D D− + −k k j 3 720 900 0yD+ − − =i k i Equate to zero the coefficients of , , :i j k i: 3 900 0, 300 Ny yD D− = = j: 0,=zD (300 N)=D j k: 1.8 1.8(300) 720 0C + − = (100 N)=C j 0: 300 100 1200 0Σ = + + − =F B j j j (800 N)=B j Free body: B: 0: (800 N) 0,BA BC BEΣ = + + + =F F F F j with (0.6 3 0.75 ) 3.15 AB BA AB FBA BA = = + −F F i j k BC BCF=F i BE BEF= −F k Substitute and equate to zero the coefficient of , , :j i k j: 3 800 N 0, 840 N, 3.315 AB ABF F + = = − 840 NABF C= i: 0.6 ( 840 N) 0 3.15 BCF − + = 160.0N=BCF T k: 0.75 ( 840 N) 0 3.15 BEF − − =− 200 NBEF T= www.elsolucionario.net www.elsolucionario.net
- 814. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 812 PROBLEM 6.39* (Continued) Free body: C: 0: (100 N) 0,CA CB CD CEΣ = + + + + =F F F F F j with ( 1.2 3 0.75 ) 3.317 AC CA AC FCA F CA = = − + −F i j k (160N)CBF = − i ( 1.8 3 ) 3,499 CE CD CD CE CE FCE F CE = − = = − −F k F F i k Substitute and equate to zero the coefficient of , , :j i k j: 3 100 N 0, 110.57 N 3.317 AC ACF F + = = − 110.6 NACF C= i: 1.2 1.8 ( 110.57) 160 0, 233.3 3.317 3.499 CE CEF F− − − − = = − 233 NCEF C= k: 0.75 3 ( 110.57) ( 233.3) 0 3.317 3.499 CDF− − − − − = 225 NCDF T= Free body: D: 0: (300 N) 0,DA DC DEΣ = + + + =F F F F j with ( 1.2 3 2.25 ) 3.937 AD DA AD FDA F DA = = − + +F i j k (225 N)DC CDF= =F k k DE DEF F= − i Substitute and equate to zero the coefficient of , , :j i k j: 3 300 N 0, 3.937 ADF + = 393.7 N,ADF = − 394 NADF C= i: 1.2 ( 393.7 N) 0 3.937 DEF − − =− 120.0 NDEF T= k: 2.25 ( 393.7 N) 225 N 0 3.937 − + = (Checks) Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. 0AEF = www.elsolucionario.net www.elsolucionario.net
- 815. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 813 PROBLEM 6.40* Solve Problem 6.39 for P = 0 and Q = (−900 N)k. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0. SOLUTION Free body: Truss: 0: 1.8 (1.8 3 ) ( )B y zC D DΣ = × + − × +M i j i k j k (0.6 3 0.75 ) ( 900N) 0+ + − × − =i j k k 1.8 1.8 1.8y zC D D+ −k k j 3 540 2700 0yD+ + − =i j i Equate to zero the coefficient of , , :i j k 3 2700 0 900 Ny yD D− = = 1.8 540 0 300 Nz zD D− + = = 1.8 1.8 0 900 Ny yC D C D+ = = − = − Thus, (900 N) (900 N) (300 N)= − = +C j D j k 0: 900 900 300 900 0Σ = − + + − =F B j j k k (600 N)=B k Free body: B: SinceB is aligned with member BE, 0,AB BCF F= = 600 NBEF T= Free body: C: 0: (900 N) 0,CA CD CEΣ = + + − =F F F F j with www.elsolucionario.net www.elsolucionario.net
- 816. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 814 PROBLEM 6.40* (Continued) ( 1.2 3 0.75 ) 3.317 AC CA AC FCA F CA = = − + −F i j k ( 1.8 3 ) 3.499 CE CD CD CE CE FCE F F CE = − = = − −F k F i k Substitute and equate to zero the coefficient of , , :j i k j: 3 900 N 0, 3.317 ACF − = 995.1 NACF = 995 NACF T= i: 1.2 1.8 (995.1) 0, 699.8 N 3.317 3.499 CE CEF F− − = = − 700 NCEF C= k: 0.75 3 (995.1) ( 699.8) 0 3.317 3.499 CDF− − − − = 375 NCDF T= Free body: D: 0: (375N) +(900 N) (300 N) 0DA DEΣ = + + + =F F F k j k with ( 1.2 3 2.25 ) 3.937 AD DA AD FDA DA = = − + +F F i j k and DE DEF F= − i Substitute and equate to zero the coefficient , , :j i k j: 3 900 N 0, 1181.1 N 3.937 AD ADF F + = = − 1181 NADF C= i: 1.2 ( 1181.1 N) 0 3.937 DEF − − − = 360 NDEF T= k: 2.25 ( 1181.1 N 375 N 300 N 0) 3.937 − + + = (Checks) Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. 0AEF = www.elsolucionario.net www.elsolucionario.net
- 817. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 815 PROBLEM 6.41* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E. SOLUTION (a) Check simple truss. (1) Start with tetrahedron BEFG. (2) Add members BD, ED, GD joining at D. (3) Add members BA, DA, EA joining at A. (4) Add members DH, EH, GH joining at H. (5) Add members BC, DC, GC joining at C. Truss has been completed: It is a simple truss. Free body: Truss: Check constraints and reactions. Six unknown reactions—ok; however, supports at A and B constrain truss to rotate about AB and support at G prevents such a rotation. Thus, Truss is completely constrained and reactions are statically determinate. Determination of reactions: 0: 11 ( ) (11 9.6 ) (10.08 9.6 ) (275 240 ) 0 A y zB B GΣ = × + + − × + − × + = M i j k i k j j k i k 11 11 11 9.6 (10.08)(275) (10.08)(240) (9.6)(275) 0 y yB B G G− + + − + − = k j k i k i j Equate to zero the coefficient of i, j, k: : 9.6 (10.08)(240) 0 252 lb+ = = −i G G ( 252 lb)= −G j : 11 (9.6)(275) 0 240 lb− − = = −j z zB B : 11 11( 252) (10.08)(275) 0, 504 lby yB B+ − − = =k (504 lb) (240 lb)= −B j k www.elsolucionario.net www.elsolucionario.net
- 818. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 816 PROBLEM 6.41* (Continued) 0: (504 lb) (240 lb) (252 lb) (275 lb) (240 lb) 0 Σ = + − − + + = F A j k j i k (275 lb) (252 lb)= − −A i j Zero-force members. The determination of these members will facilitate our solution. FB: C: Writing 0, 0, 0x y zF F FΣ = Σ = Σ = yields 0BC CD CGF F F= = = FB: F: Writing 0, 0, 0x y zF F FΣ = Σ = Σ = yields 0BF EF FGF F F= = = FB: A: Since 0,zA = writing 0zFΣ = yields 0ADF = FB: H: Writing 0yFΣ = yields 0DHF = FB: D: Since 0,AD CD DHF F F= = = we need consider only members DB, DE, and DG. Since DEF is the only force not contained in plane BDG, it must be zero. Simple reasonings show that the other two forces are also zero. 0BD DE DGF F F= = = The results obtained for the reactions at the supports and for the zero-force members are shown on the figure below. Zero-force members are indicated by a zero (“0”). (b) Force in each of the members joined at E. We already found that 0= =DE EFF F Free body: A: 0yFΣ = yields 252 lbAEF T= Free body: H: 0zFΣ = yields 240 lbEHF C= www.elsolucionario.net www.elsolucionario.net
- 819. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 817 PROBLEM 6.41* (Continued) Free body: E: 0: (240 lb) (252 lb) 0EB EGΣ = + + − =F F F k j (11 10.08 ) (11 9.6 ) 240 252 0 14.92 14.6 EGBE FF − + − + − =i j i k k j Equate to zero the coefficient of y and k: 10.08 : 252 0 14.92 BEF − − = j 373 lbBEF C= :k 9.6 240 0 14.6 EGF − + = 365 lbEGF T= www.elsolucionario.net www.elsolucionario.net
- 820. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 818 PROBLEM 6.42* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G. SOLUTION See solution to Problem 6.41 for part (a) and for reactions and zero-force members. (b) Force in each of the members joined at G. We already know that 0CG DG FGF F F= = = Free body: H: 0xFΣ = yields 275 lbGHF C= Free body: G: 0: (275 lb) (252 lb) 0GB GEΣ = + + − =F F F i j ( 10.08 9.6 ) ( 11 9.6 ) 275 252 0 13.92 14.6 − + + − + + − =BG EGF F j k i k i j Equate to zero the coefficient of i, j, k: 11 : 275 0 14.6 EGF − + = i 365 lbEGF T= 10.08 : 252 0 13.92 BGF − − = j 348 lbBGF C= 9.6 9.6 : ( 348) (365) 0 13.92 14.6 − + = k (Checks) www.elsolucionario.net www.elsolucionario.net
- 821. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 819 PROBLEM 6.43 Determine the force in members CD and DF of the truss shown. SOLUTION Reactions: 0: (12 kN)(4.8 m) (12 kN)(2.4 m) (9.6 m) 0JM BΣ = + − = 9.00 kNB = 0: 9.00 kN 12.00 kN 12.00 kN 0yF JΣ = − − + = 15.00 kNJ = Member CD: 0: 9.00 kN 0y CDF FΣ = + = 9.00 kNCDF C= Member DF: 0: (1.8 m) (9.00 kN)(2.4 m) 0C DFM FΣ = − = 12.00 kNDFF T= www.elsolucionario.net www.elsolucionario.net
- 822. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 820 PROBLEM 6.44 Determine the force in members FG and FH of the truss shown. SOLUTION Reactions: 0: (12 kN)(4.8 m) (12 kN)(2.4 m) (9.6 m) 0JM BΣ = + − = 9.00 kNB = 0: 9.00 kN 12.00 kN 12.00 kN 0yF JΣ = − − + = 15.00 kNJ = Member FG: 3 0: 12.00 kN 15.00 kN 0 5 y FGF FΣ = − − + = 5.00 kNFGF T= Member FH: 0: (15.00 kN)(2.4 m) (1.8 m) 0G FHM FΣ = − = 20.0 kNFHF T= www.elsolucionario.net www.elsolucionario.net
