4.2: Force In One Dimension

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Contributed by Benjamin Crowell
Professor (Physics) at Fullerton College

3.2.1 Momentum transfer

For every conserved quantity, we can define an associated rate of flow. An open system can have mass transferred in or out of it, and we can measure the rate of mass flow, $d{}m/d{}t$ in units of kg/s. Energy can flow in or out, and the rate of energy transfer is the power, $P=d{}E/d{}t$, measured in watts.³ The rate of momentum transfer is called force,

\[\begin{equation*} F = \frac{d{}p}{d{}t} \text{[definition of force]} . \end{equation*}\]

The units of force are $\text{kg}\!\cdot\!\text{m}/\text{s}^2$, which can be abbreviated as newtons, $1\ \text{N}=\text{kg}\!\cdot\!\text{m}/\text{s}^2$. Newtons are unfortunately not as familiar as watts. A newton is about how much force you'd use to pet a dog. The most powerful rocket engine ever built, the first stage of the Saturn V that sent astronauts to the moon, had a thrust of about 30 million newtons. In one dimension, positive and negative signs indicate the direction of the force --- a positive force is one that pushes or pulls in the direction of the positive $x$ axis.

a / Power and force are the rates at which energy and momentum are transferred.

Example 20: Walking into a lamppost
\(\triangleright\) Starting from rest, you begin walking, bringing your momentum up to 100 \(\text{kg}\!\cdot\!\text{m}/\text{s}\). You walk straight into a lamppost. Why is the momentum change of \(-100\ \text{kg}\!\cdot\!\text{m}/\text{s}\) so much more painful than the change of \(+100\ \text{kg}\!\cdot\!\text{m}/\text{s}\) when you started walking? \(\triangleright\) The forces are not really constant, but for this type of qualitative discussion we can pretend they are, and approximate \(d{} p/d{} t\) as \(\Delta{} p/\Delta{} t\). It probably takes you about 1 s to speed up initially, so the ground's force on you is \(F=\Delta{} p/\Delta{} t\approx100\ \text{N}\). Your impact with the lamppost, however, is over in the blink of an eye, say 1/10 s or less. Dividing by this much smaller \(\Delta{} t\) gives a much larger force, perhaps thousands of newtons (with a negative sign).

Example 20: Walking into a lamppost

$\triangleright$ Starting from rest, you begin walking, bringing your momentum up to 100 $\text{kg}\!\cdot\!\text{m}/\text{s}$. You walk straight into a lamppost. Why is the momentum change of $-100\ \text{kg}\!\cdot\!\text{m}/\text{s}$ so much more painful than the change of $+100\ \text{kg}\!\cdot\!\text{m}/\text{s}$ when you started walking?

$\triangleright$ The forces are not really constant, but for this type of qualitative discussion we can pretend they are, and approximate $d{} p/d{} t$ as $\Delta{} p/\Delta{} t$. It probably takes you about 1 s to speed up initially, so the ground's force on you is $F=\Delta{} p/\Delta{} t\approx100\ \text{N}$. Your impact with the lamppost, however, is over in the blink of an eye, say 1/10 s or less. Dividing by this much smaller $\Delta{} t$ gives a much larger force, perhaps thousands of newtons (with a negative sign).

This is also the principle of airbags in cars. The time required for the airbag to decelerate your head is fairly long: the time it takes your face to travel 20 or 30 cm. Without an airbag, your face would have hit the dashboard, and the time interval would have been the much shorter time taken by your skull to move a couple of centimeters while your face compressed. Note that either way, the same amount of momentum is transferred: the entire momentum of your head.

b / The airbag increases $\Delta t$ so as to reduce $F=\Delta p/\Delta t$.

Force is defined as a derivative, and the derivative of a sum is the sum of the derivatives. Therefore force is additive: when more than one force acts on an object, you add the forces to find out what happens. An important special case is that forces can cancel. Consider your body sitting in a chair as you read this book. Let the positive $x$ axis be upward. The chair's upward force on you is represented with a positive number, which cancels out with the earth's downward gravitational force, which is negative. The total rate of momentum transfer into your body is zero, and your body doesn't change its momentum.

Example 21: Finding momentum from force
\(\triangleright\) An object of mass \(m\) starts at rest at \(t=t_\text{o}\). A force varying as \(F=bt^{-2}\), where \(b\) is a constant, begins acting on it. Find the greatest speed it will ever have. \(\triangleright\) \[\begin{align} F &= \frac{dp}{dt} \\ dp &= F dt \\ p &= \int F dt + p_\text{o}\\ &= -\frac{b}{t}+p_\text{o} , \end{align}\] where \(p_\text{o}\) is a constant of integration. The given initial condition is that \(p=0\) at \(t=t_\text{o}\), so we find that \(p_\text{o}=b/t_\text{o}\). The negative term gets closer to zero with increasing time, so the maximum momentum is achieved by letting \(t\) approach infinity. That is, the object will never stop speeding up, but it will also never surpass a certain speed. In the limit \(t\rightarrow\infty\), we identify \(p_\text{o}\) as the momentum that the object will approach asymptotically. The maximum velocity is \(v=p_\text{o}/m=b/mt_\text{o}\).

Example 21: Finding momentum from force

$\triangleright$ An object of mass $m$ starts at rest at $t=t_\text{o}$. A force varying as $F=bt^{-2}$, where $b$ is a constant, begins acting on it. Find the greatest speed it will ever have.

$\triangleright$

\[\begin{align*} F &= \frac{dp}{dt} \\ dp &= F dt \\ p &= \int F dt + p_\text{o}\\ &= -\frac{b}{t}+p_\text{o} , \end{align*}\]

where $p_\text{o}$ is a constant of integration. The given initial condition is that $p=0$ at $t=t_\text{o}$, so we find that $p_\text{o}=b/t_\text{o}$. The negative term gets closer to zero with increasing time, so the maximum momentum is achieved by letting $t$ approach infinity. That is, the object will never stop speeding up, but it will also never surpass a certain speed. In the limit $t\rightarrow\infty$, we identify $p_\text{o}$ as the momentum that the object will approach asymptotically. The maximum velocity is $v=p_\text{o}/m=b/mt_\text{o}$.

Discussion Question

◊ Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities:

The collision is instantaneous.
The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.

How can two of these be ruled out based on energy or momentum considerations?

3.2.2 Newton's laws

Although momentum is the third conserved quantity we've encountered, historically it was the first to be discovered. Isaac Newton formulated a complete treatment of mechanical systems in terms of force and momentum. Newton's theory was based on three laws of motion, which we now think of as consequences of conservation of mass, energy, and momentum.

c / Isaac Newton (1643-1727).

Newton’s laws in one dimension:

Newton’s first law: If there is no force acting on an object, it stays in the same state of motion.

