4
FORCE AND MOTION
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A
force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces
that occur at a point of contact between the object and its environment. (iii) A very few forces, such as gravity and
magnetism, are long-range forces that require no contact. (iv) A force is a vector quantity, having both a magnitude
(or size) and a direction. (v) When multiple forces act on an object, the forces combine through vector addition to
r
r
give a net force Fnet = ∑ Fi .
4.2. Visualize:
r
Tension T
Weight wr
Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls.
4.3. Visualize:
r
Weight w
Normal force nr
r
Kinetic friction fk
4.4. Model: Assume friction is negligible compared to other forces.
Visualize:
r
r
Drag D
Thrust Fthrust
Weight wr
Normal force nr
4-1
4-2
Chapter 4
4.5. Visualize: Please refer to Figure Ex4.5.
Solve:
Mass is defined to be
m=
1
slope of the acceleration-versus-force graph
A larger slope implies a smaller mass. We know m2 = 0.20 kg, and we can find the other masses relative to m2 by
comparing their slopes. Thus
m1 1/ slope 1 slope 2
1
2
=
=
=
= = 0.40
m2 1/ slope 2 slope 1 5 2 5
⇒ m1 = 0.40 m2 = 0.40 × 0.20 kg = 0.08 kg
Similarly,
m3 1/ slope 3 slope 2
1
5
=
=
=
= = 2.50
m2 1/ slope 2 slope 3 2 5 2
⇒ m3 = 2.50 m2 = 2.50 × 0.20 kg = 0.50 kg
Assess: From the initial analysis of the slopes we had expected m3 > m2 and m1 < m2 . This is consistent with our
numerical answers.
4.6. Model: An object’s acceleration is linearly proportional to the net force.
Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F ∝ a
and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2 , four rubber bands will produce an
acceleration of 2.4 m/s2.
(b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F = ma,
2F = (2m)a ⇒ a = F/m = 0.6 m/s2
4.7. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the
net force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causes
acceleration. That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity
r
that measures a change of motion. Newton’s second law quantifies this idea by stating that the net force Fnet on an
object of mass m causes the object to undergo an acceleration:
r
r Fnet
a=
m
The acceleration vector and the net force vector must point in the same direction.
4.8. Visualize:
a (m/s2)
4
3
2
1
F (N)
0
0
1
2
Solve: Newton’s second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg)a, hence a = 4 m/s2 . After
repeating this procedure at various points, the above graph is obtained.
4.9. Visualize: Please refer to Figure Ex4.9.
Solve: Newton’s second law is F = ma. We can read a force and an acceleration from the graph, and hence find
the mass. Choosing the force F = 1 N gives us a = 4 m/s2. Newton’s second law yields m = 0.25 kg.
Force and Motion
4-3
4.10. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 4.3
gives us no information on textbooks, but does give the weight of a one-pound object. Place a pound weight in one
hand and the textbook on the other. The sensation on your hand is the weight of the object. The sensation from the
textbook is about five times the sensation from the pound weight. So we conclude the weight of the textbook is
about five times the weight of the one-pound object or about 25 N.
(b) According to Table 4.3, the propulsion force on a car is 5000 N. Now a bicycle (including the rider) is about
100 kg. This is about one-tenth of the mass of a car which is about 1000 kg for a compact model. The acceleration
of a bicycle is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as
follows:
1
1
5000 N
= 100 N
(mass of car) × (acceleration of car) =
10
5
50
So we would roughly estimate the propulsion force of a bicycle to be 100 N.
F( bicycle) =
4.11. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 4.3
gives us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand and
a pencil on the other hand. The sensation on your hand is the weight of the object. The sensation from the quarter is
about the same as the sensation from the pencil. So they both have about the same weight. We can estimate the
weight of the pencil to be 0.05 N.
(b) According to Table 4.3, the propulsion force on a car is 5000 N. Now the mass of a sprinter is about 100 kg. This
is about one-tenth of the mass of a car which is about 1000 kg for a compact model. The acceleration of a sprinter is
somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows:
1
1
5000 N
= 100 N
(mass of car) × (acceleration of car) =
10
5
50
So, we would roughly estimate the propulsion force of a sprinter to be 100 N.
