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Contents

Section 5.1 - Newton's third law

Section 5.2 - Classification and behavior of forces

Section 5.3 - Analysis of forces

Section 5.4 - Transmission of forces by low-mass objects

Section 5.5 - Objects under strain

Section 5.6 - Simple Machines: the pulley

Section 5.7 - Summary

Section 5.1 - Newton's third law

Section 5.2 - Classification and behavior of forces

Section 5.3 - Analysis of forces

Section 5.4 - Transmission of forces by low-mass objects

Section 5.5 - Objects under strain

Section 5.6 - Simple Machines: the pulley

Section 5.7 - Summary

\begin{margtopic}{Optional topic: Newton's third law and action at a distance}
Newton's third law is completely symmetric in the sense that
neither force constitutes a delayed response to the other.
Newton's third law does not even mention time, and the
forces are supposed to agree at any given instant. This
creates an interesting situation when it comes to noncontact
forces. Suppose two people are holding magnets, and when one
person waves or wiggles her magnet, the other person feels
an effect on his. In this way they can send signals to each
other from opposite sides of a wall, and if Newton's third
law is correct, it would seem that the signals are
transmitted instantly, with no time lag. The signals are
indeed transmitted quite quickly, but experiments with
electrically controlled magnets show that the signals do
not leap the gap instantly: they travel at the same speed as
light, which is an extremely high speed but not an infinite one.
Is this a contradiction to Newton's third law? Not really.
According to current theories, there are no true noncontact
forces. Action at a distance does not exist. Although it
appears that the wiggling of one magnet affects the other
with no need for anything to be in contact with anything,
what really happens is that wiggling a magnet creates a ripple
in the magnetic field pattern that exists even in empty space. The magnet shoves
the ripples out with a kick and receives a kick in return,
in strict obedience to Newton's third law. The ripples spread
out in all directions, and the ones that hit the other
magnet then interact with it, again obeying Newton's third law.
\end{margtopic}

Newton created the modern concept of force starting from his insight that all the effects that govern motion are interactions between two objects: unlike the Aristotelian theory, Newtonian physics has no phenomena in which an object changes its own motion.

Is one object always the “order-giver” and the other the “order-follower”? As an example, consider a batter hitting a baseball. The bat definitely exerts a large force on the ball, because the ball accelerates drastically. But if you have ever hit a baseball, you also know that the ball makes a force on the bat --- often with painful results if your technique is as bad as mine!

How does the ball's force on the bat compare with the bat's force on the ball? The bat's acceleration is not as spectacular as the ball's, but maybe we shouldn't expect it to be, since the bat's mass is much greater. In fact, careful measurements of both objects' masses and accelerations would show that \(m_{ball}a_{ball}\) is very nearly equal to \(-m_{bat}a_{bat}\), which suggests that the ball's force on the bat is of the same magnitude as the bat's force on the ball, but in the opposite direction.

Figures a and b show two somewhat more practical laboratory experiments for investigating this issue accurately and without too much interference from extraneous forces.

In experiment a, a large magnet and a small magnet are weighed separately, and then one magnet is hung from the pan of the top balance so that it is directly above the other magnet. There is an attraction between the two magnets, causing the reading on the top scale to increase and the reading on the bottom scale to decrease. The large magnet is more “powerful” in the sense that it can pick up a heavier paperclip from the same distance, so many people have a strong expectation that one scale's reading will change by a far different amount than the other. Instead, we find that the two changes are equal in magnitude but opposite in direction: the force of the bottom magnet pulling down on the top one has the same strength as the force of the top one pulling up on the bottom one.

In experiment b, two people pull on two spring scales. Regardless of who tries to pull harder, the two forces as measured on the spring scales are equal. Interposing the two spring scales is necessary in order to measure the forces, but the outcome is not some artificial result of the scales' interactions with each other. If one person slaps another hard on the hand, the slapper's hand hurts just as much as the slappee's, and it doesn't matter if the recipient of the slap tries to be inactive. (Punching someone in the mouth causes just as much force on the fist as on the lips. It's just that the lips are more delicate. The forces are equal, but not the levels of pain and injury.)

Newton, after observing a series of results such as these, decided that there must be a fundamental law of nature at work:

Forces occur in equal and opposite pairs: whenever object A exerts a force on object B, object B must also be exerting a force on object A. The two forces are equal in magnitude and opposite in direction.

Two modern, high-precision tests of the third law are described on p. 779.

In one-dimensional situations, we can use plus and minus signs to indicate the directions of forces, and Newton's third law can be written succinctly as \(F_{\text{A on B}}=-F_{\text{B on A}}\).

Figure d analyzes swimming using Newton's third law. Do a similar analysis for a sprinter leaving the starting line.

(answer in the back of the PDF version of the book)There is no cause and effect relationship between the two forces in Newton's third law. There is no “original” force, and neither one is a response to the other. The pair of forces is a relationship, like marriage, not a back-and-forth process like a tennis match. Newton came up with the third law as a generalization about all the types of forces with which he was familiar, such as frictional and gravitational forces. When later physicists discovered a new type of force, such as the force that holds atomic nuclei together, they had to check whether it obeyed Newton's third law. So far, no violation of the third law has ever been discovered, whereas the first and second laws were shown to have limitations by Einstein and the pioneers of atomic physics.

The English vocabulary for describing forces is unfortunately rooted in Aristotelianism, and often implies incorrectly that forces are one-way relationships. It is unfortunate that a half-truth such as “the table exerts an upward force on the book” is so easily expressed, while a more complete and correct description ends up sounding awkward or strange: “the table and the book interact via a force,” or “the table and book participate in a force.”

To students, it often sounds as though Newton's third law
implies nothing could ever change its motion, since the two
equal and opposite forces would always cancel. The two
forces, however, are always on two different objects, so it
doesn't make sense to add them in the first place --- we
only add forces that are acting on the same object. If two
objects are interacting via a force and no other forces are
involved, then *both* objects will accelerate --- in
opposite directions!

Mnemonics are tricks for memorizing things. For instance, the musical notes that lie between the lines on the treble clef spell the word FACE, which is easy to remember. Many people use the mnemonic “SOHCAHTOA” to remember the definitions of the sine, cosine, and tangent in trigonometry. I have my own modest offering, POFOSTITO, which I hope will make it into the mnemonics hall of fame. It's a way to avoid some of the most common problems with applying Newton's third law correctly:

\(\triangleright\) A book is lying on a table. What force is the Newton's-third-law partner of the earth's gravitational force on the book?