Newton’s second law: The sum of all the forces acting on an object determines the rate at which its momentum changes,Ftotalderpdert.

Newton’s third law: Forces occur in opposite pairs. If object A interacts with object B, then A’s force on B and B’s force on A are related byF_AB = − F_BA.

The second law is the definition of force, which we've already encountered.⁴ The first law is a special case of the second law --- if $d{}p/d{}t$ is zero, then $p=mv$ is a constant, and since mass is conserved, constant $p$ implies constant $v$. The third law is a restatement of conservation of momentum: for two objects interacting, we have constant total momentum, so $0=\frac{d}{d{}t}(p_A+p_B) =F_{BA}+F_{AB}$.

Example 22: Ia=F/m
Many modern textbooks restate Newton's second law as $a= F/ m$, i.e., as an equation that predicts an object's acceleration based on the force exerted on it. This is easily derived from Newton's original form as follows: $a=\text{d} v/\text{d} t=(\text{d} p/\text{d} t)/ m= F/ m$.

Example 23: Gravitational force related to g
As a special case of the previous example, consider an object in free fall, and let the $x$ axis point down. Then $a=+g$, and $F= ma= mg$. For example, the gravitational force on a 1 kg mass at the earth's surface is about 9.8 N. Even if other forces act on the object, and it isn't in free fall, the gravitational force on it is still the same, and can still be calculated as $mg$.

Example 24: Changing frames of reference
Suppose we change from one frame reference into another, which is moving relative to the first one at a constant velocity \(u\). If an object of mass \(m\) is moving at velocity \(v\) (which need not be constant), then the effect is to change its momentum from \(mv\) in one frame to \(mv+mu\) in the other. Force is defined as the derivative of momentum with respect to time, and the derivative of a constant is zero, so adding the constant \(mu\) has no effect on the result. We therefore conclude that observers in different inertial frames of reference agree on forces.

Example 24: Changing frames of reference

Suppose we change from one frame reference into another, which is moving relative to the first one at a constant velocity $u$. If an object of mass $m$ is moving at velocity $v$ (which need not be constant), then the effect is to change its momentum from $mv$ in one frame to $mv+mu$ in the other. Force is defined as the derivative of momentum with respect to time, and the derivative of a constant is zero, so adding the constant $mu$ has no effect on the result. We therefore conclude that observers in different inertial frames of reference agree on forces.

Using the third law correctly

If you've already accepted Galilean relativity in your heart, then there is nothing really difficult about the first and second laws. The third law, however, is more of a conceptual challenge. The first hurdle is that it is counterintuitive. Is it really true that if a fighter jet collides with a mosquito, the mosquito's force on the jet is just as strong as the jet's force on the mosquito? Yes, it is true, but it is hard to believe at first. That amount of force simply has more of an effect on the mosquito, because it has less mass.

A more humane and practical experiment is shown in figure d. 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, so the upward force of the top magnet on the bottom magnet is of the same magnitude as the downward force of the bottom magnet on the top magnet.

d / Two magnets exert forces on each other.

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. As illustrated in figure e, the fallacy arises from assuming that we can add things that it doesn't make sense to add. It only makes sense to add up forces that are acting on the same object, whereas two forces related to each other by Newton's third law are always acting on two different objects. If two objects are interacting via a force and no other forces are involved, then both objects will accelerate --- in opposite directions, as shown in figure f!

e / It doesn't make sense for the man to talk about the woman's money canceling out his bar tab, because there is no good reason to combine his debts and her assets.

f / Newton's third law does not mean that forces always cancel out so that nothing can ever move. If these two ice skaters, initially at rest, push against each other, they will both move.

Here are some suggestions for avoiding misapplication of Newton's third law:

It always relates exactly two forces, not more.
The two forces involve exactly two objects, in the pattern A on B, B on A.
The two forces are always of the same type, e.g., friction and friction, or gravity and gravity.

Directions of forces

We've already seen that momentum, unlike energy, has a direction in space. Since force is defined in terms of momentum, force also has a direction in space. For motion in one dimension, we have to pick a coordinate system, and given that choice, forces and momenta will be positive or negative. We've already used signs to represent directions of forces in Newton's third law, $F_{AB}=-F_{BA}$.

g / A swimmer doing the breast stroke pushes backward against the water. By Newton's third law, the water pushes forward on him.

There is, however, a complication with force that we were able to avoid with momentum. If an object is moving on a line, we're guaranteed that its momentum is in one of two directions: the two directions along the line. But even an object that stays on a line may still be subject to forces that act perpendicularly to the line. For example, suppose a coin is sliding to the right across a table, h, and let's choose a positive $x$ axis that points to the right. The coin's motion is along a horizontal line, and its momentum is positive and decreasing. Because the momentum is decreasing, its time derivative $dp/dt$ is negative. This derivative equals the horizontal force of friction $F_1$, and its negative sign tells us that this force on the coin is to the left.

h / A coin slides across a table. Even for motion in one dimension, some of the forces may not lie along the line of the motion.

But there are also vertical forces on the coin. The Earth exerts a downward gravitational force $F_2$ on it, and the table makes an upward force $F_3$ that prevents the coin from sinking into the wood. In fact, without these vertical forces the horizontal frictional force wouldn't exist: surfaces don't exert friction against one another unless they are being pressed together.

To avoid mathematical complication, we want to postpone the full three-dimensional treatment of force and momentum until section 3.4. For now, we'll limit ourselves to examples like the coin, in which the motion is confined to a line, and any forces perpendicular to the line cancel each other out.

Discussion Questions

◊ Criticize the following incorrect statement:
“If an object is at rest and the total force on it is zero, it stays at rest. There can also be cases where an object is moving and keeps on moving without having any total force on it, but that can only happen when there's no friction, like in outer space.”

x (m)	t (s)
10	1.84
20	2.86
30	3.80
40	4.67
50	5.53
60	6.38
70	7.23
80	8.10
90	8.96
100	9.83

Discussion question B. ”

◊ The table gives laser timing data for Ben Johnson's 100 m dash at the 1987 World Championship in Rome. (His world record was later revoked because he tested positive for steroids.) How does the total force on him change over the duration of the race?”

◊ 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.”

◊ 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?”

3.2.3 What force is not

Violin teachers have to endure their beginning students' screeching. A frown appears on the woodwind teacher's face as she watches her student take a breath with an expansion of his ribcage but none in his belly. What makes physics teachers cringe is their students' verbal statements about forces. Below I have listed several dicta about what force is not.

Force is not a property of one object.

A great many of students' incorrect descriptions of forces could be cured by keeping in mind that a force is an interaction of two objects, not a property of one object.

Incorrect statement: “That magnet has a lot of force.”