Assess: This is the same estimated number as we obtained in problem 4.10. This is reasonable since in both the
cases the propulsion force comes from a human and it probably does not matter how the human is providing that force.
F(sprinter ) =
4.12. Visualize:
r
r
r
F1 1 F 2
r
F2
r
F2
r
F1
F1
r
F3
r
r r
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction
so that the sum of all the three forces is zero.
4.13. Visualize:
r
F1
r
r
F1
r
F1 1 F2
r
r
F2
F2
r
F3
r
r r
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction
so that the sum of all three forces is zero.
4.14. Visualize:
r
r
F1
F1
r
F3
r
r
F1 1 F 2
r
F2
r
F2
r
r r
Solve: The object will be in equilibrium if F3 has the same magnitude as F1 + F2 but is in the opposite direction
so that the sum of all the three forces is zero.
4-4
Chapter 4
4.15. Visualize:
r
Tension T
r
Weight w
r
a= 0
Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force
directed down is labeled as a weight, and the force directed up is labeled as a tension. With zero net force the
acceleration is zero. So, a possible description is: “An object hangs from a rope and is at rest.” Or, “An object hanging
from a rope is moving up or down with a constant speed.”
4.16. Visualize:
Air
r
Drag D
r
Weight w
r
Thrust Fthrust
Exhaust
Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. There is a
r
r
r
force labeled w directed down, a force Fthrust directed up, and a force D directed down. So a possible description is:
“A rocket accelerates upward.”
4.17. Visualize:
r
Weight w
Normal force nr
r
Kinetic friction fk
r
Solve: The free-body
diagram shows three
r
r There is a weight force w which is down. There is a normal
r forces.
force labeled n which is up. The forces w and n are shown with vectors of the same length so they are equal in
magnitude and the net vertical force is zero. So we have an object on the ground which is not moving vertically.
r
There is also a force fk to the left. This must be a frictional force and we need to decide whether it is static or
r
kinetic friction. The frictional force is the only horizontal force so the net horizontal force must be fk . This means
there is a net force to the left producing an acceleration to the left. This all implies motion and therefore the
frictional force is kinetic. A possible description is: “A baseball player is sliding into second base.”
4.18. Visualize:
Force identification
Free-body diagram
y
nr
r
r
Fnet = 0
Normal force nr
r
Weight w
r
w
x
Force and Motion
Force identification
4.19. Visualize:
4-5
Free-body diagram
y
nr
r
r
Fnet = 0
r
Weight w
Normal force nr
x
r
w
Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not include
either of these forces. The only remaining forces are the weight and the normal force.
4.20. Visualize:
Force identification
Free-body diagram
r
Tension T
y
r
T
r
r
Fnet = 0
x
r
w
r
Weight w
r
Assess: There are only two forces on the elevator. The weight w is directed down and the tension in the cable is
directed up. Since the elevator is descending at a steady speed there is no acceleration and the two forces must be
equal in magnitude.
Force identification
4.21. Visualize:
Free-body diagram
y
nr
x
r
r
Weight w
fk
r
Normal force n r
Kinetic friction fk
r
w
r
Fnet
Assess: The problem uses the word “sliding.” Any real situation involves friction with the surface. Since we are
not told to neglect it, we show that force.
Motion diagrams
4.22. Visualize:
vr
r
ar
Fnet
(a)
r
ar
Fnet
vr
(b)
4-6
Chapter 4
Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each
r
r
step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down. Since F = ma
the force is in the same direction as the acceleration and must be directed down.
Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector
decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is also
directed down. As in part (a) the net force must be directed down.
4.23. Visualize:
Motion diagrams
vr
ar
ar
r
r
Fnet
Fnet
vr
(b)
(a)
The velocity vector in figure (a) is shown downward and to the left. So movement is downward and to the left. The
velocity vectors get successively longer which means the speed is increasing. Therefore the acceleration is
r
r
downward and to the left. By Newton’s second law F = ma , the net force must be in the same direction as the
acceleration. Thus, the net force is downward and to the left.
