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Applied Mechanics 1
• Week – 9: Chapter 13 Kinetics of Particles:
Energy and Momentum Methods
1

Course Outcomes
• 13.1 Work and Energy
• 13.2 Conservation of Energy
• 13.3 Impulse and Momentum
2

13.1A Work of a Force
• Consider a particle that moves from a point A to a neighboring point
A′ (Fig. 13.1). If r denotes the position vector corresponding to point
A, we can denote the small vector joining A and A′ by the differential
dr; the vector dr is called the displacement of the particle.
• Now, let us assume that a force F is acting on the particle. We
define the work of the force F corresponding to the
displacement dr as the quantity
• dU = F · dr…………………………(13.1)
4

Work
• We can obtain the work of F during a finite displacement of the
particle from A1 to A2 by integrating Eq. (13.1) along the path
described by the particle. This work, denoted by , is
• Work of a force
• = ……………(13.2)
5

Work of a Constant Force in Rectilinear Motion.
• When a particle moving in a straight line is acted upon by a force F
of constant magnitude and of constant direction (Fig. 13.3),
•
• = (F cosα) Δx……(13.3)
• where α = angle the force forms with the direction of motion
• Δx = displacement from A1 to A2
6

Work of the Force of Gravity
• We can obtain the work of the weight W of a body—that is, of the
force of gravity exerted on that body. Choosing the y axis upward
(Fig. 13.4), we have = 0, = −W, and = 0. This gives us
• dU = - W · dy
• = - = W - W ……..(13.4)
• = - W ( - ) = - W Δy…….(13.4’)
• where Δy is the vertical displacement from A1 to A2. The work of the
weight W is thus equal to the product of W and the vertical
displacement of the center of gravity of the body. The work is
positive when Δy < 0; that is, when the body moves down.
7

Work of the Force Exerted by a Spring
• Consider a body A attached to a fixed-point B by a spring; we
assume that the spring is undeformed when the body is at A0 (Fig.
13.5a). For a linear spring, the magnitude of the force F exerted by
the spring on body A is proportional to the deflection x of the spring
measured from the unstretched position A0.
• F = kx……(13.5)
• We can obtain the work of force F exerted by the spring during a
finite displacement of the body from A1(x = x1) to A2(x = x2) by
writing
• dU = − F dx = − kx dx
• = - = ½ k - ½ k…(13.6)
8

Work of the Force Exerted by a Spring. Continue
• Note that the work of force F
exerted by the spring on the body is
positive when x2 < x1; that is, when
the spring is returning to its
undeformed position.
• When the body is moved from x1 to
x2, the work done by the spring is
negative, because the
displacement and force are in
opposite directions.
9

Work of a Gravitational Force.
• Two particles of mass M and m separated by a distance r attract
each other with equal and opposite forces F and −F, directed along
the line joining the particles and of magnitude
• F = g M m/
• Let us assume that particle M occupies a fixed position O, while
particle m moves along the path shown in Fig. 13.6.
• We can obtain the work of force F exerted on particle m during an
infinitesimal displacement of the particle from A to A′ by multiplying
the magnitude F of the force by the radial component dr of the
displacement
• Because F is directed toward O and dr is directed away from O, the
work is negative, and we have
10

Work of a Gravitational Force… continue
• dU = - F dr = - G M m/ dr
• The work of the gravitational force F during a finite displacement
from A1(r = r1) to A2(r = r2) is therefore
• = - = G Mm/ - G Mm/ ……(13.7)
• where M is the mass of the earth. We can use this formula to
determine the work of the force exerted by the earth on a body of
mass m at a distance r from the earth’s center when r is larger than
the radius R of the earth.
11

Work of a Gravitational Force….continue
• Two particles of mass M and m separated by a distance r attract
each other with equal and opposite forces F and −F, directed along
the line joining the particles and of magnitude
• F = g M m/
• Let us assume that particle M occupies a fixed position O, while
particle m moves along the path shown in Fig. 13.6.
• We can obtain the work of force F exerted on particle m during an
infinitesimal displacement of the particle from A to A′ by multiplying
the magnitude F of the force by the radial component dr of the
displacement
• Because F is directed toward O and dr is directed away from O, the
work is negative, and we have
12

