Lecture 02.pdf

S. Tamna
ENG-204 Engineering Mechanics 2
Lecture 02

2
Outline
S. Tamna
• Introduction
• Definitions of position, displacement, velocity
and acceleration of a particle
• Solving for or or in
- Rectilinear motions
- Curvilinear motions
• Example problems
a r
v

3
Introduction
S. Tamna
Kinematic relationships are used to help us determine the trajectory
of a snowboarder completing a jump, the orbital speed of a satellite,
and accelerations during acrobatic flying.

4
Introduction
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Dynamics includes:
Kinematics: study of the geometry of motion. Relates displacement,
velocity, acceleration, and time without reference to the cause of motion.
Fdrive
Fdrag
Kinetics: study of the relations existing between the forces acting on a
body, the mass of the body, and the motion of the body. Kinetics is used
to predict the motion caused by given forces or to determine the forces
required to produce a given motion.

ENG-204 Engineering Mechanics 2 5
Introduction
S. Tamna
Particle kinetics includes:
Rectilinear motion: position, velocity, and acceleration of a particle as it
moves along a straight line.
Curvilinear motion: position, velocity, and acceleration of a particle as it
moves along a curved line in two or three dimensions.
Rectilinear motion Curvilinear motion

6
Motions of Particles
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Definition: A particle is a point mass, i.e. a point with
mass m kg associated with it.
Theoretically, a “point” is a position in space with no
dimensions, i.e. no width, no length and no height.
This means a “particle” can not exist. However, as a
concept for problem solving, it is very useful,
especially for bodies which are “small” compared with
their environment.

7
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8
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Position of
a particle
To specify the position of a particle,
we need to know where the particle is.
This means knowing how far the
particle is from a reference (fixed)
point, and in which direction. This
mean the position of a particle is a
vector.
Reference point

9
S. Tamna
O – reference point (origin)
s – distance along the path,
measuring from some fixed point.
– position vector with respect to O,
(with magnitude and direction).
r
( )
t
r = r
Position Path
O
r
s
s

10
S. Tamna
– displacement vector
0 0
( )
lim( ) lim
avg
avg
t t
t
t
t t t t
t
d
dt
 →  →





→ = + 
 
= = + 
=
= = =
r
r
r r r r
v
r
v v
is always tangent to the path.
r
v
r

r
s

r
s
Displacement

11
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Δs – displacement along the path
0
lim
Δt
s
ds
v
Δt dt
→
 = 

= = =
r
r
v
v = speed = rate of change of
distance along the path.
r
v
s
O
Velocity

12
S. Tamna
0 0
2
2
lim( ) lim
( )
avg
avg
t t
t
d
t dt
d d d
dt dt dt
 →  →

=


= = =

= =
v
a
v v
a a
r r
a
In general, is not tangent to the path
of the particle.
a

v
v
a
Acceleration
 = + 
v v v
Acceleration

13
Rectangular coordinates
S. Tamna
Position vector
2 2 2
2 2 2
( ) ( ) ( )
( )
r
r
x y z
x t y t z t
r x y z
r
r x y z
= + +
= + +
= =  = + +
=
= =
+ +
1/ 2
r i j k
r i j k
r r r
r u
r r
u
( )
t
=
r r
x
y
z
y
x
z
i
k
j
r
u
r

14
S. Tamna
1/2 2 2 2
( )
( ) [ ( ) ( ) ( ) ]
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
, ,
( )
x y z
x y z
x y z
v
d t d
t x t y t z t
dt dt
dx t d dy t d dz t d
x t y t z t
dt dt dt dt dt dt
x y z v t v t v t
v x v y v z
v v v v
v
= = = + +
= + + + + +
= + + = + +
= = =
= =  = + +
=
r
v r i j k
i j k
v i j k
v i j k i j k
v v v
v u