The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. The
velocity vector gets successively shorter which means the speed is decreasing. Therefore the acceleration is
downward and to the left. From Newton’s second law, the net force must be in the direction of the acceleration and
so it is directed downward and to the left.
4.24. Visualize:
Fx (N)
6
4
2
t (s)
0
1
2
4
3
−2
Solve: According to Newton’s second law F = ma , the force at any time is found simply by multiplying the
value of the acceleration by the mass of the object.
4.25. Visualize:
Fx (N)
0.75
0.50
0.25
t (s)
0
1
2
3
4
−0.25
Solve: According to Newton’s second law F = ma, the force at any time is found simply by multiplying the
value of the acceleration by the mass of the object.
Force and Motion
4.26. Visualize:
4-7
ax (m/s2)
1.5
1.0
0.5
t (s)
0
1
2
3
4
−0.5
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the
value of the force by the mass of the object.
4.27. Visualize:
ax (m/s2)
3.0
2.0
1.0
t (s)
0
1
2
3
4
−1.0
Solve: According to Newton’s second law F = ma, the acceleration at any time is found simply by dividing the
value of the force by the mass of the object.
4.28. Model: Use the particle model for the object.
Solve: (a) We are told that for an unknown force (call it F0) acting on an unknown mass (call it m0) the acceleration
of the mass is 10 m/s2. According to Newton’s second law, F0 = m0(10 m/s2). The force then becomes 12 F0 . Newton’s
second law gives
1
1
F0 = m0 a = m0 10 m s2
2
2
This means a is 5 m/s2.
(b) The force is F0 and the mass is now 12 m0 . Newton’s second law gives
[ (
F0 =
)]
1
m0 a = m0 (10 m s2 )
2
This means a = 20 m/s2.
(c) A similar procedure gives a = 10 m/s2.
(d) A similar procedure gives a = 2.5 m/s2.
4.29. Model: Use the particle model for the object.
Solve: (a) We are told that for an unknown force (call it F0 ) acting on an unknown mass (call it m0 ) the
acceleration of the mass is 8 m/s2 . According to Newton’s second law, F0 = m0(8 m/s2). The force then becomes
2 F0 . Newton’s second law gives
[
]
2 F0 = m0 a = 2 m0 (8 m s 2 )
This means a is 16 m/s2.
(b) The force is F0 and the mass is now 2 m0 . Newton’s second law gives
This means a = 4 m/s .
2
F0 = 2 m0 a = m0 (8 m s 2 )
4-8
Chapter 4
(c) A similar procedure gives a = 8 m/s2.
(d) A similar procedure gives a = 32 m/s2.
4.30. Visualize:
y
nr
r
r
fk
T
x
r
w
r
Fnet
ar
vr
Solve: (d) There are a normal force and a weight which are equal and opposite, so this is an object on a horizontal
surface. The description could be: “A tow truck pulls a stuck car out of the mud.”
4.31. Visualize:
y
nr
r
r
Fthrust
D
wr
x
r
r
Fnet = 0
r
ar = 0
vr
Solve: (d) There is a normal force and a weight which are equal and opposite, so this is an object on a horizontal
surface. The description of this free-body diagram could be: “A jet plane is flying at constant speed.”
4.32. Visualize:
y
nr
r
Fsp
r
fk
x
r
w
r
Fnet
ar
vr
Solve: (d) This is an object on a surface because w = n. It must be moving to the left because the kinetic friction is to
the right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic block to the left.”
4.33. Visualize:
y
x
r
w
r
Fnet
ar
Solve: (d) There is only a single force of weight. We are unable to tell the direction of motion. The description is:
“Galileo has dropped a ball from the Leaning Tower of Pisa.”