- 823. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 821 PROBLEM 6.45 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF. SOLUTION Free body: Truss: 0: 0x xFΣ = =k 0: (62.5 ft) (6000 lb)(12.5 ft) (6000 lb)(25 ft) 0 A yM kΣ = − − = 3600 lby= =k k 0: 3600 lb 6000 lb 6000 lb 0yF AΣ = + − − = 8400 lb=A We pass a section through members CE, DE, and DF and use the free body shown. 0: (15 ft) (8400 lb)(18.75 ft) (6000 lb)(6.25 ft) 0 D CEM FΣ = − + = 8000 lbCEF = + 8000 lbCEF T= 15 0: 8400 lb 6000 lb 0 16.25 y DEF FΣ = − − = 2600 lbDEF = + 2600 lbDEF T= 0: 6000 lb(12.5 ft) (8400 lb)(25 ft) (15 ft) 0 E DF M F Σ = − − = 9000 lbDFF = − 9000 lbDFF C= www.elsolucionario.net www.elsolucionario.net
- 824. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 822 PROBLEM 6.46 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH. SOLUTION See solution of Problem 6.45 for free-body diagram of truss and determination of reactions: 8400 lb=A 3600 lb=k We pass a section through members EG, FG, and FH, and use the free body shown. 0: (3600 lb)(31.25 ft) (15 ft) 0F EGM FΣ = − = 7500 lbEGF = + 7500 lbEGF T= 15 0: 3600 lb 0 16.25 y FGF FΣ = + = 3900 lbFGF = − 3900 lbFGF C= 0: (15 ft) (3600 lb)(25 ft) 0Σ = + =G FHM F 6000 lbFHF = − 6000 lbFHF C= www.elsolucionario.net www.elsolucionario.net
- 825. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 823 PROBLEM 6.47 Determine the force in members DF, EF, and EG of the truss shown. SOLUTION 3 tan 4 β = Reactions: 0= =A N Member DF: 3 0: (16 kN)(6 m) (4 m) 0 5 E DFM FΣ = + − = 40 kNDFF = + 40.0 kNDFF T= Member EF: + 0: (16 kN)sin cos 0EFF Fβ βΣ = − = 16tan 16(0.75) 12 kNEFF β= = = 12.00 kNEFF T= Member EG: 4 0: (16 kN)(9 m) (3 m) 0 5 F EGM FΣ = + = 60 kNEGF = − 60.0 kNEGF C= www.elsolucionario.net www.elsolucionario.net
- 826. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 824 PROBLEM 6.48 Determine the force in members GI, GJ, and HI of the truss shown. SOLUTION Reactions: 0= =A N Member GI: + 0: (16 kN)sin sin 0GIF Fβ βΣ = + = 16 kNGIF = − 16.00 kNGIF C= Member GJ: 4 0: (16 kN)(9 m) (3 m) 0 5 I GJM FΣ = − − = 60 kNGJF = − 60.0 kNGJF C= Member HI: 3 0: (16 kN)(9 m) (4 m) 0 5 G HIM FΣ = − + = 60 kNHIF = + 60.0 kNHIF T= www.elsolucionario.net www.elsolucionario.net
- 827. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 825 PROBLEM 6.49 Determine the force in members AD, CD, and CE of the truss shown. SOLUTION Reactions: 0: 36(2.4) (13.5) 20(9) 20(4.5) 0kM BΣ = − + + = 26.4 kN=B 0: 36 0 36 kNx x xF KΣ = − + = =K 0: 26.4 20 20 0 13.6 kNy y yF KΣ = − − + = =K 0: 36(1.2) 26.4(2.25) (1.2) 0C ADM FΣ = − − = 13.5 kNADF = − 13.5 kNADF C= 8 0: (4.5) 0 17 A CDM F Σ = = 0CDF = 15 0: (2.4) 26.4(4.5) 0 17 D CEM F Σ = − = 56.1 kNCEF = + 56.1 kNCEF T= www.elsolucionario.net www.elsolucionario.net
- 828. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 826 PROBLEM 6.50 Determine the force in members DG, FG, and FH of the truss shown. SOLUTION See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports B and K. 26.4 kN=B ; 36.0 kNx =K ; 13.60 kNy =K 0: 36(1.2) 26.4(6.75) 20(2.25) (1.2) 0F DGM FΣ = − + − = 75 kNDGF = − 75.0 kNDGF C= 8 0: 26.4(4.5) (4.5) 0 17 D FGM F Σ = − + = 56.1 kNFGF = + 56.1 kNFGF T= 15 0: 20(4.5) 26.4(9) (2.4) 0 17 G FHM F Σ = − + = 69.7 kNFHF = + 69.7 kNFHF T= www.elsolucionario.net www.elsolucionario.net
- 829. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 827 PROBLEM 6.51 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG. SOLUTION We pass a section through members AB, AG, and FG, and use the free body shown. 40 0: (6.3 ft) (1.8 kips)(14 ft) 41 (0.9 kips)(28 ft) 0 G ABM F Σ = − − = 8.20 kipsABF = + 8.20 kipsABF T= 3 0: (28 ft) (1.8 kips)(28 ft) 5 (1.8 kips)(14 ft) 0 D AGM F Σ = − + + = 4.50 kipsAGF = + 4.50 kipsAGF T= 0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0 A FGM FΣ = − − − − = 11.60 kipsFGF = − 11.60 kipsFGF C= www.elsolucionario.net www.elsolucionario.net
- 830. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 828 PROBLEM 6.52 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ. SOLUTION We pass a section through members AE, EF, and FJ, and use the free body shown. 2 2 8 0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0 8 9 F AEM F Σ = − − − = + 17.46 kipsAEF = + 17.46 kipsAEF T= 0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0A EFM FΣ = − − − − = 11.60 kipsEFF = − 11.60 kipsEFF C= 0: (8 ft) (0.9 kips)(8 ft) (1.8 kips)(20 ft) (1.8 kips)(34 ft) (0.9 kips)(48 ft) 0E FJM FΣ = − − − − − = 18.45 kipsFJF = − 18.45 kipsFJF C= www.elsolucionario.net www.elsolucionario.net
- 831. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 829 PROBLEM 6.53 Determine the force in members CD and DF of the truss shown. SOLUTION 5 tan 22.62 12 α α= = ° 5 12 sin cos 13 13 α α= = Member CD: 0: (9 m) (10 kN)(9 m) (10 kN)(6 m) (10 kN)(3 m) 0I CDM FΣ = + + + = 20 kNCDF = − 20.0 kNCDF C= Member DF: 0: ( cos )(3.75 m) (10 kN)(3 m) (10 kN)(6 m) (10 kN)(9 m) 0C DFM F αΣ = + + + = cos 48 kNDFF α = − 12 48 kN 52.0 kN 13 DF DFF F = − = − 52.0 kNDFF C= www.elsolucionario.net www.elsolucionario.net
- 832. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 830 PROBLEM 6.54 Determine the force in members CE and EF of the truss shown. SOLUTION Member CE: 0: (2.5 m) (10 kN)(3 m) (10 kN)(6 m) 0F CEM FΣ = − − = 36 kNCEF = + 36.0 kNCEF T= Member EF: 0: (6 m) (10 kN)(6 m) (10 kN)(3 m) 0I EFM FΣ = + + = 15 kNEFF = − 15.00 kNEFF C= www.elsolucionario.net www.elsolucionario.net
- 833. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 831 PROBLEM 6.55 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH. SOLUTION Reactions at supports. Because of the symmetry of the loading, 1 0, 2 x yA A O= = = total load 4.48 kN= =A O We pass a section through members FG, EG, and EH, and use the free body shown. 1.75 m Slope Slope 6 m FG FI= = 5.50 m Slope 2.4 m EG = 0: (0.6 kN)(7.44 m) (1.24 kN)(3.84 m)EMΣ = + (4.48 kN)(7.44 m) 6 (4.80 m) 0 6.25 FGF − − = 5.231 kNFGF = − 5.23 kNFGF C= 0: (5.50 m) (0.6 kN)(9.84 m)G EHM FΣ = + (1.24 kN)(6.24 m) (1.04 kN)(2.4 m) (4.48 kN)(9.84 m) 0 + + − = 5.08 kNEHF T= 5.50 1.75 0: ( 5.231 kN) 4.48 kN 0.6 kN 1.24 kN 1.04 kN 0 6.001 6.25 y EGF FΣ = + − + − − − = 0.1476 kNEGF = − 0.1476 kNEGF C= www.elsolucionario.net www.elsolucionario.net
- 834. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 832 PROBLEM 6.56 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN. SOLUTION Because of symmetry of loading, 1 load 2 =O 4.48 kN=O We pass a section through KM, LM, LN, and use free body shown. 3.84 0: (3.68 m) 4 (4.48 kN 0.6 kN)(3.6 m) 0 M LNM F Σ = + − = 3.954 kNLNF = − 3.95 kNLNF C= 0: (4.80 m) (1.24 kN)(3.84 m) (4.48 kN 0.6 kN)(7.44 m) 0 L KMM FΣ = − − + − = 5.022 kNKMF = + 5.02 kNKMF T= 4.80 1.12 0: ( 3.954 kN) 1.24 kN 0.6 kN 4.48 kN 0 6.147 4 y LMF FΣ = + − − − + = 1.963 kNLMF = − 1.963 kNLMF C= www.elsolucionario.net www.elsolucionario.net
- 835. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 833 PROBLEM 6.57 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG. SOLUTION Free body: Truss: 0: 0x xF NΣ = = 0: (200 lb)(8 ) (400 lb)(7 6 5 )+(350 lb)(4 ) (300 lb)(3 2 ) (8 ) 0NM a a a a a a a a A aΣ = + + + + + + − = 1500 lb=A 0: 1500 lb 200 lb 3(400 lb) 350 lb 3(300 lb) 150 lb 0y yF NΣ = − − − − − + = 1300 lb 1300 lbyN = =N We pass a section through DF, EF, and EG, and use the free body shown. (We apply DFF at F.) 2 2 0: (200 lb)(18 ft) (400 lb)(12 ft) (400 lb)(6 ft) (1500 lb)(18 ft) 18 (4.5 ft) 0 18 4.5 E DF M F Σ = + + − − = + 3711 lb 3710 lbDF DFF F C= − = 0: (18 ft) (400 lb)(6 ft) (400 lb)(12 ft) 0A EFM FΣ = − − = 400 lbEFF = + 400 lbEFF T= 0: (4.5 ft) (1500 lb)(18 ft)+(200 lb)(18 ft) (400 lb)(12 ft) (400 lb)(6 ft) 0F EGM FΣ = − + + = 3600 lb 3600 lbEG EGF F T= + = www.elsolucionario.net www.elsolucionario.net