13.1B Principle of Work and Energy
• Consider a particle of mass m acted upon by a force F and moving
along a path that is either rectilinear or curved (Fig. 13.7).
Expressing Newton’s second law in terms of the tangential
components of the force and of the acceleration, we have
• = m or = m dv/dt
• Where v is the speed of particle.
• We know v = ds/dt,
• = m dv/ds * ds/dt = mv dv/ds
• ds = mv ds
• Integrating from A1, where s = and v = , to A2, where s = and v = ,
we have
13

Kinetic Energy of particle
• = m = ½ m - = ½ m……..(13.8)
• The left-hand side of Eq. (13.8) represents the work U1→2 of the
force F exerted on the particle during the displacement from A1 to
A2; We define 1/2mv2 as the kinetic energy of the particle,
denoted by T. That is,
• Kinetic energy of a particle T = ½ m ……(13.9)
• Principle of work and energy = - ……(13.10)
• Rearranging the terms in Eq. (13.10) gives
• + …….(13.11)
• Because both work and kinetic energy are scalar quantities, we can
compute their sum as an ordinary algebraic sum with the work
being positive or negative according to the direction of F.
14

13.1D Power and Efficiency
• We define power as the time rate at which work is done.
• If ΔU is the work done during the time interval Δt, the average
power during that time interval is
• Average Power = ΔU / Δt
• Letting Δt approach zero, we obtain in the limit
• Power = dU / dt ……(13.12)
• Substituting the scalar product F·dr for dU, we can also write
• Power = dU / dt = F dr / dt
• Then, recalling that dr/dt represents the velocity v of the point of
application of F, we have
• Power = F·v……(13.3)
• Power is expressed in J/s; this unit is called a watt (W).
15

Mechanical Efficiency
• We defined the mechanical efficiency of a machine as the ratio of
the output work to the input work:
• η = Output work / Input work ….(13.14)
• This definition is based on the assumption that work is done at a
constant rate.
• The ratio of the output to the input work is therefore equal to the
ratio of the rates at which output, and input work are done, and we
have
• Mechanical efficiency η = Power Output / Power Input ….(13.15)
• Because of energy losses due to friction, the output work is always
smaller than the input work, and consequently, the power output is
always smaller than the power input.
16

Sample Problem 13.2
• Two blocks are joined by an inextensible cable as shown. If the
system is released from rest, determine the velocity of block A after
it has moved 2 m. Assume that the coefficient of kinetic friction
between block A and the plane is μk = 0.25 and that the pulley is
weightless and frictionless
17

FBD 13.2
18
• Define two separate systems, one for each block, and model them
as particles. As stated in the problem, assume the pulley is
weightless and frictionless.

Work and Energy for Block A.
• Denote the friction force by FA and the force exerted by the cable
by FC. Then you have (Fig. 1)
• = 200 kg
• = (200 kg)(9.81 m/s 2 ) = 1962 N
• = = = 0.25(1962 N) = 490 N
• We know +
• 0 + (2m) − (2 m) = ½
• (2 m) − (490 N)(2 m) = ½ (200 kg) .....(1)
19

Work and Energy for Block B.
• From the free-body diagram for block B (Fig. 2), you have
• = 300 kg
• = (300 kg)(9.81 m/s 2 ) = 2940 N
• We know +
• 0 + (2m) − (2 m) = ½
• (2940 m) (2 m) − (2 m) = ½ (300 kg) .....(2)
20

Velocity of block 13.2
• Now add the left-hand and right-hand sides of Eqs. (1) and (2). The
work of the forces exerted by the cable on A and B cancels out.
• This is why when solving problems using work and energy, it is
usually best to choose your system to include all the objects of
interest, so you don’t need to worry about the work of internal
forces.
• Therefore, after combining Eqs. (1) and (2) or by choosing your
system to be block A, block B, and the cable, you get
• (2940N) (2m) − (490N) (2m) = ½ (200 kg + 300 kg)
• 4900 J = ½ (500 kg)
• V = 4.43 m/s
21

