## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 11

### Forces and Newton”s Laws

**IN THIS CHAPTER**

**Summary** : A force is a push or a pull applied by one object on another object. This chapter describes the construction and use of free-body diagrams, which are key to approaching problems involving forces.

**Definitions**

A **force** is a push or a pull applied by one object and experienced by another object.

The **net force** on an object is the single force that could replace all the individual forces acting on an object and produce the same effect. Forces acting in the same direction add together to determine the net force; forces acting in opposite directions subtract to determine the net force.

**Weight** is the force of a planet on an object near that planet.

The **force of friction** is the force of a surface on an object. The friction force acts parallel to the surface. **Kinetic friction** is the friction force when something is moving along the surface and acts opposite the direction of motion. **Static friction** is the friction force between two surfaces that aren”t moving relative to one another.

The **normal force** is also the force of a surface on an object. The normal force acts perpendicular to the surface.

The **coefficient of friction** is a number that tells how sticky two surfaces are.

**Newton”s third law** says that the force of Object A on Object B is equal in amount and opposite in direction to the force of Object B on Object A.

**Newton”s second law** states that an object”s acceleration is the net force it experiences divided by its mass, and is in the direction of the net force.

**Describing Forces: Free-Body Diagrams**

A force is a push or a pull applied by one object and experienced by another object. A force in the laboratory is often measured by a spring scale, as in the preceding picture. In AP Physics 1 we have to understand two aspects of forces. First, we have to describe the application of the force: What are the objects involved, and how much force is applied and in which direction? Next, we have to connect the net force acting on an object to that object”s change in velocity.

Start with correct language: an object can “experience” a force, but an object cannot “have” a force. Don”t let yourself say, “Ball A has a bigger force than Ball B”—that means nothing. “The net force on Ball A is bigger than on Ball B” is fine, as is “The Earth pulls harder on Ball A than on Ball B.”

The canonical method of describing forces acting on an object is to draw a free-body diagram. A free-body diagram should include two elements:

(1) A labeled arrow representing each force, with each arrow beginning on the object and pointing in the direction in which the force acts

(2) A list of all the forces acting on the object, indicating the object applying the force and the object experiencing the force

On the AP Exam you”ll be asked something like, “Draw and label the forces (not components) that act on the car as it slows down.” This means “Draw a free-body diagram.”

**FACT:** Only gravitational and electrical forces can act on an object without contact.^{ }^{1}

**Example 1:** A car moving to the right on the freeway applies the brakes and skids to a stop.

Pretty much always start with the force of the Earth on the object, which is commonly known as its weight. Don”t call this force “gravity”—that”s an ambiguous term. Weight acts downward and doesn”t require any contact with the Earth in order to exist.^{ }^{2}^{ }Draw a downward arrow on the dot, label it “weight,” and in the list write “Weight: force of Earth on the car.”^{ }^{3}

Any other forces must be a result of contact with the car. What”s the car touching? It”s touching just the ground. Since the car is touching the ground, the ground exerts a normal force perpendicular to the surface. Draw an upward arrow on the dot, label it something like “*F* _{n} ,” and in the list write “*F* _{n} : The force of the ground on the car.”^{ }^{4}

Since the car is sliding along the ground, the ground exerts a force of (kinetic) friction. By definition, kinetic friction must always act in the opposite direction of motion—the car skids right along the ground, the friction force acts to the left. Draw a leftward arrow on the dot, label it “*F* _{f} ,” and in the list write “*F* _{f} : The force of the ground on the car.”^{ }^{5}

The car is not in contact with anything else, so we”re done.

*Whoa there* ! The car is moving to the right, so what about the force of its motion?

There”s no such thing as the “force of motion.” All forces must be exerted by an identifiable object; and all nongravitational and nonelectrical forces must be a result of contact. The car is not in contact with anything that pushes the car forward.

Then how is the car moving to the right?

It just is. It is critically important to focus *only* on the problem as stated. Questions about what happened before the problem started are irrelevant. Perhaps at first the car was pushed by the engine, or pulled by a team of donkeys, to start it moving; perhaps it had been in motion since the beginning of time. It doesn”t matter. All that matters is that when we tune in to the action, the car is moving right and slowing down.

If the car had been pulled by a team of donkeys to start it moving, wouldn”t we put the force of the donkeys on the car on the free-body diagram?

No, because the free-body diagram includes only forces that act *now* , not forces that acted earlier, or forces that will act in the future. If donkeys pulled the car, the force of the donkeys would appear on the free-body while the donkeys were actually pulling. After they let go and the car is slowing down, the donkeys might as well have never existed.

While it”s important to learn how to draw a free-body diagram, it”s just as important to learn how to *stop* drawing a free-body diagram. Don”t make up forces. Unless you can clearly identify the source of the force, don”t include the force.

**Exam Tip from an AP Physics Veteran**

If you see a problem involving forces, try drawing a free-body diagram for each object in the problem, or for a system including multiple objects. A free-body diagram will always be useful, even if you”re not explicitly asked to make one.