- 836. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 834 PROBLEM 6.58 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ. SOLUTION See solution of Problem 6.57 for reactions: 1500 lb=A , 1300 lb=N We pass a section through HI, GI, and GJ, and use the free body shown. (We apply HIF at H.) 2 2 6 0: (8.5 ft) (1300 lb)(24 ft) (300 lb)(6 ft) 6 4 (300 lb)(12 ft) (300 lb)(18 ft) (150 lb)(24 ft) 0 G HIM F Σ = + − + − − − = 2375.4 lb 2375 lbHI HIF F C= − = 0: (1300 lb)(18 ft) (300 lb)(6 ft) (300 lb)(12 ft) (150 lb)(18 ft) (4.5 ft) 0 I GJ M F Σ = − − − − = 3400 lb 3400 lbGJ GJF F T= + = 2 2 4 6 0: ( 2375.4 lb) 3400 lb 0 5 6 4 x GIF FΣ = − − − − = + 1779.4 lbGIF = − 1779 lbGIF C= www.elsolucionario.net www.elsolucionario.net
- 837. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 835 PROBLEM 6.59 Determine the force in members DE and DF of the truss shown when P = 20 kips. SOLUTION Reactions: 2.5P= =C K 7.5 tan 18 22.62 β β = = ° Member DE: 0: (2.5 )(6 ft) (12 ft) 0A DEM P FΣ = − = 1.25DEF P= + For 20 kips,P = 1.25(20) 25 kipsDEF = + = + 25.0 kipsDEF T= Member DF: 0: (12 ft) (2.5 )(6 ft) cos (5 ft) 0E DFM P P F βΣ = − − = 12 15 cos 22.62 (5 ft) 0DFP P F− − ° = 0.65DFF P= − For 20 kips,P = 0.65(20) 13 kipsDFF = − = − 13.00 kipsDFF C= www.elsolucionario.net www.elsolucionario.net
- 838. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 836 PROBLEM 6.60 Determine the force in members EG and EF of the truss shown when P = 20 kips. SOLUTION Reactions: 2.5 7.5 tan 6 51.34 P α α = = = = ° C K Member EG: 0: (18 ft) 2.5 (12 ft) (6 ft) (7.5 ft) 0F EGM P P P FΣ = − − + = 0.8 ;EGF P= + For 20 kips,P = 0.8(20) 16 kipsEGF = = + 16.00 kipsEGF T= Member EF: 0: 2.5 (6 ft) (12 ft) sin51.34 (12 ft) 0A EFM P P FΣ = − + ° = 0.320 ;EFF P= − For 20 kips,P = 0.320(20) 6.4 kipsEFF = − = − 6.40 kipsEFF C= www.elsolucionario.net www.elsolucionario.net
- 839. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 837 PROBLEM 6.61 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.) SOLUTION Reactions: 0: 0x xF AΣ = = 0: 12(45) 12(30) 12(15) (90) 0P yM AΣ = + + − = 12 kipsy =A 0: 12 12 12 12 0yF PΣ = − − − + = 24 kips=P 0: (12 kips)(30 ft) (16 ft) 0G EHM FΣ = − − = 22.5 kipsEHF = − 22.5 kipsEHF C= 0: 22.5 kips 0x GIF FΣ = − = 22.5 kipsGIF T= www.elsolucionario.net www.elsolucionario.net
- 840. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 838 PROBLEM 6.62 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.) SOLUTION See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A and P. 0; 12.00 kipsx y= =A A ; 24.0 kips=P 0: (16 ft) (12 kips)(15 ft) (24 kips)(30 ft) 0L HJM FΣ = − + = 33.75 kipsHJF = − 33.8 kipsHJF C= 0: 33.75 kips 0x ILF FΣ = − = 33.75 kipsILF = + 33.8 kipsILF T= www.elsolucionario.net www.elsolucionario.net
- 841. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 839 PROBLEM 6.63 Determine the force in members DG and FI of the truss shown. (Hint: Use section aa.) SOLUTION 0: (4 m) (5 kN)(3 m) 0F DGM FΣ = − = 3.75 kNDGF = + 3.75 kNDGF T= 0: 3.75 kN 0y FIF FΣ = − − = 3.75 kNFIF = − 3.75 kNFIF C= www.elsolucionario.net www.elsolucionario.net
- 842. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 840 PROBLEM 6.64 Determine the force in members GJ and IK of the truss shown. (Hint: Use section bb.) SOLUTION 0: (4 m) (5 kN)(6 m) (5 kN)(3 m) 0I GJM FΣ = − − = 11.25 kNGJF = + 11.25 kNGJF T= 0: 11.25 kN 0y IKF FΣ = − − = 11.25 kNIKF = − 11.25 kNIKF C= www.elsolucionario.net www.elsolucionario.net
- 843. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 841 PROBLEM 6.65 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading. SOLUTION Free body: Truss: 0: 0x xF FΣ = = 0: 4.8(3 ) 4.8(2 ) 4.8 2.4 (2 ) 0H yM a a a a F aΣ = + + − − = 13.20 kipsyF = + 13.20 kips=F 0: 13.20 kips 3(4.8 kips) 2(2.4 kips) 0yF HΣ = + − − = 6.00 kipsH = + 6.00 kips=H Free body: ABF: We assume that counter BG is acting. 9.6 0: 13.20 2(4.8) 0 14.6 y BGF FΣ = − + − = 5.475BGF = + 5.48 kipsBGF T= Since BG is in tension, our assumption was correct. Free body: DEH: We assume that counter DG is acting. 9.6 0: 6.00 2(2.4) 0 14.6 y DGF FΣ = − + − = 1.825DGF = + 1.825 kipsDGF T= Since DG is in tension, O.K. www.elsolucionario.net www.elsolucionario.net
- 844. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 842 PROBLEM 6.66 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading. SOLUTION Free body: Truss: 0: 0x xF FΣ = = 0: 4.8(2 ) 4.8 2.4 2.4(2 ) 0G yM F a a a a aΣ = − + + − − = 7.20yF = 7.20 kipsF = Free body: ABF: We assume that counter CF is acting. 9.6 0: 7.20 2(4.8) 0 14.6 y CFF FΣ = + − = 3.65CFF = + 3.65 kipsCFF T= Since CF is in tension, O.K. Free body: DEH: We assume that counter CH is acting. 9.6 0: 2(2.4 kips) 0 14.6 y CHF FΣ = − = 7.30CHF = + 7.30 kipsCHF T= Since CH is in tension, O.K. www.elsolucionario.net www.elsolucionario.net
- 845. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 843 PROBLEM 6.67 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters CJ and HE SOLUTION Free body: Portion ABDFEC of tower. We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ. 4 0: 2(1.2 kN)sin 20 0 1.026 kN 5 x CJ CJF F FΣ = − ° = = + Since CJ is found to be in tension, our assumption was correct. Thus, the answers are (a) CJ (b) 1.026 kN T www.elsolucionario.net www.elsolucionario.net
- 846. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 844 PROBLEM 6.68 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters IO and KN SOLUTION Free body: Portion of tower shown. We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO. 4 0: 4(1.2 kN)sin 20 0 2.05 kN 5 x IO IOF F FΣ = − ° = = + Since IO is found to be in tension, our assumption was correct. Thus, the answers are (a) IO (b) 2.05 kN T www.elsolucionario.net www.elsolucionario.net
- 847. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 845 PROBLEM 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a) Number of members: 16m = Number of joints: 10n = Reaction components: 4 20, 2 20 r m r n = + = = Thus, 2m r n+ = To determine whether the structure is actually completely constrained and determinate, we must try to find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of each truss. This is a properly supported simple truss – O.K. This is an improperly supported simple truss. (Reaction at C passes through B. Thus, Eq. 0BMΣ = cannot be satisfied.) Structure is improperly constrained. Structure (b) 16 10 4 20, 2 20 m n r m r n = = = + = = Thus, 2m r n+ = (a) (b) www.elsolucionario.net www.elsolucionario.net
- 848. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 846 PROBLEM 6.69 (Continued) We must again try to find the reactions at the supports dividing the structure as shown. Both portions are simply supported simple trusses. Structure is completely constrained and determinate. Structure (c) 17 10 4 21, 2 20 m n r m r n = = = + = = Thus, 2m r n+ > This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate. Structure is completely constrained and indeterminate. (c) www.elsolucionario.net www.elsolucionario.net