Sample Problem 13.3
• spring is used to stop a 60-kg package that is sliding on a
horizontal surface. The spring has a constant k = 20 kN/m and is
held by cables so that it is initially compressed 120 mm. The
package has a velocity of 2.5 m/s in the position shown, and the
maximum additional deflection of the spring is 40 mm. Determine
(a) the coefficient of kinetic friction between the package and the
surface, (b) the velocity of the package as it passes again through
the position shown.
22

FBD13.3
• The system is the crate, which you can model as a particle. A free-
body diagram for the crate when it is not in contact with the spring
is shown in Fig. 2.
• After it hits the spring, it has an additional force P acting on it due to
the compression of the spring (Fig. 3).
23

a. Motion from Position 1 to Position 2
• Kinetic Energy:
• Position 1. v1 = 2.5 m/s
• = ½ m = ½ (60) = 187.5 N·m = 187.5 J
• Position 2. (maximum spring deflection): v2 = 0 = 0
• Work: Friction Force F. You have (Fig. 2)
• = = = mg = (60)(9.81) = 588.6 N
• The work of F is negative and equal to
• = -Fx = −(588.6 N) (0.600 m + 0.040 m) = −(377 J)
• Spring Force P. The variable force P exerted by the spring does an
amount of negative work equal to the area under the force-
deflection curve of the spring force. You have
24

Principle of work and energy 13.3
• = = (20 kN/m)(120 mm) = (20 000 N/m)(0.120 m) = 2400 N
• = + k Δx = 2400 N + (20 kN/m)(40 mm) = 3200 N
• = -1/2 ( + ) Δx = − ½ (2400N + 3200N)(0.040 m)
• = −112.0 J
• Principle of Work and Energy. You can determine the coefficient
of kinetic friction from the expression for the principle of work and
energy in this segment of the motion.
• +
• 187.5 J − (377 J) − 112.0 J = 0
• = 0.20
25

b. Motion from Position 2 to Position 3.
• Kinetic energy:
• position 2 : v2 = 0 = 0
• Position 3: = ½ m= ½ (60)
• Work: Because the distances involved are the same, the numerical
values of the work of the friction force F and of the spring force P
are the same as before. However, the work of F is still negative,
whereas the work of P is now positive.
• = −(377 J) + 112.0 J = −75.5 J + 112.0 J = +36.5 J
• Principle of Work and Energy:
• + : 0 + 36.5 J = ½ (60)
• = 1.103 m/s
26

Sample Problem 13.7
The dumbwaiter D and its load have a combined weight of 600 lb,
whereas the counterweight C weighs 800 lb. Determine the power
delivered by the electric motor M when the dumbwaiter (a) is moving
up at a constant speed of 8 ft/s, (b) has an instantaneous velocity of 8
ft/s and an acceleration of 2.5 ft/s2, where both are directed upward.
27

FBD 13.7
• Define two separate systems, one for body C and one for body D,
and model them as particles.
• Assume the pulley is weightless and frictionless.
28

Uniform motion 13.7
• The force F exerted by the motor cable has the same direction as
the velocity of the dumbwaiter, so the power is equal to F, where =
8 ft/s. To obtain the power, you must first determine F in each of the
two given situations.
• You have = = 0; both bodies are in equilibrium (Fig. 1).
• Body C: +↑ = 0
• 2T − 800 lb = 0 T = 400 lb
• Body D: +↑ = 0
• F + T − 600 lb = 0: F = 600 lb − T = 600 lb − 400 lb = 200 lb
• F = (200 lb) (8 ft/s) = 1600 ft·lb/s
• Power = (1600 ft·lb/s) (1 hp / 550 ft.lb/s)
• Power = 2.91 hp
29

b. Accelerated Motion
• You have = 2.5 ft/s2 ↑ = -1/2 = 1.25 ft/s2 ↓
• The equations of motion are obtained using Figs. 2 and 3.
• Body C: + ↓ =
• 800 – 2T = 800/32.2 * (1.25)
• T = 384.5 lb
• Body D: + ↑ =
• F + T − 600 = 600/32.2 8 (2.5)
• F + 384.5 − 600 = 46.6 F = 262.1 lb
• F = (262.1 lb)(8 ft/s) = 2097 ft·lb/s
• Power = (2097 ft·lb/s) (1 hp / 550 ft.lb/s)
• Power: 3.81 hp
30