If the magnet is one millimeter away from a steel ball bearing, they may exert a very strong attraction on each other, but if they were a meter apart, the force would be virtually undetectable. The magnet's strength can be rated using certain electrical units $(\text{ampere}-\text{meters}^2)$, but not in units of force.

Force is not a measure of an object's motion.

If force is not a property of a single object, then it cannot be used as a measure of the object's motion.

Incorrect statement: “The freight train rumbled down the tracks with awesome force.”

Force is not a measure of motion. If the freight train collides with a stalled cement truck, then some awesome forces will occur, but if it hits a fly the force will be small.

Force is not energy.

Incorrect statement: “How can my chair be making an upward force on my rear end? It has no power!”

Power is a concept related to energy, e.g., a 100-watt lightbulb uses up 100 joules per second of energy. When you sit in a chair, no energy is used up, so forces can exist between you and the chair without any need for a source of power.

Force is not stored or used up.

Because energy can be stored and used up, people think force also can be stored or used up.

Incorrect statement: “If you don't fill up your tank with gas, you'll run out of force.”

Energy is what you'll run out of, not force.

Forces need not be exerted by living things or machines.

Transforming energy from one form into another usually requires some kind of living or mechanical mechanism. The concept is not applicable to forces, which are an interaction between objects, not a thing to be transferred or transformed.

Incorrect statement: “How can a wooden bench be making an upward force on my rear end? It doesn't have any springs or anything inside it.”

No springs or other internal mechanisms are required. If the bench didn't make any force on you, you would obey Newton's second law and fall through it. Evidently it does make a force on you!

A force is the direct cause of a change in motion

I can click a remote control to make my garage door change from being at rest to being in motion. My finger's force on the button, however, was not the force that acted on the door. When we speak of a force on an object in physics, we are talking about a force that acts directly. Similarly, when you pull a reluctant dog along by its leash, the leash and the dog are making forces on each other, not your hand and the dog. The dog is not even touching your hand.

self-check:

Which of the following things can be correctly described in terms of force?

A nuclear submarine is charging ahead at full steam.
A nuclear submarine's propellers spin in the water.
A nuclear submarine needs to refuel its reactor periodically.

(answer in the back of the PDF version of the book)

Discussion Questions

◊ Criticize the following incorrect statement: “If you shove a book across a table, friction takes away more and more of its force, until finally it stops.”

◊ You hit a tennis ball against a wall. Explain any and all incorrect ideas in the following description of the physics involved: “The ball gets some force from you when you hit it, and when it hits the wall, it loses part of that force, so it doesn't bounce back as fast. The muscles in your arm are the only things that a force can come from.”

3.2.4 Forces between solids

Conservation laws are more fundamental than Newton's laws, and they apply where Newton's laws don't, e.g., to light and to the internal structure of atoms. However, there are certain problems that are much easier to solve using Newton's laws. As a trivial example, if you drop a rock, it could conserve momentum and energy by levitating, or by falling in the usual manner.⁵ With Newton's laws, however, we can reason that $a=F/m$, so the rock must respond to the gravitational force by accelerating.

i / Static friction: the tray doesn't slip on the waiter's fingers.

Less trivially, suppose a person is hanging onto a rope, and we want to know if she will slip. Unlike the case of the levitating rock, here the no-motion solution could be perfectly reasonable if her grip is strong enough. We know that her hand's interaction with the rope is fundamentally an electrical interaction between the atoms in the surface of her palm and the nearby atoms in the surface of the rope. For practical problem-solving, however, this is a case where we're better off forgetting the fundamental classification of interactions at the atomic level and working with a more practical, everyday classification of forces. In this practical scheme, we have three types of forces that can occur between solid objects in contact:

A normal force,F_n,	is perpendicular to the surface of contact, and prevents objects from passing through each other by becoming as strong as necessary (up to the point where the objects break). “Normal” means perpendicular.
Static friction,F_s,	is parallel to the surface of contact, and prevents the surfaces from starting to slip by becoming as strong as necessary, up to a maximum value ofF_s_,max. “Static” means not moving, i.e., not slipping.
Kinetic friction,F_k,	is parallel to the surface of contact, and tends to slow down any slippage once it starts. “Kinetic” means moving, i.e., slipping.

self-check:

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 put a coin on this page, which is horizontal, gravity pulls down on the coin, but the atoms in the paper and the coin repel each other electrically, and the atoms are compressed until the repulsion becomes strong enough to stop the downward motion of the coin. We describe this complicated and invisible atomic process by saying that the paper makes an upward normal force on the coin, and the coin makes a downward normal force on the paper. (The two normal forces are related by Newton's third law. In fact, Newton's third law only relates forces that are of the same type.)

If you now tilt the book a little, static friction keeps the coin from slipping. The picture at the microscopic level is even more complicated than the previous description of the normal force. One model is to think of the tiny bumps and depressions in the coin as settling into the similar irregularities in the paper. This model predicts that rougher surfaces should have more friction, which is sometimes true but not always. Two very smooth, clean glass surfaces or very well finished machined metal surfaces may actually stick better than rougher surfaces would, the probable explanation being that there is some kind of chemical bonding going on, and the smoother surfaces allow more atoms to be in contact.

Finally, as you tilt the book more and more, there comes a point where static friction reaches its maximum value. The surfaces become unstuck, and the coin begins to slide over the paper. Kinetic friction slows down this slipping motion significantly. In terms of energy, kinetic friction is converting mechanical energy into heat, just like when you rub your hands together to keep warm. One model of kinetic friction is that the tiny irregularities in the two surfaces bump against each other, causing vibrations whose energy rapidly converts to heat and sound --- you can hear this sound if you rub your fingers together near your ear.

For dry surfaces, experiments show that the following equations usually work fairly well:

\[\begin{equation*} F_{s,max} \approx \mu_sF_n , \end{equation*}\]

and

\[\begin{equation*} F_k \approx \mu_kF_n , \end{equation*}\]

where $\mu_s$, the coefficient of static friction, and $\mu_k$, the coefficient of kinetic friction, are constants that depend on the properties of the two surfaces, such as what they're made of and how rough they are.

self-check:

When a baseball player slides in to a base, is the friction static, or kinetic?
A mattress stays on the roof of a slowly accelerating car. Is the friction static, or kinetic?
Does static friction create heat? Kinetic friction?