Force and Motion
4.34. Visualize:
4-9
y
nr
r
Fnet
r
ar
vr
fk
x
r
w
Solve: (d) There is an object on an inclined surface. The net force is down the plane so the acceleration is down
the plane. The net force includes both the frictional force and the component of the weight. The direction of the
force of kinetic friction implies that the object is moving upward. The description could be: “A car is skidding up
an embankment.”
4.35. Visualize:
y
nr
x
r
fk
r
r
Fnet
T
ar
r
w
vr
Solve: (d) There is an object on an inclined surface with a tension force down the surface. There is a small
frictional force up the surface implying that the object is sliding down the slope. A description could be: “A sled is
being pulled down a slope with a rope which is parallel to the slope.”
4.36. Visualize:
y
r
r
Fnet
Fthrust
ar
x
vr
wr
Solve: (d) There is a thrust at an angle to the horizontal and a weight. There is no normal force so the object is not
on a surface. The description could be: “A rocket is fired at an angle to the horizontal and there is no drag force.”
4.37. Visualize:
r
y
Tension T
vr
r
T
ar
x
r
r
Weight w
Fnet
r
w
Tension is the only contact force. The downward acceleration implies that w > T .
4.38. Visualize:
y
r
vr
Air resistance (drag) D
r
Fthrust
ar
r
Fnet
r
Propulsion Fthrust
r
Weight w
x
r
D
r
w
4-10
Chapter 4
4.39. Visualize:
y
r
r
nr
Drag D
Thrust Fthrust
r
r
Fthrust
D
vr
r
r
a
w
Weight wr
r
Fnet
Normal force nr
The normal force is perpendicular to the ground. The thrust force is parallel to the ground and in the direction of
acceleration. The drag force is opposite to the direction of motion.
4.40. Visualize:
y
vr
r
r
r
Weight w
n
fk
r
Fnet
Normal force nr
r
Friction f k
ar
x
r
w
The normal force is perpendicular to the hill. The frictional force is parallel to the hill.
4.41. Visualize:
y
r
Wind Fw
nr
r
v
r
r
fk
r
Weight w
ar
Fnet
r
Fw
x
Normal force nr
r
Kinetic friction fk
r
w
The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed upward
opposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the slope (that is,
there is no acceleration away from the slope) the net force must be parallel to the slope.
4.42. Visualize:
y
nr
x
vr
r
ar = 0
r
r
Weight w
r
w
Normal force nr
r
Fnet = 0
The ball is rolling, so there is no indication of a speed change (no acceleration, no net force) and no indication of
friction. Thus the weight and the normal force are the only forces.
4.43. Visualize:
y
r
vr
Drag D
ar
x
r
D
r
Weight w
The drag force due to air is opposite the motion.
r
w
r
Fnet
x
Force and Motion
4.44. Visualize:
4-11
y
r
nr
Spring force Fsp
r
r
fk
vr
ar
r
Weight w
r
Friction f k
Fsp
r
r
w
r
x
Fnet
Normal force n
The ball rests on the floor of the barrel because the weight is equal to the normal force. There is a force of the
spring to the right which causes an acceleration.
4.45. Visualize:
y
vr
ar
x
r
w
r
Weight w
r
Fnet
There are no contact forces on the rock. Weight is the only force acting on the rock.
4.46. Visualize:
y
vr
nr
r
r
a
Fnet
r
v
x
r
Weight w
vr
r
w
Normal force nr
The gymnast experiences the long range force of weight. There is also a contact force from the trampoline which is
the normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline is
decreasing her speed, so the acceleration is upward and there is a net force upward. Thus the normal force must be
larger than the weight. The actual behavior of the normal force will be complicated as it involves the stretching of
the trampoline and therefore tensions.
y
4.47. Visualize:
vr
vr
r
vr
nr Fnet
r
ar
r
Weight w
fs
Normal force nr
r
Static friction fs
x
r
w
You can see from the motion diagram that the box accelerates to the right along with the truck. According to
r
r
Newton’s second law, F = ma, there must be a force to the right acting on the box. This is friction, but not kinetic
friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents slipping. Were it
not for static friction, the box would slip off the back of the truck. Static friction acts in the direction needed to
prevent slipping. In this case, friction must act in the forward (toward the right) direction.