13.2 Conservation of Energy
31

13.2A Potential Energy
• Let’s consider again a body of weight W that moves along a curved
path from a point A1 of elevation y1 to a point A2 of elevation y2
(Fig. 13.4). Recall from Sec. 13.1A that the work done by the force
of gravity W during this displacement is
• = W - W ……..(13.4)
• That is, we obtain the work done by W by subtracting the value of
the function Wy corresponding to the second position of the body
from its value corresponding to the first position. The work of W is
independent of the actual path followed; it depends only upon the
initial and final values of the function Wy
32

Gravitational potential energy on earth
• This function is called the potential energy of the body with respect
to the force of gravity W and is denoted by Vg. We have
• Gravitational potential energy on earth
• = - Where = Wy……(13.16)
• where y is measured from an arbitrary horizontal datum where the
potential energy is zero by definition.
• Note that if (Vg)2 > (Vg)1, that is, if the potential energy increases
during the displacement (as in the case considered here), the work
U1→2 is negative.
• On the other hand, if the work of W is positive,the potential energy
decreases.
33

Gravitational potential energy in Space
• In the case of a space vehicle, however, we need to take into
consideration the variation of the force of gravity with the distance r
from the center of the earth. Using the expression obtained in Sec.
13.1A for the work of a gravitational force, we have
• = GMm/ - GMm/
• The expression that we use for the potential energy Vg when the
variation in the force of gravity cannot be neglected is
• Gravitational potential energy in Space
• = - G Mm / r…………………………………(13.17)
34

Elastic potential energy
• Consider now a body attached to a spring and moving from a
position A1, corresponding to a deflection x1 of the spring, to a
position A2, corresponding to a deflection x2 of the spring (Fig.
13.5). Recall from Sec. 13.1A that the work of the force F exerted
by the spring on the body is
• = ½ k - ½ k
• Elastic potential energy
• = - where = ½ k…..(13.18)
35

13.2C The Principle of Conservation of Energy
• Conservation of energy + = + …..(13.24)
• Formula (13.24) indicates that when a particle moves under the
action of conservative forces, the sum of the kinetic energy and
of the potential energy of the particle remains constant.
• The sum T + V is called the total mechanical energy of the particle
and is denoted by E.
• So far, we have discussed two types of potential energy:
gravitational potential energy, , and elastic potential energy, .
Therefore, another way to write Eq. (13.24) is
• + + = + + …..(13.24’)
36

Pendulum Analysis
• Consider, for example, the pendulum analyzed in Sec. 13.1C that is
released with no velocity from and allowed to swing in a vertical
plane (Fig. 13.12). Measuring the potential energy from the level
of , that is, placing our datum at , we have at
• = 0 = WL + = WL
• Recalling that at A2 the speed of the pendulum is = , we have
• = ½ m = ½ W/g () = WL = 0
• + = WL
• Thus, we can check that the total mechanical energy E = T + V of
the pendulum is the same at A1 and A2
37

13.2D Application to Space Mechanics: Motion Under a
Conservative Central Force
• Consider, a space vehicle of mass m moving under the earth’s
gravitational force. Let us assume that it begins its free flight at
point at a distance from the center of the earth with a velocity
forming an angle with the radius vector (Fig. 13.14).
• Let P be a point of the trajectory described by the vehicle; we
denote by r the distance from O to P, by v the velocity of the vehicle
at P, and by the angle formed by v and the radius vector OP.
ϕ
• Applying the principle of conservation of angular momentum about
O between and P (Sec. 12.2B), we have
• m Sin = rmv sin …..(13.25)
ϕ
38

Conservation of Force, Space Vehicle
• Recalling the expression in Eq. (13.17)
for the potential energy due to a
gravitational force, we apply the principle
of conservation of energy between and
P, obtaining
• + = +
• ½ m - G Mm/ = ½ m - G Mm/ r….(13.26)
• where M is the mass of the earth.
39