(answer in the back of the PDF version of the book)

Example 25: Maximum acceleration of a car
\(\triangleright\) Rubber on asphalt gives \(\mu_k\approx0.4\) and \(\mu_s\approx 0.6\). What is the upper limit on a car's acceleration on a flat road, assuming that the engine has plenty of power and that air friction is negligible? \(\triangleright\) This isn't a flying car, so we don't expect it to accelerate vertically. The vertical forces acting on the car should cancel out. The earth makes a downward gravitational force on the car whose absolute value is \(mg\), so the road apparently makes an upward normal force of the same magnitude, \(F_n= mg\). Now what about the horizontal motion? As is always true, the coefficient of static friction is greater than the coefficient of kinetic friction, so the maximum acceleration is obtained with static friction, i.e., the driver should try not to burn rubber. The maximum force of static friction is \(F_{s,max}=\mu_{s} F_{n} =\mu_s mg\). The maximum acceleration is \(a= F_s/ m=\mu_{s} g\approx6\ \text{m}/\text{s}^2\). This is true regardless of how big the tires are, since the experimentally determined relationship \(F_{s,max}=\mu_{s} F_n\) is independent of surface area. j / Kinetic friction: the car skids.

Example 25: Maximum acceleration of a car

$\triangleright$ Rubber on asphalt gives $\mu_k\approx0.4$ and $\mu_s\approx 0.6$. What is the upper limit on a car's acceleration on a flat road, assuming that the engine has plenty of power and that air friction is negligible?

$\triangleright$ This isn't a flying car, so we don't expect it to accelerate vertically. The vertical forces acting on the car should cancel out. The earth makes a downward gravitational force on the car whose absolute value is $mg$, so the road apparently makes an upward normal force of the same magnitude, $F_n= mg$.

Now what about the horizontal motion? As is always true, the coefficient of static friction is greater than the coefficient of kinetic friction, so the maximum acceleration is obtained with static friction, i.e., the driver should try not to burn rubber. The maximum force of static friction is $F_{s,max}=\mu_{s} F_{n} =\mu_s mg$. The maximum acceleration is $a= F_s/ m=\mu_{s} g\approx6\ \text{m}/\text{s}^2$. This is true regardless of how big the tires are, since the experimentally determined relationship $F_{s,max}=\mu_{s} F_n$ is independent of surface area.

j / Kinetic friction: the car skids.

self-check:

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

(answer in the back of the PDF version of the book).

k / 1. The cliff's normal force on the climber's feet. 2. The track's static frictional force on the wheel of the accelerating dragster. 3. The ball's normal force on the bat.

Example 26: Locomotives
Looking at a picture of a locomotive, l, we notice two obvious things that are different from an automobile. Where a car typically has two drive wheels, a locomotive normally has many --- ten in this example. (Some also have smaller, unpowered wheels in front of and behind the drive wheels, but this example doesn't.) Also, cars these days are generally built to be as light as possible for their size, whereas locomotives are very massive, and no effort seems to be made to keep their weight low. (The steam locomotive in the photo is from about 1900, but this is true even for modern diesel and electric trains.) l / Example 26. 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.

Example 26: Locomotives

Looking at a picture of a locomotive, l, we notice two obvious things that are different from an automobile. Where a car typically has two drive wheels, a locomotive normally has many --- ten in this example. (Some also have smaller, unpowered wheels in front of and behind the drive wheels, but this example doesn't.) Also, cars these days are generally built to be as light as possible for their size, whereas locomotives are very massive, and no effort seems to be made to keep their weight low. (The steam locomotive in the photo is from about 1900, but this is true even for modern diesel and electric trains.)

l / Example 26.

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.

3.2.5 Fluid friction

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.

m / The wheelbases of the Hummer H3 and the Toyota Prius are surprisingly similar, differing by only 10%. The main difference in shape is that the Hummer is much taller and wider. It presents a much greater cross-sectional area to the wind, and this is the main reason that it uses about 2.5 times more gas on the freeway.

Discussion Question

◊ Criticize the following analysis: “A book is sitting on a table. I shove it, overcoming static friction. Then it slows down until it has less force than static friction, and it stops.”

3.2.6 Analysis of forces

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.

Example 27: A barge

A barge is being pulled 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’ forward normal forces on barge	barge’s backward normal force on ropes
water’s backward fluid friction force on barge	barge’s forward fluid friction force on water
planet earth’s downward gravitational force on barge	barge’s upward gravitational force on earth
water’s upward “floating” force on barge	barge’s downward “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 in the previous section. 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 missed 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.
Whenever one solid object touches another, there will be a normal force, and possibly also a frictional force; check for both.
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.

The following is another example in which we can profit by checking against our physical intuition for what should be happening.

Example 28: Rappelling

As shown in the figure below, Cindy is rappelling down a cliff. Her downward motion is at constant speed, and she takes little hops off of the cliff, as shown by the dashed line. Analyze the forces in which she participates at a moment when her feet are on the cliff and she is pushing off.

force acting on Cindy	force related to it by Newton’s third law
planet earth’s downward gravitational force on Cindy	Cindy’s upward gravitational force on earth
ropes upward frictional force on Cindy (her hand)	Cindy’s downward frictional force on the rope
cliff’s rightward normal force on Cindy	Cindy’s leftward normal force on the cliff

The two vertical forces cancel, which is what they should be doing if she is to go down at a constant rate. The only horizontal force on her is the cliff's force, which is not canceled by any other force, and which therefore will produce an acceleration of Cindy to the right. This makes sense, since she is hopping off. (This solution is a little oversimplified, because the rope is slanting, so it also applies a small leftward force to Cindy. As she flies out to the right, the slant of the rope will increase, pulling her back in more strongly.)

I believe that constructing the type of table described in this section is the best method for beginning students. Most textbooks, however, prescribe a pictorial way of showing all the forces acting on an object. Such a picture is called a free-body diagram. It should not be a big problem if a future physics professor expects you to be able to draw such diagrams, because the conceptual reasoning is the same. You simply draw a picture of the object, with arrows representing the forces that are acting on it. Arrows representing contact forces are drawn from the point of contact, noncontact forces from the center of mass. Free-body diagrams do not show the equal and opposite forces exerted by the object itself.

Discussion Questions

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

◊ 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?”

◊ A pool ball is rebounding from the side of the pool table. Analyze the forces in which the ball participates during the short time when it is in contact with the side of the table.”

◊ 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?”

3.2.7 Transmission of forces by low-mass objects

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.

n / If we imagine dividing a taut rope up into small segments, then any segment has forces pulling outward on it at each end. If the rope is of negligible mass, then all the forces equal $+T$ or $-T$, where $T$, the tension, is a single number.

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 29).

Example 29: Types of force made by ropes

$\triangleright$ Analyze the forces in figures o/1 and o/2.

$\triangleright$ 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!

o / Example 29. The forces between the rope and other objects are normal and frictional forces.

In o/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 Cindy	force related to it by Newton’s third law
planet earth’s downward gravitational force on Cindy	Cindy’s upward gravitational force on earth
ropes upward frictional force on Cindy (her hand)	Cindy’s downward frictional force on the rope
cliff’s rightward normal force on Cindy	Cindy’s leftward normal force on the cliff

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

In figure o/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 Cindy	force related to it by Newton’s third law
planet earth’s downward gravitational force on Cindy	Cindy’s upward gravitational force on earth
ropes upward frictional force on Cindy (her hand)	Cindy’s downward frictional force on the rope
cliff’s rightward normal force on Cindy	Cindy’s leftward normal force on the cliff

(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.

