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Chapter 4
Newton’s Laws
Conceptual Problems
1 • While on a very smooth level transcontinental plane flight, your coffee cup sits motionless on your tray. Are there forces acting on the cup? If so, how do they differ from the forces that would be acting on the cup if it sat on your kitchen table at home?
Determine the Concept Yes, there are forces acting on it. They are the normal force of the table and the gravitational pull of Earth (weight). Because the cup is not accelerating relative to the ground, the forces are the same as those that would act on it if it was sitting on your table at home.
2 • You are passing another car on a highway and determine that, relative to you, the car you pass has an acceleration
r a to the west. However, the driver of the other car is maintaining a constant speed and direction relative to the road. Is the reference frame of your car an inertial one? If not, in which direction (east or west) is your car accelerating relative to the other car?
Determine the Concept No. You are in a non-inertial frame that is accelerating to the east, opposite the other car’s apparent acceleration.
3 • [ SSM ] You are riding in a limousine that has opaque windows that do not allow you to see outside. The car is on a flat horizontal plane, so the car can accelerate by speeding up, slowing down, or turning. Equipped with just a small heavy object on the end of a string, how can you use it to determine if the limousine is changing either speed or direction? Can you determine the limousine’s velocity?
Determine the Concept In the limo you hold one end of the string and suspend the object from the other end. If the string remains vertical, the reference frame of the limo is an inertial reference frame.
4 •• If only a single nonzero force acts on an object, does the object accelerate relative to all inertial reference frames? Is it possible for such an object to have zero velocity in some inertial reference frame and not in another? If so, give a specific example.
Determine the Concept An object accelerates when a net force acts on it. The fact that an object is accelerating tells us nothing about its velocity other than that it is always changing.
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Yes, the object must have acceleration relative to all inertial frames of reference. According to Newton’s first and second laws, an object must accelerate, relative to any inertial reference frame, in the direction of the net force. If there is ″only a single nonzero force,″ then this force is the net force.
Yes, the object’s velocity may be momentarily zero in some inertial reference frame and not in another. During the period in which the force is acting, the object may be momentarily at rest, but its velocity cannot remain zero because it must continue to accelerate. Thus, its velocity is always changing.
5 •• A baseball is acted upon by a single known force. From this information alone, can you tell in which direction the baseball is moving relative to some reference frame? Explain.
Determine the Concept No. Predicting the direction of the subsequent motion correctly requires additional information (knowledge of the initial velocity as well as the acceleration). While the acceleration can be obtained from the net force through Newton’s second law, the velocity can only be obtained by integrating the acceleration.
6 •• A truck moves directly away from you at constant velocity (as observed by you while standing in the middle of the road). It follows that ( a ) no forces act on the truck, ( b ) a constant force acts on the truck in the direction of its velocity, ( c ) the net force acting on the truck is zero, ( d ) the net force acting on the truck is its weight.
Determine the Concept An object in an inertial reference frame accelerates if there is a net force acting on it. Because the object is moving at constant velocity,
the net force acting on it is zero. ( c ) is correct.
7 •• Several space probes have been launched that are now far out in space Pioneer 10 , for example, was launched in the 1970s and is still moving away from the Sun and its planets. Is the mass of Pioneer 10 changing? Which of the known fundamental forces continue to act on it? Does it have a net force on it?
Determine the Concept No. The mass of the probe is constant. However, the solar system will attract the probe with a gravitational force. As the distance between Pioneer 10 and the solar system becomes larger, the magnitude of the gravitational force becomes smaller. There is a net force on the probe because no other forces act on it.
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( b ) Pulling a fish vertically upward at constant velocity while it is still in the water. The forces acting on the fish are the pull, the gravitational force (weight of the fish), and water drag forces. These forces add up to zero.
( c ) The three forces need to add vectorially to zero. An example is a picture hung by two wires.