- 849. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 847 PROBLEM 6.70 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Nonsimple truss with 4,r = 16,m = 10,n = so 20 2 ,m r n+ = = but we must examine further. FBD Sections: FBD I: 0AMΣ = 1T II: 0xFΣ = 2T I: 0xFΣ = xA I: 0yFΣ = yA II: 0EMΣ = yC II: 0yFΣ = yE Since each section is a simple truss with reactions determined, structure is completely constrained and determinate. Structure (b): Nonsimple truss with 3,r = 16,m = 10,n = so 19 2 20m r n+ = < = Structure is partially constrained. www.elsolucionario.net www.elsolucionario.net
- 850. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 848 PROBLEM 6.70 (Continued) Structure (c): Simple truss with 3,r = 17,m = 10,n = 20 2 ,m r n+ = = but the horizontal reaction forces andx xA E are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and indeterminate. www.elsolucionario.net www.elsolucionario.net
- 851. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 849 PROBLEM 6.71 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Nonsimple truss with 4,r = 12,m = 8n = so 16 2 .r m n+ = = Check for determinacy: One can solve joint F for forces in EF, FG and then solve joint E for yE and force in DE. This leaves a simple truss ABCDGH with 3, 9, 6 so 12 2r m n r m n= = = + = = Structure is completely constrained and determinate. Structure (b): Simple truss (start with ABC and add joints alphabetically to complete truss) with 4,r = 13,m = 8n = so 17 2 16r m n+ = > = Constrained but indeterminate www.elsolucionario.net www.elsolucionario.net
- 852. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 850 PROBLEM 6.71 (Continued) Structure (c): Nonsimple truss with 3,r = 13,m = 8n = so 16 2 .+ = =r m n To further examine, follow procedure in part (a) above to get truss at left. Since 1 0≠F (from solution of joint F), 1AM aFΣ = 0≠ and there is no equilibrium. Structure is improperly constrained. www.elsolucionario.net www.elsolucionario.net
- 853. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 851 PROBLEM 6.72 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a) Number of members: 12=m Number of joints: 8n = Reaction components: 3 15, 2 16 r m r n = + = = Thus, 2m r n+ < Structure is partially constrained. Structure (b) 13, 8 3 16, 2 16 = = = + = = m n r m r n Thus, 2+ =m r n To verify that the structure is actually completely constrained and determinate, we observe that it is a simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus, structure is completely constrained and determinate. www.elsolucionario.net www.elsolucionario.net
- 854. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 852 PROBLEM 6.72 (Continued) Structure (c) 13, 8 4 17, 2 16 m n r m r n = = = + = = Thus, 2m r n+ > Structure is completely constrained and indeterminate. This result can be verified by observing that the structure is a simple truss (follow lettering to check this), therefore it is rigid, and that its supports involve four unknowns. www.elsolucionario.net www.elsolucionario.net
- 855. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 853 PROBLEM 6.73 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Rigid truss with 3,r = 14,m = 8,n = so 17 2 16r m n+ = > = so completely constrained but indeterminate Structure (b): Simple truss (start with ABC and add joints alphabetically), with 3, 13, 8, so 16 2r m n r m n= = = + = = so completely constrained and determinate Structure (c): Simple truss with 3,r = 13,m = 8,n = so 16 2 ,r m n+ = = but horizontal reactions ( and )x xA D are collinear, so cannot be resolved by any equilibrium equation. Structure is improperly constrained. www.elsolucionario.net www.elsolucionario.net
- 856. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 854 PROBLEM 6.74 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): No. of members 12m = No. of joints 8 16 2n m r n= + = = No. of reaction components 4 unknows equationsr = = FBD of EH: 0HMΣ = ;DEF 0xFΣ = ;GHF 0yFΣ = yH Then ABCDGF is a simple truss and all forces can be determined. This example is completely constrained and determinate. Structure (b): No. of members 12m = No. of joints 8 15 2 16n m r n= + = < = No. of reaction components 3 unknows equationsr = < partially constrained Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents this action. www.elsolucionario.net www.elsolucionario.net
- 857. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 855 PROBLEM 6.74 (Continued) Structure (c): No. of members 13m = No. of joints 8 17 2 16n m r n= + = > = No. of reaction components 4 unknows equationsr = > completely constrained but indeterminate www.elsolucionario.net www.elsolucionario.net
- 858. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 856 PROBLEM 6.75 Determine the force in member BD and the components of the reaction at C. SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC: 2 2 (24) (10) 26 in.BD = + = 10 0: (160 lb)(30 in.) (16 in.) 0 26 C BDM F Σ = − = 780 lbBDF = + 780 lbBDF T= 24 0: (780 lb) 0 26 x xM CΣ = + = 720 lbxC = − 720 lbx =C 10 0: 160 lb (780 lb) 0 26 y yF CΣ = − + = 140.0 lbyC = − 140.0 lby =C www.elsolucionario.net www.elsolucionario.net
- 859. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 857 PROBLEM 6.76 Determine the force in member BD and the components of the reaction at C. SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC: Attaching BDF at D and resolving it into components, we write 0:CMΣ = 450 (400 N)(135 mm) (240 mm) 0 510 BDF + = 255 NBDF = − 255 NBDF C= 240 0: ( 255 N) 0 510 x xF CΣ = + − = 120.0 NxC = + 120.0 Nx =C 450 0: 400 N ( 255 N) 0 510 y yF CΣ = − + − = 625 NyC = + 625 Ny =C www.elsolucionario.net www.elsolucionario.net
- 860. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 858 PROBLEM 6.77 Determine the components of all forces acting on member ABCD of the assembly shown. SOLUTION Free body: Entire assembly: 0: (120 mm) (480 N)(80 mm) 0BM DΣ = − = 320 N=D 0: 480 N 0x xF BΣ = + = 480 Nx =B 0: 320 N 0y yF BΣ = + = 320 Ny =B Free body: Member ABCD: 0: (320 N)(200 mm) (160 mm) (320 N)(80 mm) (480 N)(40 mm) 0 AM CΣ = − − − = 120.0 N=C 0: 480 N 0x xF AΣ = − = 480 Nx =A 0: 320 N 120 N 320 N 0y yF AΣ = − − + = 120.0 Ny =A www.elsolucionario.net www.elsolucionario.net
- 861. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 859 PROBLEM 6.78 Determine the components of all forces acting on member ABD of the frame shown. SOLUTION Free body: Entire frame: 0: (300 lb) (12 ft) (450 lb)(4 ft) (6 ft) 0DM EΣ = − + − = 900 lb= +E 900 lb=E 0: 900 lb 0x xF DΣ = − = 900 lbx =D 0: 300 lb 450 lb 0y yF DΣ = − − = 750 lby =D Free body: Member ABD: We note that BC is a two-force member and that B is directed along BC. 0: (750 lb)(16 ft) (900 lb)(6 ft) (8 ft) 0AM BΣ = − − = 825 lbB = + 825 lb=B 0: 900 lb 0x xF AΣ = + = 900 lbxA = − 900 lbx =A 0: 750 lb 825 lb 0y yF AΣ = + − = 75 lbyA = + 75.0 lby =A www.elsolucionario.net www.elsolucionario.net
- 862. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 860 PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: 0: (4) (20 kips)(5) 0E xM AΣ = − − = 25 kips,xA = − 25.0 kipsx =A 0: 20 kips 0y yF AΣ = − = 20 kipsyA = 20.0 kipsy =A Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and 2 . 5 y xB B= 0: (25 kips)(4 ft) (20 kips)(10 ft) (2 ft) (5 ft) 0C x yM B BΣ = − + + = 2 100 kip ft (2 ft) (5 ft) 0 5 x xB B− ⋅ + + = 25 kipsxB = 25.0 kipsx =B 2 2 ( ) (25) 10 kips 5 5 y xB B= = = 10.00 kipsy =B 0: 25 kips 25 kips 0x xF CΣ = − − = 50 kipsxC = 50.0 kipsx =C 0: 20 kips 10 kips 0y yF CΣ = + − = 10 kipsyC = − 10.00 kipsy =C www.elsolucionario.net www.elsolucionario.net
