solucionario fundamentos de física 9na edición capitulo 4

4The Laws of Motion

CLICKER QUESTIONS

Question B1.01

Description: Understanding force and drawing out a common misconception about the persistence of forces.

Question

A baseball is struck by a bat. While the ball is in the air, what objects exert forces on the ball?

1. Earth 2. Bat 3. Air 4. Bat, air 5. Earth, bat 6. Earth, air 7. Earth, bat, air 8. There are no forces on the ball. 9. The answer depends on whether the ball is going up, going down, or at its highest point.

Commentary

Purpose: To develop the concept of force, and draw out the common misconception that a force can persist after the interaction causing it has ceased.

Discussion: Except for gravitation, all the forces you will encounter for now require direct contact with an object or substance that causes them. (Later, you will encounter a few others, such as the electric and magnetic forces that act over a distance.)

When the ball is in the air, the only object or substance touching the ball is the air, so the only forces on the ball are due to the Earth (gravitation) and the air (buoyancy and air resistance/drag).

The bat exerts a force on the ball while the bat and ball are touching, and this causes the ball’s initial motion. However, once the ball leaves the bat, this force disappears; thereafter, all changes in motion are due to the Earth and the air. Once the bat loses contact with the ball, it cannot affect its motion, and we say that the force is no longer exerted.

No forces are needed to keep something moving. An object in motion will remain in motion until forces act to change its motion. In this case, the bat gives the ball an initial velocity, and then the Earth and the air combine to give the ball an acceleration, thus changing its velocity (making it slow down and travel a curved path).

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Key Points:

• All forces are caused by interactions between two objects or substances.

• For now, the only force you need to deal with that acts over a distance is gravity. All others require physical contact between the causing object/substance and the body being acted upon.

• When contact between to objects ceases, the contact forces between them also disappear.

For Instructors Only

It is common for students to think that a force is needed to sustain motion; in some cases this conception is more specifi c, namely, that motion requires a force in the direction of motion. For this question, the concep-tion manifests itself in the belief that there is a “force of the bat” that propels the ball during fl ight. Some students believe that the force of the bat goes away when the ball reaches its highest point and the Earth takes over, causing the ball to fall.

It is often useful to discuss the forces on the ball at three separate instants: (1) as it is rising, (2) when it is at its highest point, and (3) as it is falling. Students’ answers and explanations might change.

Discussion Questions:

• What forces are being exerted on the ball and what object exerts each force?

• How do you know when a force is being exerted by one object on another?

• Do the sizes of the forces on the ball change? Do the directions of the forces change? Describe how.

• Do you have any control over the force of the bat on the ball? Can you make it larger or smaller or change its direction once the ball is fl ying through the air?

Suggestions:

Ask students if they have a way of exerting a force on an object without touching it. Invite them to move an object in the front of the room without leaving their seats and touching the item.

If Newton’s second law has been introduced, attempt to relate the forces exerted on the ball to the ball’s acceleration. See if students agree that, if air resistance and buoyancy can be neglected, the ball has a constant acceleration of 9 8. m s2 toward the Earth during its entire trajectory. If they agree, ask what they can conclude about the net force on the ball while airborne.


The Laws of Motion 121

Question B1.02

Description: Identifying forces, and targeting misconceptions about “transmitted” forces.

Texts: College Physics, Principles of Physics, Physics for Scientists & Engineers

Question

Three blocks are stacked as shown below.

m1

m2

m3

How many forces are acting on the bottom block (m3)?

1. One force 2. Two forces 3. Three forces 4. Four forces 5. Five forces 6. Six forces 7. More than six forces 8. No forces act on the block 9. Cannot be determined

Commentary

Purpose: To hone and clarify the concept of “force,” and consider strategies for identifying the forces on an object.

Discussion: Except for gravitation, all of the common forces you will see in Mechanics require “direct” contact, though sometimes the agent is the air and you might not see the contact. So, in many situations, if you can identify all of the agents touching something, you can also determine the forces being exerted.

Air exerts a buoyant force on the block, but its magnitude is small and we generally ignore this force. The only other agents touching the bottom block are the fl oor and the middle block. Thus, there are 3 forces on the bottom block: (1) gravitation, down, exerted by the Earth; (2) normal force, up, exerted by the fl oor; and (3) normal force, down, exerted by the middle block. Thus, answer (3) is preferred. Answer (4) is also defensible if you include buoyancy due to the air.

The upper block is not touching the bottom block, so it cannot exert a force on it.

Note that the force of the middle block acting on the lower block is a normal force, not a gravitational force. It turns out that this force pushing has the same strength as the total weight of the upper two blocks, but this does not mean that the force is “the weight of the two blocks.” In a free-body diagram, for instance, the force pushing down should be labeled as a normal force.


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Key Points:

• To identify the forces on a body, identify all the objects (or substances) in physical contact with the body; each can exert a contact force. Then, include any action-at-a-distance forces, such as gravity.

• Air exerts a buoyant force, but in most circumstances it is negligibly small and we ignore it.


This problem has multiple defensible answers —(3) and (4)—, and students who pick one may do so for the wrong reasons. Thus, it is crucial to have students volunteer explanations for their answers during whole-class discussion, and to interact with the reasoning rather than with the answers themselves.

A common mistake is for students to claim there are four forces: gravitation due to the Earth, normal force due to the fl oor (or table), and the two gravitational forces on the upper two blocks (somehow “transmitted” to the lower one).

Some students will say that there are 3 forces, but one will be the “weight of the upper two blocks” or the “weight of m

1 and m

2.” This seems like a correct answer, but should not be treated as valid. The third force

should be a normal force, whose value in this situation happens to be equal to the weight of the upper two blocks. If, for example, the system were accelerating, the normal force pushing down on m

3 would not be

equal to the weight of the upper two blocks.

It can help tremendously to have students analyze the wording of their own statements. Many will admit that the force pushing down on the top of the bottom block is the “gravitational force exerted on the upper two blocks.” They should be able to see that the force “on the bottom block” cannot be a force exerted “on the upper two blocks.”

This question provides a good opportunity to discuss the nature of the normal force and of “constraint forces”—forces whose value is determined procedurally, with no empirical law. Students may have prior experience that allows them to solve certain problems correctly, but they might not have seen situations in which they must actually understand the normal force.

It would be technically correct but undesirable for students to include gravitational forces between the blocks, or other infi nitesimal forces, in their count. Deciding what to include and what to neglect in prob-lem solving is a learned skill, and discussing how these decisions are made can be worthwhile.

Question B1.03 makes an effective follow-up to this. It is a similar question, but the situation includes the additional subtlety of whether a contact force should be considered as one force or split into normal and friction forces.

Discussion Questions:

• Does m3 exert a force on m

1? If not, why not? How can m

1 exert a force on m

3 without m

3 exerting a

force on m1?

• What part of m2 interacts with m

3? What part of m

1 interacts with m

3?

• Is weight a force? If so, what agent exerts the force?

• Can an object interact with another object without touching it? If so, when? If not, why not?

• Is the normal force exerted by m2 on m

3 less than, equal to, or greater than the weight of m

2?



Question B1.03

Description: Identifying the forces acting on an object.

Question

A block of mass m is on a rough surface, with a spring attached and extended. As the block moves up the incline a small distance, how many forces are exerted on the mass?

m

1. 1 force 2. 2 forces 3. 3 forces 4. 4 forces 5. 5 forces 6. 6 forces 7. 7 forces 8. More than 7 forces 9. None of the above 10. Impossible to determine

Commentary

Purpose: To develop the skill of identifying all the (relevant) forces acting on an object, and appreciate some of the ambiguities that involves.

Discussion: The question has many defensible answers. What matters is not which answer you pick, but the validity of your reasoning. The most extreme answer is “an uncountable number of forces, one for each atom outside the mass interacting with each atom inside the mass.” So (8) is not wrong, but this way of thinking is not useful: it does not help us to analyze or understand the situation.

To identify the forces on an object, we identify all the objects interacting with it. In this case, the object is touched by the plane, string, spring, and air, and the Earth pulls on it gravitationally. The plane, string, and spring each exert a contact force on the object, and the air exerts both buoyancy and drag (air resistance), so we can say there are six forces acting: answer (6). (Warning: when you study electricity and magnetism, you will encounter other forces that act “at a distance” the way that gravity does.)

However, we often separate the contact force of a surface into its two components and treat them as distinct forces (normal and friction), so we can identify seven forces on the object: normal (due to the plane), kinetic friction (also due to the plane), tension (due to the string), elastic (due to the spring), buoyancy (due to the air), air drag (also due to the air), and gravitation (due to the Earth): answer (7).

However, buoyancy and air drag are generally negligible, so to solve a physics problem we would usually ignore them and identify the other fi ve as the ones we must consider. That means answer (5) is defensible, and this would probably be the “expected” answer on a physics exam, since these are the fi ve we need


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to use to solve a typical problem. (If we treated the force of the plane as one contact force rather than as distinct friction and normal forces, we would have an argument for answer (4).)

So, one can make a plausible argument for answers (4), (5), (6), (7), or (8). What matters is not which answer you choose, but whether your reasoning is cogent and defensible.

You should not count the “net force” along with the others: the net force is the vector sum of all of the individual forces, not a separate force that needs to be considered or accounted for.

Key Points:

• To identify the forces acting on an object, identify all the objects that contact it; each one will exert a force. If the object has mass and the problem is not in “outer space,” the Earth will also exert a gravitational force.

• Some contact forces are split into “pushing” and “sliding” components and treated as two separate forces, a normal one and a frictional one.


Part of learning to do physics is learning to make the “right” assumptions—the assumptions that experts do. This is a process of acculturation as well as of understanding physical laws, and we can help students along by explicitly discussing the assumptions we make and the ways we interpret questions. This question demonstrates a powerful tactic that involves presenting students with a question with multiple defensible answers, and then discussing the various answers and demonstrating the reasoning an expert might use and the standard assumptions she might make. We strongly recommend downplaying the notion of “right” and “wrong” here in favor of cogency.

When introducing forces to your students, we suggest identifying all valid forces, even if you will routinely neglect some of them in the future.

Most experienced teachers and students consider the normal and friction forces separate and independent, even though they are simply components of one contact force exerted by the incline. Students need to understand this if the advice to “count forces by identifying all objects interacting with the body” is to be useful.

Some students will include other, “ridiculous” forces such as the gravitational attraction of the Moon, the Sun, the planets, etc. This is a good opportunity to make a distinction between “wrong,” “right,” and “right but not useful.”

Question B1.04

Description: Identifying the forces on a body.

Question

A monkey hangs on a rope. What forces act on the monkey? (Ignore forces due to the air.)

1. Friction, gravitation 2. Tension, gravitation 3. Friction, tension, gravitation 4. Normal, friction, gravitation



5. More than one answer is true. 6. None of the above 7. Cannot be determined

Commentary

Purpose: To explore the notion of a “free body” when identifying the forces on something.

Discussion: When identifying the forces on an object, we generally start with gravitation. If the object has mass and is in a gravitational fi eld (such as that produced by the Earth), then the object has weight, and therefore it has a gravitational force on it.

Gravitation is the only common force that does not require direct contact between the object and the agent exerting it. All others—tension, friction (static and kinetic), normal, spring, air resistance/drag, and buoyancy—arise from contact.

The monkey is touching only a rope, and we are told to ignore any effects due to the air, so only the rope can exert any relevant forces on the monkey.

There are many valid ways to analyze this situation. The most precise answer would be to say that the rope exerts a friction force upwards on the monkey. In other words, treating the monkey as the free body, it is friction that balances gravitation and keeps the monkey at rest. Friction requires a normal force, which in this situation would be horizontal. But there is no net normal force, so we cannot show a normal force vector on a free-body diagram for the monkey.

Another way to approach this situation is to treat that small bit of rope in the monkey’s hand as being part of the free body we are identifying forces upon, in which case the force exerted by the rope is a tension force upward. Although this is less precise than the previous answer, it is the more common.

(If forces due to the air were included, we would also identify buoyancy, and perhaps a drag force if a wind were blowing.)

Key Points:

• To identify the forces acting on a body, start with gravity if the body is in a gravitational fi eld. Then, include forces for all objects and substances that contact the body.

• There may be more than one way to describe and analyze a situation.

• The set of forces acting on a body may depend crucially on exactly how one defi nes and delineates the “body.”


This simple question can spark a lively debate among students as they try to sort out the roles of friction, normal force(s), and tension. Impassioned advocates may defend all sides of the issue, and we encourage you to let the argument unfold on its own.

It is common practice to say that the upward force on the monkey is tension, but this is not precise and therefore can confuse some students, especially if it is simply asserted without a full explanation.

Most of the issues are resolved by defi nitions, and there are many valid answers, as outlined above. If we take the system or “body” to be the monkey and only the monkey, then the tension force is not exerted. Friction and a set of normal forces are the ones making direct contact with the monkey. If we consider only


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the “net” normal force, or sum of all normal forces, then the resultant is zero, so it may be neglected. If we consider a certain small bit of rope to be part of the system as well, then tension is the only contact force on the system.

Having students understand the “right” or “best” answer to this question is not as important as exploring defi nitions for “system,” “direct contact,” “free body,” and related ideas, as well as identifying common assumptions and practice.

There are many other similar situations that arise when trying to draw free-body diagrams. For instance, have the monkey hang from a loop in the rope. It is again common practice to say that the upward force is tension, though looking at the situation more closely, we can see that the loop of rope is pushing up on the monkey’s hand, so the force is actually a normal force.

Here are some more examples:

• Replace the rope with a spring. Even if we include a small bit of the spring as part of the system, it is defensible to call the force exerted by the spring “tension,” even though it is common practice to call it a spring force.

• The monkey fl oating in water experiences a buoyant force, which is a type of normal force.

• The monkey sitting on a compressed spring experiences either a spring force or a normal force, depending on your point of view.

Question B1.05

Description: Exploring the vector nature of force.

Question

A thin wire is stretched horizontally between two walls. If a weight W is hung on the wire, what is true about the tension T in the wire?

1. T < W 2. T = W 3. T > W 4. The relationship between T and W cannot be determined.

Commentary

Purpose: To develop your understanding of the vector nature of force.

Discussion: The weight W is at rest, so the net force on it is zero, and the wire must support its weight. The wire will sag at least a little, forming a “V” shape with the weight at the lowest point. We can analyze this as two tension forces acting on the weight, one pulling up and to the left and the other pulling up and to the right. If the situation is symmetric (the weight is in the middle of the wire), each tension force has a vertical component equal to half the weight of W. The wire will also have a nonzero horizontal component. If the wire is tight enough that it stretches very little, the wire will be close to horizontal and the horizontal com-ponent will be larger than the vertical component, so T > W. (The closer to horizontal the wire is, the larger the horizontal component and thus the tension will be.)

Force is a vector, so direction and components matter. The vertical components of the tensions in the two wires combine to balance the gravitational force on the weight, and the horizontal components balance each



other. To visualize this, it can help to draw a free-body diagram for the weight, and make it as accurate as possible.

You can also get a better understanding of this situation with a simple experiment. Hold one end of a rope or string and have a friend hold the other. Pull hard enough to keep the rope straight. Now, have someone else hang a heavy object from the middle of the rope. Can you pull hard enough to straighten the rope? As you change the tension in the rope by how hard you pull, what happens to the angle of the rope? When you and your friend hold both ends of the rope together so that it hangs straight down, what fraction of the weight does each of you support?

Key Points:

• Force is a vector. When forces balance, their components balance in each direction.

• When two forces balance a third by acting from two different directions, the sum of their magnitudes will be greater than the force they are balancing.


We are assuming that the thin wire does not stretch very much. If it stretches enough, it can become far enough from the horizontal that T < W. We must admit that the wire stretches and sags at least infi nitesi-mally, however, or there would be no vertical component of the tension to balance the weight.

Students commonly say that the tension is half the weight of the hanging object. They are truly amazed when the tension is not only larger, but much larger, than the weight. They are thinking of forces as scalars, not vectors.