13 •• [SSM] Suppose a block of mass m 1 rests on a block of mass m 2 and the combination rests on a table as shown in Figure 4-33. Tell the name of the force and its category (contact versus action-at-a-distance) for each of the following forces; ( a ) force exerted by m 1 on m 2 , ( b ) force exerted by m 2 on m 1 , ( c ) force exerted by m 2 on the table, ( d ) force exerted by the table on m 2 , ( e ) force exerted by Earth on m 2. Which, if any, of these forces constitute a Newton’s third law pair of forces?
Determine the Concept
( a ) The force exerted by m 1 on m 2. Normal force, contact
( b ) The force exerted by m 2 on m 1. Normal force, contact
( c ) The force exerted by m 2 on the table.
Normal force, contact
( d ) The force exerted by the table on m 2.
Normal force, contact
( e ) The force exerted by Earth on m 2. Gravitational force, action-at-a-distance
The Newton’s third law force pairs are the two normal forces between the two blocks and the two normal forces between the table and the bottom block. The gravitational force has a third law force pair that acts on Earth and so is not in the question set.
14 •• You yank a fish you have just caught on your line upward from rest into your boat. Draw a free-body diagram of the fish after it has left the water and as it gains speed as it rises. In addition, tell the type (tension, spring, gravity, normal, friction, etc.) and category (contact versus action-at-a-distance) for each force on your diagram. Which, if any, pairs of the forces on your diagram constitute a Newton’s third law pair? Can you tell the relative magnitudes of the forces from the information given? Explain.
Newton’s Laws 299
Determine the Concept A free-body diagram showing the forces acting on the fish is shown to the right. The forces do not constitute a Newton’s 3rd law pair. A table summarizing the type and category of the forces is shown below.
fish
F stringonfish
r
F w
r (^) r Earthonfish=
Force Type Category
F stringonfish
r Tension Contact
F Earthonfish
r Gravity Action-at-a-distance
Because the fish accelerates upward, the tension force must be greater in magnitude than the gravitational force acting on the fish.
15 • If you gently set a fancy plate on the table, it will not break. However if you drop it from a height, it might very well break. Discuss the forces that act on the plate (as it contacts the table) in both these situations. Use kinematics and Newton’s second law to describe what is different about the second situation that causes the plate to break.
Determine the Concept When the plate is sitting on the table, the normal force F n acting upward on it is exerted by the table and is the same size as the gravitational force Fg acting on the plate. Hence, the plate does not accelerate. However, to slow the plate down as it hits the table requires that F n > Fg (or F n >> Fg if the table is hard and the plate slows quickly). A large normal force exerted on delicate china can easily break it.
16 •• For each of the following forces, give what produces it, what object it acts on, its direction, and the reaction force. ( a ) The force you exert on your briefcase as you hold it while standing at the bus stop. ( b ) The normal force on the soles of your feet as you stand barefooted on a horizontal wood floor. ( c ) The gravitational force on you as you stand on a horizontal floor. ( d ) The horizontal force exerted on a baseball by a bat as the ball is hit straight up the middle towards center field for a single.
Determine the Concept
( a ) The force you exert on your briefcase to hold it while standing at the bus stop:
You produce this force. It acts on the briefcase. It acts upward. The reaction force is the force the briefcase exerts on your hand.
Newton’s Laws 301
( b ) False. Action and reaction forces are equal independently of any motion of the involved objects.
19 •• An 80-kg man on ice skates is pushing his 40-kg son, also on skates, with a force of 100 N. Together, they move across the ice steadily gaining speed. ( a ) The force exerted by the boy on his father is (1) 200 N, (2) 100 N, (3) 50 N, or (4) 40 N. ( b ) How do the magnitudes of the two accelerations compare? ( c ) How do the directions of the two accelerations compare?
Determine the Concept
( a ) (2) These forces are a Newton 3rd^ law force pair, and so the force exerted by the boy on his father is 100 N.
( b ) Because the father and son move together, their accelerations will be the same.
( c ) The directions of their acceleration are the same.