Answer: Newton's third law works like “B on A, A on \(B\),” so the partner must be the book's gravitational force pulling upward on the planet earth. Yes, there is such a force! No, it does not cause the earth to do anything noticeable.

Incorrect answer: The table's upward force on the book is the Newton's-third-law partner of the earth's gravitational force on the book.

This answer violates two out of three of the commandments of POFOSTITO. The forces are not of the same type, because the table's upward force on the book is not gravitational. Also, three objects are involved instead of two: the book, the table, and the planet earth.

\(\triangleright\) The box's force on the person.

Incorrect answer: The person's force on the box is opposed by friction, and also by gravity.

This answer fails all three parts of the POFOSTITO test, the most obvious of which is that three forces are referred to instead of a pair.

◊ Solved problem: More about example 2 — problem 20

◊ Solved problem: Why did it accelerate? — problem 18

◊

When you fire a gun, the exploding gases push outward in all directions, causing the bullet to accelerate down the barrel. What third-law pairs are involved? [Hint: Remember that the gases themselves are an object.]

◊

Tam Anh grabs Sarah by the hand and tries to pull her. She tries to remain standing without moving. A student analyzes the situation as follows. “If Tam Anh's force on Sarah is greater than her force on him, he can get her to move. Otherwise, she'll be able to stay where she is.” What's wrong with this analysis?

◊

You hit a tennis ball against a wall. Explain any and all incorrect ideas in the following description of the physics involved: “According to Newton's third law, there has to be a force opposite to your force on the ball. The opposite force is the ball's mass, which resists acceleration, and also air resistance.”

One of the most basic and important tasks of physics is to classify the forces of nature. I have already referred informally to “types” of forces such as friction, magnetism, gravitational forces, and so on. Classification systems are creations of the human mind, so there is always some degree of arbitrariness in them. For one thing, the level of detail that is appropriate for a classification system depends on what you're trying to find out. Some linguists, the “lumpers,” like to emphasize the similarities among languages, and a few extremists have even tried to find signs of similarities between words in languages as different as English and Chinese, lumping the world's languages into only a few large groups. Other linguists, the “splitters,” might be more interested in studying the differences in pronunciation between English speakers in New York and Connecticut. The splitters call the lumpers sloppy, but the lumpers say that science isn't worthwhile unless it can find broad, simple patterns within the seemingly complex universe.

Scientific classification systems are also usually compromises between practicality and naturalness. An example is the question of how to classify flowering plants. Most people think that biological classification is about discovering new species, naming them, and classifying them in the class-order-family-genus-species system according to guidelines set long ago. In reality, the whole system is in a constant state of flux and controversy. One very practical way of classifying flowering plants is according to whether their petals are separate or joined into a tube or cone --- the criterion is so clear that it can be applied to a plant seen from across the street. But here practicality conflicts with naturalness. For instance, the begonia has separate petals and the pumpkin has joined petals, but they are so similar in so many other ways that they are usually placed within the same order. Some taxonomists have come up with classification criteria that they claim correspond more naturally to the apparent relationships among plants, without having to make special exceptions, but these may be far less practical, requiring for instance the examination of pollen grains under an electron microscope.

In physics, there are two main systems of classification for forces. At this point in the course, you are going to learn one that is very practical and easy to use, and that splits the forces up into a relatively large number of types: seven very common ones that we'll discuss explicitly in this chapter, plus perhaps ten less important ones such as surface tension, which we will not bother with right now.

Physicists, however, are obsessed with finding simple patterns, so recognizing as many as fifteen or twenty types of forces strikes them as distasteful and overly complex. Since about the year 1900, physics has been on an aggressive program to discover ways in which these many seemingly different types of forces arise from a smaller number of fundamental ones. For instance, when you press your hands together, the force that keeps them from passing through each other may seem to have nothing to do with electricity, but at the atomic level, it actually does arise from electrical repulsion between atoms. By about 1950, all the forces of nature had been explained as arising from four fundamental types of forces at the atomic and nuclear level, and the lumping-together process didn't stop there. By the 1960's the length of the list had been reduced to three, and some theorists even believe that they may be able to reduce it to two or one. Although the unification of the forces of nature is one of the most beautiful and important achievements of physics, it makes much more sense to start this course with the more practical and easy system of classification. The unified system of four forces will be one of the highlights of the end of your introductory physics sequence.

The practical classification scheme which concerns us now can be laid out in the form of the tree shown in figure i. The most specific types of forces are shown at the tips of the branches, and it is these types of forces that are referred to in the POFOSTITO mnemonic. For example, electrical and magnetic forces belong to the same general group, but Newton's third law would never relate an electrical force to a magnetic force.

The broadest distinction is that between contact and noncontact forces, which has been discussed in ch. 4. Among the contact forces, we distinguish between those that involve solids only and those that have to do with fluids, a term used in physics to include both gases and liquids.

It should not be necessary to memorize this diagram by rote. It is better to reinforce your memory of this system by calling to mind your commonsense knowledge of certain ordinary phenomena. For instance, we know that the gravitational attraction between us and the planet earth will act even if our feet momentarily leave the ground, and that although magnets have mass and are affected by gravity, most objects that have mass are nonmagnetic.

\(\triangleright\) A bullet, flying horizontally, hits a steel wall. What type of force is there between the bullet and the wall?

\(\triangleright\) Starting at the bottom of the tree, we determine that the force is a contact force, because it only occurs once the bullet touches the wall. Both objects are solid. The wall forms a vertical plane. If the nose of the bullet was some shape like a sphere, you might imagine that it would only touch the wall at one point. Realistically, however, we know that a lead bullet will flatten out a lot on impact, so there is a surface of contact between the two, and its orientation is vertical. The effect of the force on the bullet is to stop the horizontal motion of the bullet, and this horizontal acceleration must be produced by a horizontal force. The force is therefore perpendicular to the surface of contact, and it's also repulsive (tending to keep the bullet from entering the wall), so it must be a normal force.

Diagram i is meant to be as simple as possible while including most of the forces we deal with in everyday life. If you were an insect, you would be much more interested in the force of surface tension, which allowed you to walk on water. I have not included the nuclear forces, which are responsible for holding the nuclei of atoms, because they are not evident in everyday life.

You should not be afraid to invent your own names for types of forces that do not fit into the diagram. For instance, the force that holds a piece of tape to the wall has been left off of the tree, and if you were analyzing a situation involving scotch tape, you would be absolutely right to refer to it by some commonsense name such as “sticky force.”