Discussion Question

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

3.2.8 Work

Energy transferred to a particle

To change the kinetic energy, $K=(1/2)mv^2$, of a particle moving in one dimension, we must change its velocity. That will entail a change in its momentum, $p=mv$, as well, and since force is the rate of transfer of momentum, we conclude that the only way to change a particle's kinetic energy is to apply a force.⁶ A force in the same direction as the motion speeds it up, increasing the kinetic energy, while a force in the opposite direction slows it down.

Consider an infinitesimal time interval during which the particle moves an infinitesimal distance $d{}x$, and its kinetic energy changes by $d{}K$. In one dimension, we represent the direction of the force and the direction of the motion with positive and negative signs for $F$ and $d{}x$, so the relationship among the signs can be summarized as follows: \begin{center}

F > 0	dx>0	dK>0
F < 0	dx<0	dK>0
F > 0	dx<0	dK<0
F < 0	dx>0	dK<0

\end{center} This looks exactly like the rule for determining the sign of a product, and we can easily show using the chain rule that this is indeed a multiplicative relationship:

\[\begin{align*} d{}K &= \frac{d{}K}{d{}v}\frac{d{}v}{d{}t}\frac{d{}t}{d{}x}d{}x \text{[chain rule]} \\ &= (mv)(a)(1/v)d{}x \\ & = m\,a\,d{}x \\ & = F\,d{}x \ \text{[Newton's second law]} \\ &\text{We can verify that force multiplied by distance has units of energy:} \\ &\text{N}\!\cdot\!\text{m} = \frac{\text{kg}\!\cdot\!\text{m}/\text{s}}{\text{s}}\times\text{m} \\ &= \text{kg}\!\cdot\!\text{m}^2/\text{s}^2 \\ &= \text{J} \end{align*}\]

Example 30: A TV picture tube
\(\triangleright\) At the back of a typical TV's picture tube, electrical forces accelerate each electron to an energy of \(5\times10^{-16}\ \text{J}\) over a distance of about 1 cm. How much force is applied to a single electron? (Assume the force is constant.) What is the corresponding acceleration? \(\triangleright\) Integrating \[\begin{align} d{} K &= Fd{} x , \quad \text{we find} \quad K_{f}- K_{i}= F( x_{f}- x_{i}) \\ &\text{or} \\ \Delta K &= F\Delta x . \\ &\text{The force is} F = \Delta K/\Delta x \\ &= \frac{5\times10^{-16}\ \text{J}}{ 0.01\ \text{m}} \\ & = 5\times10^{-14}\ \text{N} . \\ &\text{This may not sound like an impressive force,} \\ &\text{but it's enough to supply an electron with a spectacular acceleration.} \\ &\text{Looking up the mass of an electron on p. 943, we find} \quad a = F/ m \\ &= 5\times10^{16}\ \text{m}/\text{s}^2 . \end{align}\]

Example 30: A TV picture tube

$\triangleright$ At the back of a typical TV's picture tube, electrical forces accelerate each electron to an energy of $5\times10^{-16}\ \text{J}$ over a distance of about 1 cm. How much force is applied to a single electron? (Assume the force is constant.) What is the corresponding acceleration?

$\triangleright$ Integrating

\[\begin{align*} d{} K &= Fd{} x , \quad \text{we find} \quad K_{f}- K_{i}= F( x_{f}- x_{i}) \\ &\text{or} \\ \Delta K &= F\Delta x . \\ &\text{The force is} F = \Delta K/\Delta x \\ &= \frac{5\times10^{-16}\ \text{J}}{ 0.01\ \text{m}} \\ & = 5\times10^{-14}\ \text{N} . \\ &\text{This may not sound like an impressive force,} \\ &\text{but it's enough to supply an electron with a spectacular acceleration.} \\ &\text{Looking up the mass of an electron on p. 943, we find} \quad a = F/ m \\ &= 5\times10^{16}\ \text{m}/\text{s}^2 . \end{align*}\]

Example 31: An air gun
{ p / A simplified drawing of an airgun. \(\triangleright\) An airgun, figure p, uses compressed air to accelerate a pellet. As the pellet moves from \(x_1\) to \(x_2\), the air decompresses, so the force is not constant. Using methods from chapter 5, one can show that the air's force on the pellet is given by \(F= bx^\text{-7/5}\). A typical high-end airgun used for competitive target shooting has \[\begin{align} x_1 &= 0.046\ \text{m} , \\ x_2 &= 0.41\ \text{m} , \\ \text{and} b &= 4.4\ \text{N}\!\cdot\!\text{m}^\text{7/5} . \end{align}\] What is the kinetic energy of the pellet when it leaves the muzzle? (Assume friction is negligible.) \(\triangleright\) Since the force isn't constant, it would be incorrect to do \(F = \Delta K/\Delta x\). Integrating both sides of the equation \(d{} K= Fd{} x\), we have \[\begin{align} \Delta K &= \int_{ x_{1}}^{ x_{2}} Fd{} x \\ &= -\frac{5 b}{2}\left( x_2^\text{-2/5} - x_1^\text{-2/5}\right) \\ &= 22\ \text{J} \end{align}\]

Example 31: An air gun

{

p / A simplified drawing of an airgun.

$\triangleright$ An airgun, figure p, uses compressed air to accelerate a pellet. As the pellet moves from $x_1$ to $x_2$, the air decompresses, so the force is not constant. Using methods from chapter 5, one can show that the air's force on the pellet is given by $F= bx^\text{-7/5}$. A typical high-end airgun used for competitive target shooting has

\[\begin{align*} x_1 &= 0.046\ \text{m} , \\ x_2 &= 0.41\ \text{m} , \\ \text{and} b &= 4.4\ \text{N}\!\cdot\!\text{m}^\text{7/5} . \end{align*}\]

What is the kinetic energy of the pellet when it leaves the muzzle? (Assume friction is negligible.)

$\triangleright$ Since the force isn't constant, it would be incorrect to do $F = \Delta K/\Delta x$. Integrating both sides of the equation $d{} K= Fd{} x$, we have

\[\begin{align*} \Delta K &= \int_{ x_{1}}^{ x_{2}} Fd{} x \\ &= -\frac{5 b}{2}\left( x_2^\text{-2/5} - x_1^\text{-2/5}\right) \\ &= 22\ \text{J} \end{align*}\]

In general, when energy is transferred by a force,⁷ we use the term work to refer to the amount of energy transferred. This is different from the way the word is used in ordinary speech. If you stand for a long time holding a bag of cement, you get tired, and everyone will agree that you've worked hard, but you haven't changed the energy of the cement, so according to the definition of the physics term, you haven't done any work on the bag. There has been an energy transformation inside your body, of chemical energy into heat, but this just means that one part of your body did positive work (lost energy) while another part did a corresponding amount of negative work (gained energy).