20 •• A girl holds a stone in her hand and can move it up or down or keep it still. True or false: ( a ) The force exerted by her hand on the rock is always the same magnitude as the weight of the stone. ( b ) The force exerted by her hand on the rock is the reaction force to the pull of gravity on the stone. ( c ) The force exerted by her hand is always the same size the force her hand feels from the stone but in the opposite direction. ( d ) If the girl moves her hand down at a constant speed, then her upward force on the stone is less than the weight of the stone. ( e ) If the girl moves her hand downward but slows the stone to rest, the force of the stone on the girl’s hand is the same magnitude as the pull of gravity on the stone.
( a ) False. If the rock is accelerating, the force the girl exerts must be greater than the weight of the stone.
( b ) False. The reaction force to the pull of gravity is the force the rock exerts on Earth.
( c ) True. These forces constitute a Newton’s third law pair.
( d ) False. If she moves the stone downward at a constant speed, the net force acting on the stone must be zero.
( e ) False. If she is slowing the stone, it is experiencing acceleration and the net force acting on it can not be zero. The force of her hand on the stone, which has the same magnitude as the force of the stone on her hand, is greater than the force of gravity on the stone.
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21 •• [SSM] A 2.5-kg object hangs at rest from a string attached to the ceiling. ( a ) Draw a free body diagram of the object, indicate the reaction force to each force drawn, and tell what object the reaction force acts on. ( b ) Draw a free body diagram of the string, indicate the reaction force to each force drawn, and tell what object each reaction force acts on. Do not neglect the mass of the string.
Determine the Concept The force diagrams will need to include forces exerted by the ceiling, on the string, on the object, and forces exerted by Earth.
( a )
2.5 kg
F bystringonobject
r
F gbyEarthon object
r
Object
Force Third-Law Pair F bystringonobject
r F byobjecton string
r
F gbyEarthon object
r F byobjecton Earth
r
( b )
String
F gbyEarthonstring
r
F byobjecton string
r
F byceilingon string
r
Force Third-Law Pair F byceilingonstring
r F bystringon ceiling
r
F byEarthon string
r F bystringon Earth
r
F byobjecton string
r F bystringon object
r
22 •• ( a ) Which of the free-body diagrams in Figure 4-34. represents a block sliding down a frictionless inclined surface? ( b ) For the correct figure, label the forces and tell which are contact forces and which are action-at-a-distance forces. ( c ) For each force in the correct figure, identify the reaction force, the object it acts on and its direction.
Determine the Concept Identify the objects in the block’s environment that are exerting forces on the block and then decide in what directions those forces must be acting if the block is sliding down the inclined plane.
( a ) Free-body diagram ( c ) is correct.
304 Chapter 4
( c ) When the box separates from the spring, the force exerted by the spring on the box goes to zero. Note that it is still true that
F n (^) byfloor F gbyEarth
r r =.
F nbyfloor
r
F gby Earth
r
24 •• Imagine yourself seated on a wheeled desk chair at your desk. Consider any friction forces between the chair and the floor to be negligible. However, the friction forces between the desk and the floor are not negligible. When sitting at rest, you decide you need another cup of coffee. You push horizontally against the desk, and the chair rolls backward away from the desk. ( a ) Draw your free-body diagram of yourself during the push and clearly indicate which force was responsible for your acceleration. ( b ) What is the reaction force to the force that caused your acceleration? ( c ) Draw the free-body diagram of the desk and explain why it did not accelerate. Does this violate Newton’s third law? Explain.
Determine the Concept In the following free-body diagrams we’ll assume that the desk is to the left and that your motion is to the right.
( a ) Newton’s third law accounts for this as follows. When you push with your hands against the desk, the desk pushes back on your hands with a force of the same magnitude but opposite direction. This force accelerates you backward.
F g
r
F by desk
r
F n
r
( b ) The reaction force to the force that caused your acceleration is the force that you exerted on the desk.
( c ) When you pushed on the desk, you did not apply sufficient force to overcome the force of friction between the desk and the floor. In terms of forces on the desk, you applied a force, and the floor applied a friction force that, when added as vectors, cancelled. The desk, therefore, did not accelerate and Newton’s third law is not violated. The forces in the diagram do not constitute a Newton’s third law pair of forces, even though they are equal in magnitude and opposite in direction.