- 863. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 861 PROBLEM 6.80 Solve Problem 6.79 assuming that the 20-kip load is replaced by a clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC at Point D. PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: 0: 0y yF AΣ = = 0: (4 ft) 100 kip ft 0E xM AΣ = − − ⋅ = 25 kipsxA = − 25.0 kipsx =A 25.0 kips=A Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and 2 . 5 y xB B= 0: (25 kips)(4 ft) (2 ft) (5 ft) 0C x yM B BΣ = + + = 2 100 kip ft (2 ft) (5 ft) 0 5 x xB B⋅ + + = 25 kipsxB = − 25.0 kipsx =B 2 2 ( 25) 10 kips; 5 5 y xB B= = − = − 10.00 kipsy =B 0: 25 kips 25 kips 0 0x x xF C CΣ = − + + = = 0: 10 kips 0y yF CΣ = + + = 10 kipsyC = − 10 kipsyC = 10.00 kips=C www.elsolucionario.net www.elsolucionario.net
- 864. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 862 PROBLEM 6.81 Determine the components of all forces acting on member ABCD when θ = 0. SOLUTION Free body: Entire assembly: 0: (200) (150 N)(500) 0BM AΣ = − = 375 NA = + 375 N=A 0: 375 N 0x xF BΣ = + = 375 NxB = − 375 Nx =B 0: 150 N 0y yF BΣ = − = 150 NyB = + 150 Ny =B Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. 0: (300) (150 N)(100) (375 N)(200) 0CM DΣ = − + = 200 ND = − 200 N=D 0: 375 375 0x xF CΣ = + − = 0xC = 0: 150 200 0y yF CΣ = + − = 50.0 NyC = + 50.0 N=C www.elsolucionario.net www.elsolucionario.net
- 865. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 863 PROBLEM 6.82 Determine the components of all forces acting on member ABCD when θ = 90°. SOLUTION Free body: Entire assembly: 0: (200) (150 N)(200) 0BM AΣ = − = 150.0 NA = + 150.0 N=A 0: 150 150 0x xF BΣ = + − = 0xB = 0: 0y yF BΣ = = 0=B Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. 0: (300) (150 N)(200) 0CM DΣ = + = 100.0 ND = − 100.0 N=D 0: 150 N 0x xF CΣ = + = 150 NxC = − 150.0 Nx =C 0: 100 N 0y yF CΣ = − = 100.0 NyC = + 100.0 Ny =C www.elsolucionario.net www.elsolucionario.net
- 866. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 864 PROBLEM 6.83 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. 0: (750 N)(80 mm) (200 mm) 0Σ = − − =E xM A 300 N 300 Nx xA = − =A 0: 300 N 0 300 N 300 Nx x x xF E EΣ = − = = =E 0: 750 N 0y y yF A EΣ = + − = (1) (a) Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. 75 mm 90 N 300 N 250 mm y y E E= = From Eq. (1): 90 N 750 N 0yA + − = 660 NyA = Thus, reactions are 300 Nx =A , 660 Ny =A 300 Nx =E , 90.0 N=Ey (b) Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. 125 mm 150 N 300 N 250 mm y y A A= = www.elsolucionario.net www.elsolucionario.net
- 867. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 865 PROBLEM 6.83 (Continued) From Eq. (1): 750 N 0 150 N 750 N 0 y y y A E E + − = + − = 600 N 600 Ny yE = =E Thus, reactions are 300 Nx =A , 150.0 Ny =A 300 Nx =E , 600 Ny =E www.elsolucionario.net www.elsolucionario.net
- 868. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 866 PROBLEM 6.84 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free-body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. 0: (750 N)(240 mm) (400 mm) 0E xM AΣ = − − = 450 N 450 Nx xA = − =A 0: 450 N 0 450 N 450 Nx x x xF E EΣ = − = = =E 0: 750 N 0y y yF A EΣ = + − = (1) (a) Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. 240 mm ; 225 N 450 N 480 mm y y E = =E From Eq. (1): 225 750 0; 525 Ny yA + − = =A Thus, reactions are 450 Nx =A , 525 Ny =A 450Nx =E , 225 Ny =E (b) Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. 160 mm 150.0 N 450 N 480 mm y y A = =A From Eq. (1): 750 N 0 150 N 750 N 0 y y y A E E + − = + − = 600 N 600 Ny yE = =E Thus, reactions are 450 Nx =A , 150.0 Ny =A 450 Nx =E , 600 Ny =E www.elsolucionario.net www.elsolucionario.net
- 869. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 867 PROBLEM 6.85 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. 0: 36 N m (0.2 m) 0Σ = − ⋅ − =E xM A 180 N 180 Nx xA = − =A 0: 180 N + 0Σ = − =x xF E 180 N 180 N= =Ex xE 0: 0y y yF A EΣ = + = (1) (a) Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. 0.075 m 54 N 180 N 0.25 m = =E y y E From Eq. (1): 54 N 0yA + = 54 N 54.0 Ny yA = − =A Thus, reactions are 180.0 Nx =A , 54.0 Ny =A 180.0 Nx =E , 54.0 N=Ey (b) Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along EC. 0.125 m 90 N 180 N 0.25 m y y A A= = www.elsolucionario.net www.elsolucionario.net
- 870. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 868 PROBLEM 6.85 (Continued) From Eq. (1): 0 90 N 0 y y y A E E + = + = 90 N 90 Ny yE = − =E Thus, reactions are 180.0 Nx =A , 90.0 Ny =A 180.0 Nx = −E , 90.0 Ny =E www.elsolucionario.net www.elsolucionario.net
- 871. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 869 PROBLEM 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. 0: 36 N m (0.4 m) 0E xM AΣ = − ⋅ − = 90 N 90.0 Nx xA = − =A 0: 90 + 0x xF EΣ = − = 90 N 90.0 Nx xE = =E 0: 0y y yF A EΣ = + = (1) (a) Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. 0.24 m ; 45.0 N 90 N 0.48 m y y E = =E From Eq. (1): 45 N 0+ =yA 45 N 45.0 Ny yA = − =A Thus, reactions are 90.0 Nx =A , 45.0 Ny =A 90.0 Nx =E , 45.0 Ny =E (b) Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. 0.16 m ; 30 N 90 N 0.48 m y y A = =A www.elsolucionario.net www.elsolucionario.net
- 872. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 870 PROBLEM 6.86 (Continued) From Eq. (1): 0 30 N 0 y y y A E E + = + = 30 N 30 Ny yE = − =E Thus, reactions are 90.0 Nx =A , 30.0 Ny =A 90.0 Nx = −E , 30.0 Ny =E www.elsolucionario.net www.elsolucionario.net
- 873. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 871 PROBLEM 6.87 Determine all the forces exerted on member AI if the frame is loaded by a clockwise couple of magnitude 1200 lb · in. applied (a) at Point D, (b) at Point E. SOLUTION Free body: Entire frame: Location of couple is immaterial. 0: (48 in.) 1200 lb in. 0HM IΣ = − ⋅ = 25.0 lbI = + (a) and (b) 25.0 lb=I We note that AB, BC, and FG are two-force members. Free body: Member AI: 20 5 tan 22.6 48 12 α α= = = ° (a) Couple applied at D. 5 0: 25 lb 0 13 yF AΣ = − + = 65.0 lbA = + 65.0 lb=A 22.6° 12 0: (65 lb)(40 in.) (20 in.) 0 13 GM CΣ = − = 120 lbC = + 120 lb=C 12 0: (65 lb) 120 lb 0 13 xF GΣ = − + + = 60.0 lbG = − 60 lb=G www.elsolucionario.net www.elsolucionario.net
- 874. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 872 PROBLEM 6.87 (Continued) (b) Couple applied at E. 5 0: 25 lb 0 13 yF AΣ = − + = 65.0 lbA = + 65.0 lb=A 22.6° 12 0: (65 lb) (40 in.) (20 in.) 1200 lb in. 0 13 GM CΣ = + − − ⋅ = 60.0 lbC = + 60.0 lb=C 12 0: (65 lb) 60 lb 0 13 xF GΣ = − + + = 0=G www.elsolucionario.net www.elsolucionario.net
- 875. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 873 PROBLEM 6.88 Determine all the forces exerted on member AI if the frame is loaded by a 40-lb force directed horizontally to the right and applied (a) at Point D, (b) at Point E. SOLUTION Free body: Entire frame: Location of 40-lb force on its line of action DE is immaterial. 0: (48 in.) (40 lb)(30 in.) 0HM IΣ = − = 25.0 lbI = + (a) and (b) 25.0 lb=I We note that AB, BC, and FG are two-force members. Free body: Member AI: 20 5 tan 22.6 48 12 α α= = = ° (a) Force applied at D. 5 0: 25 lb 0 13 yF AΣ = − + = 65.0 lbA = + 65.0 lb=A 22.6° 12 0: (65 lb)(40 in.) (20 in.) 0 13 GM CΣ = − = 120.0 lbC = + 120.0 lb=C 12 0: (65 lb) 120 lb 0 13 xF GΣ = − + + = 60.0 lbG = − 60.0 lb=G www.elsolucionario.net www.elsolucionario.net
- 876. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 874 PROBLEM 6.88 (Continued) (b) Force applied at E. 5 0: 25 lb 0 13 yF AΣ = − + = 65.0 lbA = + 65.0 lb=A 22.6° 12 0: (65 lb)(40 in.) (20 in.) (40 lb)(10 in.) 0 13 GM CΣ = − − = 100.0 lbC = + 100.0 lb=C 12 0: (65 lb) 100 lb 40 lb 0 13 xF GΣ = − + + + = 80.0 lbG = − 80.0 lb=G www.elsolucionario.net www.elsolucionario.net
