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# Force and Acceleration

By harsharan Jun 14, 2013 17605 Words
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 r highway and determine that, relative to you, the car you pass has an acceleration 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.

295

296 Chapter 4 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.

Newton’s Laws 297
8

Astronauts in apparent weightlessness during their stay on the International Space Station must carefully monitor their masses because significant loss of body mass is known to cause serious medical problems. Give an example of how you might design equipment to measure the mass of an astronaut on the orbiting space station. ••

298 Chapter 4 (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 m1 rests on a block of mass m2 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 m1 on m2, (b) force exerted by m2 on m1, (c) force exerted by m2 on the table, (d) force exerted by the table on m2, (e) force exerted by Earth on m2. Which, if any, of these forces constitute a Newton’s third law pair of forces? Determine the Concept (a) The force exerted by m1 on m2. (b) The force exerted by m2 on m1. (c) The force exerted by m2 on the table. (d) The force exerted by the table on m2. (e) The force exerted by Earth on m2. Normal force, contact Normal force, contact Normal force, contact Normal force, contact 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. Force r Fstring on fish r FEarth on fish Type Tension Category Contact r Fstring on fish

fish

r r FEarth on fish = w

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 Fn 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 Fn > Fg (or Fn >> 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.

300 Chapter 4 (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. The floor produces this force. It acts on your feet. It acts upward. The reaction force is the force your feet exert on the floor. Earth produces this force. It acts on you. It acts downward. The reaction force is the gravitational force you exert on Earth. The bat produces this force. It acts on the ball. It acts horizontally. The reaction force is the force the ball exerts on the bat.

(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.

17 •• For each case, identify the force (including its direction) that causes the acceleration. (a) A sprinter at the very start of the race. (b) A hockey puck skidding freely but slowly coming to rest on the ice. (c) A long fly ball at the top of its arc. (d) A bungee jumper at the very bottom of her descent. Determine the Concept (a) A sprinter at the very start of the race: (b) A hockey puck skidding freely but slowly coming to rest on the ice: (c) A long fly ball at the top of its arc: (d) A bungee jumper at the very bottom of her descent: 18 (a) (b) • True or false: The normal force of the block on the sprinter, in the forward direction. The frictional force by the ice on the puck, in the opposite direction to the velocity. The gravitation force by Earth on the ball, in the downward direction. The force exerted by the bungee cord on the jumper, in the upward direction.

If two external forces that are both equal in magnitude and opposite in direction act on the same object, the two forces can never be a Newton’s third law pair. The two forces of a Newton’s third law pair are equal only if the objects involved are not accelerating.

(a) True. By definition, third law pairs cannot act on the same object.

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.

302 Chapter 4 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) r Fby string on object

2.5 kg Object

r Fg by Earth on object

Force r Fby string on object r Fg by Earth on object

Third-Law Pair r Fby object on string r Fby object on Earth

(b)

r Fby ceiling on string
String

r Fg by Earth on string r Fby object on string

Force r Fby ceiling on string r Fby Earth on string r Fby object on string

Third-Law Pair r Fby string on ceiling r Fby string on Earth r Fby string on object

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.

Newton’s Laws 303 (b) Because the incline is frictionless, the force r Fn the incline exerts on the block must be normal to the surface and is a contact force. The second object capable of exerting a force on the block is Earth and its force; the gravitational force r Fg acting on the block acts directly downward and is an action-at-a-distance force. The magnitude of the normal force is less than that of the weight because it supports only a portion of the weight.

r Fn

r Fg

(c) The reaction to the normal force is the force the block exerts perpendicularly on the surface of the incline. The reaction to the gravitational force is the upward force the block exerts on Earth. 23 •• A wooden box on the floor is pressed against a compressed, horizontal spring that is attached to a wall. The horizontal floor beneath the box is frictionless. Draw the free body diagram of the box in the following cases. (a) The box is held at rest against the compressed spring. (b) The force holding the box against the spring no longer exists, but the box is still in contact with the spring. (c) When the box no longer has contact with the spring. Determine the Concept In the following free-body diagrams we’ll assume that the box is initially pushed to the left to compress the spring. (a) Note that, in the free-body diagram to the right, that r r Fn by floor = Fg by Earth and r r Fby spring = Fby hand .

r Fby hand

r Fn by floor

r Fby spring

r Fg by Earth
(b) Note that while r r r Fn by floor = Fg by Earth , Fby spring is now the net force acting on the box. As r the spring decompresses, Fby spring will become smaller.

r Fn by floor

r Fby spring

r Fg by Earth

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 r r Fn by floor = Fg by Earth .

r Fn by floor

r Fg by Earth

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.

r Fn

r Fby desk r Fg

(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

r Fby you

r Fn by floor

r Fby floor
r Fg on desk

25 ••• The same (net) horizontal force F is applied for a fixed time interval Δt to each of two objects, having masses m1 and m2, that sit on a flat, frictionless surface. (Let m1 > m2.) (a) Assuming the two objects are initially at rest, what is the ratio of their accelerations during the time interval in terms of F, m1 and m2? (b) What is the ratio of their speeds v1 and v2 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: Dividing the first of these equations by the second and simplifying yields:

a1 =

F F and a2 = m1 m2

F a1 m1 m2 = = F a2 m1 m2
v1 = a1Δt and v2 = a2 Δt

(b) Because both objects started from rest, their speeds after time Δt has elapsed are: Dividing the first of these equations by the second and simplifying yields: (c) The separation of the two objects at the end of the time interval is given by:

v1 a1Δt a1 m2 = = = v2 a2 Δt a 2 m1

Δx = Δx2 − Δx1

(1)

306 Chapter 4 Using a constant acceleration equation, express the distances traveled by the two objects when time Δt has elapsed: Substitute for Δx1 and Δx2 in equation (1) and simplify to obtain: Δx1 = 1 a1 (Δt ) 2 and 2 Δx2 = 1 a 2 (Δt ) 2

2

Δx = 1 a 2 (Δt ) − 1 a1 (Δt ) 2 2
2

2

⎛ 1 1 ⎞ 2 − ⎟ (Δt ) F⎜ ⎜m m ⎟ 1 ⎠ ⎝ 2 and, because m1 > m2, the object whose mass is m2 is ahead. = 1 2

Estimation and Approximation
26 •• Most cars have four springs attaching the body to the frame, one at each wheel position. Devise an experimental method of estimating the force constant of one of the springs using your known weight and the weights of several of your friends. Assume the four springs are identical. Use the method to estimate the force constant of your car’s springs. Picture the Problem Suppose you put in 800 lbs (or about 3600 N) of weight and the car sags several inches (or 6.00 cm). Then each spring supports about 900 N and we can use the definition of the force constant k to determine its value. The force constant is the ratio of the compressing (or stretching) force to the compression (or stretch):

k=

Fx Δx
1 4

Substitute numerical values and evaluate k:

k =

(3600 N )
6.00 cm

≈ 150 N/cm

27 •• [SSM] Estimate the force exerted on the goalie’s glove by the puck when he catches a hard slap shot for a save. Picture the Problem Suppose the goalie’s glove slows the puck from 60 m/s to zero as it recoils a distance of 10 cm. Further, assume that the puck’s mass is 200 g. Because the force the puck exerts on the goalie’s glove and the force the goalie’s glove exerts on the puck are action-and-reaction forces, they are equal in magnitude. Hence, if we use a constant-acceleration equation to find the puck’s acceleration and Newton’s second law to find the force the glove exerts on the puck, we’ll have the magnitude of the force exerted on the goalie’s glove.

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: Use a constant-acceleration equation to relate the initial and final speeds of the puck to its acceleration and stopping distance: Solving for apuck yields:

Fglove on puck = mpuck apuck

(1)

2 v 2 = v0 + 2a puck (Δx )puck

apuck =

2 v 2 − v0 2(Δx )puck

Substitute for apuck in equation (1) to obtain:

Fglove on puck =

2 mpuck v 2 − v0

2(Δx )puck

(

)

Substitute numerical values and evaluate Fglove on puck:

Fglove on puck =

(0.200 kg )(0 − (60 m/s)2 ) 2(0.10 m )

≈ 3.6 kN
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: Using a constant-acceleration equation, relate the runner’s initial and final speeds to his acceleration and the length of his slide: Substituting for a in equation (1) yields:

∑ Fx = Ffriction = ma vf2 − vi2 2Δx

(1)

vf2 = vi2 + 2aΔx ⇒ a =

Ffriction = m

vf2 − vi2 2Δx

308 Chapter 4 Assuming the player slides 2.0 m, substitute numerical values and evaluate Ffriction :

(3.0 m/s)2 − (8.0 m/s )2 Ffriction = (100 kg ) 2(2.0 m )
≈ − 1.4 kN 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: Use a constant-acceleration equation to relate the initial and final speeds of the driver to her acceleration and stopping distance:

on driver

(1)

2 v 2 = v0 + 2adriver (Δx )driver

2 v 2 − v0 = 2(Δx )driver

Substitute for adriver in equation (1) to obtain: Substitute numerical values and evaluate Fsafety harness : on driver

2 ⎛ v 2 − v0 ⎞ ⎟ Fsafety harness = mdriver ⎜ ⎟ ⎜ 2(Δx ) on driver driver ⎠ ⎝

⎛ 0 − (25 m/s )2 ⎞ ⎟ Fsafety harness = (55 kg )⎜ ⎜ 2(1.0 m ) ⎟ on driver ⎠ ⎝ ≈ − 17 kN where the minus sign indicates that the force exerted by the safety harness is in the opposite direction from the driver’s motion.