Newton’s Laws 305
F byyou
r
F by floor
r
F gon desk
r
F nby floor
r
25 ••• The same (net) horizontal force F is applied for a fixed time interval Δ t to each of two objects, having masses m 1 and m 2 , that sit on a flat, frictionless surface. (Let m 1 > m 2 .) ( a ) Assuming the two objects are initially at rest, what is the ratio of their accelerations during the time interval in terms of F , m 1 and m 2? ( b ) What is the ratio of their speeds v 1 and v 2 at the end of the time interval? ( c ) How far apart are the two objects (and which is ahead) the end of the time interval?
Picture the Problem We can apply Newton’s second law to find the ratios of the accelerations and speeds of the two objects and constant-acceleration equations to express the separation of the objects as a function of the elapsed time.
( a ) Use Newton’s second law to express the accelerations of the two objects: 1
(^1) m
F
a = and 2
(^2) m
F
a =
Dividing the first of these equations by the second and simplifying yields:
1
2
2
1 2
1 m
m
m
F
m
F
a
a = =
( b ) Because both objects started from rest, their speeds after time Δ t has elapsed are:
v (^) 1 = a 1 Δ t and v (^) 2 = a 2 Δ t
Dividing the first of these equations by the second and simplifying yields: 1
2 2
1 2
1 2
1 Δ
m
m a
a a t
a t v
v = = =
( c ) The separation of the two objects at the end of the time interval is given by:
Δ x = Δ x 2 −Δ x 1 (1)
Newton’s Laws 307
Apply Newton’s second law to the puck as it is slowed by the goalie’s glove to express the magnitude of the force the glove exerts on the puck:
F glove (^) onpuck= m puck a puck (1)
Use a constant-acceleration equation to relate the initial and final speeds of the puck to its acceleration and stopping distance:
v^2 = v 02 + 2 a puck^ (Δ^ x )puck
Solving for a puck yields:
( )puck
2 0
2 puck (^2) Δ x
v v a
Substitute for a puck in equation (1) to obtain:
( )puck
2 0
2 puck glove onpuck 2 Δ x
m v v F
Substitute numerical values and evaluate F glove on puck :
3 .6 kN
10 m
200 kg 0 60 m/s 2 gloveonpuck
F =
Remarks: The force on the puck is about 1800 times its weight.
28 •• A baseball player slides into second base during a steal attempt. Assuming reasonable values for the length of the slide, the speed of the player at the beginning of the slide, and the speed of the player at the end of the slide, estimate the average force of friction acting on the player.
Picture the Problem Let’s assume that the player’s mass is 100 kg, that he gets going fairly quickly down the base path, and that his speed is 8.0 m/s when he begins his slide. Further, let’s assume that he approaches the base at the end of the slide at 3.0 m/s. From these speeds, and the length of the slide, we can use Newton’s second law and a constant-acceleration equation to find the force due to friction (which causes the slowing down).
Apply Newton’s second law to the sliding runner:
∑ Fx = F friction = ma (1)
Using a constant-acceleration equation, relate the runner’s initial and final speeds to his acceleration and the length of his slide:
v (^) f^2 = v i^2 + 2 a Δ x ⇒ x
v v a 2 Δ
2 i
2 = f^ −
Substituting for a in equation (1) yields: (^) x
v v F m 2 Δ
2 i
2 f friction
308 Chapter 4
Assuming the player slides 2.0 m, substitute numerical values and evaluate F friction :
( ) ( ) ( ) ( )
4 kN
0 m
0 m/s 8. 0 m/s 100 kg
2 2 friction
≈ −
F =
where the minus sign indicates that the force of friction opposes the runner’s motion.
29 •• A race car skidding out of control manages to slow down to 90 km/h before crashing head-on into a brick wall. Fortunately, the driver is wearing a safety harness. Using reasonable values for the mass of the driver and the stopping distance, estimate the average force exerted on the driver by the safety harness, including its direction. Neglect any effects of frictional forces on the driver by the seat.