15
S. Tamna
2 2
2 2
1/2 2 2 2
( )
( ) ( )
( )
( ) ( )
( ) ( ) ( )
, ,
( )
x y z
x y z x y z
x x y y z z
x y z
d t d
t v v v
dt dt
d t d
t x y z
dt dt
v v v x y z a t a t a t
a v x a v y a v z
a a a a
= = = + +
= = = + +
= + + = + + = + +
= = = = = =
= =  = + +
v
a v i j k
r
a r i j k
a i j k i j k i j k
a a a
a
a
=
a u

16
Rectilinear Motions of Particles
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Consider a special case of a particle moving in a straight
line, so called “rectilinear motion.” Without loss of
generality, we can set the x-axis (with a reference point as
the origin) along the direction of motion. The distance along
the axis is measure by x.
o
x

17
S. Tamna
In this case, it is convenient to consider only the function
x(t). The position x can be both positive and negative, i.e.
vector.
( ) ( )
t x t
=
r i
o
r
i
x

18
S. Tamna
The displacement (vector) of the particle is the change of
the position.
, ( )
,
t t x x x t
x x x t t t
  
→ → =
 
 = −  = −
x
o
Δx
x'
If Δx > 0, the point x´ is on the right of x.
If Δx < 0, the point x´ is to the left of x.

19
S. Tamna
The average velocity is given by
avg
x
v
t

=

0
lim
t
x dx
v
t dt
 →

= =

The instantaneous velocity (vector) is given by
Speed, , is the magnitude of the velocity and
is a positive scalar.
v

20
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Since x is a function of time, v is also a
function of time, i.e. v = v(t).
If v > 0, the function x(t) is increasing
and the particle is moving to the right.
If v < 0, the function x(t) is decreasing
and the particle is moving to the left.
x
x
(b)
(a)
P
P
v<0
v>0

21
S. Tamna
The average acceleration is given by
)
(
)
(
, t
v
t
v
v
v
v
t
v
aavg −

=
−

=



=
2
2
0
lim ( )
t
v dv d dx d x
a
t dt dt dt dt
 →

= = = =

The instantaneous acceleration (vector) is given by

22
S. Tamna
Acceleration is a vector quantity, thus positive acceleration
may or may not mean the particle is moving faster. And
similarly, negative acceleration may or may not mean the
particle is moving slower.
( ) ( ) ( )
x x t v v t a a t
=  =  =
If a > 0, the velocity v is increasing.
If a < 0, the velocity v is decreasing.
x
x
P
a<0
v
P
P’
P’
v’
v’
v
a>0

23
S. Tamna
If v > 0 (moving to the right) and a > 0, it means the
particle is moving faster to the right and v is increasing.
If v < 0 (moving to the left) and a > 0, it means the particle
is moving slower to the left and v is increasing (becoming
less negative) while |v| is decreasing.
v > 0 (moving to the right) v < 0 (moving to the left)
x
a>0
P P’
x
P
v
P’
v’
v’
v
a>0

24
S. Tamna
2
2
dv d x
a
dt dt
= =
( )
d dv dx dv
a v x v
dt dx dt dx
adx vdv
= =  =
=
Recall

25
S. Tamna
2
( ) 4 3
8
8
x t t
dx
v t
dt
a
= +
= =
=
Sometimes, v is expressed as a function of position, i.e.
v = v(x).
2 3
4
3
4
4
2 3
4 3
8 8 4 3
(4 3)( )
8
x
x
x
x t t
v t x
dv
a v x
dx
a
−
−
−
= +  =
= = = −
= = −
=

26
S. Tamna
The special case of uniform motion
0
a =
0
0 0 0 0
0
constant
t
dx
v v
dt
x x v dt x v t
= = =
= + = +


27
S. Tamna
The special case of constant acceleration
0
0
0
c c
v t
c c
v
dv
a dv a dt
dt
dv a dt v v a t
=  =
=  − =
 
Assuming that at t = 0, v = v0 . Then
constant
c
a a
= =
0 c
v v a t
= +

28
S. Tamna
0
0 0
2
0 0 0
0
( ) ( )
1
( )
2
c c
x t
c c
x
dx
v t v a t dx v a t dt
dt
dx v a t dt x x v t a t
= = +  = +
= +  − = +
 