Some students may use experience of everyday situations to answer—for example, from hanging wet clothes on a clothesline to dry.

A demonstration can be very useful. Use spring scales to measure the tension in the rope at its two ends. Otherwise, you might have students who simply disregard what you are saying because they do not believe you.

An alternative is to do the following: Have two people hold a rope horizontally between them and tell the class that you are going to hang a gallon of water (or something equivalently heavy) from the middle of it. Ask the class how hard these people must pull to keep the rope horizontal. Many students will say that each must pull with half the weight of the hanging object. The demonstrators will fi nd it impossible, and the class will enjoy watching them try to do it anyway.

Question B2.01a

Description: Relating and distinguishing mass, weight, and density.

Question

Consider the following three items.

A. 100 g of steam on the Moon B. 10 g of water on the Earth C. 20 g of ice fl oating in water on the Earth


128 Chapter 4

Put these items in order of increasing weight:

1. A < B < C 2. A < C < B 3. B < A < C 4. B < C < A 5. C < A < B 6. C < B < A 7. None of the above 8. Impossible to determine

Commentary

Purpose: To sensitize you to the similarities and differences between weight, mass, and density.

Discussion: Weight is the total gravitational force on something; it is not the same thing as mass. If the object is far from any objects that can exert a gravitational force, we say it is “weightless,” otherwise it has weight. The empirical law for gravitation is Fg = mg, where m is the mass (e.g., in kg) and g is the gravitational constant at that point in space (in units of force per kilogram).

g is about 10 N/kg on the Earth (9.81 if you want to be picky) and about 1.7 N/kg on the Moon. In other words, near the surface of the Earth, for every kg of mass something has, its weight is 10 N. On the Moon, the same object would weigh 1.7 N—about 1/6 as large.

It is easy to be confused about the weight of steam, since it is a gas and its weight cannot be measured. However, since it has mass, it must also have weight if it is in a gravitational fi eld. Likewise, ice fl oating has weight since it has mass and is near the surface of the Earth.

“Weight” is often explained as “a scale reading.” This can be a source of confusion, since this common-sense “defi nition” only works in limited circumstances and fails for gases, fl oating solids, or accelerating bodies.

Given that g on the Moon is about 1/6 as strong as g on the Earth, 10 g of water on the Earth has the smallest weight, then 100 g of steam on the Moon, then 20 g of ice on earth (whether fl oating or not).

Key Points:

• All matter in a gravitational fi eld has weight, and the weight depends only on the mass and the strength of the gravitational fi eld at that location.

• Be careful not to confuse mass and weight, especially since the words are used interchangeably in everyday language.

• Thinking of g as a force constant in N/kg, rather than as an acceleration in m/s2, is less likely to lead to confusion.


This is the fi rst of three questions exploring weight, mass, and density. They are interrelated, and we recommend discussing them jointly after presenting and collecting answers for all three.

Many students have some chemistry background. That can be useful to tap into, since they know that all atoms have an intrinsic mass. Unfortunately, the molar mass is often referred to as “atomic weight” even though it is given in grams; this can further confuse the issue.



Some students think that there is no gravity on the Moon, since there is no atmosphere.

Unlike most textbooks, we fi nd it pedagogically advantageous to defi ne g as the “gravitational constant” or “gravitational fi eld strength” rather than the “acceleration due to gravity,” and give its units in N/kg (in analogy with, say, Hooke’s law for springs). Otherwise, students can become confused about the “ acceleration due to gravity,” sometimes looking for or expecting there to be accelerations due to other forces. A more precise phrase would be “acceleration due to gravity in the absence of all other forces” or “acceleration in freefall,” but these phrases also have limitations. “Gravitational fi eld strength” or simply “gravitational constant” do not have these limitations.

Question B2.01b


Question



Put these items in order of increasing mass:


Commentary

Purpose: To sensitize you to the similarities and differences between weight, mass, and density.

Discussion: Every atom has mass. For instance, the atomic mass (mass of 1 mole) of carbon is 12 grams. Mass represents how hard or easy something is to accelerate, and it is not affected by the local gravitational constant g.

Consider an astronaut repairing a broken satellite. On the Earth, the satellite weighs many tons, and also has a large mass. In space, however, the weight is very small and the astronaut can support its weight very easily. But the satellite remains very diffi cult to move, because it is so hard to accelerate—and once moving, it is very hard to stop again!

The quantities given are all masses, so 10 g of water (on the Earth) has the smallest mass, then 20 g of ice, then 100 g of steam.


130 Chapter 4


This is the second of three questions exploring weight, mass, and density.

The primary point of confusion for students is likely to be the mass of the steam, since it is a gas. Some students just do not understand how a gas can have mass. One way to approach this is by appealing to the microscopic view of a gas consisting of atoms, which have mass. Another is to discuss the consequences of air (or gas) having no mass: When I blow air on my hand, I would not be able to feel it; when I pump up a basketball, I would not be able to hear it; there would be no air resistance or drag, which would be great for car fuel effi ciency, but there also could be no airplanes, because there would be no lift; hurricanes would not have any devastating effects, because there would be no energy in its winds. There are many other examples as well. The manifestations of air having mass are everywhere.

Question B2.01c


Question



Put these items in order of increasing density:


Commentary

Purpose: To sensitize students to the similarities and differences between weight, mass, and density.

Discussion: The density is the ratio of mass to volume (not of weight to volume).

Some people think that solids are always more dense than fl uids. While this is often true, ice is an excep-tion, since it fl oats in water. Its density is about 90% as large as that of water. In other words, 100 milliliters of water has a mass of 100 grams, but the same volume of ice has a mass of only about 90 grams.

People often confuse density with mass and/or weight. A battleship has a very large mass (and therefore a very large weight), but since it fl oats, its average density must be smaller than that of water. A coin is very light, having a mass of only a few grams (a nickel is 5 grams), but a coin sinks in water, which means its density is larger.



Since density is defi ned in terms of the mass, not weight, gravitational fi eld strength is irrelevant; an object or substance has the same density on the moon and on the earth (unless, of course, the moon’s vacuum causes changes in the volume of the substance).

Steam (in any amount, anywhere) has the smallest density, then ice, then water. The masses and locations are not meaningful for answering this question.

Key Points:

• The density of an object or substance is its mass divided by its volume, and does not depend on gravitational fi eld strength.

• Big objects are not necessarily denser than small ones.

• Everyday language is very sloppy about distinguishing between mass, weight, and density, but physics depends on precise usage.

• If one substance fl oats on another, the fl oating one is less dense.


This is the last of three questions exploring mass, weight, and density.

In everyday usage, we often use the same terms to refer to mass, weight, and density. We might say that something is “heavy” or “light,” and usually context alone tells us whether we are referring to mass, weight, or density. Therefore, it is useful for students to think about “light” objects that sink (such as coins) and “heavy” objects that fl oat (such as battleships).

As a follow-up, you can ask students if ice fl oats in water on the Moon, and if so, whether the water line would change. This will induce students to think about their ideas of mass, weight, and density.

Question B2.02

Description: Honing and distinguishing force, weight, and mass.

Question

An astronaut fl oats inside an orbiting space station. Which of the following are true?

A. No forces act on the astronaut. B. The astronaut has no mass. C. The astronaut has no weight.

1. A only 2. B only 3. C only 4. A and B 5. A and C 6. B and C 7. All are true. 8. None are true.


132 Chapter 4

Commentary

Purpose: To distinguish the concepts of force, mass, and weight.

Discussion: Does a force act on the astronaut? Yes! The astronaut is traveling in a circle or ellipse about the Earth. Since her velocity is changing (velocity is direction and speed), the astronaut must be accelerating, which means a force must be acting on her. The Earth is exerting a gravitational force on the astronaut that keeps her in orbit, rather than sailing off into space.

Does the astronaut have mass? Yes! All matter has mass, always. Mass determines an object’s inertia: how hard or easy it is to accelerate. A fl oating object can still be diffi cult to accelerate. A fl oating coin is much easier to accelerate than a fl oating astronaut.

Does the astronaut have weight? Yes! In physics, we defi ne “weight” to be the total gravitational force on something. Since the Earth exerts a gravitational force on her, the astronaut has weight. To someone within the space station the astronaut appears to be “weightless” because the observer and the space station—the entire frame of reference—has exactly the same acceleration that she does. To be precise, we can say that an object fl oating inside an orbiting space station has an “apparent weight” of zero.

The orbiting space station is not a “proper” frame of reference, because it is accelerating. To judge whether the astronaut is accelerating, we must observe the motion from a nonaccelerating frame.

(The International Space Station orbits the Earth once every 90 minutes, and it is always within 450 km of Earth’s surface. Thus, its acceleration is about 9.5 m/s2, and a 200 lb astronaut would weigh about 190 lb. At 6 Earth radii, a 200 lb astronaut would weigh about 6 lbs. This may seem small, but it is still the largest force acting on the astronaut, and therefore it cannot be neglected.)

Key Points:

• All matter has mass, always.

• An object’s weight is the total gravitational force acting upon it. Any mass in a gravitational fi eld has weight.

• According to Newton’s fi rst law, any object that is accelerating must be experiencing a nonzero net force. An object whose direction is changing is accelerating, even if the speed is not changing.

• Objects in free-fall or orbit have only apparent, not real, weightlessness.

• Newton’s laws are not valid in an accelerating frame of reference, so such a frame is not generally useful for analyzing motion.


This is a good question for drawing out misconceptions about force, mass, weight, free-fall, and “weight-lessness.” Everyday language rarely distinguishes between free-fall and true weightlessness, so students may experience considerable confusion here.

Invoking scale readings as an operational defi nition of weight is not a good solution; a scale merely measures the normal force exerted by the scale, which only equals an object’s weight in a proper reference frame and when no other forces impact the normal force.



Question B2.03

Description: Understanding the normal force.

Question

Consider the three situations shown below. In each case two small carts are connected by a spring. A constant force F is applied to the leftmost cart in each case. In each situation the springs are compressed so that the distance between the two carts never changes.

F 2M 2M(A)

F 3M M(B)

F M 3M(C)

Which of the following statements must be true regarding the compression of the spring in each case? Assume the springs are identical.

1. Compression A = Compression B = Compression C 2. B = C < A 3. A < B = C 4. A < B < C 5. B < A < C 6. C < A < B 7. A < C < B 8. None of the above 9. Cannot be determined

Commentary

Purpose: To explore Newton’s second law and model the normal force between two objects.

Discussion: The total mass of each system is 4M and the net force on each system is F, so the acceleration of each system—and each object in each system—is the same.

Isolating the right cart in each system, the net force on it is the spring force. Note that the applied force is exerted only on the left cart, and the left cart experiences two horizontal forces, F to the right, and the spring force to the left. But the right cart experiences only one horizontal force, so it is the net force. In any one system, the spring force is the same on both carts.

Since the accelerations of the rightmost carts in all three situations are equal, the least massive cart must have the smallest net force on it, and the most massive one must have the largest net force on it. Therefore, the spring pushing the M cart to the right must be compressed the least, followed by the spring pushing the 2M cart, and fi nally, the spring pushing the 3M cart to the right.


134 Chapter 4

This arrangement of two carts connected by a spring is a useful model for the normal force, because you can “see” the forces more easily. Imagine that the carts were not connected by springs, but were only touching each other. We would replace the spring force with a normal force, but everything else about the situation would be the same. The normal force pushing the 3M cart to the right would be largest, and the normal force pushing the M cart to the right would be smallest. Also, the normal force pushing to the right in each system would be as strong as the normal force pushing to the left.

Key Points:

• Newton’s second law can be applied to a “system” of objects moving together. It can also be applied individually to each object within the system.

• Only external forces contribute to the net force on a system. In this case, the spring force does not affect the acceleration of any of the two-cart systems.

• The more a spring is compressed (or stretched), the larger the force it exerts on either end.

• A compressed spring placed between two objects is a useful model of the normal force.


Students may not be comfortable applying Newton’s second law to a multiobject system. They may not realize that they do this all the time: when we apply Newton’s laws to an extended object, we are already treating it as a collection of atoms, all of which are moving at the same velocity, and therefore, each of which has the same acceleration (ignoring microscopic motion, of course).

Students should be discouraged from calculating the accelerations, and instead be encouraged to apply qualitative reasoning. Students should be encouraged also to draw free-body diagrams for each cart and for one of the two-cart systems. Which forces are included? Only external forces should be on a free-body diagram. Which forces balance? Vertical forces balance. What is the net force in each case? The sum of the horizontal forces.

Note that we are assuming that there is no oscillatory motion, which is diffi cult to eliminate in real life. This may be a point of confusion for students.

Additional Questions:

1. How would your answer change if the force F was applied to the left on the right cart of each system?

2. Using the same three systems, assume now that each spring is compressed by the same amount d. Compare the applied forces exerted on the left cart.

3. Blocks of 3 kg and 2 kg are side-by-side on a smooth, horizontal surface. A horizontal 12-N force is applied to the 3 kg block. (a) What are the accelerations of the blocks? (b) What are the net forces on the two blocks? (c) Show that the normal forces that the two blocks exert on each other must have the same strength in order for your answers to (a) and (b) to be correct. In other words, demonstrate that Newton’s third law is satisfi ed in this situation, even though you have not used it or assumed it to be true.



Question B2.04

Description: Understanding constraint forces and reasoning with tension.

Question

Consider the three situations below, labeled A, B, and C. Ignore friction.

2 kg

2 kg

2 kg

1 kg

2 kgA B C

After each system is released from rest, how do the tensions in the strings compare?

1. A = B = C 2. B = C < A 3. A = C < B 4. A < B < C 5. A < C < B 6. B < A < C 7. B < C < A 8. C < A < B 9. C < B < A 10. Impossible to determine

Commentary

Purpose: To reason qualitatively about forces and accelerations, and to recognize that tension is a “constraint force” with a situation-dependent magnitude.

Discussion: It is common for students to think that the tension is equal to 20 N in all three cases. However, this is impossible, otherwise the net force on each hanging mass would be zero, and it would not accelerate.

In A, the 2-kg block does not move or accelerate, so the tension in the string balances gravitation, which is 20 N.

In B, both blocks speed up. This means that the tension is the string must be smaller than the weight of the hanging block, so that there is a net force acting downward on the 2kg (hanging) block.

Likewise, in C, both blocks speed up, so the tension in the string must be smaller than the weight of the hanging block.

So we need to ask, “In which case, B or C, is the acceleration larger?” because the case with the larger acceleration will have the larger net force and thus the smaller tension. Intuitively, the acceleration of the blocks in B must be largest, B would have the smallest tension. Thus, B < C < A = 20 N.


136 Chapter 4

We can also use extreme case reasoning to distinguish B and C without solving any equations. Imagine the 1kg block in B is only 1 gram (smaller than a dime), and imagine that the upper 2-kg block in C is 2 000 kg (as big as a car). In this case, the hanging block in B would fall at nearly 10 m/s/s and the tension would be almost zero. And the hanging block in C would barely accelerate at all and the tension would be just slightly less than 20 N.

Key Points:

• Tension is a “constraint” force, which means there is no formula or empirical law to determine its value. Instead, the tension adjusts to the given situation, taking on whatever (positive) value is required to keep the string a fi xed length.

• You do not have to solve these situations in order to compare the tensions. Qualitative reasoning is all you need to do the comparison. Actually solving for the tensions is much more work, and you might not even believe the result when you get it. Reasoning through to an answer can increase your understanding.

• “Extreme case reasoning”—imagining the situation with exaggerated quantities—is a useful technique for making qualitative comparisons and developing intuition for a situation.


This style of problem—comparing the value of a quantity across similar situations—is powerful for draw-ing attention to particular features and for demonstrating the utility of qualitative reasoning. The instructor should strongly resist students’ desire for quantitative calculations! Extreme case reasoning may be new and uncomfortable to students, but this is a good situation to engage them in it.

Making connections to other constraint forces, such as the normal and static friction forces, can be helpful.

Question B2.05a

Description: Understanding tension and how to fi nd it in a static situation.