On the other hand, if you are having trouble classifying a certain force, you should also consider whether it is a force at all. For instance, if someone asks you to classify the force that the earth has because of its rotation, you would have great difficulty creating a place for it on the diagram. That's because it's a type of motion, not a type of force!

A normal force, \(F_N\), is a force that keeps one solid object from passing through another. “Normal” is simply a fancy word for “perpendicular,” meaning that the force is perpendicular to the surface of contact. Intuitively, it seems the normal force magically adjusts itself to provide whatever force is needed to keep the objects from occupying the same space. If your muscles press your hands together gently, there is a gentle normal force. Press harder, and the normal force gets stronger. How does the normal force know how strong to be? The answer is that the harder you jam your hands together, the more compressed your flesh becomes. Your flesh is acting like a spring: more force is required to compress it more. The same is true when you push on a wall. The wall flexes imperceptibly in proportion to your force on it. If you exerted enough force, would it be possible for two objects to pass through each other? No, typically the result is simply to strain the objects so much that one of them breaks.

As we'll discuss in more detail later in the course, a gravitational force exists between any two things that have mass. In everyday life, the gravitational force between two cars or two people is negligible, so the only noticeable gravitational forces are the ones between the earth and various human-scale objects. We refer to these planet-earth-induced gravitational forces as weight forces, and as we have already seen, their magnitude is given by \(|F_W|=mg\).

◊ Solved problem: Weight and mass — problem 26

If you have pushed a refrigerator across a kitchen floor, you have felt a certain series of sensations. At first, you gradually increased your force on the refrigerator, but it didn't move. Finally, you supplied enough force to unstick the fridge, and there was a sudden jerk as the fridge started moving. Once the fridge was unstuck, you could reduce your force significantly and still keep it moving.

While you were gradually increasing your force, the floor's
frictional force on the fridge increased in response. The
two forces on the fridge canceled, and the fridge didn't
accelerate. How did the floor know how to respond with just
the right amount of force? Figure j shows one
possible *model* of friction that explains this
behavior. (A scientific model is a
description that we expect to be incomplete, approximate, or
unrealistic in some ways, but that nevertheless succeeds in
explaining a variety of phenomena.) Figure j/1 shows a
microscopic view of the tiny bumps and holes in the surfaces
of the floor and the refrigerator. The weight of the fridge
presses the two surfaces together, and some of the bumps in
one surface will settle as deeply as possible into some of
the holes in the other surface. In
j/2, your leftward
force on the fridge has caused it to ride up a little higher
on the bump in the floor labeled with a small arrow. Still
more force is needed to get the fridge over the bump and
allow it to start moving. Of course, this is occurring
simultaneously at millions of places on the two surfaces.

Once you had gotten the fridge moving at constant speed, you found that you needed to exert less force on it. Since zero total force is needed to make an object move with constant velocity, the floor's rightward frictional force on the fridge has apparently decreased somewhat, making it easier for you to cancel it out. Our model also gives a plausible explanation for this fact: as the surfaces slide past each other, they don't have time to settle down and mesh with one another, so there is less friction.

Even though this model is intuitively appealing and fairly successful, it should not be taken too seriously, and in some situations it is misleading. For instance, fancy racing bikes these days are made with smooth tires that have no tread --- contrary to what we'd expect from our model, this does not cause any decrease in friction. Machinists know that two very smooth and clean metal surfaces may stick to each other firmly and be very difficult to slide apart. This cannot be explained in our model, but makes more sense in terms of a model in which friction is described as arising from chemical bonds between the atoms of the two surfaces at their points of contact: very flat surfaces allow more atoms to come in contact.

Since friction changes its behavior dramatically once the
surfaces come unstuck, we define two separate types of
frictional forces. *Static friction*
is friction that occurs between surfaces that are not
slipping over each other. Slipping surfaces experience
*kinetic friction*. The forces of static and
kinetic friction, notated \(F_s\) and \(F_k\), are always
parallel to the surface of contact between the two objects.

1. When a baseball player slides in to a base, is the friction static, or kinetic?

2. A mattress stays on the roof of a slowly accelerating car. Is the friction static, or kinetic?

3. Does static friction create heat? Kinetic friction?

(answer in the back of the PDF version of the book)The maximum possible force of static friction depends on what kinds of surfaces they are, and also on how hard they are being pressed together. The approximate mathematical relationships can be expressed as follows:

\[\begin{equation*}
F_{s,max} = \mu_s F_N ,
\end{equation*}\]

where \(\mu_s\) is a unitless number, called the coefficient of static friction, which depends on what kinds of surfaces they are. The maximum force that static friction can supply, \(\mu_s F_N\), represents the boundary between static and kinetic friction. It depends on the normal force, which is numerically equal to whatever force is pressing the two surfaces together. In terms of our model, if the two surfaces are being pressed together more firmly, a greater sideways force will be required in order to make the irregularities in the surfaces ride up and over each other.

Note that just because we use an adjective such as “applied” to refer to a force, that doesn't mean that there is some special type of force called the “applied force.” The applied force could be any type of force, or it could be the sum of more than one force trying to make an object move.

The arrows in figure m show the forces of the tree trunk on the partridge. Describe the forces the bird makes on the tree.

(answer in the back of the PDF version of the book)The force of kinetic friction on each of the two objects is in the direction that resists the slippage of the surfaces. Its magnitude is usually well approximated as

\[\begin{equation*}
F_k = \mu_k F_N
\end{equation*}\]

where \(\mu_k\) is the coefficient of kinetic friction. Kinetic friction is usually more or less independent of velocity.

Can a frictionless surface exert a normal force? Can a frictional force exist without a normal force?

(answer in the back of the PDF version of the book)If you try to accelerate or decelerate your car too quickly, the forces between your wheels and the road become too great, and they begin slipping. This is not good, because kinetic friction is weaker than static friction, resulting in less control. Also, if this occurs while you are turning, the car's handling changes abruptly because the kinetic friction force is in a different direction than the static friction force had been: contrary to the car's direction of motion, rather than contrary to the forces applied to the tire.

Most people respond with disbelief when told of the experimental evidence that both static and kinetic friction are approximately independent of the amount of surface area in contact. Even after doing a hands-on exercise with spring scales to show that it is true, many students are unwilling to believe their own observations, and insist that bigger tires “give more traction.” In fact, the main reason why you would not want to put small tires on a big heavy car is that the tires would burst!