- 877. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 875 PROBLEM 6.89 Determine the components of the reactions at A and B, (a) if the 500-N load is applied as shown, (b) if the 500-N load is moved along its line of action and is applied at Point F. SOLUTION Free body: Entire frame: Analysis is valid for either parts (a) or (b), since position of 100-lb load on its line of action is immaterial. 0: (10) (100 lb)(6) 0 60 lbA y yM B BΣ = − = = + 0: 60 100 0 40 lby y yF A AΣ = + − = = + 0: 0x x xF A BΣ = + = (1) (a) Load applied at E. Free body: Member AC: Since AC is a two-force member, the reaction at A must be directed along CA. We have 40 lb 10 in. 5 in. xA = 80.0 lbx =A , 40.0 lby =A From Eq. (1): 80 0 80 lbx xB B− + = = + Thus, 80.0 lbx =B , 60.0 lby =B www.elsolucionario.net www.elsolucionario.net
- 878. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 876 PROBLEM 6.89 (Continued) (b) Load applied at F. Free body: Member BCD: Since BCD is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have, therefore, 0xB = The reaction at B is 0x =B 60.0 lby =B From Eq. (1): 0 0 0x xA A+ = = The reaction at A is 0x =A 40.0 lby =A www.elsolucionario.net www.elsolucionario.net
- 879. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 877 PROBLEM 6.90 (a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at a Point B, a force of magnitude equal to the tension in the cable should also be applied at B. SOLUTION First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged. 1 2 1 20: 0CM rT rT T TΣ = − = = (a) Replace each force with an equivalent force-couple. (b) Cut the cable and replace the forces on pulley with equivalent pair of forces at A as above. www.elsolucionario.net www.elsolucionario.net
- 880. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 878 PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE. SOLUTION Free body: 16-ft length of pipe: (500 lb/ft)(16 ft) 8 kipsW = = Force Triangle 8 kips 3 5 4 B D = = 6 kips 10 kipsB D= = Determination of CB = CD. We note that horizontal projection of horizontal projection of=BO OD CD + 3 4 ( ) 5 5 r r CD+ = Thus, 8 2(1.5 ft) 3 ft 4 CB CD r= = = = Free body: Member ABC: 0: (6 kips)( 3)A xM C h hΣ = − − 3 (6 kips)x h C h − = (1) For 9 ft,h = 9 3 (6 kips) 4 kips 9 xC − = = www.elsolucionario.net www.elsolucionario.net
- 881. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 879 PROBLEM 6.91 (Continued) Free body: Member CDE: From above, we have 4.00 kipsx =C 0: (10 kips)(7 ft) (4 kips)(6 ft) (8 ft) 0E yM CΣ = − − = 5.75 kips,yC = + 5.75 kipsy =C 3 0: 4 kips (10 kips) 0 5 x xF EΣ = − + + = 2 kips,xE = − 2.00 kipsx =E 4 0: 5.75 kips (10 kips) 0, 5 y yF EΣ = − + = 2.25 kipsy =E www.elsolucionario.net www.elsolucionario.net
- 882. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 880 PROBLEM 6.92 Solve Problem 6.91 for a frame where h = 6 ft. PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE. SOLUTION See solution of Problem 6.91 for derivation of Eq. (1). For 3 6 3 6 ft, C (6 kips) 3 kips 6 x h h h − − = = = = Free body: Member CDE: From above, we have 3.00 kipsx =C 0: (10 kips)(7 ft) (3 kips)(6 ft) (8 ft) 0E yM CΣ = − − = 6.50 kips,yC = + 6.50 kipsy =C 3 0: 3 kips (10 kips) 0 5 x xF EΣ = − + + = 3.00 kips,xE = − 3.00 kipsx =E 4 0: 6.5 kips (10 kips) 0 5 y yF EΣ = − + = 1.500 kipsyE = 1.500 kipsy =E www.elsolucionario.net www.elsolucionario.net
- 883. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 881 PROBLEM 6.93 Knowing that the pulley has a radius of 0.5 m, determine the components of the reactions at A and E. SOLUTION FBD Frame: 0: (7 m) (4.5 m)(700 N) 0A yM EΣ = − = 450 Ny =E 0: 700 N 450 N 0y yF AΣ = − + = 250 Ny =A 0: 0x x x x xF A E A EΣ = − = = Dimensions in m FBD Member ABC: 0: (1 m)(700 N) (1 m)(250 N) (3 m) 0C xM AΣ = − − = 150.0 Nx =A so 150.0 Nx =E www.elsolucionario.net www.elsolucionario.net
- 884. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 882 PROBLEM 6.94 Knowing that the pulley has a radius of 50 mm, determine the components of the reactions at B and E. SOLUTION Free body: Entire assembly: 0 : (300 N)(350 mm) (150 mm) 0E xM BΣ = − − = 700 NxB = − 700 Nx =B 0: 700 0x xF N EΣ = − + = 700 NxE = 700 Nx =E 0: 300 N 0y y yF B EΣ = + − = (1) Free body: Member ACE: 0: (700 N)(150 mm) (300 N)(50 mm) (180 mm) 0C yM EΣ = − − = 500 NyE = 500 Ny =E From Eq. (1): 500 N 300 N 0yB + − = 200 NyB = − 200 Ny =B Thus, reactions are 700 Nx =B , 200 Ny =B 700 Nx =E , 500 Ny =E www.elsolucionario.net www.elsolucionario.net
- 885. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 883 PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer. SOLUTION (a) Free body: Trailer: (We shall denote by A, B, C the reaction at one wheel.) 0: (2400 lb)(2 ft) (11ft) 0AM DΣ = − + = 436.36 lbD = 0: 2 2400 lb 436.36 lb 0yF AΣ = − + = 981.82 lbA = 982 lb=A Free body: Truck. 0: (436.36 lb)(3 ft) (2900 lb)(5 ft) 2 (9 ft) 0BM CΣ = − + = 732.83 lbC = 733 lb=C 0: 2 436.36 lb 2900 lb 2(732.83 lb) 0yF BΣ = − − + = 935.35 lbB = 935 lb=B (b) Additional load on truck wheels. Use free body diagram of truck without 2900 lb. 0: (436.36 lb)(3 ft) 2 (9 ft) 0BM CΣ = + = 72.73 lbC = − 72.7 lbCΔ = − 0: 2 436.36 lb 2(72.73 lb) 0yF BΣ = − − = 290.9 lbB = 291lbBΔ = + www.elsolucionario.net www.elsolucionario.net
- 886. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 884 PROBLEM 6.96 In order to obtain a better weight distribution over the four wheels of the pickup truck of Problem 6.95, a compensating hitch of the type shown is used to attach the trailer to the truck. The hitch consists of two bar springs (only one is shown in the figure) that fit into bearings inside a support rigidly attached to the truck. The springs are also connected by chains to the trailer frame, and specially designed hooks make it possible to place both chains in tension. (a) Determine the tension T required in each of the two chains if the additional load due to the trailer is to be evenly distributed over the four wheels of the truck. (b) What are the resulting reactions at each of the six wheels of the trailer-truck combination? PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer. SOLUTION (a) We small first find the additional reaction Δ at each wheel due to the trailer. Free body diagram: (Same Δ at each truck wheel) 0: (2400 lb)(2 ft) 2 (14 ft) 2 (23 ft) 0AMΣ = − + Δ + Δ = 64.86 lbΔ = 0: 2 2400 lb 4(64.86 lb) 0;yF AΣ = − + = 1070 lb;A = 1070 lb=A Free body: Truck: (Trailer loading only) 0: 2 (12 ft) 2 (3 ft) 2 (1.7 ft) 0DM TΣ = Δ + Δ − = 8.824 8.824(64.86 lb) 572.3 lb T T = Δ = = 572 lbT = Free body: Truck: (Truck weight only) 0: (2900 lb)(5 ft) 2 (9 ft) 0BM C′Σ = − + = 805.6 lbC′ = 805.6 lb′ =C www.elsolucionario.net www.elsolucionario.net
- 887. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 885 PROBLEM 6.96 (Continued) 0: 2 2900 lb 2(805.6 lb) 0yF B′Σ = − + = 644.4 lbB′ = 644.4 lb′ =B Actual reactions: 644.4 lb 64.86 709.2 lbB B′= + Δ = + = 709 lb=B 805.6 lb 64.86 870.46 lbC C′= + Δ = + = 870 lb=C From part a: 1070 lb=A www.elsolucionario.net www.elsolucionario.net
- 888. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 886 PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D. SOLUTION (a) Free body: Entire machine: =A Reaction at each front wheel =B Reaction at each rear wheel 0: 75(3.2 m) 100(1.2 m) 2 (4.8 m) 300(5.6 m) 0AM BΣ = − + − = 2 325 kNB = 162.5 kN=B 0: 2 325 75 100 300 0yF AΣ = + − − − = 2 150 kNA = 75.0 kN=A www.elsolucionario.net www.elsolucionario.net