Newton’s Laws 309
Remarks: The average force on the safety harness is about 32 times her weight.

Newton’s First and Second Laws: Mass, Inertia, and Force
30 • A particle is traveling in a straight line at constant speed of 25.0 m/s. Suddenly a constant force of 15.0 N acts on it, bringing it to a stop in a distance of 62.5 m. (a) What is the direction of the force? (b) Determine the time it takes for the particle to come to a stop. (c) What is its mass? Picture the Problem The acceleration of the particle, its stopping time, and its mass can be found using constant-acceleration equations and Newton’s second law. A convenient coordinate system is shown in the following diagram.

y
r F

r v0

x

(a) Because the constant force slows the particle, we can conclude that, as shown in the diagram, its direction is opposite the direction of the particle’s motion. (b) Use a constant-acceleration equation to relate the initial and final velocities of the particle to its acceleration and stopping time: v x = v0 x + a x Δt or, because vx = 0, 0 = v0 x + a x Δt ⇒ Δt = − 2 2 v x = v0 x + 2a x Δx or, because vx = 0, 2 0 = v0 x + 2a x Δx ⇒ a x = 2 − v0 x 2Δx

v0 x ax

(1)

Use a constant-acceleration equation to relate the initial and final velocities of the particle to its acceleration and stopping distance: Substituting for ax in equation (1) and simplifying yields: Substitute numerical values and evaluate Δt: (c) Apply Newton’s second law to the particle to obtain: Solving for m yields:

Δt = Δt =

2Δx v0 x 2(62.5 m ) = 5.00 s 25.0 m/s

∑ Fx = − Fnet = ma x − Fnet ax

m=

(2)

310 Chapter 4 Substitute for ax in equation (2) to obtain: Substitute numerical values and evaluate m:

m=

2ΔxFnet 2 v0 x 2(62.5 m) (15.0 N) = 3.00 kg (25.0 m/s)2

m=

An object has an acceleration of 3.0 m/s2 when a single force of 31 • magnitude F0 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/s2 under the influence of a single force of magnitude F0. 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 F0 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 Fnet = F0 = ma.

(a) Use Newton’s second law of motion to relate the acceleration of the object to the net force acting on it: When Fnet = 2F0:

a=

Fnet m

a=

2 F0 = 2a0 m

Substitute numerical values and evaluate a: (b) Let the subscripts 1 and 2 distinguish the two objects. The ratio of the two masses is found from Newton’s second law: (c) The acceleration of the composite object is the net force divided by the total mass m = m1 + m2 of the composite object: Substitute for a1 and evaluate a:

a = 2 3.0 m/s 2 = 6.0 m/s 2

(

)

m2 F0 a2 a1 3.0 m/s 2 = = = = m1 F0 a1 a2 9.0 m/s 2

1 3

a= =
a=

Fnet F0 F0 m1 = = m m1 + m2 1 + m2 m1 a1 = 3 a1 4 1+ 1 3
3 4

(3.0 m/s ) =
2

2.3 m/s 2

A tugboat tows a ship with a constant force of magnitude F1. The 32 • 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 F2 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 F1 and F2 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 Fnet = ma . Let m be the mass of the ship, a1 be the acceleration of the ship when the net force acting on it is F1, and a2 be its acceleration when the net force is F1 + F2.

Using Newton’s second law, express the net force acting on the ship when its acceleration is a1: Express the net force acting on the ship when its acceleration is a2: Divide the second of these equations by the first and solve for the ratio F2/F1: Substitute for the accelerations to determine the ratio of the accelerating forces and solve for F2 to obtain:

F1 = ma1

F1 + F2 = ma2 F1 + F2 ma2 F a = ⇒ 2 = 2 −1 F1 a1 F1 ma1 16 km/h F2 10 s = − 1 = 3 ⇒ F2 = 3F1 F1 4.0 km/h 10 s

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: 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: Substitute for ax in equation (1) to obtain:

r ∑ F Fx m= r = a ax
Δx = v0 x t + 1 a x (Δt ) 2 or, because v0x = 0,
2 2

(1)

Δx = 1 a x (Δt ) ⇒ a x = 2

2Δx (Δt )2

m=

Fx (Δt ) 2Δx

2

312 Chapter 4 Substitute numerical values and evaluate m:

(12 N )(6.0 s ) 2 m= 2(18 m )

= 12 kg

34 • A net force of (6.0 N) ˆ – (3.0 N) jˆ acts on an object of mass 1.5 kg. i r Find the acceleration a . Picture the Problem The acceleration of an object is related to its mass and the r r net force acting on it according to a = Fnet m .

Apply Newton’s second law to the object to obtain: Substitute numerical values and r evaluate a :

r r Fnet a= m
ˆ r (6.0 N ) i − (3.0 N ) ˆ j a= 1.5 kg
=

(4.0 m/s ) iˆ − (2.0 m/s ) ˆj
2 2

35 •• [SSM] A bullet of mass 1.80 × 10–3 kg moving at 500 m/s impacts a tree stump and penetrates 6.00 cm into the wood before coming to rest. (a) Assuming that the acceleration of the bullet is constant, find the force (including direction) exerted by the wood on the bullet. (b) If the same force acted on the bullet and it had the same speed but half the mass, how far would it penetrate into the wood? Picture the Problem Choose a coordinate system in which the +x direction is in the direction of the motion of the bullet and use Newton’s second law and a constant- acceleration equation to express the relationship between Fstopping and the mass of the bullet and its displacement as it is brought to rest in the block of wood.

(a) Apply Newton’s second law to the bullet to obtain: Use a constant-acceleration equation to relate the bullet’s initial and final speeds, acceleration, and stopping distance: Substitute for ax in equation (1) to obtain:

∑ Fx = Fstopping = ma x

(1)

vf2x = vi2 + 2a x Δx x or, because vfx = 0,
− vi2 x 0 = v + 2a x Δx ⇒ a x = 2Δx
2 ix

Fstopping = − m

vi2 x 2Δx

(2)

Newton’s Laws 313 Substitute numerical values and evaluate Fstopping : Fstopping = −(1.80 × 10 −3 kg ) = − 3.8 kN

(500 m/s)2 2(6.00 cm )

where the minus sign indicates that Fstopping opposes the motion of the bullet. (b) Solving equation (2) for Δx yields: For m = m′ and Δx = Δx′ : Δx = −m vi2 x 2 Fstopping vi2 x 2 Fstopping (4) (3)

Δx' = − m'

Evaluate this expression for m' = 1 m to obtain: 2 Dividing equation (4) by equation (3) yields:

vi2 x Δx' = − m 4 Fstopping −m vi2 x 4 Fstopping

Δx' = Δx

vi2 x −m 2 Fstopping

=

1 2

or Δx' = 1 Δx 2 Substitute numerical values and evaluate Δx′: Δx' =
1 2

(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.

314 Chapter 4
Picture the Problem Choose the coordinate system shown in the r diagram to the right. The force F 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 r between F and the mass of the cart-fan combination and the distance it travels in a given interval of time.

y
r Fn

r F x

r Fg
∑ Fx = F = ma x
Δx = v0 x Δt + 1 a x (Δt ) 2 or, because v0x = 0,
2 2

(a) Apply Newton’s second law to the cart-fan combination to obtain: Using a constant-acceleration equation, relate the distance the cartfan combination travels to its initial speed, acceleration, and the elapsed time: Substitute for ax in equation (1) to obtain: Substitute numerical values and evaluate F:

(1)

Δx = 1 a x (Δt ) ⇒ a x = 2

2Δx (Δt )2 (2)

F =m

2Δx (Δt )2
2(1.50 m )

F = (0.355 kg )

(4.55 s )2

= 0.05144 N = 0.0514 N (b) Solve equation (2) for Δt to obtain: Substitute numerical values and evaluate Δt: Δt = 2mΔx F 2(0.722 kg )(1.50 m ) = 6.49 s 0.05144 N

Δt =

37 •• A horizontal force of magnitude F0 causes an acceleration of 3.0 m/s2 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 4-35a and 4-35b. 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 2F0 in (b) is the positive direction and the direction of the left-most F0 in (a) is the positive direction. Find the resultant force in each case and then find the resultant acceleration.