4-12
Chapter 4
4.48. Visualize:
y
r
r
v
v
r
r
fs
v
r
a
nr
r
Fnet
r
Weight w
x
wr
Normal force nr
r
Static friction fs
You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows down.
r
r
According to Newton’s second law, F = ma, there must be a force to the left acting on the bag. This is friction, but
not kinetic friction. The bag is not sliding against across the seat. Instead, it is static friction, the force that prevents
slipping. Were it not for static friction, the bag would slide off the seat as the car stops. Static friction acts in the
direction needed to prevent slipping. In this case, friction must act in the backward (toward the left) direction.
4.49. Visualize: (a)
First contact
Loses contact
ar
ar
vr
r
Expanding
Magnified view of
ball in contact with
ground
Compressing
vr
a
Turning point
(b)
(c)
r
Weight w
Normal force nr
y
nr
r
Fnet
x
r
w
(d) The ball accelerates downward until the instant when it makes contact with the ground. Once it makes contact,
it begins to compress and to slow down. The compression takes a short but nonzero distance, as shown in the
motion diagram. The point of maximum compression is the turning point, where the ball has an instantaneous
speed of v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses contact with the
r
ground. The motion diagram shows that the acceleration vector a points upward the entire time that the ball is in
r
contact with the ground. An upward acceleration implies that there is a net upward force Fnet on the ball. The only
two forces on the ball are its weight downward and the normal force of the ground upward. To have a net force
upward requires n > w . So the ball bounces because the normal force of the ground exceeds the weight, causing a
net upward force during the entire time that the ball is in contact with the ground. This net upward force slows the
ball, turns it, and accelerates it upward until it loses contact with the ground. Once contact with the ground is lost,
the normal force vanishes and the ball is simply in free fall.
Force and Motion
4.50. Visualize: (a)
4-13
y
nr
r
Fnet = 0
x
r
r
Weight w
w
Normal force nr
You are sitting on a seat driving along to the right. Both you and the seat are moving with a constant speed. There
is a force on you due to your weight which is directed down. There is a contact force between you and the seat
which is directed up. Since you are not accelerating up or down the net vertical force on you is zero, which means
the two vertical forces are equal in magnitude. The statement of the problem gives no indication of any other
contact forces. Specifically, we are told that the seat is very slippery. We can take this to mean there is no frictional
force. So our force diagram includes only the normal force up, the weight down, and no horizontal force.
(b) The above considerations lead to the free-body diagram that is shown.
(c) The car (and therefore the seat) slows down. Does this create any new force on you? No. The forces remain the
same. This means the pictorial representation and the free-body diagram are unchanged.
(d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and the
road exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remains
unchanged. There are no horizontal forces on you. You do not slow down and you continue at an unchanged
velocity until something in the picture changes for you (for example, you fall off the seat or hit the windshield).
(e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forces
at all. According to Newton’s first law if the net force on you is zero, then you continue to move in a straight line
with a constant velocity. That is what happens to you when the car slows down. You continue to move forward
with a constant velocity. The statement that you are “thrown forward” is misleading and incorrect. To be “thrown”
there would need to be a net force on you and there is none. It might be correct to say that the car has been “thrown
backward” leaving you to continue onward (until you part company with the seat).
(f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional force
between the seat and you. This force is certainly horizontal (parallel to the surface of the seat). Is the frictional
force directed forward (in the direction of motion) or backward? The car is slowing down and you are staying on
the bench. That means you are slowing down with the bench. Your velocity to the right is decreasing (you are
moving right and slowing down) so you are accelerating to the left. By Newton’s second law that means the force
producing the acceleration must be to the left. That force is the force of static friction and it is shown on the freebody diagram below. Of course, when the car accelerates (increases in speed to the right) and you accelerate with
it, then your acceleration is to the right and the frictional force must be to the right.
y
nr
r
fs
x
r
w
r
Weight w
Normal force nr
r
Friction fs
r
Fnet