Assuming that at t = 0, x = x0 . Then
2
0 0
1
2
c
x x v t a t
= + +

29
S. Tamna
0 0
2 2
0 0
1
( ) ( )
2
x v
c
x v
c
a dx vdv
a x x v v
=
− = −
 
Recall adx = vdv, then
2 2
0 0
2 ( )
c
v v a x x
= + −

30
S. Tamna
v = v(t)
Given v = v(t), and initially at t = 0, x = x0, then
0 0
0 0
0
( )
( ) ( )
( )
area under the curve from 0
x t
x
t
dv df
a a t
dt dt
dx
v t dx vdt dx v t dt
dt
x x v t dt
x x v -t t
= = =
=  =  =
= +
= + →
 


31
S. Tamna
a = a(t)
Given a = a(t), x(0) = x0, v(0) = v0, a(0) = a0.
0 0
0 0
0
( ) ( )
( ) ( )
( ) area under curve from 0
v t
v
t
dv
a t dv adt dv a t dt
dt
v t v a t dt
v t v a t t
=  =  =
= +
= + − →
 


32
S. Tamna
a = a(t)
0 0
0 0
0
( )
( )
area under - curve from 0 .
x t
x
t
dx
v dx vdt dx v t dt
dt
x x v t dt
x x v t t
=  =  =
= +
= + →
 


33
S. Tamna
a = a(x)
Given a = a(x), and v = v0 at x = x0.
0 0
0
0
2 2 2
0
2 2
0
( )
1 1 1
( )
2 2 2
2 ( )
v
x v
x v
v
x
x
dv
a v a dx vdv a x dx vdv
dx
a x dx vdv v v v
v v a x dx
=  =  =
= = = −
= +
 


34
S. Tamna
a = a(x)
0
0
( )
( )
( )
t x
x
dx dx
v v x dt
dt v x
dx
dt
v x
= =  =
=
 

35
S. Tamna
a = a(v)
Given a = a(v), and v = v0 at x = x0
0 0
0 ( ) ( )
( )
t v v
v v
dv dv
a dt
dt a
dv dv
dt t
a v a v
v v t
=  =
=  =
=
  

36
S. Tamna
a = k(v)
Alternatively,
0 0 0
0
( ) ( )
( )
x v v
x v v
dv v
a v dx dv
dx a
v v
dx dv x x dv
a v a v
v v x
=  =
=  − =
=
  

37
Sample Problem 11.1
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3 2
( ) 6 15 40
a) time when 0?
b) position and distance traveled at that time?
c) acceleration?
d) distance traveled from 4 to 6?
x t t t t
v
t t
= − − +
=
= =

38
Sample Problem 11.1
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3 2
2
6 15 40
3 12 15
6 12
x t t t
dx
v t t
dt
dv
a t
dt
= − − +
= = − −
= = −
(m)
x
(m/s)
v
2
(m/s )
a

39
Sample Problem 11.1
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3
2
2
2
3 12 15 0
(3 3)( 5) 0 1, 5
5 6(5) 15(5) 40
Time at zero velocity = 5 s
( 5) 60
Displacement (5) (0) 60 40
100 m
6(5) 1
m
Distance travelled 100 m
( 5) 18 m
2 /s
x t
a t
v t t
t t t
x x
= = −
= − − =
 + − =  = −
=
=
= =
− − +
= − = − −
= −
= −
(m)
x
(m/s)
v
2
(m/s )
a

40
Sample Problem 11.1
S. Tamna
3 2
3 2
Distance from 4 s to 6 s
(6) 6 6(6) 15(6) 40 50
(5) 60
(4) 4 6
distance
18 m
(4) 15(4) 40 52
[ (4) (5)] [ (5) (6)]
t t
x
x
x
x x x x
=
= =
= − − + = −
= −
= − − + =
→
=
−
→ +
(m)
x
(m/s)
v
2
(m/s )
a