Although many people expect that friction would be proportional to surface area, such a proportionality would make predictions contrary to many everyday observations. A dog's feet, for example, have very little surface area in contact with the ground compared to a human's feet, and yet we know that a dog can often win a tug-of-war with a person.

The reason a smaller surface area does not lead to less friction is that the force between the two surfaces is more concentrated, causing their bumps and holes to dig into each other more deeply.

Find the direction of each of the forces in figure o.

(answer in the back of the PDF version of the book)The reason locomotives are built to be so heavy is for traction. The upward normal force of the rails on the wheels, \(F_N\), cancels the downward force of gravity, \(F_W\), so ignoring plus and minus signs, these two forces are equal in absolute value, \(F_N=F_W\). Given this amount of normal force, the maximum force of static friction is \(F_s=\mu_s F_N=\mu_s F_W\). This static frictional force, of the rails pushing forward on the wheels, is the only force that can accelerate the train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed. The coefficient of static friction for steel on steel is about 1/4, so no locomotive can pull with a force greater than about 1/4 of its own weight. If the engine is capable of supplying more than that amount of force, the result will be simply to break static friction and spin the wheels.

The reason this is all so different from the situation with a car is that a car isn't pulling something else. If you put extra weight in a car, you improve the traction, but you also increase the inertia of the car, and make it just as hard to accelerate. In a train, the inertia is almost all in the cars being pulled, not in the locomotive.

The other fact we have to explain is the large number of driving wheels. First, we have to realize that increasing the number of driving wheels neither increases nor decreases the total amount of static friction, because static friction is independent of the amount of surface area in contact. (The reason four-wheel-drive is good in a car is that if one or more of the wheels is slipping on ice or in mud, the other wheels may still have traction. This isn't typically an issue for a train, since all the wheels experience the same conditions.) The advantage of having more driving wheels on a train is that it allows us to increase the weight of the locomotive without crushing the rails, or damaging bridges.

Try to drive a nail into a waterfall and you will be confronted with the main difference between solid friction and fluid friction. Fluid friction is purely kinetic; there is no static fluid friction. The nail in the waterfall may tend to get dragged along by the water flowing past it, but it does not stick in the water. The same is true for gases such as air: recall that we are using the word “fluid” to include both gases and liquids.

Unlike kinetic friction between solids, fluid friction increases rapidly with velocity. It also depends on the shape of the object, which is why a fighter jet is more streamlined than a Model T. For objects of the same shape but different sizes, fluid friction typically scales up with the cross-sectional area of the object, which is one of the main reasons that an SUV gets worse mileage on the freeway than a compact car.

◊

A student states that when he tries to push his refrigerator, the reason it won't move is because Newton's third law says there's an equal and opposite frictional force pushing back. After all, the static friction force is equal and opposite to the applied force. How would you convince him he is wrong?

◊

Kinetic friction is usually more or less independent of velocity. However, inexperienced drivers tend to produce a jerk at the last moment of deceleration when they stop at a stop light. What does this tell you about the kinetic friction between the brake shoes and the brake drums?

◊

Some of the following are correct descriptions of types of forces that could be added on as new branches of the classification tree. Others are not really types of forces, and still others are not force phenomena at all. In each case, decide what's going on, and if appropriate, figure out how you would incorporate them into the tree.

sticky force | makes tape stick to things |

opposite force | the force that Newton’s third law says relates to every force you make |

flowing force | the force that water carries with it as it flows out of a hose |

surface tension | lets insects walk on water |

horizontal force | a force that is horizontal |

motor force | the force that a motor makes on the thing it is turning |

canceled force | a force that is being canceled out by some other force |

Newton's first and second laws deal with the total of all the forces exerted on a specific object, so it is very important to be able to figure out what forces there are. Once you have focused your attention on one object and listed the forces on it, it is also helpful to describe all the corresponding forces that must exist according to Newton's third law. We refer to this as “analyzing the forces” in which the object participates.

A barge is being pulled to the right along a canal by teams of horses on the shores. Analyze all the forces in which the barge participates.

force acting on barge | force related to it by Newton’s third law |

ropes’ normal forces on barge,→ | barge’s normal force on ropes,← |

water’s fluid friction force on barge,← | barge’s fluid friction force on water,→ |

planet earth’s gravitational force on barge,↓ | barge’s gravitational force on earth,↑ |

water’s “floating” force on barge,↑ | barge’s “floating” force on water,↓ |

Here I've used the word “floating” force as an example of a sensible invented term for a type of force not classified on the tree on p. 156. A more formal technical term would be “hydrostatic force.”

Note how the pairs of forces are all structured as “A's force on B, B's force on A”: ropes on barge and barge on ropes; water on barge and barge on water. Because all the forces in the left column are forces acting on the barge, all the forces in the right column are forces being exerted by the barge, which is why each entry in the column begins with “barge.”

Often you may be unsure whether you have forgotten one of the forces. Here are three strategies for checking your list:

- See what physical result would come from the forces you've found so far. Suppose, for instance, that you'd forgotten the “floating” force on the barge in the example above. Looking at the forces you'd found, you would have found that there was a downward gravitational force on the barge which was not canceled by any upward force. The barge isn't supposed to sink, so you know you need to find a fourth, upward force.
- Another technique for finding missing forces is simply to go through the list of all the common types of forces and see if any of them apply.
- Make a drawing of the object, and draw a dashed boundary line around it that separates it from its environment. Look for points on the boundary where other objects come in contact with your object. This strategy guarantees that you'll find every contact force that acts on the object, although it won't help you to find non-contact forces.

\(\triangleright\) We analyze the forces:

force acting on Fifi | force related to it by Newton’s third law |

planet earth’s gravitational
force | Fifi’s gravitational force on earth, ↑ |

belt’s kinetic frictional
force | Fifi’s kinetic frictional force on belt, ← |

belt’s normal
force | Fifi’s normalforce on belt, ↓ |

Checking the analysis of the forces as described on p. 164:

(1) The physical result makes sense. The left-hand column consists of forces \(\downarrow\rightarrow\uparrow\). We're describing the time when she's moving horizontally on the belt, so it makes sense that we have two vertical forces that could cancel. The rightward force is what will accelerate her until her speed matches that of the belt.

(2) We've included every relevant type of force from the tree on p. 156.

(3) We've included forces from the belt, which is the only object in contact with Fifi.