- 889. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 887 PROBLEM 6.97 (Continued) (b) Free body: Motor unit: 0: (1 m) 2 (2.8 m) 300(3.6 m) 0DM C BΣ = + − = 1080 5.6C B= − (1) Recalling 162.5 kN, 1080 5.6(162.5) 170 kNB C= = − = 170.0 kN=C 0: 170 0x xF DΣ = − = 170.0 kNx =D 0: 2(162.5) 300 0y yF DΣ = − − = 25.0 kNy =D www.elsolucionario.net www.elsolucionario.net
- 890. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 888 PROBLEM 6.98 Solve Problem 6.97 assuming that the 75-kN load has been removed. PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D. SOLUTION (a) Free body: Entire machine: =A Reaction at each front wheel =B Reaction at each rear wheel 0: 2 (4.8 m) 100(1.2 m) 300(5.6 m) 0AM BΣ = − − = 2 375 kNB = 187.5 kN=B 0: 2 375 100 300 0yF AΣ = + − − = 2 25 kNA = 12.50 kN=A (b) Free body: Motor unit: See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With 187.5 kN,B = we have 1080 5.6(187.5) 30 kNC = − = 30.0 kN=C 0: 30 0x xF DΣ = − = 30.0 kNx =D 0: 2(187.5) 300 0y yF DΣ = − − = 75.0 kNy =D www.elsolucionario.net www.elsolucionario.net
- 891. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 889 PROBLEM 6.99 For the frame and loading shown, determine the components of all forces acting on member ABE. SOLUTION FBD Frame: 0: (1.8 m) (2.1m)(12 kN) 0E yM FΣ = − = 14.00 kNy =F 0: 14.00 kN 12 kN 0y yF EΣ = − + − = 2 kNyE = 2.00 kNy =E FBD member BCD: 0: (1.2 m) (12 kN)(1.8 m) 0 18.00 kNB y yM C CΣ = − = = But C is ⊥ ACF, so 2 ; 36.0 kNx y xC C C= = 0: 0 36.0 kNx x x x xF B C B CΣ = − + = = = 36.0 kNxB = on BCD 0: 18.00 kN 12 kN 0 6.00 kNy y yF B BΣ = − + − = = on BCD On ABE: 36.0 kNx =B 6.00 kNy =B FBD member ABE: 0: (1.2 m)(36.0 kN) (0.6 m)(6.00 kN) (0.9 m)(2.00 kN) (1.8 m)( ) 0 A x M E Σ = − + − = 23.0 kNx =E 0: 23.0 kN 36.0 kN 0x xF AΣ = − + − = 13.00 kNx =A 0: 2.00 kN 6.00 kN 0y yF AΣ = − + − = 4.00 kNy =A www.elsolucionario.net www.elsolucionario.net
- 892. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 890 PROBLEM 6.100 For the frame and loading shown, determine the components of all forces acting on member ABE. PROBLEM 6.99 For the frame and loading shown, determine the components of all forces acting on member ABE. SOLUTION FBD Frame: 0: (1.2 m)(2400 N) (4.8 m) 0F yM EΣ = − = 600 Ny =E FBD member BC: 4.8 8 5.4 9 y x xC C C= = 0: (2.4 m) (1.2 m)(2400 N) 0 1200 NC y yM B BΣ = − = = On ABE: 1200 Ny =B 0: 1200 N 2400 N 0 3600 Ny y yF C CΣ = − + − = = so 9 4050 N 8 x y xC C C= = 0: 0 4050 Nx x x xF B C BΣ = − + = = on BC On ABE: 4050 Nx =B www.elsolucionario.net www.elsolucionario.net
- 893. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 891 PROBLEM 6.100 (Continued) FBD member AB0E: 0: (4050 N) 2 0A xM a aEΣ = − = 2025 NxE = 2025 Nx =E 0 : (4050 2025) N 0x xF AΣ = − + − = 2025 Nx =A 0: 600 N 1200 N 0y yF AΣ = + − = 1800 Ny =A www.elsolucionario.net www.elsolucionario.net
- 894. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 892 PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body: Entire frame: 0: (12 in.) (360 lb)(15 in.) (240 lb)(33 in.) 0AM EΣ = − − = 1110 lbE = + 1110 lb= +E 0: 1110 lb 0x xF AΣ = + = 1110 lbxA = − 1110 lbx =A 0: 360 lb 240 lb 0y yF AΣ = − − = 600 lbyA = + 600 lby =A Free body: Member CDE: 0: (1110 lb)(24 in.) (12 in.) 0C xM DΣ = − = 2220 lbxD = + www.elsolucionario.net www.elsolucionario.net
- 895. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 893 PROBLEM 6.101 (Continued) Free body: Member ABD: From above: 2220 lbx =D 0: (18 in.) (600 lb)(6 in.) 0B yM DΣ = − = 200 lbyD = + 200 lby =D 0: 2220 lb 1110 lb 0x xF BΣ = + − = 1110 lbxB = − 1110 lbx =B 0: 200 lb 600 lb 0y yF BΣ = + + = 800 lbyB = − 800 lby =B www.elsolucionario.net www.elsolucionario.net
- 896. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 894 PROBLEM 6.102 Solve Problem 6.101 assuming that the 360-lb load has been removed. PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body diagram of entire frame. 0: (12 in.) (240 lb)(33 in.) 0AM EΣ = − = 660 lbE = + 660 lb=E 0: 660 lb 0x xF AΣ = + = 660 lbxA = − 660 lbx =A 0: 240 lb 0y yF AΣ = − = 240 lbyA = + 240 lby =A Free body: Member CDE: 0: (660 lb)(24 in.) (12 in.) 0C xM DΣ = − = 1320 lbxD = + www.elsolucionario.net www.elsolucionario.net
- 897. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 895 PROBLEM 6.102 (Continued) Free body: Member ABD: From above: 1320 lbx =D 0: (18 in.) (240 lb)(6 in.) 0B yM DΣ = − = 80 lbyD = + 80.0 lby =D 0: 1320 lb 660 lb 0x xF BΣ = + − = 660 lbxB = − 660 lbx =B 0: 80 lb 240 lb 0y yF BΣ = + + = 320 lbyB = − 320 lby =B www.elsolucionario.net www.elsolucionario.net
- 898. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 896 PROBLEM 6.103 For the frame and loading shown, determine the components of the forces acting on member CDE at C and D. SOLUTION Free body: Entire frame: 0: 25 lb 0y yM AΣ = − = 25 lbyA = 25 lby =A 0: (6.928 2 3.464) (25 lb)(12 in.) 0F xM AΣ = + × − = 21.651lbxA = 21.65 lby =A 0: 21.651lb 0xF FΣ = − = 21.651lbF = 21.65 lb=F Free body: Member CDE: 0: (4 in.) (25 lb)(10 in.) 0C yM DΣ = − = 62.5 lbyD = + 62.5 lby =D 0: 62.5 lb 25 lb 0y yF CΣ = − + − = 37.5 lbyC = + 37.5 lby =C Free body: Member ABD: 0: (3.464 in.) (21.65 lb)(6.928 in.)B xM DΣ = + (25 lb)(4 in.) (62.5 lb)(2 in.) 0− − = 21.65 lbxD = + Return to free body: Member CDE: From above: 21.65 lbxD = + 21.7 lbx =D 0: 21.65 lbx xF CΣ = − 21.65 lbxC = + 21.7 lbx =C www.elsolucionario.net www.elsolucionario.net
- 899. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 897 PROBLEM 6.104 For the frame and loading shown, determine the components of the forces acting on member CFE at C and F. SOLUTION Free body: Entire frame: 0: (40 lb)(13 in.) (10 in.) 0D xM AΣ = + = 52 lb,xA = − 52 lbx =A Free body: Member ABF: 0: (52 lb)(6 in.) (4 in.) 0B xM FΣ = − + = 78 lbxF = + Free body: Member CFE: From above: 78.0 lbx =F 0: (40 lb)(9 in.) (78 lb)(4 in.) (4 in.) 0C yM FΣ = − − = 12 lbyF = + 12.00 lby =F 0: 78 lb 0x xF CΣ = − = 78 lbxC = + 78.0 lbx =C 0: 40 lb 12 lb 0; 28 lby y yF C CΣ = − + + = = + 28.0 lby =C www.elsolucionario.net www.elsolucionario.net
- 900. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 898 PROBLEM 6.105 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body: Entire frame: 0: (500) (3 kN)(600) (2 kN)(1000) 0AM FΣ = − − = 7.60 kNF = + 7.60 kN=F 0: 7.60 N 0,x xF AΣ = + = 7.60 NxA = − 7.60 kNx =A 0: 3 kN 2 kN 0y yF AΣ = − − = 5 kNyA = + 5.00 kNy =A Free body: Member CDE: 0: (400) (3 kN)(200) (2 kN)(600) 0C yM DΣ = − − = 4.50 kNyD = + Free body: Member ABD: From above: 4.50 kNy =D 0: (200) (4.50 kN)(400) (5 kN)(400) 0B xM DΣ = − − = 19 kNxD = + 19.00 kNx =D 0: 19 kN 7.60 kN 0x xF BΣ = + − = 11.40 kNxB = − 11.40 kNx =B 0: 5 kN 4.50 kN 0y yF BΣ = + − = 0.50 kNyB = − 0.500 kNy =B Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 901. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 899 PROBLEM 6.106 Solve Problem 6.105 assuming that the 3-kN load has been removed. PROBLEM 6.105 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body: Entire frame: 0: (500) (2 kN)(1000) 0AM FΣ = − = 4 kNF = + 4.00 kN=F 0: 4 kN 0,x xF AΣ = + = 4 kNxA = − 4.00 kNx =A 0: 2 kN 0,y yF AΣ = − = 2 kNyA = + 2.00 kNy =A Free body: Member CDE: 0: (400) (2 kN)(600) 0C yM DΣ = − = 3.00 kNyD = + Free body: Member ABD: From above: 3.00 kNy =D 0: (200) (3 kN)(400) (2 kN)(400) 0B xM DΣ = − − = 10.00 kNxD = + 10.00 kNx =D 0: 10 kN 4 kN 0x xF BΣ = + − = 6 kNxB = − 6.00 kNx =B 0: 2 kN 3 kN 0y yF BΣ = + − = 1kNyB = + 1.000 kNy =B Dimensions in mm www.elsolucionario.net www.elsolucionario.net