Picture the Problem Assume a crush distance of 1.0 m at 90 km/h (25 m/s) and a driver’s mass of 55 kg. We can use a constant-acceleration equation (the definition of average acceleration) to find the acceleration of the driver and Newton’s second law to find the force exerted on the driver by the seat belt.
Apply Newton’s second law to the driver as she is brought to rest by her safety harness:
driver driver on driver
F safety (^) harness = m a (1)
Use a constant-acceleration equation to relate the initial and final speeds of the driver to her acceleration and stopping distance:
v^2 = v 02 + 2 a driver (Δ x )driver
Solving for a driver yields:
( )driver
2 0
2 driver 2 Δ x
v v a
Substitute for a driver in equation (1) to obtain: (^) ( ) ⎟⎟ ⎠
driver
2 0
2 driver ondriver safety harness 2 Δ x
v v F m
Substitute numerical values and evaluate ondriver
F safety (^) harness: (^ )^
( ) ( )
17 kN
- 0 m
0 25 m/s 55 kg
2
on driver safetyharness
F =
where the minus sign indicates that the force exerted by the safety harness is in the opposite direction from the driver’s motion.
310 Chapter 4
Substitute for ax in equation (2) to obtain: 2 0
(^2) net v x
xF m
Substitute numerical values and evaluate m : (^) ( 25.0m/s) 3.^00 kg
2 (62.5 m)(15.0N) m = 2 =
31 • An object has an acceleration of 3.0 m/s^2 when a single force of magnitude F 0 acts on it. ( a ) What is the magnitude of its acceleration when the magnitude of this force is doubled? ( b ) A second object has an acceleration of 9.0 m/s 2 under the influence of a single force of magnitude F 0. What is the ratio of the mass of the second object to that of the first object? ( c ) If the two objects are glued together to form a composite object, what acceleration magnitude will a single force of magnitude F 0 acting on the composite object produce?
Picture the Problem The acceleration of an object is related to its mass and the net force acting on it by F net (^) = F 0 = ma.
( a ) Use Newton’s second law of motion to relate the acceleration of the object to the net force acting on it:
m
F
a = net
When F net = 2 F 0 : 0
(^2 0 2) a m
F
a = =
Substitute numerical values and evaluate a :
a = 2 ( 3.0 m/s^2 ) = 6.0m/s^2
( b ) Let the subscripts 1 and 2 distinguish the two objects. The ratio of the two masses is found from Newton’s second law:
3
1 2
2
2
1 0 1
0 2 1
2 9.0m/s
3.0 m/s = = = = a
a F a
F a m
m
( c ) The acceleration of the composite object is the net force divided by the total mass m = m 1 + m 2 of the composite object: (^431) 3
1
1
2 1
0 1 1 2
net 0
a
a
m m
F m m m
F
m
F
a
Substitute for a 1 and evaluate a : (^) ( 2 ) 2 4 a =^3 3. 0 m/s = 2. 3 m/s
32 • A tugboat tows a ship with a constant force of magnitude F 1. The increase in the ship’s speed during a 10-s interval is 4.0 km/h. When a second tugboat applies an additional constant force of magnitude F 2 in the same direction, the speed increases by 16 km/h during a 10-s interval. How do the
Newton’s Laws 311
magnitudes of F 1 and F 2 compare? (Neglect the effects of water resistance and air resistance.)
Picture the Problem The acceleration of an object is related to its mass and the net force acting on it by F net (^) = ma. Let m be the mass of the ship, a 1 be the
acceleration of the ship when the net force acting on it is F 1 , and a 2 be its acceleration when the net force is F 1 + F 2.
Using Newton’s second law, express the net force acting on the ship when its acceleration is a 1 :
F 1 = ma 1
Express the net force acting on the ship when its acceleration is a 2 :
F 1 + F 2 = ma 2
Divide the second of these equations by the first and solve for the ratio F 2 / F 1 : 1
2 1
1 2 ma
ma F
F F
1
2 1
a
a F
F
Substitute for the accelerations to determine the ratio of the accelerating forces and solve for F 2 to obtain:
10 s
- 0 km/h
10 s
16 km/h
1
F
F
⇒ F 2 = 3 F 1
33 • A single constant force of magnitude 12 N acts on a particle of mass m. The particle starts from rest and travels in a straight line a distance of 18 m in 6.0 s. Find m.