Work in general

I derived the expression $Fd{}x$ for one particular type of kinetic-energy transfer, the work done in accelerating a particle, and then defined work as a more general term. Is the equation correct for other types of work as well? For example, if a force lifts a mass $m$ against the resistance of gravity at constant velocity, the increase in the mass's gravitational energy is $d{}(mgy)=mgd{}y=Fd{}y$, so again the equation works, but this still doesn't prove that the equation is always correct as a way of calculating energy transfers.

q / The black box does work by reeling in its cable.

Imagine a black box⁸, containing a gasoline-powered engine, which is designed to reel in a steel cable, exerting a certain force $F$. For simplicity, we imagine that this force is always constant, so we can talk about $\Delta{}x$ rather than an infinitesimal $d{}x$. If this black box is used to accelerate a particle (or any mass without internal structure), and no other forces act on the particle, then the original derivation applies, and the work done by the box is $W=F\Delta{}x$. Since $F$ is constant, the box will run out of gas after reeling in a certain amount of cable $\Delta{}x$. The chemical energy inside the box has decreased by $-W$, while the mass being accelerated has gained $W$ worth of kinetic energy.⁹

Now what if we use the black box to pull a plow? The energy increase in the outside world is of a different type than before; it takes the forms of (1) the gravitational energy of the dirt that has been lifted out to the sides of the furrow, (2) frictional heating of the dirt and the plowshare, and (3) the energy needed to break up the dirt clods (a form of electrical energy involving the attractions among the atoms in the clod). The box, however, only communicates with the outside world via the hole through which its cable passes. The amount of chemical energy lost by the gasoline can therefore only depend on $F$ and $\Delta{}x$, so it is the same $-W$ as when the box was being used to accelerate a mass, and thus by conservation of energy, the work done on the outside world is again $W$.

This is starting to sound like a proof that the force-times-distance method is always correct, but there was one subtle assumption involved, which was that the force was exerted at one point (the end of the cable, in the black box example). Real life often isn't like that. For example, a cyclist exerts forces on both pedals at once. Serious cyclists use toe-clips, and the conventional wisdom is that one should use equal amounts of force on the upstroke and downstroke, to make full use of both sets of muscles. This is a two-dimensional example, since the pedals go in circles. We're only discussing one-dimensional motion right now, so let's just pretend that the upstroke and downstroke are both executed in straight lines. Since the forces are in opposite directions, one is positive and one is negative. The cyclist's total force on the crank set is zero, but the work done isn't zero. We have to add the work done by each stroke, $W=F_1\Delta{}x_1+F_2\Delta{}x_2$. (I'm pretending that both forces are constant, so we don't have to do integrals.) Both terms are positive; one is a positive number multiplied by a positive number, while the other is a negative times a negative.

This might not seem like a big deal --- just remember not to use the total force --- but there are many situations where the total force is all we can measure. The ultimate example is heat conduction. Heat conduction is not supposed to be counted as a form of work, since it occurs without a force. But at the atomic level, there are forces, and work is done by one atom on another. When you hold a hot potato in your hand, the transfer of heat energy through your skin takes place with a total force that's extremely close to zero. At the atomic level, atoms in your skin are interacting electrically with atoms in the potato, but the attractions and repulsions add up to zero total force. It's just like the cyclist's feet acting on the pedals, but with zillions of forces involved instead of two. There is no practical way to measure all the individual forces, and therefore we can't calculate the total energy transferred.

To summarize, $\sum{F_jd{}x_j}$ is a correct way of calculating work, where $F_j$ is the individual force acting on particle $j$, which moves a distance $d{}x_j$. However, this is only useful if you can identify all the individual forces and determine the distance moved at each point of contact. For convenience, I'll refer to this as the work theorem. (It doesn't have a standard name.)

There is, however, something useful we can do with the total force. We can use it to calculate the part of the work done on an object that consists of a change in the kinetic energy it has due to the motion of its center of mass. The proof is essentially the same as the proof on p. 161, except that now we don't assume the force is acting on a single particle, so we have to be a little more delicate. Let the object consist of $n$ particles. Its total kinetic energy is $K=\sum_{j=1}^n{(1/2)m_jv_j^2}$, but this is what we've already realized can't be calculated using the total force. The kinetic energy it has due to motion of its center of mass is

\[\begin{align*} K_{cm} &= \frac{1}{2}m_{total}v_{cm}^2 . \end{align*}\]

r / The wheel spinning in the air has $K_{cm}=0$. The space shuttle has all its kinetic energy in the form of center of mass motion, $K=K_{cm}$. The rolling ball has some, but not all, of its energy in the form of center of mass motion, $K_{cm}\lt K$. (Space Shuttle photo by NASA)

Figure r shows some examples of the distinction between $K_{cm}$ and $K$. Differentiating $K_{cm}$, we have

\[\begin{align*} d{}K_{cm} &= m_{total}v_{cm}d{}v_{cm} \\ &= m_{total}v_{cm} \frac{d{}v_{cm}}{d{}t}\frac{d{}t}{d{}x_{cm}}d{}x_{cm} \text{[chain rule]} \\ &= m_{total}\frac{d{}v_{cm}}{d{}t}d{}x_{cm} \text{[$d{}t/d{}x_{cm}=1/v_{cm}$]} \\ &= \frac{d{}p_{total}}{d{}t}d{}x_{cm} \text{[$p_{total}=m_{total}v_{cm}$]} \\ &= F_{total}d{}x_{cm} \end{align*}\]

I'll call this the kinetic energy theorem --- like the work theorem, it has no standard name.

Example 32: An ice skater pushing off from a wall
The kinetic energy theorem tells us how to calculate the skater's kinetic energy if we know the amount of force and the distance her center of mass travels while she is pushing off. The work theorem tells us that the wall does no work on the skater, since the point of contact isn't moving. This makes sense, because the wall does not have any source of energy.

Example 32: An ice skater pushing off from a wall

The kinetic energy theorem tells us how to calculate the skater's kinetic energy if we know the amount of force and the distance her center of mass travels while she is pushing off.

The work theorem tells us that the wall does no work on the skater, since the point of contact isn't moving. This makes sense, because the wall does not have any source of energy.

Example 33: Absorbing an impact without recoiling?
\(\triangleright\) Is it possible to absorb an impact without recoiling? For instance, if a ping-pong ball hits a brick wall, does the wall “give” at all? \(\triangleright\) There will always be a recoil. In the example proposed, the wall will surely have some energy transferred to it in the form of heat and vibration. The work theorem tells us that we can only have an energy transfer if the distance traveled by the point of contact is nonzero.