41
Problem 11.23
S. Tamna
+y
2
2
2
8 0
3 0.1
3 0.1
(3 0.1 )
v y
dv
a v v
dy
v
dv dy
v
v
dv dy
v
= = −
=
−
=
−
 
A bowling ball is dropped from a boat so that it strikes the surface of a
lake with a speed of 8 m/s. Assuming the ball experiences a downward
acceleration of a = 3 − 0.1v2 when in the water, determine the velocity of
the ball when it strikes the bottom of the lake. (a and v expressed in m/s2
and m/s respectively)
10 m

42
Problem 11.23
S. Tamna
2
8 0
2
8
2
2
2 0.2
2 0.2
1
0.2
(3 0.1 )
ln(3 0.1 )
ln(3 0.1 ) ln(3 6.4) 0.2
(3 0.1 )
ln 0.2
(3 6.4)
(3 0.1 ) 3.4
10(3 3.4 )
For 5.88
1 , m/s
0
v y
v
y
y
v
dv dy
v
v y
v
v y
v
y
v e
v e
y
−
−
−
=
−
− =
− − − = −
−
= −
−
− = −
= +
=  =
 

43
Problem 11.38
S. Tamna
2
0
2
2
0
1 1
2 2
constant, 5.4s
35 (5.4)
2.4 m/s
12
2.4(
.96 m/s
5.4)
c
c
c c
c
a a t
x v t a t a
v
v
v a t
a
 =
 =
= = =
= +  =
= + =
A sprinter in a 100-m race accelerates uniformly for the first 35 m
and then runs with constant velocity. If the sprinter’s time for the
first 35 m is 5.4 s, determine (a) his acceleration, (b) his final
velocity, (c) his time for the race.

44
Problem 11.38
S. Tamna
0, (100 35) 65
65
5.02 s
12.9
Total time 5.4 5.02 10. 2
6
4 s
a x
x x
v t
t v
t
= + =
= = − =
=  =
= =
A sprinter in a 100-m race accelerates uniformly for the first 35 m
and then runs with constant velocity. If the sprinter’s time for the
first 35 m is 5.4 s, determine (a) his acceleration, (b) his final
velocity, (c) his time for the race.

45
Summary
S. Tamna
0
0 0
constant
v v
x x v t
= =
= +
constant
c
a a
=
=
0
2
0 0
2 2
0 0
( )
1
( )
2
2 ( )
c
c
c
v t v a t
x t x v t a t
v v a x x
= +
= + +
= + −
0
a =

46
Summary
S. Tamna
( )
v v t
=
0 0
( )
t
dv
a
dt
x x v t dt
=
= + 
( )
v v x
=
0 ( )
x
x
dv
a v
dx
dx
t
v x
=
= 

47
Summary
S. Tamna
( )
a a t
= 0 0
( ) ( )
t
v t v a t dt
= + 
( )
a a x
=
0
2 2
0 ( )
x
x
v v a x dx
= + 
( )
a a v
=
0
0
0
( )
( )
v
v
v
v
dv
t
a v
v
x x dv
a v
=
= +


( )
dv
a t
dt
=
( )
dv
v a x
dx
=
( )
dv
v a v
dx
=
( )
dv
a v
dt
=

48
Work problem 11.9
S. Tamna
v0 = 24.5 m/s
t = 8.17 s
The brakes of a car are applied, causing it to slow down at a rate of 3
m/s2. Knowing that the car stops in 100 m, determine (a) how fast the
car was traveling immediately before the brakes were applied, (b) the
time required for the car to stop.