Question

Two blocks are arranged as shown and kept at rest by holding the 1-kg block in place.

1 kg

1.2 kg

The tension in the string is closest to:

1. 9 N 2. 10 N 3. 11 N



4. 12 N 5. 13 N 6. 10 N at the left end; 12 N at the right 7. 10 N in the left segment; 11 N in the middle segment; 12 N in the right segment 8. Smoothly varying from 10 N by the left block to 12 N by the right block 9. None of the above 10. Impossible to determine

Commentary

Purpose: To clarify the concept of tension, how it can be determined, and the assumptions we often make when working with it.

Discussion: Let’s assume as little as possible, and see where it leads us. The block on the right weighs about 12 N, so we know that the tension at that end of the string is 12 N also, since the block is not accel-erating. Traveling up the string, we expect the tension to become slightly larger, as we take into account the weight of the string: a point in the string must support the weight of the block and all the string hanging below it. If we neglect the string’s weight, the tension must be 12 N everywhere in the rightmost segment of the string.

The pulley on the right has no rotational acceleration, so the tension pulling down on it must be equal to the tension pulling left on it. This is true whether or not the pulley has mass.

Moving along the string to the left, a real string (i.e., with mass) would sag in the middle. This one doesn’t sag, so the tension is 12 N everywhere in the middle segment of the string.

The pulley on the left also has no rotational acceleratino, so the tension at the top of the leftmost segment is 12 N.

Traveling down the left string, the tension would get slightly smaller if the string had mass, since lower parts support less weight. If the string’s weight is negligible, then the tension is 12 N everywhere along the leftmost segment.

Thus, assuming the string is massless, the tension everywhere in it is 12 N.

Note that the weight of the 1-kg block is about 10 N, so a downward force of 2 N is needed to keep it at rest.

If the 1.2-kg block were supported at rest instead of the 1-kb block, the tension in the string would be 10 N everywhere.

Key Points:

• There is no straightforward law or equation for calculating the tension in a string. The tension will take on whatever value is required to keep the objects it is attached to from pulling farther apart.

• If we don't approximate strings as massless, the tension in them is not the same everywhere, and this makes calculations much more diffi cult.

• If a string passes over a pulley, the tension in each part of the string must be the same or the pulley will have a rotational acceleration.


138 Chapter 4


This is the fi rst of two related questions using the same situation.

Instructors often assert or assume that the tension is the same everywhere in a string without letting students see the implications of a nonmassless string or nonmassless pulleys. As a rule, students have diffi culty separating the consequences of the approximation from more general principles. This can be particularly hard to overcome later when we address massive pulleys and strings.

Students might not recognize or register the critical feature that the 1-kg block is held in place, but think that the system is moving or accelerating.

As a follow-up, try asking the question again with the other block held in place. To check for understanding, you can also ask students to explain how the 1-kg block can remain at rest with a tension force of 12 N pulling up on it.

Question B2.05b

Description: Understanding tension and how to fi nd it in a dynamic situation.

Question

Two blocks are arranged as shown and released from rest.

1 kg

1.2 kg

The tension in the string is closest to:

1. 9 N 2. 10 N 3. 11 N 4. 12 N 5. 13 N 6. 10 N at the left end; 12 N at the right 7. 10 N in the left segment; 11 N in the middle segment; 12 N in the right segment 8. Smoothly varying from 10 N by the left block to 12 N by the right block 9. None of the above 10. Impossible to determine

Commentary

Purpose: To help you to understand tension, especially in a dynamic situation.



Discussion: If the string is massless, the tension must be the same everywhere in a segment (between a block and pulley, or between two pulleys). Otherwise, some massless sub-portion of the segment would have a nonzero net force on it (more tension on one end than the other), and its acceleration would be infi nite.

When a string passes over a pulley, the pulley will have a rotational acceleration if the tensions are different on the two sides. If the pulley is massless, the rotational acceleration would be infi nite, so with a massless pulley the tension is always the same on either side. We won’t have the tools to handle pulleys with mass until we discuss rotational dynamics, so for now we’ll generally assume all pulleys are massless. So, in this question we can say the tension in the string will be the same everywhere, from one block all the way to the other.

The 1.2-kg block accelerates downward, so the net force on it must also point downward, which means the tension in the string is less than 12 N. The 1-kg block accelerates upward, so the net force on it must point upward as well, which means the tension in the string must be larger than 10 N. So, we know that 10 N < T < 12 N.

Without solving for the exact values of the accelerations and the tension, it is impossible to give a precise value for the tension, but 11 N is certainly a reasonable approximation. 11 N cannot be exactly correct, because then the accelerations of the two blocks would be different, which is impossible if the string remains taut and does not stretch. A tension of 11 N on the 1-kg block yields an acceleration of 1 m s2 upwards, while the same tension on the 1.2-kg block yields an acceleration of about 0 8. m s2 downwards. Therefore, the tension must be slightly smaller than 11 N, and the acceleration must be something near 0 9. m s2.

Note that tension is a “constraint” force, which means its value adjusts to take on whatever value is required to enforce a physical constraint of the system—in this case, that the string maintains a constant length and the blocks stay the same distance apart (along the string). This means the two blocks also have the same speed at all times, as well as the same magnitudes of acceleration. (In this situation, the directions of their velocities are opposite each other, as are the directions of their accelerations.)

Key Points:

• Tension is a constraint force, and will have whatever value is required to keep the objects at either end of the string from moving farther apart than the length of the string. There is no one equation for calculating the tension.

• The tension is the same everywhere in a section of massless string.

• The tension is the same in the string on either side of a massless pulley, even if the pulley is experiencing rotational acceleration.

• In many problems, you can estimate a numerical value fairly well without a detailed calculation to detrermine its precise value.


This is the second of two questions using the same situation.

This question may engender—or rather, reveal—much confusion, since students often believe that T = mg is a formula or empirical law for the tension. They do not appreciate that tension adjusts to the physical situation. Thus, most students’ answers will arise out of an attempt to reconcile confl icting ideas, and the resulting distribution of answers can appear random.


140 Chapter 4

Some students will attempt to solve the problem and compute a numerical value for the tension. We recommend discouraging this—for example, by giving students very little time to determine and submit an answer. Students should be encouraged to reason toward and estimate an answer, as that better develops conceptual understanding and physical intuition.

A nice follow-up or exam question is to put identical blocks on either end of the string and indicate that the system is in motion. Students who are still confused may not be able to see that since the blocks are not accelerating, the tension in the string is equal to the (common) weight of each block.

Question B1.06

Description: Reasoning with forces about self-propelled motion.

Question

A car accelerates down a straight highway. Which of the free-body diagrams shown below best represents the forces on the car?

Fg

FN

Fmotor

Fmotor

FN

Fg

FmotorFair

FNFN

FgFg

FfFf

Fair

Fair

(1) (2)

(3)(5) None of the above (6) Cannot be determined

(4)

Commentary

Purpose: To hone your understanding of forces by exploring the motion of a self-propelled vehicle.

Discussion: Many people, when asked “What makes a car accelerate?” will answer “the motor.” The motor is certainly an essential feature of this situation, but having a motor is not the whole story.

Imagine that the car is on ice. The motor can spin just as much as it does on a dry road, but the car does not move. Why? Because friction is needed for motion. On the road, the friction force of the roadway on the wheels prevents the wheels from spinning, so it acts in the direction the car moves: the direction of the net force.



Think about your own movement. You are a self-propelled vehicle, just like a car. You convert energy and push on the ground in order to go. If the ground is slippery, you don’t move. The direction you are pushing on the ground is “backwards” (assuming you want to move forward), so the ground is pushing in the opposite direction, or “forward.” Thus, friction provides the external force to make you go forward.

Therefore, the horizontal forces are friction in the direction of motion and air resistance opposite the direction of motion. Note that the friction is “static,” because the tires generally do not slip during normal operation. If the tires slip because the accelerator pedal is pushed too hard, the friction force will be kinetic but will still act in the direction of the car’s motion.

It’s not clear exactly what the “force of the motor” might be—presumably something to do with the drive shaft. Any such force is an internal force, and therefore cannot affect the motion of the car and should not be included in a free-body diagram. Only external forces are included in a free-body diagram.

Key Points:

• Only external forces belong on free-body diagrams.

• For self-propelled objects, an internal mechanism may provide the energy, but the friction force of a surface acting on the object provides the force to make the object accelerate.

• Even though the object as a whole is moving, the friction force is static if the part of the object contacting the ground is stationary at the moment of contact.


Self-propelled motion is very common. We literally do it every day, whether we drive a car or not, but it is poorly understood. Many students have trouble understanding how the friction force can be exerted in the direction of motion, because all of their experience is with kinetic friction opposing the motion of objects.

Students will need patient help sorting through these ideas, but the time and effort is usually worth it, since it will deepen students’ appreciation of many critical concepts: internal vs. external forces, static friction, Newton’s second and third laws, etc.

To help students see that the force they exert while walking is “backwards,” stand on a large board rest-ing on marbles or a similar frictionless surface. When you start walking, the board will fl y backwards (and you will not move much!). This can be done as a thought experiment as well; most students will intuitively understand that the board must move backward.

Question B2.06

Description: Understanding constraint forces and determining the static friction force.

Question

A mass of 5 kg sits at rest on an incline making an angle of 30° to the horizontal. If µs = 0 7. , what is the friction force on the block?


142 Chapter 4

1. 43.3 N, down the incline 2. 25 N, up the incline 3. 10 N, down the incline 4. 30.3 N, up the incline 5. None of the above

Commentary

Purpose: To understand constraint forces and how to determine the value of the static friction force.

Discussion: Static friction is a constraint force: a force that isn’t defi ned by a formula, but rather takes on whatever value is necessary (within limits) to satisfy Newton’s second law. The normal force and tension force are other constraint forces. We sometimes call them procedural forces because one must resort to a procedure, rather than a specifi c equation, to determine their value.

A common mistake is to think that the magnitude of the friction force is the coeffi cient of friction µ times the normal force N (which must itself be determined procedurally, though that is straightforward here). That would be correct for kinetic friction, where f Nk k= µ ; however, here we are dealing with static friction, where f Ns s≤ µ . Because of the inequality, it only gives us an upper limit on the magnitude; the actual value will be whatever is required to keep the block stationary on the plane, i.e., to satisfy Newton’s second law with zero acceleration.

If you work out the numbers, (2) turns out to be the best answer: the component of the block’s weight paral-lel to the plane is 25 N (downhill), and no other forces have a parallel component, so the static friction force must be 25 N up the plane. You don’t even need to determine the normal force, unless you want to check whether friction is strong enough here to hold the block in place. (If the angle were larger than about 35°, it would not be.)

Key Points:

• Constraint forces like normal, tension, and static friction forces have no force law and cannot generally be calculated from a formula. Rather, they take on whatever value is required to satisfy a physical constraint on the object (such as “the block doesn’t slide” or “the block doesn’t fall through the plane”).

• The “equation” for static friction, f Ns s≤ µ , only gives the maximum possible value of the force, not the actual value.


Choice (1) is the magnitude of the normal force, but pointing down the incline rather than up. (2) is correct. (3) is a random distracter. (4) will be popular: it is the maximum static friction force µs N( ), which students will choose if they interpret it as a force law rather than a maximum limit.

This problem can strengthen the common misconception that N = mg cos(θ ) is a general formula for deter-mining the normal force. In order to combat that and enhance students’ appreciation of constraint forces, we suggest following this question with one in which the normal force must be determined procedurally and is not equal to mg cos(θ ). During discussion, you might ask students what the normal force on the block in this problem is, and then ask them what it would be if a second, smaller block were perched on top of it, or if a string were pulling vertically upward on it.



Question B2.07a

Description: Exploring friction.

Question

Two blocks, having the same mass but different sizes, slide with the same constant speed on a smooth surface, then move onto a surface having friction coeffi cient µk.

M1

M2

Which stops in the shorter time?

1. M1

2. M2

3. Both stop in the same time. 4. Cannot be determined

Commentary

Purpose: To develop your understanding of friction (as we normally model it).

Discussion: We are told that even though the two blocks have different sizes, they have the same mass. Since the normal force balances the gravitational force in this situation, this means that the normal forces exerted by the fl oor are the same on both blocks.

In this course we use a model of friction in which the force does not depend on the size of objects or on the contact area between them. It depends only on the normal force and the coeffi cient of kinetic friction: F

kf = µkFN

. Therefore, both blocks experience the same force of friction, which is equal to the net force. Since they also have the same mass, they must have the same acceleration. Since they start at the same speed, they must take the same amount of time to stop.

We are ignoring air resistance, which would affect the two blocks differently due to their different sizes. However, since you are told that the blocks are moving with the same constant speed before moving into the rough region, you know that air resistance is negligibly small.

Other models of friction are possible, and might take into account other factors and features. For most situations, though, our simple model works quite well.

Key Points:

• The friction force, as we model it, depends only on the coeffi cient of friction and the normal force.

• We generally ignore air resistance (drag). In this case, we can reason that air resistance must be negligible in order to fi t the given description.

• An object’s size (volume) does not affect its acceleration; only its mass and the net force upon it matter.


144 Chapter 4


This is the fi rst of two similar questions. This one is intended merely to clarify the common model of kinetic friction. Students might have intuitive models that contradict the (oversimplifi ed but standard) model generally taught. Dealing with this issue should not take too much time.

Introductory physics is often taught as “revealed truth.” We prefer to make it explicit that everything in physics is a model of reality, with some sphere of utility and some degree of incompleteness or inaccuracy.

Students who include air resistance can validly choose the larger mass (M2) as stopping fi rst. However, you

should note that including air resistance is not consistent with the description, because in that case the two blocks could not be moving at constant speed, even on a frictionless surface.

The term “smooth” is used synonymously with “frictionless.” Some students might not be aware of this usage.

Question B2.07b

Description: Exploring friction: role of object’s mass in slowing down.

Question

Two blocks, M1 > M

2, having the same speed, move from a frictionless surface onto a surface having

friction coeffi cient µk.

M1

M2

Which stops in the shorter time?

1. M1

2. M2

3. Both stop in the same time. 4. Cannot be determined

Commentary

Purpose: To explore the effect an object’s mass has on acceleration due to friction.

Discussion: Note that the block that appears smaller in the fi gure has the larger mass.

As discussed in the previous question, the friction force on an object is proportional to its mass and the coeffi cient of friction. The net force on each block is due to friction alone. Thus, the net force on each is proportional to the block’s mass. According to Newton’s second law, each block’s acceleration is equal to the net force on it divided by its mass. Thus, the acceleration of each block must be the same: the mass “cancels out” since it appears in both the numerator (from the empirical law for friction) and the denominator (from Newton’s second law).

If the blocks have the same acceleration and same initial speed, they stop in the same amount of time.



Key Points:

• In many situations, the force on an object is proportional to its mass, so mass cancels out and the resulting acceleration does not depend on mass.


This is the second of two similar questions. In this one, the masses differ. This question gets at the com-mon phenomenon, often confusing to students, in which mass cancels out of the calculation for an object’s dynamical behavior. Students should learn to understand this and be able to see it qualitatively, rather than fi nding that sometimes the variable m mysteriously disappears from their algebra and sometimes it doesn’t.

Some students may carelessly assume that the block drawn larger in the fi gure is the more massive. Atten-tion to detail is important in physics!

It is common for students to apply incomplete reasoning, either arguing that the heavier mass experiences a larger friction force and thus slows more rapidly, or that the heavier mass has more inertia and thus slows more gradually. Some students may guess the correct answer without being able to justify it; they should be encouraged to formulate an argument.

A good follow-up discussion question is “Can you think of any other situations for which an object’s accel-eration is independent of its mass?”