Example 33: Absorbing an impact without recoiling?

$\triangleright$ Is it possible to absorb an impact without recoiling? For instance, if a ping-pong ball hits a brick wall, does the wall “give” at all?

$\triangleright$ There will always be a recoil. In the example proposed, the wall will surely have some energy transferred to it in the form of heat and vibration. The work theorem tells us that we can only have an energy transfer if the distance traveled by the point of contact is nonzero.

Example 34: Dragging a refrigerator at constant velocity
The fridge's momentum is constant, so there is no net momentum transfer, and the total force on it must be zero: your force is canceling the floor's kinetic frictional force. The kinetic energy theorem is therefore true but useless. It tells us that there is zero total force on the refrigerator, and that the refrigerator's kinetic energy doesn't change. The work theorem tells us that the work you do equals your hand's force on the refrigerator multiplied by the distance traveled. Since we know the floor has no source of energy, the only way for the floor and refrigerator to gain energy is from the work you do. We can thus calculate the total heat dissipated by friction in the refrigerator and the floor. Note that there is no way to find how much of the heat is dissipated in the floor and how much in the refrigerator.

Example 34: Dragging a refrigerator at constant velocity

The fridge's momentum is constant, so there is no net momentum transfer, and the total force on it must be zero: your force is canceling the floor's kinetic frictional force. The kinetic energy theorem is therefore true but useless. It tells us that there is zero total force on the refrigerator, and that the refrigerator's kinetic energy doesn't change.

The work theorem tells us that the work you do equals your hand's force on the refrigerator multiplied by the distance traveled. Since we know the floor has no source of energy, the only way for the floor and refrigerator to gain energy is from the work you do. We can thus calculate the total heat dissipated by friction in the refrigerator and the floor.

Note that there is no way to find how much of the heat is dissipated in the floor and how much in the refrigerator.

Example 35: Accelerating a cart
If you push on a cart and accelerate it, there are two forces acting on the cart: your hand's force, and the static frictional force of the ground pushing on the wheels in the opposite direction. Applying the work theorem to your force tells us how to calculate the work you do. Applying the work theorem to the floor's force tells us that the floor does no work on the cart. There is no motion at the point of contact, because the atoms in the floor are not moving. (The atoms in the surface of the wheel are also momentarily at rest when they touch the floor.) This makes sense, because the floor does not have any source of energy. The kinetic energy theorem refers to the total force, and because the floor's backward force cancels part of your force, the total force is less than your force. This tells us that only part of your work goes into the kinetic energy associated with the forward motion of the cart's center of mass. The rest goes into rotation of the wheels.

Example 35: Accelerating a cart

If you push on a cart and accelerate it, there are two forces acting on the cart: your hand's force, and the static frictional force of the ground pushing on the wheels in the opposite direction.

Applying the work theorem to your force tells us how to calculate the work you do.

Applying the work theorem to the floor's force tells us that the floor does no work on the cart. There is no motion at the point of contact, because the atoms in the floor are not moving. (The atoms in the surface of the wheel are also momentarily at rest when they touch the floor.) This makes sense, because the floor does not have any source of energy.

The kinetic energy theorem refers to the total force, and because the floor's backward force cancels part of your force, the total force is less than your force. This tells us that only part of your work goes into the kinetic energy associated with the forward motion of the cart's center of mass. The rest goes into rotation of the wheels.

Discussion Questions

◊ Criticize the following incorrect statement: “A force doesn't do any work unless it's causing the object to move.”

◊ To stop your car, you must first have time to react, and then it takes some time for the car to slow down. Both of these times contribute to the distance you will travel before you can stop. The figure shows how the average stopping distance increases with speed. Because the stopping distance increases more and more rapidly as you go faster, the rule of one car length per 10 m.p.h. of speed is not conservative enough at high speeds. In terms of work and kinetic energy, what is the reason for the more rapid increase at high speeds?

s / Discussion question B.

3.2.9 Simple Machines

Conservation of energy provided the necessary tools for analyzing some mechanical systems, such as the seesaw on page 85 and the pulley arrangements of the homework problems on page 120, but we could only analyze those machines by computing the total energy of the system. That approach wouldn't work for systems like the biceps/forearm machine on page 85, or the one in figure t, where the energy content of the person's body is impossible to compute directly. Even though the seesaw and the biceps/forearm system were clearly just two different forms of the lever, we had no way to treat them both on the same footing. We can now successfully attack such problems using the work and kinetic energy theorems.

Example 36: Constant tension around a pulley
t / The force is transmitted to the block. \(\triangleright\) In figure t, what is the relationship between the force applied by the person's hand and the force exerted on the block? \(\triangleright\) If we assume the rope and the pulley are ideal, i.e., frictionless and massless, then there is no way for them to absorb or release energy, so the work done by the hand must be the same as the work done on the block. Since the hand and the block move the same distance, the work theorem tells us the two forces are the same. Similar arguments provide an alternative justification for the statement made in section 3.2.7 that show that an idealized rope exerts the same force, the tension, anywhere it's attached to something, and the same amount of force is also exerted by each segment of the rope on the neighboring segments. Going around an ideal pulley also has no effect on the tension. This is an example of a simple machine, which is any mechanical system that manipulates forces to do work. This particular machine reverses the direction of the motion, but doesn't change the force or the speed of motion.

Example 36: Constant tension around a pulley

t / The force is transmitted to the block.

$\triangleright$ In figure t, what is the relationship between the force applied by the person's hand and the force exerted on the block?

$\triangleright$ If we assume the rope and the pulley are ideal, i.e., frictionless and massless, then there is no way for them to absorb or release energy, so the work done by the hand must be the same as the work done on the block. Since the hand and the block move the same distance, the work theorem tells us the two forces are the same.

Similar arguments provide an alternative justification for the statement made in section 3.2.7 that show that an idealized rope exerts the same force, the tension, anywhere it's attached to something, and the same amount of force is also exerted by each segment of the rope on the neighboring segments. Going around an ideal pulley also has no effect on the tension.

This is an example of a simple machine, which is any mechanical system that manipulates forces to do work. This particular machine reverses the direction of the motion, but doesn't change the force or the speed of motion.

Example 38: Inclined plane and wedge
v / An inclined plane. In figure v, the force applied by the hand is equal to the one applied to the load, but there is a mechanical advantage compared to the force that would have been required to lift the load straight up. The distance traveled up the inclined plane is greater by a factor of 1/sin \(\theta\), so by the work theorem, the force is smaller by a factor of sin \(\theta\), and we have M.A.=1/sin \(\theta\). The wedge, w, is similar. w / A wedge.

Example 38: Inclined plane and wedge

v / An inclined plane.