- 902. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 900 PROBLEM 6.107 Determine the reaction at F and the force in members AE and BD. SOLUTION Free body: Entire frame: 0: (9 in.) (450 lb)(24 in.) 0C yM FΣ = − = 1200 lbyF = 1200 lby =F Free body: Member DEF: 0: (1200 lb)(4.5 in.) (18 in.) 0J xM FΣ = − = 300 lbxF = 300 lbx =F 3 0: (24 in.) (12 in.) 0 5 D x AEM F F Σ = − − = 10 10 (300 lb) 3 3 AE xF F= − = − 1000 lbAEF = − 1000 lbAEF C= 4 4 0: 1200 lb ( 1000 lb) 0 5 5 y BDF FΣ = + − − = 5 (1200 lb) 1000 lb 500 lb 4 BDF = − = + 500 lbBDF T= www.elsolucionario.net www.elsolucionario.net
- 903. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 901 PROBLEM 6.108 For the frame and loading shown, determine the reactions at A, B, D, and E. Assume that the surface at each support is frictionless. SOLUTION Free body: Entire frame: 0: (1000 lb)sin30 0xF A BΣ = − + ° = 500 0A B− + = (1) 0: (1000 lb)cos30 0yF D EΣ = + − ° = 866.03 0D E+ − = (2) Free body: Member ACE: 0: (6 in.) (8 in.) 0CM A EΣ = − + = 3 4 E A= (3) Free body: Member BCD: 0: (8 in.) (6 in.) 0CM D BΣ = − + = 3 4 D B= (4) Substitute E and D from Eqs. (3) and (4) into Eq. (2): 3 3 866.06 0 4 4 1154.71 0 A B A B − + − = + − = (5) From Eq. (1): 500 0A B− + = (6) Eqs. (5) (6):+ 2 654.71 0A − = 327.4 lbA = 327 lb=A Eqs. (5) (6):− 2 1654.71 0B − = 827.4 lbB = 827 lb=B From Eq. (4): 3 (827.4) 4 D = 620.5 lbD = 621lb=D From Eq. (3): 3 (327.4) 4 E = 245.5 lbE = 246 lb=E www.elsolucionario.net www.elsolucionario.net
- 904. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 902 PROBLEM 6.109 The axis of the three-hinge arch ABC is a parabola with the vertex at B. Knowing that P = 112 kN and Q = 140 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB. SOLUTION Free body: Segment AB: 0: (3.2 m) (8 m) (5 m) 0A x yM B B PΣ = − − = (1) 0.75 (Eq. 1): (2.4 m) (6 m) (3.75 m) 0x yB B P− − = (2) Free body: Segment BC: 0: (1.8 m) (6 m) (3 m) 0C x yM B B QΣ = + − = (3) Add Eqs. (2) and (3): 4.2 3.75 3 0xB P Q− − = (3.75 3 )/4.2xB P Q= + (4) From Eq. (1): 3.2 (3.75 3 ) 8 5 0 4.2 yP Q B P+ − − = ( 9 9.6 )/33.6yB P Q= − + (5) given that 112 kN and 140 kN.P Q= = (a) Reaction at A. Considering again AB as a free body, 0: 0; 200 kNx x x x xF A B A BΣ = − = = = 200 kNx =A 0: 0y y yF A P BΣ = − − = 112 kN 10 kN 0yA − − = 122 kNyA = + 122.0 kNy =A www.elsolucionario.net www.elsolucionario.net
- 905. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 903 PROBLEM 6.109 (Continued) (b) Force exerted at B on AB. From Eq. (4): (3.75 112 3 140)/4.2 200 kNxB = × + × = 200 kNx =B From Eq. (5): ( 9 112 9.6 140)/33.6 10 kNyB = − × + × = + 10.00 kNy =B www.elsolucionario.net www.elsolucionario.net
- 906. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 904 PROBLEM 6.110 The axis of the three-hinge arch ABC is a parabola with the vertex at B. Knowing that P = 140 kN and Q = 112 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB. SOLUTION Free body: Segment AB: 0: (3.2 m) (8 m) (5 m) 0A x yM B B PΣ = − − = (1) 0.75 (Eq. 1): (2.4 m) (6 m) (3.75 m) 0x yB B P− − = (2) Free body: Segment BC: 0: (1.8 m) (6 m) (3 m) 0C x yM B B QΣ = + − = (3) Add Eqs. (2) and (3): 4.2 3.75 3 0xB P Q− − = (3.75 3 )/4.2xB P Q= + (4) From Eq. (1): 3.2 (3.75 3 ) 8 5 0 4.2 yP Q B P+ − − = ( 9 9.6 )/33.6yB P Q= − + (5) given that 140 kN and 112 kN.P Q= = (a) Reaction at A. 0: 0; 205 kNx x x x xF A B A BΣ = − = = = 205 kNx =A 0: 0y y yF A P BΣ = − − = 140 kN ( 5.5 kN) 0yA − − − = 134.5 kNyA = 134.5 kNy =A www.elsolucionario.net www.elsolucionario.net
- 907. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 905 PROBLEM 6.110 (Continued) (b) Force exerted at B on AB. From Eq. (4): (3.75 140 3 112)/4.2 205 kNxB = × + × = 205 kNx =B From Eq. (5): ( 9 140 9.6 112)/33.6 5.5 kNyB = − × + × = − 5.50 kNy =B www.elsolucionario.net www.elsolucionario.net
- 908. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 906 PROBLEM 6.111 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Member FBDs: I II FBD I: 1 0: 0 2 2 B y AF AF yM aC a F F CΣ = − = = FBD II: 1 0: 0 2 2 D y EH EH yM aC a F F C= − = = FBDs combined: 1 1 1 1 0: 0 2 2 2 2 2 2 G AF EH y yM aP a F a F P C CΣ = − − = = + 2 y P C = 2 so 2 AFF P C= 2 2 EHF P T= FBD I: 1 1 1 0: 0 0 2 22 2 2 y AF BG y BG P P F F F P C F PΣ = + − + = + − + = 0BGF = FBD II: 1 1 1 0: 0 0 2 22 2 2 y y DG EH DG P P F C F F FΣ = − + − = − + − = 2DGF P C= www.elsolucionario.net www.elsolucionario.net
- 909. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 907 PROBLEM 6.112 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Member FBDs: I II FBD I: 0: 2 0 2I y x y xM aC aC aP C C PΣ = + − = + = FBD II: 0: 2 0 2 0J y x y xM aC aC C CΣ = − = − = Solving, ; as shown. 2 4 x y P P C C= = FBD I: 1 0: 0 2 2 x BG x BG xF F C F CΣ = − + = = 2 BG P F C= 1 2 0: 0 2 42 y AF P F F P P Σ = − + + = 4 AF P F C= FBD II: 1 0: 0 2 2 x x DG DG xF C F F CΣ = − + = = 2 DG P F C= 1 2 0: 0 2 4 2 42 y y EH EH P P P F C P F F Σ = − + + = = − = − 4 EH P F T= www.elsolucionario.net www.elsolucionario.net
- 910. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 908 PROBLEM 6.113 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Member FBDs: I II From FBD I: 3 0: 0 3 2 2 2 J x y x y a a a M C C P C C PΣ = + − = + = FBD II: 3 0: 0 3 0 2 2 K x y x y a a M C C C CΣ = − = − = Solving, ; as drawn. 2 6 x y P P C C= = FBD I: 1 2 0: 0 2 62 B y AG AG yM aC a F F C PΣ = − = = = 2 6 AGF P C= 1 1 2 2 0: 0 2 6 22 2 x AG BF x BF AG xF F F C F F C P PΣ = − + − = = + = + 2 2 3 BFF P C= FBD II: 1 2 0: 0 2 62 D EH y EH yM a F aC F C PΣ = + = = − = − 2 6 EHF P T= 1 1 2 2 0: 0 2 6 22 2 x x DI EH DI EH xF C F F F F C P PΣ = − + = = + = − + 2 3 DIF P C= www.elsolucionario.net www.elsolucionario.net
- 911. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 909 PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link. SOLUTION We consider members ABC and CDE: Free body: CDE: 0: (4 ) (2 ) 0J x yM C a C aΣ = + = 2y xC C= − (1) Free body: ABC: 0: (2 ) (4 ) (3 ) 0K x yM C a C a P aΣ = + − = Substituting for Cy from Eq. (1): (2 ) 2 (4 ) (3 ) 0x xC a C a P a− − = 1 1 2 2 2 x yC P C P P = − = − − = + 1 0: 0 2 x BG xF F CΣ = + = 1 2 2 , 2 2 BG x P F C P = − = − − = + 2 BG P F T= www.elsolucionario.net www.elsolucionario.net
- 912. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 910 PROBLEM 6.114 (Continued) 1 0: 0 2 y AF BG yF F F P CΣ = − − − + = 1 22 2 AF P P F P P= − − + = − 2 AF P F C= Free body: CDE: 1 0: 0 2 y DG yF F CΣ = − = 2 2DG yF C P= = + 2DGF P T= 1 0: 0 2 x EH x DGF F C FΣ = − − − = 1 1 2 2 22 EH P F P P = − − − = − 2 EH P F C= www.elsolucionario.net www.elsolucionario.net
- 913. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 911 PROBLEM 6.115 Solve Problem 6.112 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.112 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Free body: Member ABC: 0: (2 ) ( ) 0J y xM C a C aΣ = + = 2x yC C= − Free body: Member CDE: 00: (2 ) ( ) 0K y xM C a C a MΣ = − − = 0(2 ) ( 2 )( ) 0y yC a C a M− − − = 0 4 y M C a = ᭠ 2 :x yC C= − 0 2 x M C a = − ᭠ 0 0: 0; 0 22 2 x x MD D F C a Σ = + = − = 0 2 M D a = 0 2 DG M F T a = 00: ( ) ( ) 0D yM E a C a MΣ = − + = 0 0( ) ( ) 0 4 M E a a M a − + = 03 4 M E a = − 03 4 EH M F C a = Return to free body of ABC: 0 0: 0; 0 22 2 x x MB B F C a Σ = + = − = 0 2 M B a = 0 2 BG M F T a = 0 0: ( ) ( ); ( ) ( ) 0 4 B y M M A a C a A a a a Σ = + + = 0 4 M A a = − 0 4 AF M F C a = www.elsolucionario.net www.elsolucionario.net
- 914. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 912 PROBLEM 6.116 Solve Problem 6.114 assuming that the force P is replaced by a clockwise couple of moment M0 applied at the same point. PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading sh