Newton’s Laws 315 (a) Apply Newton’s second law to the object to obtain: The magnitude of the net force is given by: Substitute for Fnet in equation (1) and simplify to obtain: Substitute the numerical value of a0 and evaluate a: (b)The magnitude of net force is given by: Fnet = Fx2 + Fy2 =

a=

Fnet m

(1)

Fnet = Fx2 + Fy2 = 2F0 2 F0 = 2a0 m
2 3.0 m/s 2 = 4.2 m/s 2

a =
a =

(

)

(− F0 sin 45°)2 + (2 F0 + F0 cos 45°)2
a = 2.80

= 2.80 F0

Substitute for Fnet in equation (1) and simplify to obtain: Substitute the numerical value of a0 and evaluate a:

F0 ma = 2.80 0 = 2.80a0 m m

a = 2.80 3.0 m/s 2 = 8.4 m/s 2

(

)

38 •• Al and Bert stand in the middle of a large frozen lake (frictionless surface). Al pushes on Bert with a force of 20 N for 1.5 s. Bert’s mass is 100 kg. Assume that both are at rest before Al pushes Bert. (a) What is the speed that Bert reaches as he is pushed away from Al? (b) What speed does Al reach if his mass is 80 kg? Picture the Problem The speed of either Al or Bert can be obtained from their accelerations; in turn, they can be obtained from Newtons second law applied to each person. The free-body diagrams to the right show the forces acting on Al and Bert. The forces that Al and Bert exert on each other are action-and-reaction forces. y

r Fn, Bert

r Fn, Al

r FAl on Bert

r FBert on Al

x

r mBert g

r m Al g

(a) Apply ∑ Fx = ma x to Bert:

− FAl on Bert = mBert aBert ⇒ aBert = − 20 N = −0.200 m/s 2 100 kg

− FAl on Bert mBert

Substitute numerical values and evaluate aBert:

a Bert =

316 Chapter 4 Using a constant-acceleration equation, relate Bert’s speed to his initial speed, speed after 1.5 s, and acceleration: Substitute numerical values and evaluate Bert’s speed at the end of 1.5 s: (b) From Newton's 3rd law, an equal but oppositely directed force acts on Al while he pushes Bert. Because the ice is frictionless, Al speeds off in the opposite direction. Apply ∑ Fx = ma x to Al: Solving for Al’s acceleration yields: Substitute numerical values and evaluate aAl,x: Using a constant-acceleration equation, relate Al’s speed to his initial speed, speed after 1.5 s, and acceleration: Substitute numerical values and evaluate Al’s speed at the end of 1.5 s: v x = v0 x + aBert,x Δt

v x = 0 + (− 0.200 m/s 2 )(1.5 s ) = − 0.30 m/s

∑F

x , Al

= FBert on Al,x = mAl a Al,x

aAl,x =

FBert on Al,x mAl

a Al,x =

20 N = 0.250 m/s 2 80 kg

v x = v0 x + aAl,x Δt

v x (1.5 s ) = 0 + (0.250 m/s 2 )(1.5 s ) = 0.38 m/s

39 •• If you push a block whose mass is m1 across a frictionless floor with a horizontal force of magnitude F0, the block has an acceleration of 12 m/s2. If you push on a different block whose mass is m2 with a horizontal force of magnitude F0, its acceleration is 3.0 m/s2. (a) What acceleration will a horizontal force of magnitude F0 give to a single block with mass m2 – m1? (b) What acceleration will a horizontal force of magnitude F0 give to a single block with mass m2 + m1? Picture the Problem The free-body diagrams show the forces acting on the two blocks. We can apply Newton’s second law to the forces acting on the blocks and eliminate F to obtain a relationship between the masses. Additional applications of Newton’s second law to the sum and difference of the masses will lead us to values for the accelerations of these combinations of mass.

Newton’s Laws 317
y y

r Fn,1

r Fn,2
r F0

m1

x

m2

r F0

x

r m1 g

r m2 g

(a) Apply blocks:

∑F

x

= ma x to the two

∑F

1, x

= F0 = m1a1, x

and ∑ F2,x = F0 = m2a2,x m2 = a1, x a2, x m1

Eliminate F0 between the two equations and solve for m2: Substitute numerical values to obtain: Express the acceleration of an object whose mass is m2 – m1 when the net force acting on it is F0: Substitute numerical values and evaluate ax: (b) Express the acceleration of an object whose mass is m2 + m1 when the net force acting on it is F0: Substitute numerical values and evaluate ax:

m2 =

12 m/s 2 m1 = 4.0 m1 3.0 m/s 2 F0 F0 F = = 0 = 1 a1, x 3 m2 − m1 4m1 − m1 3m1

ax =

a x = 1 12 m/s 2 = 4.0 m/s 2 3

(

)

ax =

F0 F0 F = = 0 = 1 a1, x 5 m2 + m1 4m1 + m1 5m1

a x = 1 12 m/s 2 = 2.4 m/s 2 5

(

)

40 •• To drag a 75.0-kg log along the ground at constant velocity, your tractor has to pull it with a horizontal force of 250 N. (a) Draw the free body diagram of the log. (b) Use Newton’s laws to determine the force of friction on the log. (c) What is the normal force of the ground on the log? (d) What horizontal force must you exert if you want to give the log an acceleration of 2.00 m/s2 assuming the force of friction does not change. Redraw the log’s free body diagram for this situation. Picture the Problem Because the velocity is constant, the net force acting on the log must be zero. Choose a coordinate system in which the positive x direction is the direction of motion of the log and apply Newton’s second law to the log.

318 Chapter 4 (a) The free-body diagram shows the forces acting on the log when it is being dragged in the +x direction at constant velocity. r Fby ground r Fg

y
r Fn
r

Fpull

x

(b) Apply

∑F

x

= ma x to the log

∑ Fx = Fpull − Fby ground = ma x = 0 or Fby ground = Fpull Fby ground = Fpull = 250 N

when it is moving at constant speed:

Substitute for Fpull and evaluate the force of friction Fby ground: (c) Apply to obtain:

∑F

y

= ma y to the log

∑F

y

= Fn − Fg = ma y = 0

or Fn = Fg
Fn = mg Fn = (75.0 kg )(9.81 m/s 2 ) = 736 N

Because the gravitational force is given by Fg = mg : Substitute numerical values and evaluate Fn: (d) The free-body diagram shows the forces acting on the log when it is accelerating in the positive x direction.

y
r Fn
r

r Fby ground r Fg

Fpull

x

Apply

∑F

x

= ma x to the log when

∑ Fx = Fpull − Fby ground = ma x

it is accelerating to the right: Solving for Fpull yields: Fpull = ma x + Fby ground

Newton’s Laws 319 Substitute numerical values and evaluate Fpull : Fpull = (75.0 kg ) 2.00 m/s 2 + 250 N = 400 N

(

)

r A 4.0-kg object is subjected to two constant forces, F = (2.0 N) ˆ i 1 r i + (–3.0 N) jˆ and F = (4.0 N) ˆ – (11 N) jˆ . The object is at rest at the origin at

41

••

2

time t = 0. (a) What is the object’s acceleration? (b) What is its velocity at time t = 3.0 s? (c) Where is the object at time t = 3.0 s? Picture the Problem The acceleration can be found from Newton’s second law. Because both forces are constant, the net force and the acceleration are constant; hence, we can use the constant-acceleration equations to answer questions concerning the motion of the object at various times.

(a) Apply Newton’s second law to the 4.0-kg object to obtain:

r r r r Fnet F1 + F2 = a= m m

r Substitute numerical values and simplify to evaluate a :

ˆ ˆ ˆ r (2.0 N ) i + (− 3.0 N ) ˆ + (4.0 N ) i + (− 11 N ) ˆ (6.0 N ) i + (− 14 N ) ˆ j j j a= = 4.0 kg 4.0 kg ˆ = 1.5 m/s 2 i + − 3.5 m/s 2 ˆ j

(

) (

)

(b) Using a constant-acceleration equation, express the velocity of the object as a function of time:

r r r v = v 0 + at

r Substitute numerical values and evaluate v (3.0 s ) :

r ˆ ˆ v (3.0 s ) = 1.5 m/s 2 i + − 3.5 m/s 2 ˆ (3.0 s ) = (4.5 m/s ) i + (− 10.5 m/s ) ˆ j j =

[(

) (

)]

(4.5 m/s) iˆ + (− 11m/s) ˆ j
r r r r r v0 + v r = v av t = t = 1 vt 2 2

(c) Express the position of the object in terms of its average velocity:

r Substitute for v and evaluate this expression at t = 3.0 s:

r r (3.0 s ) = =

1 2

[(4.5 m/s)iˆ + (− 10.5 m/s) ˆj ](3.0 s) = (6.75 m)iˆ + (− 15.8 m) ˆj (6.8 m ) iˆ + (− 16 m ) ˆ j

320 Chapter 4

Mass and Weight
42 • On the moon, the acceleration due to gravity is only about 1/6 of that on Earth. An astronaut, whose weight on Earth is 600 N, travels to the lunar surface. His mass, as measured on the moon, will be (a) 600 kg, (b) 100 kg, (c) 61.2 kg, (d) 9.81 kg, (e) 360 kg. Picture the Problem The mass of the astronaut is independent of gravitational fields and will be the same on the moon or, for that matter, out in deep space.

Express the mass of the astronaut in terms of his weight on Earth and the gravitational field at the surface of Earth: 43 •

m=

wearth 600 N = = 61.2 kg g earth 9.81 N/kg

and (c) is correct.

Find the weight of a 54-kg student in (a) newtons and (b) pounds.

Picture the Problem The weight of an object is related to its mass and the gravitational field through Fg = mg.

(a) The weight of the student is:

w = mg = (54 kg )(9.81 N/kg ) = 530 N = 5.3 ×10 2 N

(b) Convert newtons to pounds:

w=

530 N = 119 lb ≈ 1.2 × 10 2 lb 4.45 N/lb

44

Find the mass of a 165-lb engineer in kilograms.

Picture the Problem The mass of an object is related to its weight and the gravitational field.