49
Work problem 11.27
S. Tamna
2
3 1
0.0525m/s
6.17s
a
t t
= −
− =
Experimental data indicate that in a region downstream of a given
louvered supply vent the velocity of the emitted air is defined by v =
0.18v0/x, where v and x are expressed in m/s and meters, respectively, and
v0 is the initial discharge velocity of the air. For v0 = 3.6 m/s, determine
(a) the acceleration of the air at x = 2 m, (b) the time required for the air to
flow from x =1 to x = 3 m.

50
Work problem 11.33
S. Tamna
2
2.0 m/s
60.0 m/s
B
a
v
=
=
An airplane begins its take-off run at A with zero velocity and a
constant acceleration a. Knowing that it becomes airborne 30 s later
at B and that the distance AB is 900 m, determine (a) the acceleration
a (b) the take-off velocity vB.

51
Work problem 11.34
S. Tamna
2
0.417m/s
18.0km/h
a
v
= −
=
A motorist is traveling at 54 km/h when she observes that a traffic
light 240 m ahead of her turns red. The traffic light is timed to stay
red for 24 s. If the motorist wishes to pass the light without stopping
just as it turns green again, determine (a) the required uniform
deceleration of the car, (b) the speed of the car as it passes the light.

52
Work problem 11.37
S. Tamna
2.40 m
2.06 s
D
d
t
=
=
A small package is released from rest at A and moves along the skate
wheel conveyor ABCD. The package has a uniform acceleration of
4.8 m/s2 as it moves down sections AB and CD, and its velocity is
constant between B and C. If the velocity of the package at D is 7.2
m/s, determine (a) the distance d between C and D, (b) the time
required for the package to reach D.

53
Relative motion
S. Tamna
xA = the position of A
xB = the position of B
/
/
as seen f
relative position of with respect to
= position of
relative position of with respect to
= position of
ro
as
m
A B A B
B A B A
A B
A B
x x x
x
B
x B A
x
=
=
−
= −
=
seen from A
/ /
( )
A B A B B A B A
x x x x x x
= − = − − = −
/
A B
x
o x
B
A
A
x /
B A
x
B
x

54
Relative motion
S. Tamna
/
/
/
/
is to the right of
see at the position to the right
is to the left of
see at the position to the le t
0
0
f
A B
A
B
A B
B
A A B
B A x
A B
B A
x
x
x 





/
A B
x
o x
B
A
A
x /
B A
x
B
x

55
Relative motion
S. Tamna
vA = the velocity of A
vB = the velocity of B
/
/
as seen f
relative velocity of with respect to
= velocity of
relative velocity of with respect to
= velocity of
ro
as
m
A B A B
B A B A
A B
A B
v v v
v
B
v B A
v
=
=
−
= −
=
seen from A
o x
B
A
A
x /
B A
x
B
x
B
v
A
v

56
Relative motion
S. Tamna
/
/
( )
A B A B
A B A B
dx d x x
v v v
dt dt
−
= = = −
/
/
see moving in to the right
see moving in to the
0
lef
0 t
A B
A B
B A
B A
v
v




o x
B
A
A
x /
B A
x
B
x
B
v
A
v
/ /
A B B A
v v
= −

57
Relative motion
S. Tamna
aA = the acceleration of A
aB = the acceleration of B
/
/
as seen fr
relative acceleration of w.r.t.
= acceleration of
relative acceleration of w.r.t.
= acceleration of
om
as s
A B A B
B A B A
A B
A B
a a a
a B A
B
a a
= − =
− =
=
een from A
o x
B
A
A
x /
B A
x
B
x
B
a
A
a
/ /
A B B A
a a
= −

58
Dependent motion
S. Tamna
In many cases, motions of two or more particles are related
by certain constraints. For example, there is an inextensible
rope linking two particles.
xB
xA
Datum
Datum
O

59
Dependent motion
S. Tamna
xA measured from C.
xB measured from D.
constant
0
A B
A B
A B
A B
A B
A B
x x CD l
dx dx
v v
dt dt
v v
dv dv
a a
dt dt
+ + = =
+ = + =
= −
= −  = −
vA
vB
vA
vB
Actual motion