QUICK QUIZZES

1. Newton’s second law says that the acceleration of an object is directly proportional to the resultant (or net) force acting on. Recognizing this, consider the given statements one at a time: (a) True—If the resultant force on an object is zero (either because no forces are present or the vector sum of the forces present is zero), the object can still move with constant velocity. (b) False—An object that remains at rest has zero acceleration. However, any number of external forces could be acting on it, provided that the vector sum of these forces is zero. (c) True—When a single force acts, the resultant force cannot be zero and the object must accelerate. (d) True—When an object accelerates, a set containing one or more forces with a nonzero resultant must be acting on it. (e) False—Many external forces could be acting on an object with zero acceleration, provided that the vector sum of these forces is zero. (f ) False—If the net force is in the positive x-direction, the acceleration will be in the positive x-direction. However, the velocity of an object does not have to be in the same direction as its acceleration (consider the motion of a projectile).

2. (b). The newton is a unit of weight, and the quantity (or mass) of gold that weighs 1 newton is m = ( )1 N g. Because the value of g is smaller on the Moon than on the Earth, someone possessing 1 newton of gold on the Moon has more gold than does a person having 1 newton of gold on Earth.

3. (c) and (d). Newton’s third law states that the car and truck will experience equal magnitude (but oppositely directed) forces. Newton’s second law states that acceleration is inversely proportional to mass when the force is constant. Thus, the lower mass vehicle (the car) will experience the greater acceleration.


146 Chapter 4

4. (c). In case (i), the scale records the tension in the rope attached to its right end. The section of rope in the man’s hands has zero acceleration, and hence, zero net force acting on it. This means that the tension in the rope pulling to the left on this section must equal the force F the man exerts toward the right on it. The scale reading in this case will then be F. In case (ii), the person on the left can be modeled as simply holding the rope tightly while the person on the right pulls. Thus, the person on the left is doing the same thing that the wall does in case (i). The resulting scale reading is the same whether there is a wall or a person holding the left side of the scale.

5. (c). The tension in the rope has a vertical component that supports part of the total weight of the woman and sled. Thus, the upward normal force exerted by the ground is less than the total weight.

6. (b). Friction forces are always parallel to the surfaces in contact, which, in this case, are the wall and the cover of the book. This tells us that the friction force is either upward or downward. Because the tendency of the book is to fall downward due to gravity, the friction force must be in the upward direction.

7. (b). The static friction force between the bottom surface of the crate and the surface of the truck bed is the net horizontal force on the crate that causes it to accelerate. It is in the same direction as the acceleration, toward the east.

8. (b). It is easier to attach the rope and pull. In this case, there is a component of your applied force that is upward. This reduces the normal force between the sled and the snow. In turn, this reduces the friction force between the sled and the snow, making it easier to move. If you push from behind, with a force having a downward component, the normal force is larger, the friction force is larger, and the sled is harder to move.

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. Newton’s second law gives the net force acting on the crate as

F fknet2N kg m s N= − = ( )( ) =95 0 60 0 1 20 72 0. . . .

This gives the kinetic friction force as fk = 23 0. N, so choice (a) is correct.

2. Since the pedestal is in equilibrium, the normal force pressing upward on its base supports the total weight of the man and pedestal. Therefore,

or

n F F m m gg g= ( ) + ( ) = +( ) =

man pedestal man pedestal 97.. .0 9 80 951kg m s N2( )( ) =

showing that (e) is the correct choice.

3. The tension, Fupper , in the vine at the point above the upper monkey must support the weight

of both monkeys i.e., upper singlemonkey

F Fg= ( )⎛⎝⎜

⎞⎠⎟

2 . However, the tension in the vine between the two

monkeys supports only the weight of the lower monkey F Fglower singlemonkey

= ( )⎛⎝⎜

⎞⎠⎟, meaning that

F Fupper lower = 2 and choice (d) is correct.



4. Because the block has zero vertical acceleration, Newton’s second law says that

ΣF F mg ny = −( )° − + =sin .30 0 0

or

n = − ( ) −( )° + ( )( ) =70 0 30 0 8 00 9 80. sin . . .N kg m s2 ++ + =35 0 78 4 113. .N N N

and we see that (c) is correct.

5. From Newton’s law of universal gravitation, the force Earth exerts on an object on its surface is F GM m R mgg E E= =2 , or the acceleration of gravity at Earth’s surface is g GM RE E= 2 . If both the mass and radius of the Earth should double, so ′ =M ME E2 and ′ =R RE E2 , the acceleration of gravity at the surface would then be

′ = ′′( )

=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

g GM

RG

M

RG

M

RE

E

E

E

E

E2 2 2

2

4

1

2== = =g

2

9 80

24 90

..

m sm s

22

meaning that (b) is the correct answer.

6. When the crate is held in equilibrium on the incline as shown in the sketch, Newton’s second law requires that Σ ΣF Fx y= = 0. From ΣFx g

xs= + − =F f

�� 0, the magnitude of the friction force

equals the component of gravitational force acting down the incline, or choice (e) is correct. Note that f f

� �s s sn= =

maxµ only

when the crate is on the verge of starting to slide.

7. According to Newton’s third law, the force of reaction (in this case, the force exerted on the rock by the glass) is always equal in magnitude to the force of action (in this case, the force the rock exerts on the glass). Thus, (b) is the correct choice.

8. The box will accelerate in the direction of the resultant force acting on it. The only horizontal forces present are the force exerted by the manager and the friction force. Since the acceleration is given to be in the direction of the force applied by the manager, the magnitude of this force must exceed that of the opposing friction force, however the friction force is not necessarily zero. Thus, choice (b) is correct.

9. Choose a coordinate system with the positive x-direction being east and the positive y-direction being north. Then the components of the four given forces are as follows:

A A B B

C

x y x y

x

= + = = = +

= −

40 0 0 50

7

N, N,

00 0 0 90N, NC D Dy x y= = = −,

The components of the resultant (or net) force are R A B C Dx x x x x= + + + = −30 N and R A B C Dy y y y y= + + + = −40 N. Therefore, the magnitude of the net force acting on the object is

R R Rx y= + = −( ) + −( ) =2 2 2 230 40 50N N N

or choice (a) is correct.

�y

�x

n

fs

Fg

→

→

→

qq

�y

�x

n

fs

Fg

→

→

→

qq


148 Chapter 4

10. Constant velocity means zero acceleration. From Newton’s second law, ΣF a��

= m , so the total (or resultant) force acting on the object must be zero if it moves at constant velocity. This means that (d) is the correct choice.

11. An object in equilibrium has zero acceleration a�

=( )0 , so both the magnitude and direction of the object’s velocity must be constant. Also, Newton’s second law states that the net force acting on an object in equilibrium is zero. The only untrue statement among the given choices is (d), untrue because the value of the velocity’s constant magnitude need not be zero.

12. As the trailer leaks sand at a constant rate, the total mass of the vehicle (truck, trailer and remain-ing sand) decreases at a steady rate. Then, with a constant net force present, Newton’s second law states that the magnitude of the vehicle’s acceleration a F mnet=( ) will steadily increase. Choice (b) is the correct answer.

13. When the truck accelerates forward, its natural tendency is to slip from beneath the crate, leaving the crate behind. However, friction between the crate and the bed of the truck acts in such a man-ner as to oppose this relative motion between truck and crate. Thus, the friction force acting on the crate will be in the forward horizontal direction and tend to accelerate the crate forward. The crate will slide only when the coeffi cient of static friction is inadequate to prevent slipping. The correct response to this question is (c).

14. The mass of an object is the same at all locations in space (e.g., on Earth, the Moon, or space station). However, the gravitational force the object experiences weight, w mg=( ) does vary, depending on the acceleration of gravity g at the object’s current location in space. It is the gravi-tation attraction of the Earth that holds the space station (and all its contents, including astro-nauts) in orbit around Earth. The only correct choice listed is (b).

15. Assuming that the cord connecting m m1 2and has constant length, the two masses are a fi xed dis-tance (measured along the cord) apart. Thus, their speeds must always be the same, which means that their accelerations must have equal magnitudes. The magnitude of the downward accelera-tion of m2 is given by Newton’s second law as

aF

m

m g T

mg

T

mgy

22

2

2 2

= = − = −⎛⎝⎜

⎞⎠⎟

<Σ

where T is the tension in the cord, and downward has been chosen as the positive direction. Therefore, the only correct statement among the listed choices is (a).

16. Only forces that act on the object should be included in the free-body diagram of the object. In this case, these forces are (1) the gravitational force (acting vertically downward), (2) the normal force (perpendicular to the incline) exerted on the object by the incline, and (3) the friction force exerted on the object by the incline, and acting up the incline to oppose the motion of the object down the incline). Choices (d) and (f) are forces exerted on the incline by the object. Choice (b) is the resultant of forces (1), (2), and (3) listed above, and its inclusion in the free-body diagram would duplicate information already present. Thus, correct answers to this question are (b), (d), and (f ).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. w = mg and g decreases with altitude. Thus, to get a good buy, purchase it in Denver. If gold were sold by mass, it would not matter where you bought it.

4. If it has a large mass, it will take a large force to alter its motion even when fl oating in space. Thus, to avoid injuring himself, he should push it gently toward the storage compartment.



6. The barbell always exerts a downward force on the lifter equal in magnitude to the upward force that she exerts on the barbell. Since the lifter is in equilibrium, the magnitude of the upward force exerted on her by the scale (that is, the scale reading) equals the sum of her weight and the downward force exerted by the barbell. As the barbell goes through the bottom of the cycle and is being lifted upward, the scale reading exceeds the combined weights of the lifter and the barbell. At the top of the motion and as the barbell is allowed to move back downward, the scale reading is less than the combined weights. If the barbell is moving upward, the lifter can declare she has thrown it just by letting go of it for a moment. Thus, the case is included in the previous answer.

8. The net force acting on the object decreases as the resistive force increases. Eventually, the resistive force becomes equal to the weight of the object, and the net force goes to zero. In this condition, the object stops accelerating, and the velocity stays constant. The rock has reached its terminal velocity.

10. While the engines operate, their total upward thrust exceeds the weight of the rocket, and the rocket experiences a net upward force. This net force causes the upward velocity of the rocket to increase in magnitude (speed). The upward thrust of the engines is constant, but the remain-ing mass of the rocket (and hence, the downward gravitational force or weight) decreases as the rocket consumes its fuel. Thus, there is an increasing net upward force acting on a diminishing mass. This yields an acceleration that increases in time.

12.

mg→

R→

R→

T→

mg→

R

f

→

→

n→

(a) (b) (c)

mg→

In the free-body diagrams given above, R��

represents a force due to air resistance, T��

is a force due to the thrust of the rocket engine, n

� is a normal force, f

� is a friction force, and the forces

labeled mg�

are gravitational forces.

14. If the brakes lock, the car will travel farther than it would travel if the wheels continued to roll, because the coeffi cient of kinetic friction is less than that of static friction. Hence, the force of kinetic friction is less than the maximum force of static friction.

PROBLEM SOLUTIONS

4.1 w = ( )⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞24 448

tons2000 lbs

1 ton

N

1 lb

.⎠⎠ = ×2 104 N

4.2 From v v= +0 at, the acceleration given to the football is

atav

2m s

sm s=

−= − =

v v0 10 0

0 2050

.

Then, from Newton’s second law, we fi nd

F m aav av

2kg m s 2 N= = ( )( ) =0 50 50 5.


150 Chapter 4

4.3 (a) ΣF max x= = ( )( ) =6 0 2 0. .kg m s 12 N2

(b) aF

mxx= = =Σ 12 N

4.0 kg3.0 m s2

4.4 (a) Action: The hand exerts a force to the right on the spring. Reaction: The spring exerts an equal magnitude force to the left on the hand. Action: The wall exerts a force to the left on the spring. Reaction: The spring exerts an equal magnitude force to the right on the wall. Action: Earth exerts an downward gravitational force on the spring. Reaction: The spring exerts an equal magnitude gravitational force upward on the Earth.

(b) Action: The handle exerts a force upward to the right on the wagon. Reaction: The wagon exerts an equal magnitude force downward to the left on the handle. Action: Earth exerts an upward contact force on the wagon. Reaction: The wagon exerts an equal magnitude downward contact force on the Earth. Action: Earth exerts an downward gravitational force on the wagon. Reaction: The wagon exerts an equal magnitude gravitational force upward on the Earth.

(c) Action: The player exerts a force upward to the left on the ball. Reaction: The ball exerts an equal magnitude force downward to the right on the player. Action: Earth exerts an downward gravitational force on the ball. Reaction: The ball exerts an equal magnitude gravitational force upward on the Earth.

(d) Action: M exerts a gravitational force to the right on m. Reaction: m exerts an equal magnitude gravitational force to the left on M.

(e) Action: The charge + Q exerts an electrostatic force to the right on the charge −q. Reaction: The charge −q exerts an equal magnitude electrostatic force to the left on the charge + Q.

(f ) Action: The magnet exerts a force to the right on the iron. Reaction: The iron exerts an equal magnitude force to the left on the magnet.

4.5 The weight of the bag of sugar on Earth is

w mgE E= = ( )⎛⎝

⎞⎠ =5 00 22 2. .lbs

4.448 N

1 lbN

If gM is the free-fall acceleration on the surface of the Moon, the ratio of the weight of an object on the Moon to its weight when on Earth is

w

w

mg

mg

g

gM

E

M

E

M

E

= =

so

w wg

gM EM

E

=⎛⎝⎜

⎞⎠⎟

.

Hence, the weight of the bag of sugar on the Moon is

wM = ( )⎛⎝

⎞⎠ =2

1

63 712.2 N N.

On Jupiter, its weight would be

w wg

gJ EJ

E

=⎛⎝⎜

⎞⎠⎟

= ( )( ) =22 2 2 64 58 7. . .N N

The mass is the same at all three locations, and is given by

mw

gE

E

= =( )( ) =5 00 4 448

9 802 27

. .

..

lb N lb

m skg2



4.6 aF

m= = ×

×= × −Σ 7 5 10

1 5 105 0 10

5

72.

..

N

kgm s2

and v v= +0 at gives

ta

=−

= −×

⎛⎝−

v v02

80 0

5 0 10

0 278

1

km h

m s

m s

km h2.

.⎜⎜

⎞⎠⎟

⎛⎝

⎞⎠ =1 min

s.4 min

607

4.7 Summing the forces on the plane shown gives

ΣF ma fx x= ⇒ − = ( )( )N kg m s210 0 20 2 0. .

From which, f = 9 6. N .

4.8 (a) The sphere has a larger mass than the feather. Hence, the sphere experiences a larger gravitational force F mgg = than does the feather.

(b) The time of fall is less for the sphere than for the feather. This is because air resistance affects the motion of the feather more than that of the sphere.

(c) In a vacuum, the time of fall is the same for the sphere and the feather. In the absence of air resistance, both objects have the free-fall acceleration g.

(d) In a vacuum, the total force on the sphere is greater than that on the feather. In the absence of air resistance, the total force is just the gravitational force, and the sphere weighs more than the feather.

4.9 The vertical acceleration of the salmon as it goes from v0 3 0y = . m s (underwater) to vy = 6 0. m s (after moving upward 1.0 m or 2/3 of its body length) is

ayy

y y=−

=( ) − ( )

( )v v2

02 2

2

6 0 3 0

2 1 00∆. .

.

m s m s

m

2

== 13 5. m s2

Applying Newton’s second law to the vertical leap of this salmon having a mass of 61 kg, we fi nd

ΣF ma F mg may y y= ⇒ − =

or

F m a gy= +( ) = ( ) +⎛⎝

⎞⎠ = ×61 13 5 9 8 1 4 1kg

m

s

m

s2 2. . . 003 N

4.10 The acceleration of the bullet is given by

ax

=−

( ) =( ) −

( )v v2

02 2

2

320 0

2 0 82∆m s

m.

Then,

ΣF ma= = ×( ) ( )( )

⎡

⎣⎢

⎤

⎦⎥−5 0 10

320

2 0 823

2

..

kgm s

m== ×3 1 102. N

F�10 N

a�2.0 m/s2

f→ F�10 N

a�2.0 m/s2

f→


152 Chapter 4

4.11 (a) From the second law, the acceleration of the boat is

a

F

m= = −

=

Σ 2 000 1 800

0 200

N N

1 000 kg

m s2.