Convert the weight of the man into newtons: Calculate the mass of the man from his weight and the gravitational field:

165 lb = (165 lb )(4.45 N/lb ) = 734 N

m=

w 734 N = = 74.8 kg g 9.81 N/kg

45 •• [SSM] To train astronauts to work on the moon, where the acceleration due to gravity is only about 1/6 of that on Earth, NASA submerges them in a tank of water. If an astronaut, who is carrying a backpack, air conditioning unit, oxygen supply, and other equipment, has a total mass of 250 kg, determine the following quantities. (a) her weight, including her backpack, etc., on Earth, (b) her weight on the moon, (c) the required upward

Newton’s Laws 321 buoyancy force of the water during her training for the moon’s environment on Earth. Picture the Problem We can use the relationship between weight (gravitational force) and mass, together with the given information about the acceleration due to gravity on the moon, to find the astronaut’s weight on Earth and on the moon.

(a) Her weight on Earth is the product of her mass and the gravitational field at the surface of Earth: Substitute numerical values and evaluate w: (b) Her weight on the moon is the product of her mass and the gravitational field at the surface of the moon: Substitute for her weight on Earth and evaluate her weight on the moon: (c) The required upward buoyancy force of the water equals the difference between her weight on Earth and on the moon:

wEarth = mg

wEarth = (250 kg ) (9.81 m/s 2 ) = 2.453 kN = 2.45 kN wmoon = mg moon = 1 mg = 1 wearth 6 6

wmoon =

1 6

(2.453 kN ) =

409 N

wbuoyancy = wEarth − wmoon = 2.45 kN − 0.41 kN = 2.04kN

46 •• It is the year 2075 and space travel is common. A physics professor brings his favorite teaching demonstration with him to the moon. The apparatus consists of a very smooth horizontal (frictionless) table and an object to slide on it. On Earth, when the professor attaches a spring (spring constant 50 N/m) to the object and pulls horizontally so the spring stretches 2.0 cm, the object accelerates at1.5 m/s2. (a) Draw the free-body diagram of the object and use it and Newton’s laws to determine the object’s mass. (b) What would the object’s acceleration be under identical conditions on the moon? Picture the Problem The forces acting on the object are the normal force exerted by the table, the gravitational force exerted by Earth, and the force exerted by the stretched spring.

322 Chapter 4 (a) The free-body diagram shown to the right assumes that the spring has been stretched to the right. Hence the force that the spring exerts on the object is to the left. Note that the +x direction has been chosen to be in the same direction as the force exerted by the spring. Apply ∑ Fx = ma x to the object to obtain: The force exerted by the spring on the object is given by:

y
r Fn

x

r Fs

r Fg
∑ Fx = Fs = ma x ⇒ m = Fs ax (1)

Fs = kΔx where Δx is the amount by which the spring has been stretched or compressed and k is the force constant. m= kΔx ax

Substituting for Fs in equation (1) yields: Substitute numerical values and evaluate m:

m=

(50 N/m )(2.0 cm) =
1.5 m/s 2

0.67 kg

(b) Because the object’s mass is the same on the moon as on Earth and the force exerted by the spring is the same, its acceleration on the moon would be the same as on Earth.

Free-Body Diagrams: Static Equilibrium
47 • A 35.0-kg traffic light is supported by two wires as in Figure 4-36. (a) Draw the light’s free-body diagram and use it to answer the following question qualitatively: Is the tension in wire 2 greater than or less than the tension in wire 1? (b) Prove your answer by applying Newton’s laws and solving for the two tensions. Picture the Problem Because the traffic light is not accelerating, the net force r r r acting on it must be zero; i.e., T1 + T2 + Fg = 0.

Newton’s Laws 323 (a) Construct a free-body diagram showing the forces acting on the support point: r T2
60°

y

30°

r T1

x

r Fg

Apply point:

∑F

x

= ma x to the support

T1 cos 30° − T2 cos 60° = ma x = 0

Solve for T2 in terms of T1:

cos 30° T1 = 1.732T1 cos 60° Thus T2 is greater than T1. T2 =

(1)

(b) Apply

∑F

y

= max to the

T1 sin 30° + T2 sin 60° − Fg = ma y = 0

support point: Substitute for T2 to obtain: Solving for T1 gives: 1 2

T1 +

3 2

(1.732T1 ) − mg = 0

T1 =

mg 2.000

Substitute numerical values and evaluate T1: From equation (1) we have:

T1 =

(35.0 kg )(9.81 m/s2 ) =
2.000

172 N

T2 = 1.732T1 = (1.732 )(172 N ) = 298 N

48 • A 42.6-kg lamp is hanging from wires as shown in Figure 4-37. The ring has negligible mass. The tension T1 in the vertical wire is (a) 209 N, (b) 418 N, (c) 570 N, (d) 360 N, (e) 730 N. Picture the Problem From the figure, it is clear that T1 supports the full weight of the lamp. Draw a free-body diagram showing the forces acting on the lamp and apply ∑ Fy = 0 . y r T1

m

x

r Fg

Apply obtain:

∑F

y

= 0 to the lamp to

∑F

y

= T1 − Fg = 0

324 Chapter 4 Solve for T1 and substitute for Fg to obtain: Substitute numerical values and evaluate T1: T1 = Fg = mg T1 = (42.6 kg ) 9.81 m/s 2 = 418 N and (b) is correct.

(

)

[SSM] In Figure 4-38a, a 0.500-kg block is suspended at the 49 •• midpoint of a 1.25-m-long string. The ends of the string are attached to the ceiling at points separated by 1.00 m. (a) What angle does the string make with the ceiling? (b) What is the tension in the string? (c) The 0.500-kg block is removed and two 0.250-kg blocks are attached to the string such that the lengths of the three string segments are equal (Figure 4-38b). What is the tension in each segment of the string? Picture the Problem The free-body diagrams for Parts (a), (b), and (c) are shown below. In both cases, the block is in equilibrium under the influence of the forces and we can use Newton’s second law of motion and geometry and trigonometry to obtain relationships between θ and the tensions.

(a) and (b)

(c)

(a) Referring to the free-body diagram for Part (a), use trigonometry to determine θ : (b) Noting that T = T′, apply ∑ Fy = ma y to the 0.500-kg block and solve for the tension T: Substitute numerical values and evaluate T:

θ = cos −1 ⎜ ⎜

⎛ 0.50 m ⎞ ⎟ = 36.9° = 37° 0.625 m ⎟ ⎝ ⎠

2T sin θ − mg = 0 because a = 0 and mg T= 2 sin θ
T=

(0.500 kg )(9.81m/s 2 ) =
2sin36.9°

4.1 N

Newton’s Laws 325 (c) The length of each segment is: 1.25 m = 0.41667 m 3 d= 1.00 m − 0.41667m 2 = 0.29167 m

Find the distance d:

Express θ in terms of d and solve for its value:

θ = cos −1 ⎜ ⎜

⎞ d ⎟ ⎟ ⎝ 0.417 m ⎠

⎛ 0.2917 m ⎞ = cos −1 ⎜ ⎜ 0.4167 m ⎟ = 45.57° ⎟ ⎝ ⎠ Apply

∑F

y

= ma y to the

0.250-kg block: Substitute numerical values and evaluate T3:

T3 sin θ − mg = 0 ⇒ T3 =
T3 =

mg sin θ

(0.250 kg )(9.81m/s 2 ) = 3.434 N
sin45.57°

= 3.4 N

Apply

∑F

x

= ma x to the

0.250-kg block and solve for the tension T2: Substitute numerical values and evaluate T2: By symmetry:

T3 cos θ − T2 = 0 since a = 0. and T2 = T3 cosθ
T2 = (3.434 N )cos 45.57° = 2.4 N

T1 = T3 = 3.4 N

50 •• A ball weighing 100 N is shown suspended from a system of cords (Figure 4-39). What are the tensions in the horizontal and angled cords?

326 Chapter 4
Picture the Problem The suspended body is in equilibrium under the r r influence of the forces Thor , T45 , and r r r r Fg . That is, Thor + T45 + Fg = 0. Draw the y

r T45 r Thor

free-body diagram of the forces acting on the knot just above the 100-N body. Choose a coordinate system with the positive x direction to the right and the positive y direction upward. Apply the conditions for translational equilibrium to determine the tension in the horizontal cord. Apply ∑ Fy = ma y to the knot: Solving for T45 yields:

45°

x

r Fg

∑ Fy = T45 sin 45° − Fg = ma y = 0 T45 = Fg sin 45° = 100 N = 141 N sin 45°

Apply ∑ Fx = ma x to the knot: Solving for Thor gives:

∑ Fx = T45 cos 45° − Thor = ma x = 0 Thor = T45 cos 45° = (141 N ) cos 45° = 100 N

51 •• [SSM] r A 10-kg object on a frictionless table is subjected to two r horizontal forces, F and F , with magnitudes F1 = 20 N and F2 = 30 N, as shown 1 2 r in Figure 4-40. Find the third force F that must be applied so that the object is in 3 static equilibrium. Picture the Problem The acceleration of any object is directly proportional to the net force acting on it. Choose a coordinate system in which the positive x direction r is the same as that of F1 and the positive y direction is to the right. Add the two forces to determine the net force and then use Newton’s second law to find the r acceleration of the object. If F3 brings the system into equilibrium, it must be true r r r that F3 + F1 + F2 = 0.

r r r Express F3 in terms of F1 and F2 :

r r r F3 = − F1 − F2

(1)

Newton’s Laws 327

r r Express F1 and F2 in unit vector notation: r ˆ F1 = (20 N) i and

r ˆ ˆ F2 = {(−30 N) sin 30°}i + {(30 N) cos 30°} ˆ = (−15 N)i + (26 N) ˆ j j

r r Substitute for F1 and F2 in equation (1) and simplify to obtain: r ˆ ˆ F3 = −(20 N ) i − (−15 N)i + (26 N) ˆ = j

[

]

(− 5.0 N ) iˆ + (− 26 N ) ˆ j

52 •• For the systems to be in equilibrium in Figure 4-41a, Figure 4-41b, and Figure 4-41c find the unknown tensions and masses. Picture the Problem The free-body diagrams for the systems in equilibrium are shown below. Apply the conditions for translational equilibrium to find the unknown tensions.