The purpose of the analysis is to let us set up equations containing enough information to solve the problem. Using the generalization of Newton's second law given on p. 133, we use the horizontal force to determine the horizontal acceleration, and separately require the vertical forces to cancel out.

Let positive \(x\) be to the right. Newton's second law gives

\[\begin{equation*}
(\rightarrow) a=F_k/m
\end{equation*}\]

Although it's the horizontal motion we care about, the only way to find \(F_k\) is via the relation \(F_k=\mu_k F_N\), and the only way to find \(F_N\) is from the \(\uparrow\downarrow\) forces. The two vertical forces must cancel, which means they have to be of equal strength:

\[\begin{equation*}
(\uparrow\downarrow) F_N - mg = 0 .
\end{equation*}\]

Using the constant-acceleration equation \(a=\Delta v/\Delta t\), we have

\[\begin{align*}
\Delta t &= \frac{\Delta v}{a} \\
&= \frac{v_b-v_\text{o}}{\mu_k mg/m}\\
&= \frac{v_b-v_\text{o}}{\mu_k g} .
\end{align*}\]

The units check out:

\[\begin{equation*}
\text{s} = \frac{\text{m}/\text{s}}{\text{m}/\text{s}^2} ,
\end{equation*}\]

where \(\mu_k\) is omitted as a factor because it's unitless.

We should also check that the dependence on the variables makes sense. If Fifi puts on her rubber ninja booties, increasing \(\mu_k\), then dividing by a larger number gives a smaller result for \(\Delta t\); this makes sense physically, because the greater friction will cause her to come up to the belt's speed more quickly. The dependence on \(g\) is similar; more gravity would press her harder against the belt, improving her traction. Increasing \(v_b\) increases \(\Delta t\), which makes sense because it will take her longer to get up to a bigger speed. Since \(v_\text{o}\) is subtracted, the dependence of \(\Delta t\) on it is the other way around, and that makes sense too, because if she can land with a greater speed, she has less speeding up left to do.

◊

In the example of the barge going down the canal, I referred to a “floating” or “hydrostatic” force that keeps the boat from sinking. If you were adding a new branch on the force-classification tree to represent this force, where would it go?

◊

The earth's gravitational force on you, i.e., your weight, is always equal to \(mg\), where \(m\) is your mass. So why can you get a shovel to go deeper into the ground by jumping onto it? Just because you're jumping, that doesn't mean your mass or weight is any greater, does it?

You're walking your dog. The dog wants to go faster than you do, and the leash is taut. Does Newton's third law guarantee that your force on your end of the leash is equal and opposite to the dog's force on its end? If they're not exactly equal, is there any reason why they should be approximately equal?

If there was no leash between you, and you were in direct contact with the dog, then Newton's third law would apply, but Newton's third law cannot relate your force on the leash to the dog's force on the leash, because that would involve three separate objects. Newton's third law only says that your force on the leash is equal and opposite to the leash's force on you,

\[\begin{equation*}
F_{yL} = - F_{Ly} ,
\end{equation*}\]

and that the dog's force on the leash is equal and opposite to its force on the dog

\[\begin{equation*}
F_{dL} = - F_{Ld} .
\end{equation*}\]

Still, we have a strong intuitive expectation that whatever force we make on our end of the leash is transmitted to the dog, and vice-versa. We can analyze the situation by concentrating on the forces that act on the leash, \(F_{dL}\) and \(F_{yL}\). According to Newton's second law, these relate to the leash's mass and acceleration:

\[\begin{equation*}
F_{dL} + F_{yL} = m_La_L .
\end{equation*}\]

The leash is far less massive then any of the other objects involved, and if \(m_L\) is very small, then apparently the total force on the leash is also very small, \(F_{dL}\) + \(F_{yL}\approx 0\), and therefore

\[\begin{equation*}
F_{dL}\approx - F_{yL} .
\end{equation*}\]

Thus even though Newton's third law does not apply directly to these two forces, we can approximate the low-mass leash as if it was not intervening between you and the dog. It's at least approximately as if you and the dog were acting directly on each other, in which case Newton's third law would have applied.

In general, low-mass objects can be treated approximately as if they simply transmitted forces from one object to another. This can be true for strings, ropes, and cords, and also for rigid objects such as rods and sticks.

If you look at a piece of string under a magnifying glass as you pull on the ends more and more strongly, you will see the fibers straightening and becoming taut. Different parts of the string are apparently exerting forces on each other. For instance, if we think of the two halves of the string as two objects, then each half is exerting a force on the other half. If we imagine the string as consisting of many small parts, then each segment is transmitting a force to the next segment, and if the string has very little mass, then all the forces are equal in magnitude. We refer to the magnitude of the forces as the tension in the string, \(T\).

The term “tension” refers only to internal forces within the string. If the string makes forces on objects at its ends, then those forces are typically normal or frictional forces (example 8).

\(\triangleright\) In all cases, a rope can only make “pulling” forces, i.e., forces that are parallel to its own length and that are toward itself, not away from itself. You can't push with a rope!

In w/1, the rope passes through a type of hook, called a carabiner, used in rock climbing and mountaineering. Since the rope can only pull along its own length, the direction of its force on the carabiner must be down and to the right. This is perpendicular to the surface of contact, so the force is a normal force.

force acting on Fifi | force related to it by Newton’s third law |

planet earth’s gravitational
force | Fifi’s gravitational force on earth, ↑ |

belt’s kinetic frictional
force | Fifi’s kinetic frictional force on belt, ← |

belt’s normal
force | Fifi’s normalforce on belt, ↓ |

(There are presumably other forces acting on the carabiner from other hardware above it.)

In figure w/2, the rope can only exert a net force at its end that is parallel to itself and in the pulling direction, so its force on the hand is down and to the left. This is parallel to the surface of contact, so it must be a frictional force. If the rope isn't slipping through the hand, we have static friction. Friction can't exist without normal forces. These forces are perpendicular to the surface of contact. For simplicity, we show only two pairs of these normal forces, as if the hand were a pair of pliers.

force acting on Fifi | force related to it by Newton’s third law |

planet earth’s gravitational
force | Fifi’s gravitational force on earth, ↑ |

belt’s kinetic frictional
force | Fifi’s kinetic frictional force on belt, ← |

belt’s normal
force | Fifi’s normalforce on belt, ↓ |

(There are presumably other forces acting on the person as well, such as gravity.)