(b) The distance moved is

∆x t at= + = + ( )( ) =v02 21

20

1

20 200 10 0. .m s s 10.02 mm

(c) The fi nal velocity is v v= + = + ( )( ) =0 0 0 200 10 0 2 00at . . .m s s m s2 .

4.12 (a) Choose the positive y-axis in the forward direction. We resolve the forces into their components as

Force x-component y-component

400 N 200 N 346 N

450 N –78.1 N 443 N

Resultant ΣFx = 122 N ΣFy = 790 N

The magnitude and direction of the resultant force is

F F FR x y= ( ) + ( ) =Σ Σ2 2799 N θ =

⎛

⎝⎜⎞

⎠⎟= °−tan 1 Σ

ΣF

Fx

y

8 77. to right of y-axis

Thus, �FR = °799 8 77N at to the right of the forw. aard direction .

(b) The acceleration is in the same direction as �FR and has magnitude

aF

mR= = =799

0N

3 000 kg.266 m s2

4.13 (a) At terminal speed, the magnitude of the air resistance force is equal to the weight of the skydiver F mgdrag =( ) . Therefore,

kF mg= = =

( )( )drag

terminal2

2kg m s

5v v2

65 0 9 80. .

55.0 m skg m

( )=2 0 211.

(b) At the speed v v= 12 terminal, the upward force due to air resistance is less than the downward

gravitational force acting on the skydiver, leaving a net downward force of magnitude F mg Fnet drag= − . Thus, the skydiver will have a downward acceleration of magnitude

aF

m

mg k

mg

k

m= = − = − ⎛

⎝⎞⎠

⎛⎝

⎞⎠

=

net terminalv v2 2

2

9 8. 000 211

65 0

55 0

2

2

m skg m

kg

m s2 −⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠ =.

.

.77 35. m s2

f→

F→

�y

� x f �1 800 NF�2 000 N

f→

F→

�y

� x f �1 800 NF�2 000 N

F 1�

400

N

F2

�450 N

�y

�x

30°

10°

F 1�

400

N

F2

�450 N

�y

�x

30°

10°



4.14 (a) With the wind force being horizontal, the only vertical force acting on the object is its own weight, mg. This gives the object a downward acceleration of

aF

m

mg

mgy

y= = − = −Σ

The time required to undergo a vertical displacement ∆y h= − , starting with initial vertical velocity v0 0y = , is found from ∆y t a ty y= +v0

12

2 as

− = − =hg

t th

g0

2

22 or

(b) The only horizontal force acting on the object is that due to the wind, so ΣF Fx = and the horizontal acceleration will be

aF

m

F

mxx= =Σ

(c) With v0 0x = , the horizontal displacement the object undergoes while falling a vertical distance h is given by ∆x t a tx x= +v0

12

2 as

∆xF

m

h

g

Fh

mg= + ⎛

⎝⎞⎠

⎛⎝⎜

⎞⎠⎟

=01

2

22

(d) The total acceleration of this object while it is falling will be

a a a F m g F m gx y= + = ( ) + −( ) = ( ) +2 2 2 2 2 2

4.15 Starting with v0 0y = and falling 30 m to the ground, the velocity of the ball just before it hits is

v v1 02 2 0 2 9 80 30 24= − + = − + −( ) −( ) = −y ya y∆ . m s m m2 ss

On the rebound, the ball has vy = 0 after a displacement ∆y = +20 m. Its velocity as it left the ground must have been

v v22 2 0 2 9 80 20 20= + − = + − −( )( ) = +y ya y∆ . m s m m s2

Thus, the average acceleration of the ball during the 2.0-ms contact with the ground was

atav

m s m s

s= − =

+ − −( )×

= + ×−v v2 1

3

20 24

2 0 102 2

∆ .. 1104 m s upward2

The average resultant force acting on the ball during this time interval must have been

F manet av

2kg m s= = ( ) + ×( ) = + ×0 50 2 2 10 1 1 104 4. . . NN

or

F��

net N upward= ×1 1 104.


154 Chapter 4

4.16 Since the two forces are perpendicular to each other, their resultant is

FR = ( ) + ( ) =180 390 4302 2N N N, at θ = ⎛⎝

⎞⎠ = °−tan

N

NN of E1 390

18065 2.

Thus,

aF

mR= = =4

1 5930 N

270 kgm s2.

or

�a = °1 59. m s at 65.2 N of E2 .

4.17 (a) Since the burglar is held in equilibrium, the tension

in the vertical cable equals the burglar’s weight of 600 N .

Now, consider the junction in the three cables:

ΣFy = 0 , giving T2 37 0 600 0sin . ° − =N

or

T2

600

37 0997=

°=N

N in the inclined cablsin .

ee

Also, ΣFx = 0 which yields T T2 137 0 0cos . ° − = , or

T1 997 37 0 796= ( ) ° =N N in the horizontalcos . ccable

(b) If the left end of the originally horizontal cable was attached to a point higher up the wall, the tension in this cable would then have an upward component. This upward component would support part of the weight of the cat burglar, thus

decreasing the tension in the cable on the rright

4.18 Using the reference axis shown in the sketch at the right, we see that

ΣF T Tx = ° − ° =cos . cos .14 0 14 0 0

and

ΣF T T Ty = − ° − ° = − °sin . sin . sin .14 0 14 0 2 14 0

Thus, the magnitude of the resultant force exerted on the tooth by the wire brace is

R F F T Tx y= ( ) + ( ) = + − °( ) = °Σ Σ2 2 20 2 14 0 2 14 0sin . sin .

or

R = ( ) ° =2 18 0 14 0 8 71. sin . .N N

x

y

w�600 N

37.0°T1→

T2→

x

y

w�600 N

37.0°T1→

T2→

y

T

T

x14°

14°

y

T

T

x14°

14°



4.19 From ΣFx = 0, T T1 230 0 60 0 0cos . cos .° − ° =

or T T2 11 73= ( ). [1]

Then, ΣFy = 0 becomes

T T1 130 0 1 73 60 0 150 0sin . . sin .° + ( ) ° − =N

which gives T1 75 0= . N in the right side cable .

Finally, Equation [1] above gives T2 130= N in the left side cable .

4.20 If the hip exerts no force on the leg, the system must be in equilibrium with the three forces shown in the free-body diagram.

Thus, ΣFx = 0 becomes

w2 110 40cos cosα = ( ) °N [1]

From ΣFy = 0, we fi nd

w2 220 110 40sin sinα = − ( ) °N N [2]

Dividing Equation [2] by Equation [1] yields

α = − ( ) °( ) °

⎛⎝⎜

⎞−tansin

cos1 220 110 40

110 40

N N

N ⎠⎠⎟= °61

Then, from either Equation [1] or [2], w221 7 10= ×. N .

4.21 (a) Free-body diagrams of the two blocks are shown at the right. Note that each block experiences a downward gravitational force

Fg = ( )( ) =3 50 9 80 34 3. . .kg m s N2

Also, each has the same upward acceleration as the elevator, in this case ay = +1 60. m s2.

Applying Newton’s second law to the lower block:

ΣF ma T F may y g y= ⇒ − =2

or

T F mag y2 34 3 3 50 1 60 39 9= + = + ( )( ) =. . . .N kg m s N2

Next, applying Newton’s second law to the upper block:

ΣF ma T T F may y g y= ⇒ − − =1 2

or

T T F mag y1 2 39 9 34 3 3 50 1 60= + + = + + ( ). . . .N N kg m s22 79.8 N( ) =

T1→T2

→

y

x30.0°60.0°

150 N

T1→T2

→

y

x30.0°60.0°

150 N

w2→

x

y

40° α

220 N

110 N w2→

x

y

40° α

220 N

110 N

Upper Blockm�3.50 kg

Fg

T1

T2

Lower Blockm�3.50 kg

Fg

T2

Upper Blockm�3.50 kg

Fg

T1

T2

Lower Blockm�3.50 kg

Fg

T2

continued on next page


156 Chapter 4

(b) Note that the tension is greater in the upper string, and this string will break fi rst as the acceleration of the system increases. Thus, we wish to fi nd the value of ay when T1 85 0= . N. Making use of the general relationships derived in (a) above gives:

T T F ma F ma F ma F mag y g y g y g y1 2 2 2= + + = +( ) + + = +

or

aT F

myg=

−= − ( )

( ) =1 2

2

85 0 2 34 32 34

. ..

N N

2 3.50 kgm s2

4.22 (a) Free-body diagrams of the two blocks are shown at the right. Note that each block experiences a downward gravitational force F mgg = .

Also, each has the same upward acceleration as the elevator, a ay = + .

Applying Newton’s second law to the lower block:

ΣF ma T F may y g y= ⇒ − =2 or T mg ma m g a2 = + = +( )

Next, applying Newton’s second law to the upper block:

ΣF ma T T F may y g y= ⇒ − − =1 2

or

T T F ma mg ma mg ma mg ma Tg y1 2 12 2= + + = +( ) + + = +( ) =

(b) Note that T T1 22= , so the upper string breaks first as the acceleration of the system

increases.

(c) When the upper string breaks, both blocks will be in free-fall with a g= − . Then, using the results of part (a), T m g a m g g2 0= +( ) = −( ) = and T T1 22 0= = .

4.23 m = 1 00. kg and mg = 9 80. N

α = ⎛⎝

⎞⎠ = °−tan

..1 0 200

0 458m

25.0 m

Since ay = 0, require that ΣF T T mgy = + − =sin sinα α 0,

giving 2T mgsinα = ,

or

T = =9 80613

. N

2sinN

α

Upper Blockmass�m

Fg

T1

T2

Lower Blockmass�m

Fg

T2

Upper Blockmass�m

Fg

T1

T2

Lower Blockmass�m

Fg

T2

aa

mg→T→

T→

25.0 m 25.0 m

0.200 maa

mg→T→

T→

25.0 m 25.0 m

0.200 m



4.24 The resultant force exerted on the boat by the people is

2 600 30 0 1 04 103N N( )[ ]° = ×cos . . in the forward direction

If the boat moves with constant velocity, the total force acting on it must be zero. Hence, the resistive force exerted on the boat by the water must be

�f = ×1 04 103. N in the rearward direction

4.25 The forces on the bucket are the tension in the rope and the weight of the bucket,

mg = ( )( ) =5 0 9 80 49. .kg m s N2 . Choose the positive direction upward

and use Newton’s second law:

ΣF may y=

T − = ( )( )49 5 0 3 0N kg m s2. .

T = 6 N4

4.26 (a) From Newton’s second law, we fi nd the acceleration as

aF

mxx= = =Σ 10

0 33N

30 kgm s2.

To fi nd the distance moved, we use

∆x t a tx x= + = + ( )( ) =v02 21

20

1

23 0 1 50.33 m s s m2 . .

(b) If the shopper places her 30 N 3.1 kg( ) child in the cart, the new acceleration will be

aF

mxx= = =Σ

total

2N

33 kgm s

100 30.

and the new distance traveled in 3.0 s will be

∆ ′ = + ( )( ) =x 01

23 0 1 420.30 m s s m2 . .

4.27 (a) The average acceleration is given by

atav

2m s m s

sm s=

−= − = −

v v0 5 00 20 0

4 003 75

∆. .

..

The average force is found from Newton’s second law as

F maav av2kg m s N= = ( ) −( ) = − ×2 000 3 75 7 50 103. .

(b) The distance traveled is

x t= ( ) = +⎛⎝

⎞⎠ ( ) =vav

m s m ss∆ 5 00 20 0

24 00 50 0

. .. . m

→mg

→a

→T

→mg

→a

→T

F �10 NF �10 N


158 Chapter 4

4.28 Let m1 10 0= . kg, m2 5 00= . kg, and θ = °40 0. .

Applying the second law to each object gives

m a m g T1 1= − [1]

and

m a T m g2 2= − sinθ [2]

Adding these equations yields

m a m a m g T T m g am m

m1 2 1 21 2+ = − + − = −

orsinθ

11 2+⎛⎝⎜

⎞⎠⎟m

g

so,

a =− ( ) °⎛

⎝⎜⎞⎠⎟

10 0 5 00 40 0

15 09 8

. . sin .

..

kg kg

kg00 4 43m s m s2 2( ) = .

Then, Equation [1] yields

T m g a= −( ) = ( ) −( )⎡⎣ ⎤⎦ =1 10 0 9 80 4 43 53 7. . . .kg m s2 N

4.29 (a) The resultant external force acting on this system, consisting of all three blocks having a total mass of 6.0 kg, is 42 N directed horizontally toward the right. Thus, the acceleration produced is

aF

m= = =Σ 42

7 0N

6.0 kgm s horizontally to th2. ee right

(b) Draw a free body diagram of the 3.0-kg block and apply Newton’s second law to the horizontal forces acting on this block:

ΣF max x= ⇒ 42 3 0 7 0N kg m s2− = ( )( )T . .

and therefore T = 21 N .

(c) The force accelerating the 2.0-kg block is the force exerted on it by the 1.0-kg block. Therefore, this force is given by F ma= = ( )( )2 0 7 0. .kg m s2 , or �F = 14 N horizontally to the right .

4.30 The acceleration of the mass down the incline is given by

∆x t a t= +v021

2 or 0 80 0

1

20 50 2. .m s= + ( )a

This gives

a = ( )( )

=2 0 80

0 506 42

.

..

N

sm s2

Thus, the net force directed down the incline is F ma= = ( )( ) =2 0 6 4 13. .kg m s N2 .

q

m2g→m1g→

m2 m1

n→

a→

a→T→ T

→

q

m2g→m1g→

m2 m1

n→

a→

a→T→ T

→



4.31 (a) Assuming frictionless pulleys, the tension is uniform through the entire length of the rope. Thus, the tension at the point where the rope attaches to the leg is the same as that at the 8.00-kg block. Part (a) of the sketch at the right gives a free-body diagram of the suspended block. Recognizing that the block has zero acceleration, Newton’s second law gives

ΣF T mgy = − = 0

or

T mg= = ( )( ) =8 00 9 80 78 4. . .kg m s N2

(b) Part (b) of the sketch above gives a free-body diagram of the pulley near the foot. Here, F is the magnitude of the force the foot exerts on the pulley. By Newton’s third law, this is the same as the magnitude of the force the pulley exerts on the foot. Applying the second law gives:

ΣF T T F max x= + ° − = =cos .70 0 0

or

F T= +( )° = ( ) +( )° =1 70 0 78 4 1 70 0 105cos . . cos .N N

4.32 (a)

m1g→

m1 m2

m2g→

P→

F→

n2→

n1→

P→

(b) Note that the blocks move on a horizontal surface with ay = 0. Thus, the net vertical force acting on each block and on the combined system of both blocks is zero. The net horizontal force acting on the combined system consisting of both m m1 2and is ΣF F P P Fx = − + = .

(c) Looking at just m1, ΣFy = 0 as explained above, while ΣF F Px = − .

(d) Looking at just m2, we again have ΣFy = 0, while ΣF Px = + .

(e) For m1: ΣF ma F P m ax x= ⇒ − = 1

For m2: ΣF ma P m ax x= ⇒ = 2

(f ) Substituting the second of the equations found in (e) above into the fi rst gives the following:

m a F P F m a m m a F1 2 1 2= − = − +( ) =or and a F m m= +( )1 2

Then substituting this result into the second equation from (e), we have

P m a mF

m m= =

+⎛⎝⎜

⎞⎠⎟2 2

1 2

or Pm

m mF=

+⎛⎝⎜

⎞⎠⎟

2

1 2

(a) (b)

T

mg

8.00 kgF

T

T

70.0°

(a) (b)

T

mg

8.00 kgF

T

T

70.0°



160 Chapter 4

(g) Realize that applying the force to m2 rather than m1 would have the effect of interchanging the roles of m m1 2and . We may easily fi nd the results for that case by simply interchanging the labels m m1 2and in the results found in (f ) above. This gives

a F m m= +( )2 1 (the same result as in the fi rst case) and

Pm

m mF=

+⎛⎝⎜

⎞⎠⎟

1

2 1

We see that

the contact force, , is larger in this caP sse because m m1 2> .