(a)

(b)

(c)

(a) Apply ∑ Fx = 0 and ∑ Fy = 0 to the knot above the suspended mass to obtain: Solving equation (1) for T1 yields:

∑ Fx = T1 cos 60° − 30 N = 0 and ∑ Fy = T1 sin 60° − T2 = 0 T1 = 30 N = 60 N cos 60°

(1) (2)

Solving equation (2) for T2 yields:

T2 = T1 sin 60° = (60 N )sin 60° = 51.96 N = 52 N

Because T2 is the weight of the object whose mass is m:

T2 = Fg = mg ⇒ m =

T2 g

328 Chapter 4 Substitute numerical values and evaluate m: (b) Apply ∑ Fx = 0 and ∑ Fy = 0 to the knot above the suspended mass to obtain: Solving the first of these equations for T1 yields: m= 51.96 N = 5.3 kg 9.81 m/s 2

∑ Fx = (80 N )cos 60° − T1 sin 60° = 0 and ∑ Fy = (80 N )sin 60° − T2 − T1 cos 60° = 0 T1 =

(80 N )cos 60° = 46.19 N
sin 60°

= 46 N

Solving the second of these equations for T2 yields: Substitute numerical values and evaluate T2: Because T2 is the weight of the object whose mass is m: Substitute numerical values and evaluate m: (c) Apply ∑ Fx = 0 and ∑ Fy = 0 to the knot above the suspended mass to obtain: Solving the first of these equations for T1 yields: Solving the second of these equations for T1 yields: Substitute numerical values and evaluate T1 and T3:

T2 = (80 N )sin 60° − T1 cos 60°

T2 = (80 N )sin 60° − (46.19 N ) cos 60° = 46.19 N = 46 N T2 = Fg = mg ⇒ m = m= T2 g

46.19 N = 4.7 kg 9.81 m/s 2

∑ Fx = −T1 cos 60° + T3 cos 60° = 0 and ∑ Fy = T1 sin 60° + T3 sin 60° − Fg = 0 T1 = T3 Fg 2 sin 60°

T1 = T3 =
T1 = T3 =

=

mg 2 sin 60°

(6.0 kg )(9.81 m/s 2 ) = 33.98 N
2 sin 60°

= 34 N

Because T2 = Fg:

T2 = (6.0 kg ) 9.81 m/s 2 = 58.9 N = 59 N

(

)

Newton’s Laws 329 Because the effect of the pulley is to change the direction T1 acts, T1 is the weight of the object whose mass is m: Substitute numerical values and evaluate m: T1 = mg ⇒ m =

T1 g

m=

33.98 N = 3.5 kg 9.81 m/s 2

53 •• Your car is stuck in a mud hole. You are alone, but you have a long, strong rope. Having studied physics, you tie the rope tautly to a telephone pole and pull on it sideways, as shown in Figure 4-42. (a) Find the force exerted by the rope on the car when the angleθ is 3.00º and you are pulling with a force of 400 N but the car does not move. (b) How strong must the rope be if it takes a force of 600 N to move the car whenθ is 4.00º? Picture the Problem Construct the free-body diagram for that point in the rope at which you exert the r force F and choose the coordinate system shown in the free-body diagram. We can apply Newton’s second law to the rope to relate the tension to F. r T1

y

r T2

θ
r F

θ

x

(a) Noting that T1 = T2 = T and that the car’s acceleration is zero, apply ∑ Fy = ma y to the car: Substitute numerical values and evaluate T: (b) Proceed as in Part (a) to obtain:

2T sin θ − F = ma y = 0 ⇒ T =

F 2 sin θ

T=

400 N = 3.82 kN 2 sin 3.00° 600 N = 4.30 kN 2 sin 4.00°

T=

54 ••• Balloon arches are often seen at festivals or celebrations; they are made by attaching helium-filled balloons to a rope that is fixed to the ground at each end. The lift from the balloons raises the structure into the arch shape. Figure 4-43a shows the geometry of such a structure: N balloons are attached at equally spaced intervals along a massless rope of length L, which is attached to two supports at its ends. Each balloon provides a lift force F. The horizontal and vertical coordinates of the point on the rope where the ith balloon is attached are xi and yi, and Ti is the tension in the ith segment. (Note segment 0 is the segment between the point of attachment and the first balloon, and segment N is the segment between the last balloon and the other point of attachment). (a) Figure 4-43b shows a free-body diagram for the ith balloon. From this diagram, show that the horizontal component of the force Ti (call it TH) is the same for all the

330 Chapter 4 string segments. (b) By considering the vertical component of the forces, use Newton’s laws to derive the following relationship between the tension in the ith and (i – 1)th segments: Ti–1 sin θi–1 – Ti sin θi = F (b) Show that tanθ0 = –tan θN+1 = NF/2TH. (c) From the diagram and the two expressions above, i−1 L show that tan θi = (N – 2i)F/2TH and that xi = ∑ cosθ j , N + 1 j= 0 L i −1 yi = ∑ sin θ j . N + 1 j =0

Picture the Problem (a) Applying ∑ Fx = 0 to the balloon shown in Figure 4-43

(b) will show that the horizontal component of the force Ti is the same for all string segments. In Part (b) we can apply ∑ Fy = 0 to obtain the given expression

for F. In (c) we can use a symmetry argument to find an expression for tan θ0. Finally, in (d) we can use our results obtained in (a) and (b) to express xi and yi.

(a) Applying

∑F

x

= 0 to the

Ti cos θ i − Ti−1 cos θ i−1 = 0

balloon shown in Figure 4-43 (b) gives: Because TH = Ti cos θ i : TH − Ti−1 cosθ i−1 = 0 ⇒ TH = Ti−1 cos θ i−1 independently of the string segment. F + Ti sin θ i − Ti −1 sin θ i −1 = 0

(b) Apply

∑F

y

= 0 to the balloon

shown in Figure 4-43 (b) gives: Solving for F yields: (c) By symmetry, each support must balance half of the force acting on the entire arch. Therefore, the vertical component of the force on the support must be NF/2. The horizontal component of the tension must be TH. Express tanθ0 in terms of NF/2 and TH: By symmetry, θN+1 = − θ0. Therefore, because the tangent function is odd: F = Ti −1 sin θ i −1 − Ti sin θ i

tan θ 0 =

NF 2 NF = 2TH TH

tan θ 0 = − tan θ N +1 =

NF 2TH

Newton’s Laws 331 (d) Using TH = Ti cosθi = Ti−1cosθi−1, divide both sides of the result in (b) by TH and simplify to obtain: Using this result, express tan θ1: F Ti −1 sin θ i −1 Ti sin θ i = − TH Ti −1 cos θ i −1 Ti cos θ i = tan θ i −1 − tan θ i tan θ1 = tan θ 0 − F TH

Substitute for tan θ0 from (a):

tan θ1 =

NF F F − = (N − 2) 2TH TH 2TH

Generalize this result to obtain:

tan θ i =

(N − 2i )
L N +1

F 2TH

Express the length of rope between two balloons: Express the horizontal coordinate of the point on the rope where the ith balloon is attached, xi, in terms of xi−1 and the length of rope between two balloons: Sum over all the coordinates to obtain:

l between =
balloons

xi = xi −1 +

L cos θ i −1 N +1

xi =

L i −1 ∑ cosθ j N + 1 j =0 L i −1 ∑ sin θ j N + 1 j =0

Proceed similarly for the vertical coordinates to obtain:

yi =

55 ••• (a) Consider a numerical solution to Problem 54. Write a spreadsheet program to make a graph of the shape of a balloon arch. Use the following parameters: N = 10 balloons each providing a lift force F = 1.0 N and each attached to a rope length L = 11 m, with a horizontal component of tension TH = 10 N. How far apart are the two points of attachment? How high is the arch at its highest point? (b) Note that we haven’t specified the spacing between the supports—it is determined by the other parameters. Vary TH while keeping the other parameters the same until you create an arch that has a spacing of 8.0 m between the supports. What is TH then? As you increase TH, the arch should get flatter and more spread out. Does your spreadsheet model show this?