If a rope goes over a pulley or around some other object, then the tension throughout the rope is approximately equal so long as the pulley has negligible mass and there is not too much friction. A rod or stick can be treated in much the same way as a string, but it is possible to have either compression or tension.

◊

When you step on the gas pedal, is your foot's force being transmitted in the sense of the word used in this section?

A string lengthens slightly when you stretch it. Similarly, we have already discussed how an apparently rigid object such as a wall is actually flexing when it participates in a normal force. In other cases, the effect is more obvious. A spring or a rubber band visibly elongates when stretched.

Common to all these examples is a change in shape of some kind: lengthening, bending, compressing, etc. The change in shape can be measured by picking some part of the object and measuring its position, \(x\). For concreteness, let's imagine a spring with one end attached to a wall. When no force is exerted, the unfixed end of the spring is at some position \(x_o\). If a force acts at the unfixed end, its position will change to some new value of \(x\). The more force, the greater the departure of \(x\) from \(x_o\).

Back in Newton's time, experiments like this were considered cutting-edge research, and his contemporary Hooke is remembered today for doing them and for coming up with a simple mathematical generalization called Hooke's law:

\[\begin{multline*}
F \approx k(x-x_o) . \shoveright{\text{[force required to stretch a spring; valid}}\\
\text{for small forces only]}
\end{multline*}\]

Here \(k\) is a constant, called the spring constant, that depends on how stiff the object is. If too much force is applied, the spring exhibits more complicated behavior, so the equation is only a good approximation if the force is sufficiently small. Usually when the force is so large that Hooke's law is a bad approximation, the force ends up permanently bending or breaking the spring.

Although Hooke's law may seem like a piece of trivia about springs, it is actually far more important than that, because all solid objects exert Hooke's-law behavior over some range of sufficiently small forces. For example, if you push down on the hood of a car, it dips by an amount that is directly proportional to the force. (But the car's behavior would not be as mathematically simple if you dropped a boulder on the hood!)

◊ Solved problem: Combining springs — problem 14

◊ Solved problem: Young's modulus — problem 16

◊

A car is connected to its axles through big, stiff springs called shock absorbers, or “shocks.” Although we've discussed Hooke's law above only in the case of stretching a spring, a car's shocks are continually going through both stretching and compression. In this situation, how would you interpret the positive and negative signs in Hooke's law?

Even the most complex machines, such as cars or pianos, are
built out of certain basic units called
*simple machines*. The following are some of
the main functions of simple machines:

transmitting a force: The chain on a bicycle transmits a force from the crank set to the rear wheel.

changing the direction of a force: If you push down on a seesaw, the other end goes up.

changing the speed and precision of motion: When you make the “come here” motion, your biceps only moves a couple of centimeters where it attaches to your forearm, but your arm moves much farther and more rapidly.

changing the amount of force: A lever or pulley can be used to increase or decrease the amount of force.

You are now prepared to understand one-dimensional simple machines, of which the pulley is the main example.

\(\triangleright\) To use Newton's first law, we need to pick an
object and consider the sum of the forces on it. Since our
goal is to relate the tension in the part of the cable
attached to the stump to the tension in the part attached to
the tractor, we should pick an object to which both those
cables are attached, i.e., the pulley itself. The tension in a string or cable remains
approximately constant as it passes around an idealized pulley.
^{1}
There are
therefore two leftward forces acting on the pulley, each
equal to the force exerted by the tractor. Since the
acceleration of the pulley is essentially zero, the forces
on it must be canceling out, so the rightward force of the
pulley-stump cable on the pulley must be double the force
exerted by the tractor. Yes, Farmer Bill knows what he's talking about.

*repulsive* — describes a force that tends to push the two
participating objects apart

*attractive* — describes a force that tends to pull the two
participating objects together

*oblique* — describes a force that acts at some other angle, one
that is not a direct repulsion or attraction

*normal force* — the force that keeps two objects from
occupying the same space

*static friction* — a friction force between surfaces that are
not slipping past each other

*kinetic friction* — a friction force between surfaces that are
slipping past each other

*fluid* — a gas or a liquid

*fluid friction* — a friction force in which at least one of the
object is is a fluid

*spring constant* — the constant of proportionality between
force and elongation of a spring or other object under strain

\(F_N\) — a normal force

\(F_s\) — a static frictional force

\(F_k\) — a kinetic frictional force

\(\mu_s\) — the coefficient of static friction; the constant of proportionality between the maximum static frictional force and the normal force; depends on what types of surfaces are involved

\(\mu_k\) — the coefficient of kinetic friction; the constant of proportionality between the kinetic frictional force and the normal force; depends on what types of surfaces are involved

\(k\) — the spring constant; the constant of proportionality between the force exerted on an object and the amount by which the object is lengthened or compressed

{}

Newton's third law states that forces occur in equal and opposite pairs. If object A exerts a force on object B, then object B must simultaneously be exerting an equal and opposite force on object A. Each instance of Newton's third law involves exactly two objects, and exactly two forces, which are of the same type.

There are two systems for classifying forces. We are presently using the more practical but less fundamental one. In this system, forces are classified by whether they are repulsive, attractive, or oblique; whether they are contact or noncontact forces; and whether the two objects involved are solids or fluids.

Static friction adjusts itself to match the force that is trying to make the surfaces slide past each other, until the maximum value is reached,

\[\begin{equation*}
F_{s,max} = \mu_s F_N .
\end{equation*}\]

Once this force is exceeded, the surfaces slip past one another, and kinetic friction applies,

\[\begin{equation*}
F_k = \mu_k F_N .
\end{equation*}\]

Both types of frictional force are nearly independent of surface area, and kinetic friction is usually approximately independent of the speed at which the surfaces are slipping. The direction of the force is in the direction that would tend to stop or prevent slipping.

A good first step in applying Newton's laws of motion to any physical situation is to pick an object of interest, and then to list all the forces acting on that object. We classify each force by its type, and find its Newton's-third-law partner, which is exerted by the object on some other object.

When two objects are connected by a third low-mass object, their forces are transmitted to each other nearly unchanged.