4.33 Taking the downward direction as positive, applying Newton’s second law to the falling person yields ΣF mg f may y= − = , or

a g

f

my = − = −⎛⎝⎜

⎞⎠⎟

=9 80100

8 6. .m sN

80 kgm s2 2

Then, v vy y ya y202 2= + ( )∆ gives the velocity just before hitting the net as

v vy y ya y= + ( ) = + ( )( ) =02 2 0 2 8 6 30 23∆ . m s m m s2

4.34 (a) First, consider a system consisting of the two blocks combined, with mass m m1 2+ . For this system, the only external horizontal force is the tension in cord A pulling to the right. The tension in cord B is a force one part of our system exerts on another part of our system, and is therefore an internal force.

Applying Newton’s second law to this system (including only external forces, as we should) gives

ΣF ma T m m ax A= ⇒ = +( )1 2 [1]

Now, consider a system consisting of only m2. For this system, the tension in cord B is an external force since it is a force exerted on block 2 by block 1 (which is not part of this system). Applying Newton’s second law to this system gives

ΣF ma T m ax B= ⇒ = 2 [2]

Comparing Equations [1] and [2], and realizing that the acceleration is the same in both cases (see Part (b) below), the following is clear:

Cord A exerts a larger force on block 1 thann cord B exerts on block 2

(b) Since cord B connecting the two blocks is taut and unstretchable, the two blocks stay a fi xed distance apart, and the velocities of the two blocks must be equal at all times. Thus, the rates at which the velocities of the two blocks change in time are equal, or the two blocks must have equal accelerationss .

(c) Yes. Block 1 exerts a forward force on Cord B, so Newton’s third law tells us that Cord B exerts a force of equal magnitude in the backward direction on Block 1.

m22

m11

B A

a

m22

m11

B A

a



4.35 (a) When the acceleration is upward, the total upward force T must exceed the total

downward force w mg= = ( )( ) = ×1 500 9 80 1 47 104kg m s N2. . .

(b) When the velocity is constant, the acceleration is zero. The total upward force T and the total downward force w must be equal in magnitude .

(c) If the acceleration is directed downward, the total downward force w must exceed the total upward force T.

(d) ΣF ma T mg may y y= ⇒ = + = ( ) +1 500 9 80 2 50kg m s m s2 2. .(( ) = ×1 85 104. N

Yes , T w> .

(e) ΣF ma T mg may y y= ⇒ = + = ( ) +( ) =1 500 9 80 0 1 47kg m s2. . ×× 104 N

Yes , T w= .

(f ) ΣF ma T mg may y y= ⇒ = + = ( ) −1 500 9 80 1 50kg m s m s2 2. .(( ) = ×1 25 104. N

Yes , T w< .

4.36 Note that if the cord connecting the two blocks has a fi xed length, the accelerations of the blocks must have equal magnitudes, even though they differ in directions. Also, observe from the diagrams, we choose the positive direction for each block to be in its direction of motion.

First consider the block moving along the horizontal. The only force in the direction of movement is T. Thus,

ΣF max x= ⇒ T a= ( )5 00. kg [1]

Next consider the block which moves verti-cally. The forces on it are the tension T and its weight, 98.0 N.

ΣF may y= ⇒ 98 0 10 0. .N kg− = ( )T a [2]

Equations [1] and [2] can be solved simultaneously to give

98 0 5 00 10 0. . .N kg kg− ( ) = ( )a a or a = =98 0. N

15.0 kg6.53 m s2

and

T = ( )( ) =5 00 2 7. .kg 6.53 m s 3 N2

5.00 kg 10.0 kg

mg�98.0 N

�x

�yT→

T→

(a) (b)

5.00 kg 10.0 kg

mg�98.0 N

�x

�yT→

T→

(a) (b)


162 Chapter 4

4.37 Trailer300 kg

Car1000 kg

nT→

Rcar→

nc→

nc→

wT→

wc→

T→

T→

F→

F→

q

Choose the +x direction to be horizontal and forward with the +y vertical and upward. The common acceleration of the car and trailer then has components of a ax y= + =2 15 0. m s and2 .

(a) The net force on the car is horizontal and given by

ΣF F T m ax x( ) = − = = ( )( ) =car car

2kg m s1 000 2 15 2 1. . 55 103× N forward

(b) The net force on the trailer is also horizontal and given by

ΣF T m ax x( ) = + = = ( )trailer trailer

2kg m s300 2 15.(( ) = 645 N forward

(c) Consider the free-body diagrams of the car and trailer. The only horizontal force acting on the trailer is T = 645 N forward, and this is exerted on the trailer by the car. Newton’s third law then states that the force the trailer exerts on the car is 645 N toward the rear .

(d) The road exerts two forces on the car. These are F ncand shown in the free-body diagram of the car.

From part (a),

F T= + × = + × = + ×2 15 10 645 2 15 10 2 80 103 3 3. . .N N N N

Also, ΣF n w m ay c c y( ) = − = =car car 0, so n w m gc c= = = ×car N9 80 103.

The resultant force exerted on the car by the road is then

R F nccar N N= + = ×( ) + ×( ) =2 2 3 2 3 22 80 10 9 80 10 1 0. . . 22 104× N

at θ = = ( ) = °− −tan ( ) tan . .1 1 3 51 74 1n Fc above the horizontal and forward. Newton’s third law then states that the resultant force exerted on the road by the car is

1 02 104. × N at 74.1 below the horizontal an° dd rearward .

4.38 First, consider the 3.00-kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes

T a− = ( )29 4 3 00. .N kg [1]

The forces on the falling 5.00-kg mass are its weight and T, and its acceleration

has the same magnitude as that of the rising mass. Choosing the positive direction as downward for this mass, gives

49 0 5 00. .N kg− = ( )T a [2]

RisingMass

FallingMass

m1 � 3.00 kg

w1 � 29.4 N w2 � 49.0 N

m2 � 5.00 kg

T→

T→

��

RisingMass

FallingMass

m1 � 3.00 kg

w1 � 29.4 N w2 � 49.0 N

m2 � 5.00 kg

T→

T→

��




(a) Solving Equation [2] for a and substituting into [1] gives

T T− =⎛⎝⎜

⎞⎠⎟

−( )29 43 00

49 0..

.Nkg

5.00 kgN or 1 60 58 8. .T = N

and the tension is T = 36 8. N .

(b) Equation [2] then gives the acceleration as

a = − =49 0 36 82 44

. ..

N N

5.00 kgm s2

(c) Consider the 3.00-kg mass. We have

∆y t a ty y= + = + ( )( ) =v02 21

20

1

22 44 1 00 1 22. . .m s s2 m

4.39 When the block is on the verge of moving, the static friction force has a magnitude f f ns s s= ( ) =

maxµ .

Since equilibrium still exists and the applied force is 75 N, we have

ΣF fx s= − =75 0N or fs( ) =max

N75

In this case, the normal force is just the weight of the crate, or n mg= . Thus, the coeffi cient of static friction is

µss sf

n

f

mg=

( )=

( )= ( )( )

max max2

N

20 kg m s

75

9 80.== 0 38.

After motion exists, the friction force is that of kinetic friction, f nk k= µ .

Since the crate moves with constant velocity when the applied force is 60 N, we fi nd that ΣF fx k= − =60 0N or fk = 60 N. Therefore, the coeffi cient of kinetic friction is

µkk kf

n

f

mg= = = ( )( ) =6

9 800 31

0 N

20 kg m s2..

4.40 (a) The static friction force attempting to prevent motion may reach a maximum value of

f n m gs s s( ) = = = ( )( )( ) =max

2kg m sµ µ1 1 0 50 10 9 80. . 449 N

This exceeds the force attempting to move the system, w m g2 2 39= = N. Hence, the system remains at rest and the acceleration is a = 0 .

(b) Once motion begins, the friction force retarding the motion is

f n m gk k k= = = ( )( )( ) =µ µ1 1 0 30 10 9 80 29. .kg m s N2

This is less than the force trying to move the system, w m g2 2= . Hence, the system gains speed at the rate

aF

m

m g m g

m mk= = −

+=

−net

total

kg k2 1

1 2

4 0 0 30 10µ . . gg m s

kg kgm s

22( )⎡⎣ ⎤⎦( )

+=

9 80

4 0 100 70

.

..


164 Chapter 4

4.41 (a) Since the crate has constant velocity, a ax y= = 0.

Applying Newton’s second law:

ΣF F f max k x= ° − = =cos .20 0 0 or fk = ( ) ° =300 20 0 282N Ncos .

and

ΣF n F wy = − ° − =sin .20 0 0 or n = ( ) ° + = ×300 20 0 1 000 1 10 103N N Nsin . .

The coeffi cient of friction is then

µkkf

n= =

×=2

1 10 103

82 N

N0.256

.

(b) In this case, ΣF n F wy = + ° − =sin .20 0 0, so n w F= − ° =sin .20 0 897 N.

The friction force now becomes f nk k= = ( )( ) =µ 0 256 897 230. N N.

Therefore, ΣF F f ma w g ax k x x= ° − = =cos . ( )20 0 and the acceleration is

aF f g

wk=

° −( )=

( ) ° −[ ]cos . cos .20 0 300 20 0 230 9N N ...

80

1 0000 509

m s

Nm s

22( )

=

4.42 (a) atx

x x=−

= − = −v v0 6 00 12 0

5 001 20

. .

..

m s m s

sm s2

(b) From Newton’s second law, ΣF f ma f max k x k x= − = = −, or .

The normal force exerted on the puck by the ice is n mg= , so the coeffi cient of friction is

µkkf

n

m

m= =

− −( )( ) =

1 20

9 800 122

.

..

m s

m s

2

2

(c) ∆x t txx x= ( ) =

+⎛⎝

⎞⎠ = +⎛

⎝⎞v

v vav

m s m s0

2

6 00 12 0

2

. .⎠⎠ ( ) =5 00 45 0. .s m

4.43 When the load on the verge of sliding forward on the bed of the slowing truck, the rearward directed static friction force has its maximum value

f n m gs s s( ) =max load= µ µ

Since slipping is not yet occurring, this single horizontal force must give the load an acceleration equal to that the truck.

Thus, ΣF ma m g m ax x s= ⇒ − =load load truckµ , or a gstruck = −µ .

fs→

n→

→mload g

fs→

n→

→mload g




(a) If slipping is to be avoided, the maximum allowable rearward acceleration of the truck is seen to be a gstruck = −µ , and v vx x xa x2

02 2= + ( )∆ gives the minimum stopping distance as

∆xa g

x x

s

( ) =−

( ) =min

max

0

2 202

02v v

truck µ

If v0 12 0 500x s= =m s and µ . , then ∆x( ) =( )

( )( ) =min

.

. ..

12 0

2 0 500 9 8014 7

2m s

m sm

2

(b) Examining the calculation of Part (a) shows that neither mass is necessary .

4.44 (a) The free-body diagram of the crate is shown at the right. Since

the crate has no vertical acceleration ay =( )0 , we see that

ΣF ma n mgy y= ⇒ − = 0 or n mg=

The only horizontal force present is the friction force exerted on the crate by the truck bed. Thus,

ΣF ma f max x= ⇒ =

If the crate is not to slip (i.e., the static case is to prevail), it is necessary that the required friction force not exceed the maximum possible static friction force, f ns s( ) =

maxµ . From

this, we fi nd the maximum allowable acceleration as

af

m

f

m

n

m

mg

mgs s s

smaxmax max .= =

( )= =

( ) = = (µ µµ 0 350))( ) =9 80 3 43. .m s m s2 2

(b) Once slipping has started, the kinetic friction case prevails and f f nk k= = µ . The acceleration of the crate in this case will be

af

m

n

m

mg

mgk k

k= = =( ) = = ( )( ) =µ µ

µ 0 320 9 80 3 1. . .m s2 44 m s2

4.45 The acceleration of the system is found from

∆y t aty= +v021

2 or 1 00 0

1

21 20 2. .m s= + ( )a

which gives a = 1 39. m s2.

Using the free body diagram of m2, the second law gives

5 00 9 80 5 00 1. . .kg m s kg .39 m s2 2( )( ) − = ( )( )T or T = 42 1. N

Then applying the second law to the horizontal motion of m1

42 1 10 0 1. .N kg .39 m s2− = ( )( )f or f = 28 2. N

Since n m g= =1 98 0. N, we have

µk

f

n= = =28.2 N

N98 00 287

..

m2g→m1g→T→f

→

T→n ��m1g→ →

a→

a→m1 m2

m2g→m1g→T→f

→

T→n ��m1g→ →

a→

a→m1 m2


166 Chapter 4

4.46 (a) Since the puck is on a horizontal surface, the normal force is vertical. With ay = 0, we see that

ΣF ma n mgy y= ⇒ − = 0 or n mg=

Once the puck leaves the stick, the only horizontal force is a friction force in the negative x-direction (to oppose the motion of the puck). The acceleration of the puck is

a

F

m

f

m

n

m

mg

mgx

x k k kk= = − = − =

− ( ) = −Σ µ µµ

(b) Then v vx x xa x202 2= + ( )∆ gives the horizontal displacement of the puck before coming to

rest as

∆xa g g

x x

x k k

=−

=−−( ) =

v v v v202

02

02

2

0

2 2µ µ

4.47 The crate does not accelerate perpendicular to the incline. Thus,

ΣF ma n F mg⊥ ⊥= = ⇒ = +0 cosθ

The net force tending to move the crate down the incline is ΣF mg fs|| sin= −θ , where fs is the force of static friction between the crate and the incline. If the crate is in equilibrium, then mg fssinθ − = 0, so f Fs g= sinθ .

But, we also know f n F mgs s s≤ = +( )µ µ θcos

Therefore, we may write mg F mgssin cosθ µ θ≤ +( ), or,

F mgs

≥ −⎛⎝⎜

⎞⎠⎟

= ( )( )sincos . .

siθµ

θ 3 00 9 80kg m s2 nn .

.cos . .

35 0

0 30035 0 32 1

° −⎛⎝

⎞⎠° = N

4.48 (a) Find the normal force n�

on the 25.0 kg box:

ΣF ny = + ( ) ° − =80.0 N Nsin .25 0 245 0

or n = 211 N.

Now fi nd the friction force, f, as

f nk= = ( ) =µ 0 300 211 63 4. .N N

From the second law, we have ΣF max = , or

80 0 25 0 63 4. cos . .N N 25.0 kg( ) ° − = ( )a

which yields a = 0 366. m s2 .

F→ n

→

fs→

mg→

q � 35.0°

F→ n

→

fs→

mg→

q � 35.0°

F�80.0 N

mg�245 N

f→

n→

25.0°x

y

F�80.0 N

mg�245 N

f→

n→

25.0°x

y




(b) When the box is on the incline,

ΣF ny = + ( ) ° − ( ) ° =80.0 N Nsin . cos .25 0 245 10 0 0

giving n = 207 N.

The friction force is f nk= = ( ) =µ 0 300 207 62 2. .N N.

The net force parallel to the incline is then

ΣFx = ( ) ° − ( ) ° −80 0 25 0 245 10 0 62 2. cos . sin . .N N N == −32.3 N

Thus,

aF

mx= = − = −Σ 32 3

1 29.

.N

25.0 kgm s2

or 1 29. m s down the incline2 .

4.49 (a) The object will fall so that ma mg b a mg b m= − = −v v, ( )or where the downward direction is taken as positive. Equilibrium a =( )0 is reached when

v v= = =( )( )

=terminal

2kg m s

15 kg s

mg

b

50 9 8033

.mm s

(b) If the initial velocity is less than 33 m/s, then a ≥ 0 and 33 m/s is the largest velocity attained by the object. On the other hand, if the initial velocity is greater than 33 m/s, then a ≤ 0 and 33 m/s is the smallest velocity attained by the object. Note also that if the initial velocity is 33 m/s, then a = 0 and the object continues falling with a constant speed of 33 m/s.