332 Chapter 4
Picture the Problem (a) A spreadsheet program is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Content/Formula C7 (\$B\$2−2*B7)/(2*\$B\$4) D7 E7 F7 G7 F8 G8 SIN(ATAN(C7)) COS(ATAN(C7)) 0.000 0.000 F7+\$B\$1/(\$B\$2+1)*E7 G7+\$B\$1/(\$B\$2+1)*D7

Algebraic Form (N − 2i ) F 2TH

( ) cos(tan θ )
sin tan −1 θ i
−1
i

0 0 L cos θ i −1 xi −1 + N +1 L cos θ i −1 yi −1 + N +1 E F G

1 2 3 4 5 6 7 8 9 10 14 15 19 20

A L= N= F= TH=

B C 11 m 10 1 N 3.72 N i 0 1 2 3 5 6 10 11 tanθi 1.344 1.075 0.806 0.538

D

sinθi 0.802 0.732 0.628 0.474

cosθi 0.597 0.681 0.778 0.881

xi 0.000 0.543 1.162 1.869

yi 0.000 0.729 1.395 1.966

0.000 0.000 1.000 3.548 2.632 −0.269 −0.260 0.966 4.457 2.632 −1.344 −0.802 0.597 7.462 0.729 8.005 0.000

Newton’s Laws 333 (b) A horizontal component of tension 3.72 N gives a spacing of 8 m. At this spacing, the arch is 2.63 m high, tall enough for someone to walk through. A graph of yi as a function of xi for the conditions specified follows: 3.0 2.5 2.0 y i, m 1.5 1.0 0.5 0.0 0 1 2 3 4 x i, m 5 6 7 8 9

The spreadsheet graph shown below shows that changing the horizontal component of tension to 5 N broadens the arch and decreases its height as predicted by our mathematical model. 2.5

2.0

1.5 y i, m 1.0 0.5 0.0 0 1 2 3 4 x i, m 5 6 7 8 9

334 Chapter 4

Free-Body Diagrams: Inclined Planes and the Normal Force
56 • A large box whose mass is 20.0 kg rests on a frictionless floor. A mover pushes on the box with a force of 250 N at an angle 35.0º below the horizontal. Draw the box’s free body diagram and use it to determine the acceleration of the box. Picture the Problem The free-body diagram shows the forces acting on the box as the man pushes it across the frictionless floor. We can apply Newton’s second law to the box to find its acceleration. r Fg

y
r Fn

θ

x

r F

Apply

∑F

x

= ma x to the box:

F cos θ = ma x ⇒ a x = ax =

F cos θ m 10.2 m/s 2

Substitute numerical values and evaluate ax:

(250 N )cos35.0° =
20.0 kg

57 • A 20-kg box rests on a frictionless ramp with a 15.0º slope. The mover pulls on a rope attached to the box to pull it up the incline (Figure 4-44). If the rope makes an angle of 40.0º with the horizontal, what is the smallest force F the mover will have to exert to move the box up the ramp? Picture the Problem The free-body diagram shows the forces acting on the box as the man pushes it up the frictionless incline. We can apply Newton’s second law to the box to determine the smallest force that will move it up the incline at constant speed.

y
r Fn
r F

φ

x

r Fg θ

Letting Fmin = F, apply

∑F

x

= ma x

Fmin cos(φ − θ ) − Fg sin θ = 0

to the box as it moves up the incline with constant speed: Solve for Fmin to obtain: Fmin = cos (φ − θ ) Fg sin θ

Newton’s Laws 335 Because Fg = mg: Fmin = mg sin θ cos (φ − θ )

Substitute numerical values and evaluate Fmin:

(20.0 kg )(9.81m/s 2 )sin 15° Fmin = cos(40.0° − 15.0°) = 56.0 N

58 • In Figure 4-45, the objects are attached to spring scales calibrated in newtons. Give the readings of the balance(s) in each case, assuming that both the scales and the strings are massless. Picture the Problem The balance(s) indicate the tension in the string(s). Draw free-body diagrams for each of these systems and apply the condition(s) for equilibrium.

(a)
y

(b)
r T
hook

r balance T'

r T

x

r Fg

(c)

(d)

y
r T r T x r Fg
r balance T' r T

x

(a) Apply obtain:

∑F

y

= 0 to the hook to

∑F

y

= T − Fg = 0

or, because Fg = mg, T = mg
T = (10 kg ) 9.81 m/s 2 = 98 N

Substitute numerical values and evaluate T: (b) Apply obtain:

(

)

∑F

x

= 0 to the balance to

∑F

x

= T − T' = 0

or, because T ′= mg, T = T' = mg

336 Chapter 4 Substitute numerical values and evaluate T and T ′: (c) Apply T = T' = (10 kg ) 9.81m/s 2 = 98 N

(

)

∑F

y

= 0 to the suspended

∑F

y

= 2T − Fg = 0

object to obtain:

or, because Fg = mg, 2T − mg = 0 ⇒ T = 1 mg 2
T=
1 2

Substitute numerical values and evaluate T: (d) Apply obtain:

(10 kg )(9.81m/s 2 ) =
= T −T '= 0

49 N

∑F

x

= 0 to the balance to

∑F

x

or, because T ′= mg, T = T ' = mg
T = T ' = (10 kg )(9.81 m/s 2 ) = 98 N

Substitute numerical values and evaluate T and T ′:

Remarks: Note that (a), (b), and (d) give the same answers … a rather surprising result until one has learned to draw FBDs and apply the conditions for translational equilibrium. 59 •• A box is held in position on a frictionless incline by a cable (Figure 446). (a) If θ = 60º and m = 50 kg, find the tension in the cable and the normal force exerted by the incline. (b) Find the tension as a function of θ and m, and check your result for plausibility in the special cases when θ = 0º and θ = 90º. Picture the Problem Because the box is held in place (is in equilibrium) by the forces acting on it, we know that r r r T + Fn + Fg = 0

y

Choose a coordinate system in which the +x direction is in the direction of r T and the +y direction is in the r direction of F n . Apply Newton’s second law to the block to obtain r r expressions for T and Fn . (a) Apply

r Fn

x r T

θ
r Fg

∑F

x

= 0 to the box:

T − Fg sin θ = 0 or, because Fg = mg, T − mg sin θ = 0 ⇒ T = mg sin θ (1)

Newton’s Laws 337 Substitute numerical values and evaluate T: T = (50 kg )(9.81 m/s 2 )sin 60° = 0.42 kN Fn − Fg cosθ = 0 or, because Fg = mg, Fn − mg cosθ = 0 ⇒ Fn = mg cosθ Fn = (50 kg ) (9.81m/s 2 )cos 60° = 0.25 kN T = mg sin θ T90° = mg sin 90° = mg

Apply

∑F

y

= 0 to the box:

Substitute numerical values and evaluate Fn: (b) The tension as a function of θ and m is given by equation (1): For θ = 90°:

… a result we

know to be correct. For θ = 0°:
T0° = mg sin 0° = 0 … a result we

know to be correct.
60 •• A horizontal force of 100 N pushes a 12-kg block up a frictionless incline that makes an angle of 25º with the horizontal. (a) What is the normal force that the incline exerts on the block? (b) What is the acceleration of the block? Picture the Problem Draw a free-body diagram for the box. Choose a coordinate system in which the positive x-axis is parallel to the inclined plane and the positive y-axis is in the direction of the normal force the incline exerts on the block. Apply Newton’s second law of motion to find the normal force Fn that the incline exerts on the block and the acceleration a of the block. y

r Fn

x
25° 100 N

r 25° Fg

(a) Apply

∑F

y

= ma y to the block:

Fn − Fg cos 25° − (100 N )sin 25° = 0 or, because Fg = mg, Fn − mg cos 25° − (100 N )sin 25° = 0 Fn = mg cos 25° + (100 N )sin 25°

Solving for Fn yields:

338 Chapter 4 Substitute numerical values and evaluate Fn:
Fn = (12 kg ) 9.81 m/s 2 cos25° + (100 N )sin 25° = 1.5 × 10 2 N

(

)

(b) Apply block:

∑F

x

= ma x to the

(100 N )cos 25° − Fg sin 25° = ma
or, because Fg = mg, (100 N )cos 25° − mg sin 25° = ma a=

Solve for a to obtain:

(100 N )cos 25° − g sin 25°
m

Substitute numerical values and evaluate a:

a=

(100 N )cos 25° − (9.81m/s 2 )sin 25°
12 kg

= 3.4 m/s 2
61 •• [SSM] A 65-kg student weighs himself by standing on a scale mounted on a skateboard that is rolling down an incline, as shown in Figure 4-47. Assume there is no friction so that the force exerted by the incline on the skateboard is normal to the incline. What is the reading on the scale if θ = 30º? Picture the Problem The scale reading (the boy’s apparent weight) is the force the scale exerts on the boy. Draw a free-body diagram for the boy, choosing a coordinate system in which the positive x-axis is parallel to and down the inclined plane and the positive y-axis is in the direction of the normal force the incline exerts on the boy. Apply Newton’s second law of motion in the y direction. y

r Fn

x
30°

r Fg

Apply

∑F

y

= ma y to the boy to find

Fn − Fg cos 30° = 0 or, because Fg = mg, Fn − mg cos 30° = 0 Fn = mg cos 30°

Fn. Remember that there is no acceleration in the y direction: Solving for Fn yields: Substitute numerical values and evaluate Fn:

Fn = (65 kg ) (9.81m/s 2 )cos 30° = 0.55 kN

Newton’s Laws 339
62 •• A block of mass m slides across a frictionless floor and then up a frictionless ramp (Figure 4-48). The angle of the ramp is θ and the speed of the block before it starts up the ramp is v0. The block will slide up to some maximum height h above the floor before stopping. Show that h is independent of θ by deriving an expression for h in terms of v0 and g. Picture the Problem The free-body diagram for the block sliding up the incline is shown to the right. Applying Newton’s second law to the forces acting in the x direction will lead us to an expression for ax. Using this expression in a constant-acceleration equation will allow us to express h as a function of v0 and g. y

r Fn
x

θ
h = Δx sin θ
2 2 v x = v0 x + 2a x Δx or, because vx = 0,

rθ Fg

The height h is related to the distance Δx traveled up the incline: Using a constant-acceleration equation, relate the final speed of the block to its initial speed, acceleration, and distance traveled: Substituting for Δx in equation (1) yields: Apply

(1)

2 − v0 x 0 = v + 2a x Δx ⇒ Δx = 2a x

2 0x

h=

2 − v0 x sin θ 2a x

(2)

∑F

x

= ma x to the block and

− Fg sin θ = ma x

solve for its acceleration: Because Fg = mg : Substitute for ax in equation (2) and simplify to obtain: − mg sin θ = ma x ⇒ a x = − g sin θ 2 ⎛ v0 x ⎞ v2 ⎟ sin θ = 0 x h=⎜ ⎜ 2 g sin θ ⎟ 2g ⎠ ⎝ which is independent of the ramp’s angle θ.