Objects under strain always obey Hooke's law to a good approximation, as long as the force is small. Hooke's law states that the stretching or compression of the object is proportional to the force exerted on it,

\[\begin{equation*}
F \approx k(x-x_\text{o}) .
\end{equation*}\]

\begin{homeworkforcelabel}{oldlady}{1}{}{1}A little old lady and a pro football player collide head-on. Compare their forces on each other, and compare their accelerations. Explain. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{earthaccelup}{1}{}{2}The earth is attracted to an object with a force equal and opposite to the force of the earth on the object. If this is true, why is it that when you drop an object, the earth does not have an acceleration equal and opposite to that of the object? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{gravityvsnormal}{1}{}{3}When you stand still, there are two forces acting on you, the force of gravity (your weight) and the normal force of the floor pushing up on your feet. Are these forces equal and opposite? Does Newton's third law relate them to each other? Explain. \end{homeworkforcelabel}

*In problems 4-8, analyze the forces using a table in the
format shown in section 5.3. Analyze the forces in which the
italicized object participates.*

\begin{homeworkforcelabel}{magnetundercar}{1}{}{4}Some people put a spare car key in a little magnetic
*box* that they stick under the chassis of their car.
Let's say that the box is stuck directly underneath a horizontal
surface, and the car is parked.
(See instructions above.)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{nonormal}{1}{}{5}Analyze two examples of *objects* at rest relative
to the earth that are being kept from falling by forces
other than the normal force. Do not use objects in outer
space, and do not duplicate problem 4 or 8.
(See instructions above.)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{row}{1}{}{6}A *person* is rowing a boat, with her feet braced.
She is doing the part of the stroke that propels the boat,
with the ends of the oars in the water (not the part where
the oars are out of the water).
(See instructions above.)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{cowcrush}{1}{}{7}A *farmer* is in a stall with a cow when the cow
decides to press him against the wall, pinning him with his
feet off the ground. Analyze the forces in which the
farmer participates.
(See instructions above.)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{airplaneforces}{1}{}{8}A propeller *plane* is cruising east at constant
speed and altitude.
(See instructions above.)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{elevator}{1}{}{9}Today's tallest buildings are really not that much
taller than the tallest buildings of the 1940's. One big
problem with making an even taller skyscraper is that every
elevator needs its own shaft running the whole height of the
building. So many elevators are needed to serve the
building's thousands of occupants that the elevator shafts
start taking up too much of the space within the building.
An alternative is to have elevators that can move both
horizontally and vertically: with such a design, many
elevator cars can share a few shafts, and they don't get in
each other's way too much because they can detour around
each other. In this design, it becomes impossible to hang
the cars from cables, so they would instead have to ride on
rails which they grab onto with wheels. Friction would keep
them from slipping. The figure shows such a frictional
elevator in its vertical travel mode. (The wheels on the
bottom are for when it needs to switch to horizontal
motion.)

(a) If the coefficient of static friction between
rubber and steel is \(\mu_s\), and the maximum mass of the car
plus its passengers is \(M\), how much force must there be
pressing each wheel against the rail in order to keep the
car from slipping? (Assume the car is not accelerating.)(answer check available at lightandmatter.com)

(b)
Show that your result has physically reasonable behavior
with respect to \(\mu_s\). In other words, if there was less
friction, would the wheels need to be pressed more firmly or
less firmly? Does your equation behave that way?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{atwood}{1}{}{10}Unequal masses \(M\) and \(m\) are suspended from a pulley
as shown in the figure.

(a) Analyze the forces in which mass \(m\) participates, using
a table in the format shown in section 5.3. [The forces in
which the other mass participates will of course be similar,
but not numerically the same.]

(b) Find the magnitude of the accelerations of the two
masses. [Hints: (1) Pick a coordinate system, and use
positive and negative signs consistently to indicate the
directions of the forces and accelerations. (2) The two
accelerations of the two masses have to be equal in
magnitude but of opposite signs, since one side eats up rope
at the same rate at which the other side pays it out. (3)
You need to apply Newton's second law twice, once to each
mass, and then solve the two equations for the unknowns: the
acceleration, \(a\), and the tension in the rope, \(T\).] (answer check available at lightandmatter.com)

(c) Many people expect that in the special case of \(M=m\),
the two masses will naturally settle down to an equilibrium
position side by side. Based on your answer from part b,
is this correct?

(d) Find the tension in the rope, \(T\).(answer check available at lightandmatter.com)

(e) Interpret your equation from part d in the special case where
one of the masses is zero. Here “interpret” means to figure out
what happens mathematically, figure out what should happen
physically, and connect the two.

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{tugboat}{1}{}{11}A tugboat of mass \(m\) pulls a ship of mass \(M\),
accelerating it. The speeds are low enough that you can
ignore fluid friction acting on their hulls, although there
will of course need to be fluid friction acting on
the tug's propellers.

(a) Analyze the forces in which the tugboat participates,
using a table in the format shown in section 5.3. Don't
worry about vertical forces.

(b) Do the same for the ship.

(c) If the force acting on the tug's
propeller is \(F\), what is the tension, \(T\), in the cable
connecting the two ships? [Hint: Write down two equations,
one for Newton's second law applied to each object. Solve
these for the two unknowns \(T\) and \(a\).](answer check available at lightandmatter.com)

(d) Interpret your answer in the special cases of \(M=0\) and \(M=\infty\).
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{kineticstrongerthanstatic}{1}{}{12}Someone tells you she knows of a certain type of Central American earthworm whose skin, when rubbed on polished diamond, has \(\mu_k>\mu_s\). Why is this not just empirically unlikely but logically suspect? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{three-pulleys}{1}{}{13}In the system shown in the figure, the pulleys on the left and right are fixed, but the pulley in the center can move to the left or right. The two masses are identical. Show that the mass on the left will have an upward acceleration equal to \(g/5\). Assume all the ropes and pulleys are massless and frictionless.

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{combine-springs}{1}{}{14}The figure shows two different ways of combining a
pair of identical springs, each with spring constant \(k\). We
refer to the top setup as parallel, and the bottom one as a
series arrangement.

(a) For the parallel arrangement,
analyze the forces acting on the connector piece on the
left, and then use this analysis to determine the equivalent
spring constant of the whole setup. Explain whether the
combined spring constant should be interpreted as being
stiffer or less stiff.

(b) For the series arrangement,
analyze the forces acting on each spring and figure
out the same things.(solution in the pdf version of the book)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{combinesprings}{1}{}{15}Generalize the results of problem 14 to the case where the two spring constants are unequal. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{youngs-modulus}{1}{}{16}(a) Using the solution of problem 14, which is given
in the back of the book, predict how the spring constant of
a fiber will depend on its length and cross-sectional area.