4.50 (a) The force of friction is found as f n mgk k= = ( )µ µ .

Choose the positive direction of the x-axis in the direction of motion and apply Newton’s second law. We have

ΣF f max x= − = or af

mgx k= − = −µ

From v v202 2= + ( )a x∆ , with v v= = =0 50 0 13 90, . .km h m s, we fi nd

0 13 9 22= ( ) + −( )( ). m s µk g x∆ or ∆x

gk

=( )13 9

2

2. m s

µ [1]

With µk = 0 100. , this gives

∆x =( )

( )( ) =13 9

2 0 100 9 80

2.

. .

m s

m s98.6 m

2

(b) With µk = 0 600. , Equation (1) above gives

∆x =( )

( )( ) =13 9

2 0 600 9 806 4

2.

. ..

m s

m s1 m

2

F �80.0 N

x

y

245

N

n→

f→

25.0°

10.0°

F �80.0 N

x

y

245

N

n→

f→

25.0°

10.0°

f��bv→

→

mg→

v→m

f��bv→

→

mg→

v→m


168 Chapter 4

4.51 (a) ∆x t a t a tx x= + = +v012

2 12

20 gives

ax

tx =( ) =

( )( )

=2 2 2 00

1 501 782 2

∆ .

..

m

sm s2

(b) Considering forces parallel to the incline, Newton’s second law yields

ΣF fx k= ( ) ° − = ( )( )29 4 30 0 3 00 1. sin . .N kg .78 m s2

or fk = 9 37. N.

Perpendicular to the plane, we have equilibrium, so

ΣF ny = − ( ) ° =29 4 30 0 0. cos .N or n = 25 5. N

Then,

= =9.37 N

25.5 Nµk

f

nk = 0 368.

(c) From part (b) above, fk = 9 37. N .

(d) Finally, v v202 2= + ( )a xx ∆ gives

v v= + ( ) = + ( )( ) =02 2 0 2 1 78 2 00 2 67a xx ∆ . . .m s m m s2

4.52 When the minimum force �F is used, the block tends to slide down

the incline so the friction force, �f s is directed up the incline.

While the block is in equilibrium, we have

ΣF F fx s= ° + − ( ) ° =cos . sin .60 0 1 60 0 09.6 N [1]

and

ΣF n Fy = − ° − ( ) ° =sin . cos .60 0 1 60 0 09.6 N [2]

For minimum F (impending motion), f f n ns s s= ( ) = = ( )max

µ 0 300. [3]

Equation [2] gives n F= +0 866 9 80. . N [4]

(a) Equation [3] becomes f F Fs = +( ) = +0 300 0 866 9 80 0 260 2 94. . . . .N N, so

Equation [1] gives 0 500 0 260 2 94 17 0 0. . . .F F+ + − =N N , or F = 18.5 N .

(b) Finally, Equation [4] gives the normal force, n = ( ) + =0 866 18 5 9 80 5 8. . . .N N 2 N .

w�mg�29.4 N

f k�m knn

→

30.0°

30.0°

�x

�y

w�mg�29.4 N

f k�m knn

→

30.0°

30.0°

�x

�y

mg�19.6 N

n→

fs→

F→

60.0°

60.0°�y

�x

m

mg�19.6 N

n→

fs→

F→

60.0°

60.0°�y

�x

m



4.53 First, taking downward as positive, apply Newton’s second law to the 12.0-kg block:

ΣF g T ay = ( ) − = ( )12 0 12 0. .kg kg

or

T a= ( ) −( )12 0 9 80. .kg m s2 [1]

For the 7.00 kg block, we have

ΣF n⊥ = ⇒ = ( ) ° =0 68 6 37 0 54 8N N. cos . .

and

f nk= = ( )( ) =µ 0 250 54 8 13 7. . .N N

Taking up the incline as the positive direction and applying Newton’s second law to the 7.00-kg block gives ΣF T f ax = − − ( ) ° = ( )68 6 37 0 7. sin .N .00 kg , or

7 13 7 41 3.00 kg N N( ) = − −a T . . [2]

Substituting Equation [1] into [2] yields

7 12 0 62 7.00 kg kg N+( ) =. .a or a = 3 30. m s2

4.54 (a) Both objects start from rest and have accelerations of the same magnitude, a. This magnitude can be determined by applying ∆y t a ty y= +v0

12

2 to the motion of m1:

ay

t=

( ) = ( )( )

=2 2 1 00

4 000 1252 2

∆ .

..

m

sm s2

(b) Consider the free-body diagram of m1 and apply Newton’s second law:

ΣF ma T m g m ay y= ⇒ − = +( )1 1

or

T m g a= +( ) = ( ) +( ) =1 4 00 9 80 0 125 39. . . .kg m s m s2 2 77 N

(c) Considering the free-body diagram of m2:

ΣF ma n m gy y= ⇒ − =2 0cosθ or n m g= 2 cosθ ,

so n kg m s N2= ( )( ) ° =9 00 9 80 40 0 67 6. . cos . . .

ΣF ma m g T f m ax x k= ⇒ − − = +( )2 2sinθ . Then, f m g a Tk = −( ) −2 sinθ , or

fk = ( ) ( ) −⎡⎣ ⎤9 00 9 80 0 125. . sin .kg m s 40.0° m s2 2

⎦⎦ − =39 7 15 9. .N N

The coeffi cient of kinetic friction is

µkkf

n= = =15 9

0 235.

.N

67.6 N

w�118 Nw�68.6 N

T→

T→

f→

n→

37.0°

12.0 kg 7.00 kg�y

�x

w�118 Nw�68.6 N

T→

T→

f→

n→

37.0°

12.0 kg 7.00 kg�y

�x

n→

fk→

T→

T→

m1

→m2g

→m1g

m2

q

�y

�x

n→

fk→

T→

T→

m1

→m2g

→m1g

m2

q

�y

�x


170 Chapter 4

4.55

22.0° 22.0°

w�170 lb

Free-Body Diagram of Person Free-Body Diagram of Crutch Tip

ntip→

f→

F2→

F1→

22.0°

�y �y

�x�x

F�

45.8

lb

nground� w/2�85.0 lb

From the free-body diagram of the person, ΣF F Fx = ( )° − ( )° =1 222 0 22 0 0sin . sin . , which gives or F F F1 2= = .

Then, ΣF Fy = ° + − =2 22 0 85 0 170 0cos . . lbs lbs yields F = 45.8 lb.

(a) Now consider the free-body diagram of a crutch tip.

ΣF fx = − ( ) ° =45 8 22 0 0. sin .lb

or

f = 1 lb7 2.

ΣF ny = − ( ) ° =tip lb45 8 22 0 0. cos .

which gives ntip lb= 42 5. .

For minimum coeffi cient of friction, the crutch tip will be on the verge of slipping,

so f f ns s= ( ) =max tipµ

and µs

f

n= = =

tip

lb

42.5 lb

17 20 404

..

(b) As found above, the compression force in each crutch is

F F F1 2= = = 45.8 lb

4.56 The acceleration of the ball is found from

ay

=−

( ) =( ) −

( ) =v v2

02

2

20 0 0

2 1 50133

∆.

.

m s

mm s

22

From the second law, ΣF F w may y= − = , so

F w may= + = + ( )( ) =1 47 0 150 1 21 5. . .N kg 33 m s N2

w�1.47 N

m�0.150 kg

F→

w�1.47 N

m�0.150 kg

F→



4.57 (a) T1→

T2→

T1→

T2→

a4→

a2→

n→

f→

a1→

1.00 kg4.00 kg

m4m1

2.00 kg

m2

→m4g

→m1g →m2g

(b) Note that the suspended block on the left, m4, is heavier than that on the right, m2. Thus, if the system overcomes friction and moves, the center block will move right to left with each block’s acceleration being in the directions shown above .

First, consider the center block, m1, which has no vertical acceleration. Then,

ΣF n m gy = − =1 0 or n m g= = ( )( ) =1 1 00 9 80 9 80. . .kg m s N2

This means the friction force is

f nk= = ( )( ) =µ 0 350 9 80 3 43. . .N N

Assuming the cords do not stretch, the speeds of the three blocks must always be equal. Thus, the magnitudes of the blocks’ accelerations must have a common value, a.

a a a4 1 2

�� = = = a

Taking the indicated direction of the acceleration as the positive direction of motion for each block, we apply Newton’s second law to each block as follows

For m4: m g T m a4 1 4− = or T m g a g a1 4 4 00= −( ) = ( ) −( ). kg [1]

For m1: T T f m a1 2 1− − = or T T a1 2 1 00 3 43− = ( ) +. .kg N [2]

For m2: T m g m a2 2 2− = or T m g a g a2 2 2 00= +( ) = ( ) +( ). kg [3]

Substituting Equations [1] and [3] into Equation [2], and solving for a yields the following:

4 00 2 00 1 00 3. . . .kg kg kg( ) −( ) − ( ) +( ) = ( ) +g a g a a 443

4 00 2 00 9 80 3 43

4

N

kg kg m s N2

a =−( )( ) −. . . .

.000 2 00 1 002 31

kg kg kgm s2

+ +=

. ..



172 Chapter 4

(c) Using this result in Equations [1] and [3] gives the tensions in the two cords as

T g a1 4 00 4 00 9 80 2 31= ( ) −( ) = ( ) −( ) =. . . .kg kg m s2 330 0. N

and

T g a2 2 00 2 00 9 80 2 31= ( ) +( ) = ( ) +( ) =. . . .kg kg m s2 224 2. N

(d) From the fi nal calculation in part (b), observe that if the friction force had a value of zero (rather than 3.53 N), the acceleration of the system would increase in magnitude. Then, observe from Equations [1] and [3] that this would mean T1 would decrease while T2 would increase .

4.58 The sketch at the right gives an edge view of the sail (heavy line) as seen from above. The velocity of the wind, v

�wind, is directed to

the east and the force the wind exerts on the sail is perpendicular to the sail. The magnitude of this force is

Fsail wind

N

m s=

⎛⎝⎜

⎞⎠⎟ ⊥

550 v�

where v�

wind ⊥ is the component of the wind velocity perpendicular

to the sail.

When the sail is oriented at 30° from the north–south line and the wind speed is vwind knots= 17 , we have

Fsail wind

N

m s

N

m s=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟⊥

550 550 17v�

kknotsm s

knot( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥° =0 514

130 4

.cos .22 103× N

The eastward component of this force will be counterbalanced by the force of the water on the keel of the boat. Before the sailboat has signifi cant speed (that is, before the drag force develops), its acceleration is provided by the northward component of F

��sail. Thus, the initial acceleration is

am

= =×( ) °

=F��

sailnorth

N

800 kg

4 2 10 302 6

3. sin. m s2

4.59 (a) The horizontal component of the resultant force exerted on the light by the cables is

R Fx x= = ( ) ° − ( ) ° =Σ 60 0 45 0 60 0 45 0 0. cos . . cos .N N

The resultant y component is

R Fy y= = ( ) ° + ( ) ° =Σ 60 0 45 0 60 0 45 0 84. sin . . sin .N N ..9 N

Hence, the resultant force is 84.9 N vertically upward .

(b) The forces on the traffi c light are the weight, directed downward, and the 84.9-N vertically upward force exerted by the cables. Since the light is in equilibrium, the resultant of these forces must be zero. Thus, w = 84.9 N .

north

east

vwind→

Fsail→

30° 30°

north

east

vwind→

Fsail→

30° 30°

60.0 N60

.0 N

�y

�x

45.0° 45.0°

60.0 N60

.0 N

�y

�x

45.0° 45.0°



4.60 (a) For the suspended block, ΣF Ty = − =50 0. N 0, so the tension in the rope is T = 50 0. N. Then, considering the horizontal forces on the 100-N block, we fi nd

ΣF T fx s= − = 0 or f Ts = = 50 0. N

(b) If the system is on the verge of slipping, f f ns s s= ( ) =max

µ . Therefore, the minimum acceptable coeffi cient of friction is

µssf

n= = =50 0

0 500.

.N

100 N

(c) If µk = 0 250. , then the friction force acting on the 100-N block is

f nk k= = ( )( ) =µ 0 250 100 25 0. .N N

Since the system is to move with constant velocity, the net horizontal force on the 100-N block must be zero, or ΣF T f Tx k= − = − =25 0 0. N . The required tension in the rope is T = 25 0. N. Now, considering the forces acting on the suspended block when it moves with constant velocity, ΣF T wy = − = 0, giving the required weight of this block as

w T= = 25 0. N .

4.61 On the level surface, the normal force exerted on the sled by the ice equals the total weight, or n = 600 N. Thus, the friction force is

f nk= = ( )( ) =µ 0 050 600 30. N N

Hence, Newton’s second law yields ΣF f max x= − = , or

af

m

f

w gx = − = − =− ( )( )

= −30 9 80

0 49N m s

600 Nm s

2.. 22

The distance the sled travels on the level surface before coming to rest is

∆xa

x x

x

=−

=− ( )−( ) =

v v202 2

2

0 7 0

2 0 4950

.

.

m s

m sm

2

4.62 Consider the vertical forces acting on the block:

ΣF n may y= ( ) ° − − = =85 0 55 0 39 2 0. sin . .N N

so the normal force is n = − =69 6 39 2 30 4. . .N N N.

Now, consider the horizontal forces:

ΣF f max k x= ( ) ° − = = ( )85 0 55 0 4 00 6 00. cos . . .N kg m ss2( ) or

fk = ( ) ° − =85 0 55 0 24 0 24 8. cos . . .N N N

The coeffi cient of kinetic friction is then µkkf

n= = =24 8

0 814.

.N

30.4 N.

55.0°

85.0 N

mg�39.2 N

fk→ n

→

4.00 kg

55.0°

85.0 N

mg�39.2 N

fk→ n

→

4.00 kg


174 Chapter 4

4.63 (a) The force that accelerates the box is the friction force between the box and the truck bed.

(b) The maximum acceleration the truck can have before the box slides is found by considering the maximum static friction force the truck bed can exert on the box:

f n mgs s s( ) = = ( )max

µ µ

Thus, from Newton’s second law,

af

m

mg

mgs s

smaxmax 2m s=

( )=

( ) = = ( )( )µµ 0 300 9 80. . == 2 94. m s2

4.64 Let m m m1 2 3= = =5.00 kg, 4.00 kg, and 3.00 kg. Let T1 be the tension in the string between m m1 2and , and T2 the tension in the string between m m2 3and .

(a) We may apply Newton’s second law to each of the masses.

for m1: m a T m g1 1 1= − [1]

for m2 : m a T m g T2 2 2 1= + − [2]

for m3 : m a m g T3 3 2= − [3]

Adding these equations yields m m m a m m m g1 2 3 1 2 3+ +( ) = − + +( ) , so

am m m

m m mg=

− + ++ +

⎛⎝⎜

⎞⎠⎟

=⎛⎝

1 2 3

1 2 3

2 00. kg

12.0 kg⎜⎜⎞⎠⎟

( ) =9 80 1 63. .m s m s2 2

(b) From Equation [1], T m a g1 1 5 00 11 4 57 2= +( ) = ( )( ) =. . .kg m s N2 , and

from Equation [3], T m g a2 3 3 00 8 17 24 5= −( ) = ( )( ) =. . .kg m s N2

4.65 When an object of mass m is on this frictionless incline, the only force acting parallel to the incline is the parallel component of weight, mg sinθ directed down the incline. The acceleration is then

aF

m

mg

mg= = = = ( ) ° =� sin

sin . sin . .θ θ 9 80 35 0 5 62m s2 m s2 (directed down the incline)

(a) Taking up the incline as positive, the time for the sled projected up the incline to come to rest is given by

t

a=

−= −

−=

v v0 0 5 00

5 620 890

.