Free-Body Diagrams: Elevators
63 • [SSM] (a) Draw the free body diagram (with accurate relative force magnitudes) for an object that is hung by a rope from the ceiling of an elevator that is ascending but slowing. (b) Repeat Part (a) but for the situation in which the

340 Chapter 4 elevator is descending and speeding up. (c) Can you tell the difference between the two diagrams? Explain why the diagrams do not tell anything about the object’s velocity. Picture the Problem (a) The free body diagram for an object that is hung by a rope from the ceiling of an ascending elevator that is slowing down is shown to the right. Note that because Fg > T, the net force acting on the object is downward; as it must be if the object is slowing down as it is moving upward.

r T

r Fg
r T

(b) The free body diagram for an object that is hung by a rope from the ceiling of an elevator that is descending and speeding up is shown to the right. Note that because Fg > T, the net force acting on the object is downward; as it must be if the object is speeding up as it descends.

r Fg

(c) No, there is no difference. In both cases the acceleration is downward. You can only tell the direction of the acceleration, not the direction of the velocity. 64 • A 10.0-kg block is suspended from the ceiling of an elevator by a cord rated to withstand a tension of 150 N. Shortly after the elevator starts to ascend, the cord breaks. What was the minimum acceleration of the elevator when the cord broke? Picture the Problem The free-body diagram shows the forces acting on the 10.0-kg block as the elevator accelerates upward. Apply Newton’s second law to the block to find the minimum acceleration of the elevator required to break the cord. y

r T

r Fg

Apply

∑F

y

= ma y to the block:

T − Fg = ma y or, because Fg = mg, T − mg = ma y ay = T − mg T = −g m m

Solve for ay to determine the minimum breaking acceleration:

Newton’s Laws 341 Substitute numerical values and evaluate ay: ay = 150 N − 9.81m/s 2 = 5.19 m/s 2 10.0 kg

65 •• A 2.0-kg block hangs from a spring scale calibrated in newtons that is attached to the ceiling of an elevator (Figure 4-49).What does the scale read when (a) the elevator is ascending with a constant speed of 30 m/s, (b) the elevator is descending with a constant speed of 30 m/s, (c) the elevator is ascending at 20 m/s and gaining speed at a rate of 3.0 m/s2? (d) Suppose that from t = 0 to t = 5.0 s, the elevator ascends at a constant speed of 10 m/s. Its speed is then steadily reduced to zero during the next 4.0 s, so that it is at rest at t = 9.0 s. Describe the reading of the scale during the interval 0 < t < 9.0 s. Picture the Problem The free-body diagram shows the forces acting on the 2-kg block as the elevator ascends at a constant velocity. Because the acceleration of the elevator is zero, the block is in equilibrium under the r r Apply influence of T and mg. Newton’s second law of motion to the block to determine the scale reading. y

r T

2.0 kg
r Fg

(a) Apply ∑ Fy = ma y to the block to obtain:

T − Fg = ma y or, because Fg = mg, T − mg = ma y T − mg = 0 ⇒ T = mg T = (2.0 kg ) (9.81 m/s 2 ) = 20 N

(1)

For motion with constant velocity, ay = 0 and: Substitute numerical values and evaluate T: (b) As in Part (a), for constant velocity, ay = 0. Hence:

T − mg = 0 and T = (2.0 kg ) (9.81 m/s 2 ) = 20 N
T = mg + ma y = m(g + a y ) (2)

(c) Solve equation (1) for T and simplify to obtain: Because the elevator is ascending and its speed is increasing, we have ay = 3.0 m/s2. Substitute numerical values and evaluate T:

T = (2.0 kg )(9.81m/s 2 + 3.0 m/s 2 ) = 26 N

342 Chapter 4 (d) During the interval 0 < t < 5.0 s, ay = 0. Hence: Using its definition, calculate a for 5.0 s < t < 9.0 s: Substitute in equation (2) and evaluate T: T0→5.0 s = 20 N

a=

Δv 0 − 10 m/s = = −2.5 m/s 2 Δt 4.0 s

T5 s→9 s = (2.0 kg ) 9.81 m/s 2 − 2.5m/s 2 = 15 N

(

)

Free-Body Diagrams: Several Objects and Newton’s Third Law 66 •• Two boxes of mass m1 and m2 connected by a massless string are being pulled along a horizontal frictionless surface, as shown in Figure 4-50. (a) Draw the free body diagram of both boxes separately and show that T1 T2 = m1 (m1 + m2 ) . (b) Is this result plausible? Explain. Does your answer make sense both in the limit that m2 m1 >> 1 and in the limit that m2 m1 > m2 these equations become: Divide the numerators and denominators if equations (3) and (4) by m2 to obtain:

a = g and T = 2m2 g … as expected.
⎞ ⎛ m1 −1 ⎟ ⎜ m ⎟ g and T = 2m1 g a=⎜ 2 ⎟ ⎜ m1 m1 +1 +1⎟ ⎜ m2 ⎠ ⎝ m2

For m1 mg/sin30°. Under this condition, there would be a net force in the y direction and the block would accelerate up the wedge. With an acceleration less than g/tan30°, the block would accelerate down the wedge. 93 ••• [SSM] The masses attached to each side of an ideal Atwood’s machine consist of a stack of five washers, each of mass m, as shown in Figure 465. The tension in the string is T0. When one of the washers is removed from the left side, the remaining washers accelerate and the tension decreases by 0.300 N. (a) Find m. (b) Find the new tension and the acceleration of each mass when a second washer is removed from the left side. Picture the Problem Because the system is initially in equilibrium, it follows that T0 = 5mg. When one washer is removed from the left side the remaining washers on the left side will accelerate upward (and those on the right side downward) in response to the net force that results. The free-body diagrams show the forces under this unbalanced condition. Applying Newton’s second law to each collection of washers will allow us to determine both the acceleration of the system and the mass of a single washer. y

r T

r T'

r 4m g
y

r 5m g

(a) Apply washers:

∑F

y

= ma y to the rising

T − 4mg = (4m )a y 5mg − T = (5m )a y

(1)

Noting that T = T ′, apply to the descending masses:

∑F

y

= ma y

(2)

Eliminate T between these equations to obtain: Use this acceleration in equation (1) or equation (2) to obtain: Expressing the difference ΔT between T0 and T yields:

ay = 1 g 9 40 mg 9
9 ΔT 40 mg ⇒ m = 5 g 9

T=

ΔT = 5mg −

Newton’s Laws 377 Substitute numerical values and evaluate m: (b) Proceed as in (a) to obtain: Add these equations to eliminate T and solve for ay to obtain: Substitute numerical values and evaluate ay: Eliminate ay in either of the motion equations and solve for T to obtain: Substitute numerical values and evaluate T: m=

9 5

(0.300 N )

9.81 m/s 2

= 55.0 g

T − 3mg = 3ma y and 5mg − T = 5ma y ay = 1 g 4
ay =
1 4

(9.81m/s ) =
2

2.45 m/s 2

T = 15 mg 4
T = 15 (0.05505 kg ) 9.81 m/s 2 4 = 2.03 N

(

)

94 •• Consider the ideal Atwood’s machine in Figure 4-65. When N washers are transferred from the left side to the right side, the right side drops 47.1 cm in 0.40 s. Find N. Picture the Problem The free-body diagram represents the Atwood’s machine with N washers moved from the left side to the right side. Application of Newton’s second law to each collection of washers will result in two equations that can be solved simultaneously to relate N, a, and g. The acceleration can then be found from the given data. y

r T

r T'

r (5 − N )mg
y

r (5 + N )mg

Apply

∑F

y

= ma y to the rising

T − (5 − N )mg = (5 − N )ma y

washers: Noting that T = T ′, apply ∑ Fy = ma y to the descending washers: Add these equations to eliminate T:

(5 + N )mg − T = (5 + N )ma y

(5 + N )mg − (5 − N )mg = (5 − N )ma y + (5 + N )ma y

378 Chapter 4 Solving for N yields:
N= 5a y g
2

Using a constant-acceleration equation, relate the distance the washers fell to their time of fall:

Δy = v0 y Δt + 1 a y (Δt ) 2 or, because v0y = 0,
2

Δy = 1 a y (Δt ) ⇒ a y = 2

2Δy (Δt )2

Substitute numerical values and evaluate ay: Substitute in the expression for N:

ay =

2(0.471 m ) = 5.89 m/s 2 2 (0.40 s )