Picture the Problem The mass of the particle is related to its acceleration and the net force acting on it by Newton’s second law of motion. Because the force is constant, we can use constant-acceleration formulas to calculate the acceleration. Choose a coordinate system in which the + x direction is the direction of motion of the particle.
The mass is related to the net force and the acceleration by Newton’s second law: x
x a
F
m =
a
F
r
r (1)
Because the force is constant, the acceleration is constant. Use a constant-acceleration equation to relate the displacement of the particle to it’s acceleration:
Δ x = v 0 xt + 21 ax (Δ t )^2 or, because v 0 x = 0,
( ) 2 2 Δ x^1 a Δ t = (^) x ⇒ ( Δ)^2
t
x a (^) x =
Substitute for a (^) x in equation (1) to obtain:
( ) x
F t m x 2 Δ
Newton’s Laws 313
Substitute numerical values and evaluate F stopping : (^ )
( ) ( )
8 kN
00 cm
500 m/s
- 80 10 kg
2 3 stopping
= −
F = − × −
where the minus sign indicates that F stopping opposes the motion of the bullet.
( b ) Solving equation (2) for Δ x yields: stopping
2 i 2
F
v x = − m x (3)
For m = m ′ and Δ x = Δ x ′ :
stopping
2 i 2
F
v x' = − m' x
Evaluate this expression for m' = 21 m to obtain: stopping
2 i 4
F
v x' = − m x (4)
Dividing equation (4) by equation (3) yields:
2 1
stopping
2 i
stopping
2 i
F
v m
F
v m
x
x' x
x
or Δ x' = 21 Δ x
Substitute numerical values and evaluate Δ x ′:
Δ x' = 21 ( 6. 00 cm) = 3. 00 cm
36 •• A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.55 s to travel a distance of 1.50 m. The mass of the cart plus fan is 355 g. Assume that the cart travels with constant acceleration. ( a ) What is the net force exerted on the cart-fan combination? ( b ) Mass is added to the cart until the total mass of the cart-fan combination is 722 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.50 m now? Ignore the effects due to friction.
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Picture the Problem Choose the coordinate system shown in the
diagram to the right. The force F
r acting on the cart-fan combination is the consequence of the fan blowing air to the left. We can use Newton’s second law and a constant-acceleration equation to express the relationship
between F
r and the mass of the cart-fan combination and the distance it travels in a given interval of time.
F g
r
F
r
F n
r
y
x
( a ) Apply Newton’s second law to the cart-fan combination to obtain:
∑ Fx = F = max (1)
Using a constant-acceleration equation, relate the distance the cart- fan combination travels to its initial speed, acceleration, and the elapsed time:
Δ x = v 0 x Δ t + 21 ax (Δ t )^2 or, because v 0 x = 0,
Δ x = 21 ax (Δ t )^2 ⇒ ( Δ)^2
t
x a (^) x =
Substitute for ax in equation (1) to obtain: (^) ( Δ )^2
t
x F = m (2)
Substitute numerical values and evaluate F : (^ )^
( ) ( )
05144 N 0. 0514 N
55 s
2 1.50m
- 355 kg 2
F =
( b ) Solve equation (2) for Δ t to obtain: (^) F
m x t
Substitute numerical values and evaluate Δ t :
( )( )
- 49 s
- 05144 N
2 0.722kg 1. 50 m Δ t = =
37 •• A horizontal force of magnitude F 0 causes an acceleration of 3.0 m/s 2 when it acts on an object of mass m sliding on a frictionless surface. Find the magnitude of the acceleration of the same object in the circumstances shown in Figure 4-35 a and 4-35 b.
Picture the Problem The acceleration of an object is related to its mass and the net force acting on it through Newton’s second law. Choose a coordinate system in which the direction of 2 F 0 in ( b ) is the positive direction and the direction of the left-most F 0 in ( a ) is the positive direction. Find the resultant force in each case and then find the resultant acceleration.