Sample Problem 13.8
• A 20-lb collar slides without friction along a vertical rod as shown.
The spring attached to the collar has an undeformed length of 4 in.
and a spring constant of 3 lb/in. If the collar is released from rest in
position 1, determine its velocity after it has moved 6 in. to position
2.
40

FBD 13.8
• For your system, choose the collar and the spring. You can treat
the collar as a particle
41

Position 1… 13.8
• Conservation of energy: Applying the principle of conservation of
energy between positions 1 and 2 gives
• + + = + +
• Position 1:
• Potential Energy: The elongation of the linear spring (Fig. 1) is
• = 8 in. − 4 in. = 4 in.
• This gives = ½ k = ½ (3) = 24 in.lb = 2 ft.lb
• Choosing the datum as shown, you have = 0.
• Kinetic Energy: Because the velocity at position 1 is zero, = 0.
42

Position 2… 13.8
• Position 2:
• Potential Energy: The elongation of the linear spring (Fig. 1) is
• = 10 in. − 4 in. = 6 in.
• This gives = ½ k = ½ (3) = 54 in.lb = 4.5 ft.lb
• = w = (20)(-6) = -120 in.lb = -10 ft.lb
• Kinetic Energy: = ½ m = ½ (20/32.2) = 0.311
• Conservation of Energy. Substituting into Eq. (1) gives
• + + = + +
• 0 + 0 + 2 = 0.311 + (-10) + (4.5)
• = 4.91 ft/s
43

Sample Problem 13.9
• A 2.5-lb collar is attached to a spring and slides along a smooth
circular rod in a vertical plane. The spring has an undeformed
length of 4 in. and a spring constant k. The collar is at rest at point
C and is given a slight push to the right. Knowing that the maximum
velocity of the collar is achieved as it passes through point A,
determine (a) the spring constant k, (b) the force exerted by the rod
on the collar at point A.
44

FBD 13.9
• For the conservation of energy portion of the problem, model the
collar as a particle and use it and the spring as your system. When
using Newton’s second law, use the collar as your system
45

Solution 13.9
• Conservation of Energy. Position 1 is when the collar is at point
C, and position 2 is when it is at point A (Fig. 1).
• Applying conservation of energy between positions 1 and 2 gives
• + + = + + ….(1)
• Position 1: Because the system starts from rest, = 0, and because
the spring has an unstretched length of 4 in., you know = 0. Putting
the datum at A gives.
• = (2.5 lb)(7/12 ft) = 1.4583 ft·lb
46

Position 2…. 13.9
• From geometry, the distance from the pin to A is = 7.6 in.
• Therefore, the elongation of the linear spring (Fig. 1) is x2 = 7.616
in. − 4 in. = 3.616 in. = 0.3013 ft.
• You know = 0 because the datum is at position 2. You also know
• = ½ k = ½ k = 0.04539k
• = ½ m = ½ (2.5/32.2) = 0.03882
• Substituting these expressions into Eq. (1) gives
• 0 + 1.4585 + 0 = 0.03882 + 0 + 0.04539 k….(2)
47

Spring constant k…. 13.9
• You have two unknowns in this equation, so you need another
equation. In the problem statement, you are also given that the
collar has a maximum velocity at point A.
• Therefore, the tangential acceleration must be zero at A, and you
should use Newton’s second law to get additional equation.
• A free-body diagram and kinetic diagram for the collar at position 2
are shown in Fig. 2. Applying Newton’s second law in the t direction
gives
• +↑Σ = 0 = k sin θ − W
• k(0.3013 ft)(3/7.616) − 2 lb = 0
• k = 16.85 lb/ft
48

• Substituting this value of k into Eq. (2) gives = 3.597 ft/s. Applying
Newton’s second law in the n direction gives
• + ← Σ = m
• k cos θ – N = m
• Solving for N and substituting in values provides
• N = k cos θ - m
=(16.85 lb/ft)(0.3013 ft)(7/7.616) – {(2.5lb/32.2ft/s2)(3.597 ft/s2)}/(7/12
ft)
• N = 4.19 lb
Force Exerted by the Rod. …. 13.9
49