In figure v, the force applied by the hand is equal to the one applied to the load, but there is a mechanical advantage compared to the force that would have been required to lift the load straight up. The distance traveled up the inclined plane is greater by a factor of 1/sin $\theta$, so by the work theorem, the force is smaller by a factor of sin $\theta$, and we have M.A.=1/sin $\theta$. The wedge, w, is similar.

w / A wedge.

Example 39: Archimedes' screw
In one revolution, the crank travels a distance \(2\pi{} b\), and the water rises by a height \(h\). The mechanical advantage is \(2\pi{} b/ h\). x / Archimedes' screw

Example 39: Archimedes' screw

In one revolution, the crank travels a distance $2\pi{} b$, and the water rises by a height $h$. The mechanical advantage is $2\pi{} b/ h$.

x / Archimedes' screw

3.2.10 Force related to interaction energy

In section 2.3, we saw that there were two equivalent ways of looking at gravity, the gravitational field and the gravitational energy. They were related by the equation $d{}U=mgd{}r$, so if we knew the field, we could find the energy by integration, $U=\int{mgd{}r}$, and if we knew the energy, we could find the field by differentiation, $g=(1/m)d{}U/d{}r$.

The same approach can be applied to other interactions, for example a mass on a spring. The main difference is that only in gravitational interactions does the strength of the interaction depend on the mass of the object, so in general, it doesn't make sense to separate out the factor of $m$ as in the equation $d{}U=mgd{}r$. Since $F=mg$ is the gravitational force, we can rewrite the equation in the more suggestive form $d{}U=Fd{}r$. This form no longer refers to gravity specifically, and can be applied much more generally. The only remaining detail is that I've been fairly cavalier about positive and negative signs up until now. That wasn't such a big problem for gravitational interactions, since gravity is always attractive, but it requires more careful treatment for nongravitational forces, where we don't necessarily know the direction of the force in advance, and we need to use positive and negative signs carefully for the direction of the force.

In general, suppose that forces are acting on a particle --- we can think of them as coming from other objects that are “off stage” --- and that the interaction between the particle and the off-stage objects can be characterized by an interaction energy, $U$, which depends only on the particle's position, $x$. Using the kinetic energy theorem, we have $d{}K=Fd{}x$. (It's not necessary to write $K_{cm}$, since a particle can't have any other kind of kinetic energy.) Conservation of energy tells us $d{}K+d{}U=0$, so the relationship between force and interaction energy is $d{}U=-Fd{}x$, or

\[\begin{equation*} F = -\frac{d{}U}{d{}x} \text{[relationship between force and interaction energy]} . \end{equation*}\]

Example 40: Force exerted by a spring
\(\triangleright\) A mass is attached to the end of a spring, and the energy of the spring is \(U=(1/2) kx^2\), where \(x\) is the position of the mass, and \(x=0\) is defined to be the equilibrium position. What is the force the spring exerts on the mass? Interpret the sign of the result. \(\triangleright\) Differentiating, we find \[\begin{align} F &= -\frac{d{} U}{d{} x} \\ &= - kx . \end{align}\] If \(x\) is positive, then the force is negative, i.e., it acts so as to bring the mass back to equilibrium, and similarly for \(x\lt0\) we have \(F>0\). Most books do the \(F=- kx\) form before the \(U=(1/2) kx^2\) form, and call it Hooke's law. Neither form is really more fundamental than the other --- we can always get from one to the other by integrating or differentiating.

Example 40: Force exerted by a spring

$\triangleright$ A mass is attached to the end of a spring, and the energy of the spring is $U=(1/2) kx^2$, where $x$ is the position of the mass, and $x=0$ is defined to be the equilibrium position. What is the force the spring exerts on the mass? Interpret the sign of the result.

$\triangleright$ Differentiating, we find

\[\begin{align*} F &= -\frac{d{} U}{d{} x} \\ &= - kx . \end{align*}\]

If $x$ is positive, then the force is negative, i.e., it acts so as to bring the mass back to equilibrium, and similarly for $x\lt0$ we have $F>0$.

Most books do the $F=- kx$ form before the $U=(1/2) kx^2$ form, and call it Hooke's law. Neither form is really more fundamental than the other --- we can always get from one to the other by integrating or differentiating.

Example 41: Newton's law of gravity
\(\triangleright\) Given the equation \(U=- Gm_1 m_{2}/ r\) for the energy of gravitational interactions, find the corresponding equation for the gravitational force on mass \(m_2\). Interpret the positive and negative signs. \(\triangleright\) We have to be a little careful here, because we've been taking \(r\) to be positive by definition, whereas the position, \(x\), of mass \(m_2\) could be positive or negative, depending on which side of \(m_1\) it's on. For positive \(x\), we have \(r= x\), and differentiation gives \[\begin{align} F &= -\frac{d{} U}{d{} x} \\ &= - Gm_1 m_{2}/ x^2 . \end{align}\] As in the preceding example, we have \(F\lt0\) when \(x\) is positive, because the object is being attracted back toward \(x=0\). When \(x\) is negative, the relationship between \(r\) and \(x\) becomes \(r=- x\), and the result for the force is the same as before, but with a minus sign. We can combine the two equations by writing \[\begin{equation} \| F\| = Gm_1 m_{2}/ r^2 , \end{equation}\] and this is the form traditionally known as Newton's law of gravity. As in the preceding example, the \(U\) and \(F\) equations contain equivalent information, and neither is more fundamental than the other.

Example 41: Newton's law of gravity

$\triangleright$ Given the equation $U=- Gm_1 m_{2}/ r$ for the energy of gravitational interactions, find the corresponding equation for the gravitational force on mass $m_2$. Interpret the positive and negative signs.

$\triangleright$ We have to be a little careful here, because we've been taking $r$ to be positive by definition, whereas the position, $x$, of mass $m_2$ could be positive or negative, depending on which side of $m_1$ it's on.

For positive $x$, we have $r= x$, and differentiation gives

\[\begin{align*} F &= -\frac{d{} U}{d{} x} \\ &= - Gm_1 m_{2}/ x^2 . \end{align*}\]

As in the preceding example, we have $F\lt0$ when $x$ is positive, because the object is being attracted back toward $x=0$.

When $x$ is negative, the relationship between $r$ and $x$ becomes $r=- x$, and the result for the force is the same as before, but with a minus sign. We can combine the two equations by writing

\[\begin{equation*} | F| = Gm_1 m_{2}/ r^2 , \end{equation*}\]

and this is the form traditionally known as Newton's law of gravity. As in the preceding example, the $U$ and $F$ equations contain equivalent information, and neither is more fundamental than the other.

Example 42: Equilibrium
I previously described the condition for equilibrium as a local maximum or minimum of $U$. A differentiable function has a zero derivative at its extrema, and we can now relate this directly to force: zero force acts on an object when it is at equilibrium.

Contributors

Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.