⎛ 5.89 m/s 2 ⎞ N = 5⎜ ⎟ ⎜ 9.81m/s 2 ⎟ = 3 ⎠ ⎝

95 •• Blocks of mass m and 2m are on a horizontal frictionless surface (Figure 4-66). The blocks are connected by a string. In addition, forces F1 and F2 are applied to the blocks as shown. (a) If the forces shown are constant, find the tension in the connecting string. (b) If the forces vary with time as F1 = Ct and F2 = 2Ct, where C = 5.00 N/s and t is time, find the time t0 at which the tension in the string equals 10.0 N. Picture the Problem Draw the freebody diagram for the block of mass m and apply Newton’s second law to obtain the acceleration of the system and then the tension in the rope connecting the two blocks. y

r Fn r F1

m
r Fg

r T

x

(a) Apply

∑F
x

x

= ma x to the block

T − F1 = ma x F2 − F1 = (m + 2m )a = 3ma x ax = F2 − F1 3m

(1)

of mass m: Apply

∑F

= ma x to both blocks:

Solving for a gives:

Substitute for ax in the equation (1) to obtain:

⎛ F − F1 ⎞ T − F1 = m⎜ 2 ⎟ ⎝ 3m ⎠

Newton’s Laws 379 Solving for T yields: (b) Substitute for F1 and F2 in the equation derived in Part (a): Substitute numerical values and evaluate t = t0: T=
1 3

(F2 + 2F1 )
3T 4C

T = 1 (2Ct + 2(Ct )) = 4 Ct ⇒ t = 3 3 t0 = 3(10.0 N ) = 1.50 s 4(5.00 N/s )

97 •• Elvis Presley, has supposedly been sighted numerous times after he passed away on August 16, 1977. The following is a chart of what Elvis’ weight would be if he were sighted on the surface of other objects in our solar system. Use the chart to determine: (a) Elvis’ mass on Earth, (b) Elvis’ mass on Pluto, and (c) the acceleration due to gravity on Mars. (d) Compare the free-fall acceleration on Pluto to the free-fall acceleration on the moon.

planet Mercury Venus Earth Mars Jupiter Saturn Pluto Moon

Elvis’s weight (N) 431 1031 1133 431 2880 1222 58 191

Picture the Problem Elvis’ mass is the ratio of his weight on a given planet to the acceleration of gravity on that planet. In (b), (c), and (d) we can use the relationship between the gravitational force (weight) acting on an object, its mass (independent of location), and the local value of the free-fall acceleration.

(a) Elvis’ mass on Earth is given by:

m=

wEarth g Earth 1131 N = 115.3 kg = 115 kg 9.81 m/s 2

Substitute numerical values and evaluate m:

m=

(b) Because his mass is independent of his location, Elvis’ mass on Pluto is 115 kg.

380 Chapter 4 (c) The free-fall acceleration on Mars is the ratio of the weight of an object on Mars to the mass of the object: Substitute numerical values for Elvis and evaluate gMars: (d) The free-fall acceleration on Pluto is given by: The free-fall acceleration on the moon is given by: Divide equation (1) by equation (2) and simplify to obtain: g Mars = won Mars m

g Mars =

431 N = 3.74 m/s 2 115.3 kg won Pluto m won moon m (1)

g Pluto = g moon =

(2)

won Pluto g Pluto w = m = on Pluto g moon won moon won moon m

Substituting numerical values yields:

g Pluto 58 N = = 0.30 g moon 191 N or g Pluto = 0.30 g moon

97 ••• As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. When you wake up you are 30.0 m from the nearest shore. The ice is so slippery (i.e. frictionless) that you can not seem to get yourself moving. You realize that you can use Newton’s third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 595 N. (a) What direction should you throw your boot so that you will most quickly reach the shore? (b) If you throw your 1.20-kg boot with an average force of 420 N, and the throw takes 0.600 s (the time interval over which you apply the force), what is the magnitude of the average force that the boot exerts on you? (Assume constant acceleration.) (c) How long does it take you to reach shore, including the short time in which you were throwing the boot? Picture the Problem The diagram shown below summarizes the information about your trip to the shore and will be helpful in solving Part (c) of the problem.

Newton’s Laws 381

0 t0 = 0 x0 = 0 v0 = 0

a01

1 t1 = 0.600 s x1 v1 = vcoasting

a12 = 0

2 t2 = ? x2 = 30.0 m v 2 = vcoasting

(a) You should throw your boot in the direction away from the closest shore. (b) The magnitude of the average force you exert on the boot equals the magnitude of the average force the boot exerts on you: (c) The time required for you to reach the shore is the sum of your travel time while accelerating and your travel time while coasting: Use a constant-acceleration equation to relate your displacement Δx01 to your acceleration time Δt01: Apply Newton’s second law to express your acceleration during this time interval: Substitute numerical values and evaluate a01: Substitute numerical values in equation (2) and evaluate Δx01: Your coasting time is the ratio of your displacement while coasting to your speed while coasting: Fav, on you = 420 N

Δt total = Δt 01 + Δt12 or, because Δt01 = 0.600 s, Δt total = 0.600 s + Δt12 Δx01 = 1 a01 (Δt 01 ) 2
2

(1) (2)

a01 =

Fnet F F g = av = av m w/ g w

a01 =

(420 N )(9.81 m/s 2 ) = 6.925 m/s 2
595 N
1 2

Δx01 =

(6.925 m/s )(0.600 s)
2

2

= 1.246 m Δx12 v1 or, because Δx12 = 30.0 m − Δx01 , 30.0 m − Δx01 Δt12 = (3) v1 Δt12 =

382 Chapter 4 Use a constant-acceleration equation to relate your terminal speed (your speed after the interval of acceleration) to your acceleration and displacement during this interval: Substitute for v1 in equation (3) to obtain: Substituting for Δtcoasting in equation (1) yields: v1 = v0 + a01Δt 01 or, because v0 = 0, v1 = a01Δt 01

Δt12 =

30.0 m − Δx01 a01Δt 01 30.0 m − Δx01 a01Δt01

Δt total = 0.600 s +

Substitute numerical values and evaluate Δttotal: Δt total = 0.600 s +

(

30.0 m − 1.246 m = 7.52 s 6.925 m/s 2 (0.600 s )

)

98 ••• The pulley of an ideal Atwood’s machine is given an upward acceleration a, as shown in Figure 4-67. Find the acceleration of each mass and the tension in the string that connects them. Picture the Problem Because a constant-upward acceleration has the same effect as an increase in the acceleration due to gravity, we can use the result of Problem 82 (for the tension) with a replaced by a + g. The application of Newton’s second law to the object whose mass is m2 will connect the acceleration of this body to tension from Problem 82. y

r T

x
r r Fg,2 = m2 g

In Problem 82 it is given that, when the support pulley is not accelerating, the tension in the rope and the acceleration of the masses are related according to: Replace a with a + g:

T=

2m1m2 g m1 + m2

T=

2m1m2 (a + g ) m1 + m2 T − m2 g m2

Apply

∑F

y

= ma y to the object

whose mass is m2:

T − m2 g = m2 a 2 ⇒ a 2 =

Newton’s Laws 383 Substitute for T and simplify to obtain: The expression for a1 is the same as for a2 with all subscripts interchanged (note that a positive value for a1 represents acceleration upward): a2 =

(m1 − m2 )g + 2m1a
m1 + m2

a1 =

(m2 − m1 )g + 2m2a
m1 + m2

99 ••• You are working for an automotive magazine and putting a certain new automobile (mass 650 kg) through its paces. While accelerating from rest, its onboard computer records its speed as a function of time as follows:

v (m/s): 0 10 20 30 40 50 t (s): 0 1.8 2.8 3.6 4.9 6.5 (a) Using a spreadsheet, find the average acceleration over each 1.8-s time interval, and graph the velocity versus time and acceleration versus time for this car. (b) Where on the graph of velocity versus time is the net force on the car highest and lowest? Explain your reasoning. (c) What is the average net force on the car over the whole trip? (d) From the graph of velocity versus time, estimate the total distance covered by the car. Picture the Problem

(a) A spreadsheet program is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell C1 D5 Algebraic Form m 1 2 Δt E5 Δv (C5−C4)/D5 Δt ma F5 \$D\$1*E5 F10 (F5+F6+F7+F8+F9)/5 Fave A 1 2 3 4 5 6 7 8 B C D m= 650 E kg F Content/Formula 650 (B5−B4)/2

t (s) v (m/s) tmidpt (s) a (m/s2) F = ma (N) 0.0 0 1.8 10 0.90 5.56 3611 2.8 20 2.30 10.00 6500 3.6 30 3.20 12.50 8125 4.9 40 4.25 7.69 5000

384 Chapter 4 9 10 11 6.5 50 5.70 6.25 4063 5460

A graph of velocity as a function of time follows:
60 50 40

v , m/s

30 20 10 0 0 1 2 3 4 5 6 7

t, s

A graph of acceleration as a function of time is shown below: 14 12 10
2

a , m/s

8 6 4 2 0 0 1 2 3 4 5 6

t, s

(b) Because the net force is lowest where the acceleration is lowest, we can see from the graph of velocity versus time that its slope (the acceleration) is smallest in the interval from 0 to 1.8 s. Because the net force is highest where the acceleration is highest, we can see from the graph of velocity versus time that its slope (the acceleration) is greatest in the interval from 2.8 s to 3.6 s.

Newton’s Laws 385 (c) From the table we see that: (d) The distance covered by the car is the area under the graph of velocity versus time. Because the graph of speed versus time is approximately linear, we can estimate the total distance covered by the car by finding the area of the triangular region under it. Fave ≈ 5500 N

Atriangle = 1 bh = 2

1 2

(6.5 s )(50 m/s)

≈ 160 m

386 Chapter 4

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