• A sphere of mass m = 0.6 kg is attached to an elastic cord of
constant k = 100 N/m, which is undeformed when the sphere is
located at the origin O. The sphere may slide without friction on the
horizontal surface and in the position shown its velocity vA has a
magnitude of 20 m/s. Determine (a) the maximum and minimum
distances from the sphere to the origin O, (b) the corresponding
values of its speed.
Sample Problem 13.11
50

• Choose the sphere, which can be modeled as a particle, as your
system
FBD 13.11
51

• At point B, where the distance from O is maximum (Fig. 1), the
velocity of the sphere is perpendicular to OB and the angular
momentum is m. A similar property holds at point C, where the
distance from O is minimum. Expressing conservation of angular
momentum between A and B, you have
• m sin60 = m
• (0.5)(0.6)(20) sin 60 = (0.6)
• = 8.66 / …….(1)
Solution 13.11
52

• At Point A: = ½ m = ½ (0.6) = 120 J
= ½ k = ½ (100) = 12.5 J
• At Point B: = ½ m = ½ (0.6)= 0.3
= ½ k = ½ (100) = 50
• Apply the principle of conservation of energy between points A and
B:
• + = +
• 120 + 12.5 = 0.3 + 50 …..(2)
Conservation of energy 13.11
53

a. Maximum and Minimum Values of Distance: Substituting for
from Eq. (1) into Eq. (2) and solving for , you obtain
• = 2.468 or 0.1824
• = 1.571 m , = 0.427m
b. Corresponding Values of Speed: Substituting the values obtained
for rm and r ′m into Eq. (1), you have
• = 8.66/ 1.571 = 5.51m/s
• = 8.66/0.427 = 20.3 m/s
Value of speed 13.11
54

• Consider a particle of mass m acted upon by a force F we can
express Newton’s second law in the form
• F = d/dt (mv)………………(13.27)
• where mv is the linear momentum of the particle. Multiplying both
sides of Eq. (13.27) by dt and integrating from a time t1 to a time t2,
we have
• F dt = d (mv)
• = m - m
• Moving mv1 to the left side of this equation gives us
• + m= m…….(13.28)
13.3A Principle of Impulse and Momentum
56

• The integral in Eq. (13.28) is a vector known as the linear impulse,
or simply the impulse, of the force F during the interval of time
considered.
• = = i + j + k .....(13.29)
• In the case of a force F of constant magnitude and direction, the
impulse is represented by the vector F(t2 − t1), which has the same
direction as F.
• Eq. (13.28) states that when a particle is acted upon by a force F
during a given time interval, we can obtain the final momentum mv2
of the particle by adding vectorially its initial momentum mv1 and
the impulse of the force F during the time interval considered.
Impulse
57

• Principle of impulse and momentum:
• m = m…….(13.30)
• To obtain an analytic solution, it is thus necessary to replace Eq.
(13.30) with the corresponding component equations. Note that
whereas kinetic energy and work are scalar quantities, momentum
and impulse are vector quantities.
• + =
• + =
• + = ……….(13.31)
• When several forces act on a particle, we must consider the
impulse of each of the forces. We have
• m = m……(13.32)
Principle of impulse and momentum
58

• When a problem involves two or more particles, we can consider
each particle separately and write Eq. (13.32) for each particle. We
can also add vectorially the momenta of all the particles and the
impulses of all the forces involved. We then have
• m = m……(13.33)
• Conservation of linear momentum : m = m….(13.34)
• For two particles A and B, this is
• + = +
• where and represent the velocities of the bodies at the second
time. This equation says that the total momentum of the particles is
conserved.
Conservation of linear momentum
59

• A force acting on a particle during a very short time interval but
large enough to produce a definite change in momentum is called
an impulsive force. The resulting motion is called an impulsive
motion.
• For example, when a baseball is struck, the contact between bat
and ball takes place during a very short time interval Δt.
• But the average value of the force Favg exerted by the bat on the
ball is very large, and the resulting impulse FavgΔt is large enough
to change the sense of motion of the ball (Fig. 13.19).
13.3B Impulsive Motion
60