(b) The constant of proportionality is called the Young's modulus, \(E\), and typical values of the
Young's modulus are about \(10^{10}\) to \(10^{11}\). What
units would the Young's modulus have in the SI (meter-kilogram-second) system? (solution in the pdf version of the book)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{bondstretching}{1}{}{17}This problem depends on the results of problems 14 and 16,
whose solutions are in the back of the book. When atoms
form chemical bonds, it makes sense to talk about the spring
constant of the bond as a measure of how “stiff” it is. Of
course, there aren't really little springs --- this is just
a mechanical model. The purpose of this problem is to
estimate the spring constant, \(k\), for a single bond in a
typical piece of solid matter. Suppose we have a fiber, like
a hair or a piece of fishing line, and imagine for
simplicity that it is made of atoms of a single element
stacked in a cubical manner, as shown in the figure, with a
center-to-center spacing \(b\). A typical value for \(b\) would
be about \(10^{-10}\ \text{m}\).

(a) Find an equation for \(k\) in terms
of \(b\), and in terms of the Young's modulus, \(E\), defined in
problem 16 and its solution.

(b) Estimate \(k\) using the
numerical data given in problem 16.

(c) Suppose you could
grab one of the atoms in a diatomic molecule like \(\text{H}_2\) or
\(\text{O}_2\), and let the other atom hang vertically below it. Does
the bond stretch by any appreciable fraction due to gravity?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{third-law-partners}{1}{}{18}(solution in the pdf version of the book) In each case, identify the force that causes the acceleration, and give its Newton's-third-law partner. Describe the effect of the partner force. (a) A swimmer speeds up. (b) A golfer hits the ball off of the tee. (c) An archer fires an arrow. (d) A locomotive slows down. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{dogsled}{1}{}{19}Ginny has a plan. She is going to ride her sled while
her dog Foo pulls her, and she holds on to his leash.
However, Ginny hasn't taken physics,
so there may be a problem: she may slide right off the sled
when Foo starts pulling.

(a) Analyze all the forces in which
Ginny participates, making a table as in section 5.3.

(b) Analyze all the forces in which the sled participates.

(c) The sled has mass \(m\), and Ginny has mass \(M\). The
coefficient of static friction between the sled and the snow
is \(\mu_1\), and \(\mu_2\) is the corresponding quantity for
static friction between the sled and her snow pants. Ginny
must have a certain minimum mass so that she will not slip
off the sled. Find this in terms of the other three
variables.(answer check available at lightandmatter.com)

(d) Interpreting your equation from part c,
under what conditions will there be no physically realistic solution for \(M\)? Discuss
what this means physically.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{box-uphill-partners}{1}{}{20}(solution in the pdf version of the book) Example 2 on page 153 involves a person pushing a box up a hill. The incorrect answer describes three forces. For each of these three forces, give the force that it is related to by Newton's third law, and state the type of force. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{tractormadness}{1}{}{21}Example 9 on page 170 describes a force-doubling setup involving a pulley. Make up a more complicated arrangement, using two pulleys, that would multiply the force by four. The basic idea is to take the output of one force doubler and feed it into the input of a second one. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{scaleoscillation}{1}{}{22}Pick up a heavy object such as a backpack or a chair, and stand on a bathroom scale. Shake the object up and down. What do you observe? Interpret your observations in terms of Newton's third law. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{skidmarks}{1}{}{23}A cop investigating the scene of an accident measures the length \(L\) of a car's skid marks in order to find out its speed \(v\) at the beginning of the skid. Express \(v\) in terms of \(L\) and any other relevant variables.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{baseballparadox}{1}{}{24}The following reasoning leads to an apparent paradox; explain what's wrong with the logic. A baseball player hits a ball. The ball and the bat spend a fraction of a second in contact. During that time they're moving together, so their accelerations must be equal. Newton's third law says that their forces on each other are also equal. But \(a=F/m\), so how can this be, since their masses are unequal? (Note that the paradox isn't resolved by considering the force of the batter's hands on the bat. Not only is this force very small compared to the ball-bat force, but the batter could have just thrown the bat at the ball.) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{deleted}{1}{}{25}This problem has been deleted. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{waterbottle-in-space}{1}{}{26}(solution in the pdf version of the book) (a) Compare the mass of a one-liter water bottle on
earth, on the moon, and in interstellar space.

(b) Do the
same for its weight.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{coasting-skater}{1}{}{27}An ice skater builds up some speed, and then coasts across the ice passively in
a straight line. (a) Analyze the forces, using
a table the format shown in section 5.3.

(b) If his initial speed is \(v\), and the
coefficient of kinetic friction is \(\mu_k\), find the maximum theoretical distance he can glide before
coming to a stop. Ignore air resistance.(answer check available at lightandmatter.com)

(c) Show that your answer to part b has the right units.

(d) Show that your answer to part b depends on the variables in a way that makes
sense physically.

(e) Evaluate your answer numerically for \(\mu_k=0.0046\), and a
world-record speed of \(14.58\) m/s. (The coefficient of friction was measured by
De Koning et al., using special skates worn by real speed skaters.)(answer check available at lightandmatter.com)

(f) Comment on whether
your answer in part e seems realistic. If it doesn't, suggest possible reasons why.

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{crevasse}{1}{}{28}Mountain climbers with masses \(m\) and \(M\) are roped together while crossing
a horizontal glacier when a vertical crevasse opens up under the climber with mass \(M\).
The climber with mass \(m\) drops down on the snow and tries to stop by digging
into the snow with the pick of an ice ax. Alas, this story does not have a
happy ending, because this doesn't provide enough friction to stop. Both \(m\) and \(M\)
continue accelerating, with \(M\) dropping down into the crevasse and \(m\) being dragged
across the snow, slowed only by the kinetic friction with coefficient \(\mu_k\) acting
between the ax and the snow. There is no significant friction between the rope and the
lip of the crevasse.

(a) Find the acceleration \(a\).(answer check available at lightandmatter.com)

(b) Check the units of your result.

(c) Check the dependence of your equation on the variables. That means that for each
variable, you should determine what its effect on \(a\) should be physically, and then
what your answer from part a says its effect would be mathematically.
\end{homeworkforcelabel}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

[1] This was asserted in section 5.4 without proof.
Essentially it holds because of symmetry. E.g., if the U-shaped piece of rope in figure y
had unequal tension in its two legs, then this would have to be caused by some asymmetry between clockwise and counterclockwise
rotation. But such an asymmetry can only be caused by friction or inertia, which we assume don't exist.