..

m s

m ss2

The distance the sled travels up the incline in this time is

∆s t t= =+⎛

⎝⎞⎠ = +⎛

⎝⎞⎠ ( ) =v

v vav

m ss0

2

0 5 00

20 890

.. 22 22. m

(b) The time required for the fi rst sled to return to the bottom of the incline is the same as the time needed to go up, that is, t = 0 890. s. In this time, the second sled must travel down the entire 10.0 m length of the incline. The needed initial velocity is found from

∆s t at= +v012

2 as

v0 2

10 0 5 62 0 890= − = − −

−( )∆s

t

at . . .m

0.890 s

m s2 ssm s

( )= −

28 74.

or 8 74. m s down the incline .



4.66 Before she enters the water, the diver is in free-fall, with an acceleration of 9 80. m s2 downward. Taking downward as the positive direction, her velocity when she reaches the water is given by

v v= + ( ) = + ( )( ) =02 2 0 2 9 80 10 0 14 0a y∆ . . .m s m m s2

This is also her initial velocity for the 2.00 s after hitting the water. Her average acceleration during this 2.00 s interval is

atav

2m s

sm s=

−=

− ( ) = −v v0 0 14 0

2 007 00

.

..

Continuing to take downward as the positive direction, the average upward force by the water is found as ΣF F mg may = + =av av , or

kg m s m sav av2 2F m a g= −( ) = ( ) −( ) −70 0 7 00 9 80. . .⎡⎡⎣ ⎤⎦ = − ×1 18 103. N

or

N upwardaF v = ×1 18 103.

4.67 (a) Free-body diagrams for the two blocks are given at the right. The coeffi cient of kinetic friction for aluminum on steel is µ1 0 47= . while that for copper on steel is µ2 0 36= . . Since ay = 0 for each block,

n w1 1= and n w2 2 30 0= °cos . .

Thus,

f n1 1 1 0 47 19 6 9 21= = ( ) =µ . . .N N

and

f n2 2 2 0 36 58 8 30 0 18 3= = ( ) ° =µ . . cos . .N N

For the aluminum block:

ΣF ma T f m a T f max x= ⇒ − = +( ) = +or1 1

giving

T a= + ( )9 21 2 00. .N kg [1]

For the copper block:

ΣF ma Tx x= ⇒ ( ) ° − − =58.8 N Nsin . . .30 0 18 3 6 00 kkg( )a

or

11 1 6 00. .N kg− = ( )T a [2]

Substituting Equation [1] into Equation [2] gives

11 1 9 21 2 00 6 00. . . .N N kg kg− − ( ) = ( )a a

or

a = =1 890 236

..

N

8.00 kgm s2

(b) From Equation [1] above, T = + ( )( ) =9 21 2 00 0 236 9 68. . . . N kg m s N2 .

30°

Aluminum2.00 kg

Copper6.00 kg

w1�19.6 N

w2�58.8 N

T→

T→

a→

a→

n1→

n2→

f1→

f2→

�x

30°

Aluminum2.00 kg

Copper6.00 kg

w1�19.6 N

w2�58.8 N

T→

T→

a→

a→

n1→

n2→

f1→

f2→

�x


176 Chapter 4

4.68 In the vertical direction, we have

ΣF T mgy = ° − =cos .4 0 0, or Tmg=

°cos .4 0

In the horizontal direction, the second law becomes:

ΣF T max = ° =sin .4 0

so

aT

mg= ° = ° =sin .

tan . .4 0

4 0 0 69 m s2

4.69 Figure 1 is a free-body diagram for the system consisting of both blocks. The friction forces are

f n m gk k1 1 1= = ( )µ µ and f m gk2 2= ( )µ

For this system, the tension in the connecting rope is an internal force and is not included in second law calculations. The second law gives

ΣF f f m m ax = − − = +( )50 1 2 1 2N

which reduces to

am m

gk=+

−50

1 2

N µ [1]

Figure 2 gives a free-body diagram of m1 alone. For this system, the tension is an external force and must be included in the second law. We fi nd: ΣF T f m ax = − =1 1 , or

T m a gk= +( )1 µ [2]

(a) If the surface is frictionless, µk = 0. Then, Equation [1] gives

am m

=+

− = =500

50

301 7

1 2

N N

kgm s2.

and Equation [2] yields T = ( ) +( ) =10 1 7 0 17kg m s N2. .

(b) If µk = 0 10. , Equation [1] gives the acceleration as

a = − ( )( ) =50

300 10 9 80 0 69

N

kgm s m s2 2. . .

while Equation [2] gives the tension as

T = ( ) + ( )( )⎡⎣ ⎤⎦ =10 0 0 10 9 80 17kg .69 m s m s N2 2. .

4.0°�y

�x

T→

mg→

4.0°�y

�x

T→

mg→

50 N

a→

n1→

f1→

f2→

m1g→

m2g→

n2→

m1 m2

50 N

a→

n1→

f1→

f2→

m1g→

m2g→

n2→

m1 m2

Figure 1

Figure 2



4.70 (a)

(b) No. In general, the static friction force is less than the maximum value of f ns s( ) =max

µ . It is equal to this maximum value only when the coin is on the verge of slipping, or at the

critical angle θc. For θ θ µ≤ ≤ ( ) =c s sf f n,max

.

(c) Recognize that when the y axis is chosen perpendicular to the incline as shown above, ay = 0 and we fi nd

ΣF n mg may y= − = =cosθ 0 or n mg= cosθ

Also, when static conditions still prevail, but the coin is on the verge of slipping, we have a f f n mgx c s s s c= = = ( ) = =0, , cos

maxandθ θ µ µ θ . Then, Newton’s second law becomes

ΣF mg mg max c s c x= − = =sin cosθ µ θ 0

and

µ θ θs c cmg mgcos sin= yielding µ θθ

θsc

cc= =sin

costan

(d) Once the coin starts to slide, kinetic conditions prevail and the friction force is

f f n mgk k k= = =µ µ θcos

At θ θ θ= ′ <c c , the coin slides with constant velocity, and ax = 0 again. Under these conditions, Newton’s second law gives

ΣF mg mg max c k c x= ′ − ′ = =sin cosθ µ θ 0

and

µ θ θk c cmg mgcos sin′ = ′

yielding

µ θθ

θkc

cc= ′

′= ′sin

costan


178 Chapter 4

4.71 (a) When the pole exerts a force downward and toward the rear on the lakebed, the lakebed exerts an oppositely directed force of equal magnitude, F, on the end of the pole.

As the boat fl oats on the surface of the lake, its vertical acceleration is ay = 0 . Thus, Newton’s second law gives the magnitude of the buoyant force, FB , as

ΣF F F mgy B= + − =cosθ 0

and, with θ = °35 0. ,

F mg FB = − = ( )( ) − ( )cos . cosθ 370 9 80 240 35kg m s N2 ..0°

or

FB = × =3 43 10 3 433. .N kN

(b) Applying Newton’s second law to the horizontal motion of the boat gives

ΣF F max x= − =sinθ F��

drag or ax = ( ) ° − =240 35 0 47 50 244

N N

370 kgm s2sin . .

.

After an elapsed time t = 0 450. s, v vx x xa t= +0 gives the velocity of the boat as

vx = + ( )( ) =0 857 0 244 0 450 0 967. . . .m s m s s m s2

(c) If angle q increased while the magnitude of F��

remained constant, the vertical component of this force would decrease. The buoyant force would have to increase to support more of the weight of the boat and its contents. At the same time, the horizontal component of F

��

would increase, which would increase the acceleration of the boat.

4.72 (a) n→

→

mr

mbg

mbT→

T→

F→

(b) Applying Newton’s second law to the rope yields

ΣF ma F T m a T F mx x r r= ⇒ − = = −or aa [1]

Then, applying Newton’s second law to the block, we fi nd

ΣF ma T m a F m a m ax x b r b= ⇒ = − =or

which gives

aF

m mb r

=+




(c) Substituting the acceleration found above back into Equation [1] gives the tension at the left end of the rope as

T F m a F mF

m mF

m m m

m mr rb r

b r r

b r

= − = −+

⎛⎝⎜

⎞⎠⎟

=+ −

+⎛

⎝⎜⎞

⎠⎠⎟

or

Tm

m mFb

b r

=+

⎛⎝⎜

⎞⎠⎟

(d) From the result of (c) above, we see that as mr approaches zero, T approaches F. Thus,

the tension in a cord of negligible mass is constant along its length .

4.73 Choose the positive x-axis to be down the incline and the y-axis perpendicular to this as shown in the free-body diagram of the toy. The acceleration of the toy then has components of

a aty xx= = = + = +0

30 0

6 005 00and

m s

sm s

∆∆v .

.. 22

Applying the second law to the toy gives the following:

(a) ΣF mg ma max x x= = ⇒ =sin sinθ θ mmg a gx= ,

and

θ =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠

− −sin sin.1 1 5 00a

gx m s

9.80 m s

2

2 ⎟⎟ = °30 7.

(b) ΣF T mg may y= − = =cosθ 0, or

T mg= = ( )( ) ° =cos . . cos . .θ 0 100 9 80 30 7 0 843kg m s2 N

4.74 The sketch at the right gives the free-body diagram of the person. The scale simply reads the magnitude of the normal force exerted on the student by the seat. From Newton’s second law, we obtain

ΣF ma n mgy y= ⇒ − ° ==0 cos .30 0 0

or

n mg= = ( ) ° =cos cos .θ 200 30 0 173lb lb

q q

a→

mg→

T→

�y

m

�x q q

a→

mg→

T→

�y

m

�x


180 Chapter 4

4.75 The acceleration the car has as it is coming to a stop is

ax

=−

( ) =− ( )( ) = −

v v202 2

2

0 35

2 1 0000 61

∆m s

mm s2.

Thus, the magnitude of the total retarding force acting on the car is

F m aw

ga= =

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

8 8200 61

N

9.80 m sm s2

2.(( ) = ×5 5 102. N

4.76 (a) In the vertical direction, we have

ΣF w may y= ( ) ° − = =8 000 65 0 0N sin .

so w = ( ) ° = ×8 000 65 0 7 25 103N Nsin . . .

(b) Along the horizontal, Newton’s second law yields

ΣF maw

gax x x= ( ) ° = =

⎛⎝⎜

⎞⎠⎟

8 000 65 0N cos .

or

ag

wx =( )⎡⎣ ⎤⎦° =

( )(8 000 65 0 9 80 8 000N m s N2cos . . )) °×

=cos .

..

65 0

7 25 104 573 N

m s2

4.77 Since the board is in equilibrium, ΣFx = 0, and we see that the normal forces must have the same magnitudes on both sides of the board. Also, if the minimum normal forces (compression forces) are being applied, the board is on the verge of slipping and the friction force on each side is f f ns s= ( ) =

maxµ .

The board is also in equilibrium in the vertical direction, so

ΣF f w fw

y = − = =2 02

, or

The minimum compression force needed is then

nf w

s s

= = =( ) =

µ µ2

95 5

272 0

..

N

0.663N

65.0°

8 000 N

w�mg→ →

65.0°

8 000 N

w�mg→ →

w � 95.5 N

n→

n→

f→

f→

w � 95.5 N

n→

n→

f→

f→



4.78 Consider the two free-body diagrams, one of the penguin alone and one of the combined system consisting of penguin plus sled.

The normal force exerted on the penguin by the sled is

n w m g1 1 1= =

and the normal force exerted on the combined system by the ground is

n w m g2 130= = =total total N

The penguin is accelerated forward by the static friction force exerted on it by the sled. When the penguin is on the verge of slipping, this acceleration is

af

m

n

m

m g

mgs s

smaxmax=

( )= =

( )= = ( )1

1

1

1

1

1

0 700µ µ

µ . 99 80 6 86. .m s m s2 2( ) =

Since the penguin does not slip on the sled, the combined system must have the same acceleration as the penguin. Applying Newton’s second law to thiss system gives

ΣF F f m ax = − =2 total max or F f m a ww

gak= + = ( ) +

⎛⎝⎜

⎞⎠⎟2 total max total

totalmaµ xx

which yields

F = ( )( ) +⎛⎝⎜

⎞⎠⎟

0 100 130130

9 806.

.N

N

m s.86 m s2

22 N( ) = 104

w1�70.0 N wtotal�130 N

f1→

f2→

F→

n1→ n2

→

w1�70.0 N wtotal�130 N

f1→

f2→

F→

n1→ n2

→


182 Chapter 4

4.79 First, we will compute the needed accelerations:

(1) Before it starts to move:

ay = 0

(2) During the fi rst 0.80 s:

aty

y y=−

= − =v v0 1 2 0

0 801 5

.

..

m s

sm s2

(3) While moving at constant velocity:

ay = 0

(4) During the last 1.5 s:

aty

y y=−

= − = −v v0 0 1 2

10 80

..

m s

.5 sm s2

The spring scale reads the normal force the scale exerts on the man. Applying Newton’s second law to the vertical motion of the man gives:

ΣF n mg may y= − = or n m g ay= +( ) (a) When ay = 0,

n = ( ) +( ) = ×72 9 80 0 7 1 102kg m s N2. .

(b) When ay = 1 5. m s2,

n = ( ) +( ) = ×72 9 80 1 5 8 1 102kg m s m s N2 2. . .

(c) When ay = 0,

n = ( ) +( ) = ×72 9 80 0 7 1 102kg m s N2. .

(d) When ay = −0 80. m s2,

n = ( ) −( ) = ×72 9 80 0 80 6 5 102kg m s m s N2 2. . .



4.80 The friction force exerted on the mug by the moving tablecloth is the only horizontal force the mug experiences during this process. Thus, the horizontal acceleration of the mug will be

af

mk

mugmug

2N

kgm s= = =0 100

0 2000 500

.

..

The cloth and the mug both start from rest (v x0 0= ) at time t = 0. Then, at time t > 0, the horizontal displacements of the mug and cloth are given by ∆x t a tx x= +v0

12

2 as

∆x t tmug2 2m s m s= + ( ) = ( )0 0 500 0 2501

22 2. .

and

∆x t tcloth2 2m s m s= + ( ) = ( )0 3 00 1 501

22 2. .

In order for the edge of the cloth to slip under the mug, it is necessary that ∆ ∆x xcloth mug m= + 0 300. , or

1 50 0 250 0 3002 2. . .m s m s m2 2( ) = ( ) +t t

The elapsed time when this occurs is

t =−( ) =0 300

1 50 0 2500 490

.

. ..

m

m ss2

At this time, the mug has moved a distance of

∆xmug2m s s m= ( )( ) = × =−0 250 0 490 6 00 10 6 02 2. . . . 00 cm

4.81 (a) Consider the fi rst free-body diagram in which Chris and the chair are treated as a combined system. The weight of this system is wtotal N= 480 , and its mass is

m w gtotal total kg= = 49 0.

Taking upward as positive, the acceleration of this system is found from Newton’s second law as

ΣF T w m ay y= − =2 total total

Thus,

ay = ( ) − = +2 250 4800 408

N N

49.0 kgm s2.

or

0 408. m s upward2

wtotal�320 N � 160 N wChris�320 N

T �250 N

T �250 N T �250 N

a→

n→

wtotal�320 N � 160 N wChris�320 N

T �250 N

T �250 N T �250 N

a→

n→



184 Chapter 4

(b) The downward force that Chris exerts on the chair has the same magnitude as the upward normal force exerted on Chris by the chair. This is found from the free-body diagram of Chris alone as

ΣF T n w m ay y= + − =Chris Chris

so

n m a w Ty= + −Chris Chris

Hence,

n =⎛⎝⎜

⎞⎠⎟

( ) + −3200 408 320 250

N

9.80 m sm s N2

2. NN N= 83 3.

4.82 Let R��

represent the horizontal force of air resistance. Since the helicopter and bucket move at constant velocity, a ax y= = 0. The second law then gives

ΣF T mgy = ° − =cos .40 0 0 or Tmg=

°cos .40 0

Also,

ΣF T Rx = ° − =sin .40 0 0 or R T= °sin .40 0

Thus,

Rmg=

°⎛⎝

⎞⎠ ° = ( )( )

cos .sin . .

40 040 0 620 9 80kg m s2 ttan . .40 0 5 10 103° = × N

40.0°

R

T→

w�m→g→

�y

�x


solucionario fundamentos de física 9na edición capitulo 4

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