• Impulse–Momentum principale for impulsive motion
• m Δt = m…..(13.35)
• Note that the method of impulse and momentum is particularly
effective in analyzing the impulsive motion of a particle, because it
involves only the initial and final velocities of the particle and the
impulses of the forces exerted on the particle.
Impulse–Momentum principale for impulsive motion
61

• In order to determine the weight of a freight train of 40 identical
boxcars, an engineer attaches a dynamometer between the train
and the locomotive. The train starts from rest, travels over a
straight, level track, and reaches a speed of 30 mi/h after three
minutes. During this time interval, the average reading of the
dynamometer is 120 tons. Knowing that the effective coefficient of
friction in the system is 0.03 and air resistance is negligible,
determine (a) the weight of the train (in tons), (b) the coupling force
between boxcars A and B.
62

• Choose the system to be the 40 boxcars behind the engine. An
impulse–momentum diagram for this system is shown in Fig. 1,
where F is the dynamometer force.
FBD 13.14
63

• Apply the principle of impulse and momentum
• m = m
• You can obtain scalar equations by using Fig. 1 and looking at the x
and y directions.
• + ↑ y components: Nt – Wt= 0 N=W
• → + x components: 0 + − μk N t= m
• 0 + (120 ton)(2000 lb/ton)(180 s) − 0.03 (W) (180 s) = ( W/32.2
ft/s2) (30 mi/h) (1h/3600 s) (5280ft/mi)
• Solving for W, you obtain
• W = 6.384 × lb = 3190 tons
Solution 13.14
64

• You need to define a new system where the force of interest is an
external force.
• Therefore, choose car A to be your system and define FA as the
coupling force between cars A and B. The impulse–momentum
diagram for this system is shown in Fig. 2.
• Because all the cars weigh the same amount, the weight of A is WA
= W/40 = 159,600 lb.
• Applying impulse–momentum in the y direction gives you NA = WA.
Considering the x direction,
• → + x components: 0 + − μk t - t =
• Substituting in numbers and solving for gives
• = 117.0 tons
Coupling Force Between Cars A and B. 13.14
65

• A 10-kg package drops from a chute into a 25-kg cart with a
velocity of 3 m/s. The cart is initially at rest and can roll freely.
Determine (a) the final velocity of the cart, (b) the impulse exerted
by the cart on the package, (c) the fraction of the initial energy lost
in the impact.
66

• Choose the package and the cart to be your system and assume
that both can be treated as particles. The impulse–momentum
diagram for this system is shown in Fig. 1.
• Note that a vertical impulse occurs between the cart and the
ground, because the cart is constrained to move horizontally.
FBD 13.17
67

• Apply the principle of impulse and momentum.
• m = m
a. Package and Cart. Applying this principle in the x direction gives
• → + x components: cos30 + 0 =
• (10 kg)(3 m/s) cos 30° = (10 kg + 25 kg)
• = 0.742 m/s→
• In Fig. 1, the force between the package and the cart is not shown
because it is internal to the defined system.
• To determine this force, you need a new system; that is, just the
package by itself. The impulse–momentum diagram for the
package alone is shown in Fig. 2.
Solution 13.17
68

• The package moves in both x and y directions, so write the
conservation of momentum equation for each component of the
motion.
• → + x components: -m= m
• (10 kg) (3 m/s) cos 30° + = (10 kg) (0.742 m/s)
• = -18.56 N.s
• + ↑ y components: - = 0
• − (10 kg) (3 m/s) sin 30° + = 0
• = +15 N.s
• The impulse exerted on the package is
• = 23.9 N·s 38.9°
⦩
b. Impulse–Momentum Principle: Package
69

• The initial and final energies are
• = ½ = ½ (10 kg) = 45 J
• = ½ ( = ½ (10 kg + 25kg) = 9.63 J
• The fraction of energy lost is
• ( - ) / = (45-9.63)/45 = 0.786
c. Fraction of Energy Lost.
70

References
• Vector Mechanics for Engineers_ Statics and Dynamics (2019,
McGraw Hill) by Beer , Johnston, Mazurek, Cornwell.
71

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