60
Dependent motion
S. Tamna
H
The string is inextensible and has a constant
length l.
constant
[ ] [ ( )] [ ]
2 constant
con
( )
2
2 0 2
2 0 2
st n
2
a t
B B A
B A
B A
B A B A
B A B A
FB FB HD CH
FB HD CH
x x x l
x x l
x x
v v v v
a a a a
− + − + − =
+ = +
+ =
+ =
=

+
+ =
+ 
+
= −
= = −

61
Dependent motion
S. Tamna
constant
2 2
2 2 0
2 2 0
C B A
C B A
C B A
x x x l
v v v
a a a
+ + =
+
+ +
=
+ =
=

62
Problem 11.53
S. Tamna
Slider block A moves to the left with a
constant velocity of 6 m/s.
Determine
(a) the velocity of block B,
(b) the velocity of portion D of the cable,
(c) the relative velocity of portion C of
the cable with respect to portion D.

63
Problem 11.53
S. Tamna
yB
xA
/
1
2
2 m/s (moving up)
3 constant
2 m/s (moving down)
8 m/s (moving
3 0
3
6
0
0 6 m/s
6 2 up)
B A
A
B A B
A
B D B D
A C A C C
C
B
D B
C D D
v
v v
y x
v
v v v
v
y y l v v
x y l v v v
v v
v
= −
= − =
=
+ =
+ =  = −
=  
+ =  + =

+ =  + =  = −
= − = − −
−

64
Work problem 11.40
S. Tamna
aA = 1.563 m/s2
aB = 3.13 m/s2
In a boat race, boat A is leading boat B by 50 m and both boats are
traveling at a constant speed of 180 km/h. At t = 0, the boats
accelerate at constant rates. Knowing that when B passes A, t = 8 s
and vA = 225 km/h, determine (a) the acceleration of A, (b) the
acceleration of B.

65
Work problem 11.47
S. Tamna
The elevator shown in the figure moves downward with a constant
velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the
velocity of the counterweight W, (c) the relative velocity of the cable C
with respect to the elevator, (d) the relative velocity of the counterweight
W with respect to the elevator.
/
/
8.0 m/s
4.0 m/s
12.0 m/s
8.0 m/s
C
W
C E
W E
= 
= 
= 
= 
v
v
v
v

66
Work problem 11.49
S. Tamna
An athlete pulls handle A to the left with a constant velocity of 0.5
m/s. Determine (a) the velocity of the weight B, (b) the relative
velocity of weight B with respect to the handle A.
vB = 0.125 m/s
vB/A = 0.5154 m/s 14

67
Work problem 11.51
S. Tamna
Slider block B moves to the right with a constant velocity of 300
mm/s. Determine (a) the velocity of slider block A, (b) the velocity of
portion C of the cable, (c) the velocity of portion D of the cable, (d)
the relative velocity of portion C of the cable with respect to slider
block A.
/
0.2 m/s , 0.6 m/s
0.2 m/s , 0.4 m/s
A C
D C A
= → = →
=  = →
v v
v v

68
Work problem 11.54
S. Tamna
/
16.67mm/s
16.67mm/s
L
B L
= 
= 
v
v
The motor M reels in the cable at a constant rate of 100 mm/s.
Determine (a) the velocity of load L, (b) the velocity of pulley B with
respect to load L.

69
Work problem 11.58
S. Tamna
Block B moves downward with a constant velocity of 20 mm/s. At t
= 0, block A is moving upward with a constant acceleration, and its
velocity is 30 mm/s. Knowing that at t = 3 s slider block C has
moved 57 mm to the right, determine (a) the velocity of slider block
C at t = 0, (b) the accelerations of A and C, (c) the change in position
of block A after 5 s.
2
2
10.0 mm/s
2.0 mm/s
6.0 mm/s
175 mm
C
A
C
A
= →
= 
= →
 = 
v
a
a
y

70
S. Tamna
Thank You

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