**Determining the Net Force**

To determine the net force on an object, treat each direction separately. Add forces that point in the same direction; subtract forces that point in opposite directions. Or, if you know the acceleration in a direction, use *F* _{net} = *ma* .

**FACT:** When an object moves along a surface, the acceleration in a direction perpendicular to that surface must be zero. Therefore, the net force perpendicular to the surface is also zero.

In Example 1, the net force horizontally is equal to the force of friction, because that”s the only force acting in the horizontal direction—there”s no other force to add or subtract. Vertically, the net force is equal to the normal force minus the weight. But since the car is moving along the surface, the vertical acceleration and the vertical net force on the car are zero.

We can conclude, then, that the normal force on the car is equal to the car”s weight. This isn”t a general fact, though—a normal force is *not* always equal to an object”s weight. If more vertical forces are acting, or if the surface is changing speed vertically (as in an elevator), the normal force can be different from the weight.

**FACT:** The kinetic friction force is equal to the coefficient of kinetic friction times the normal force.

A good AP question might describe a second car, identical in mass and initial speed to the car in Example 1, but on a wet freeway. The question might ask you to explain why this second car skids to a stop over a longer distance.

The coefficient of friction is a property of the surfaces in contact. Here, since a wet road is less “sticky” than a dry road, the coefficient of friction has decreased. But since the second car is identical to the first, its weight and thus the normal force of the surface on the car is the same as before. Therefore, by the equation *F* _{f} = *μF* _{n} , the new car experiences a smaller force of friction.

With a smaller net force on the second car, its acceleration is also smaller by *F* _{net} = *ma* . Then the distance traveled during the skid depends on the car”s acceleration by the kinematics equation (3), . Take the final speed *v* _{f}_{ }to zero and solve for Δ*x* to see that acceleration *a* is in the denominator of the equation. Thus, a smaller acceleration means a larger distance to stop.

**Static and Kinetic Friction**

You may have learned that the coefficient of friction takes two forms: **static** and **kinetic** friction. Use the coefficient of static friction if something is stationary, and the coefficient of kinetic friction if the object is moving. The equation for the force of friction is essentially the same in either case: *F* _{f}_{ }= *μF* _{N}_{ }.

The only strange part about static friction is that the coefficient of static friction is a *maximum* value. Think about this for a moment—if a book just sits on a table, it doesn”t need any friction to stay in place. But that book won”t slide if you apply a very small horizontal pushing force to it, so static friction can act on the book. To find the maximum coefficient of static friction, find out how much horizontal pushing force will just barely cause the book to move; then use *F* _{f}_{ }= *μF* _{N}_{ }.

**Newton”s Third Law**

**FACT:** The force of Object A on Object B is equal in amount and opposite in direction to the force of Object B on Object A. These two forces, which act on different objects, are called Newton”s third law companion forces.^{ }^{6}

In Example 1, then, what”s the Newton”s third law companion force to the normal force? It”s tempting to say, “Oh, the weight.” After all, the weight is equal to the normal force and is opposite in direction to the normal force. But that”s wrong.

To find the companion force, look at the description of the force in the free-body diagram, and reverse the objects applying and experiencing the force. The normal force is the force of the ground on the car, and that acts upward. Therefore, the third law companion force is the force of the car on the ground, acting downward.

**FACT:** If the net force has both a vertical and a horizontal component, use the Pythagorean theorem to determine the magnitude of the net force, and use the tangent function to determine the direction of the net force.

In most AP problems, though, the net force will be zero in one or both directions. In Example 1, the magnitude of the net force is equal to only the magnitude of the friction force, because the vertical forces must subtract to zero.

If this were an AP problem, chances are it would ask about the connection between the net force and the change in the object”s speed. We”ll revisit this example later.

**Forces at Angles**

A force at an angle is drawn on a free-body diagram just like any other force. But when you”re ready to do any analysis on the free-body diagram, start by breaking the angled force into components.

**Example 2:** A pair of fuzzy dice is hanging by a string from your rearview mirror, as shown in the preceding figure. You speed up from a stoplight. During the acceleration, the dice do not move vertically; the string makes an angle of *θ* = 22° with the vertical. The dice have mass 0.10 kg.

No matter what this problem ends up asking, you

What about the force of the car on the dice?

What about it? The car is not in contact with the dice, the string is. It”s the string, not the car, applying the force to the dice.

**Determining the Net Force**

It”s likely that you”d be asked to determine the net force on these dice. But the tension acts both up *and* to the right. How do you deal with the vertical and horizontal forces, then?

**FACT:** When a force acts at an angle *θ* measured from the horizontal:

- The vertical component of that force is equal to the amount of the force itself times sin
*θ*. - The horizontal component of that force is equal to the amount of the force itself times cos
*θ*.

Before you do any further work to find the net force, break all individual forces into horizontal and vertical components as best as you can. Here, the weight is already vertical. The tension becomes two separate components: *T*sin 68° goes in the vertical direction, and *T* cos 68° goes in the horizontal direction. Now you”re ready to answer any possible problem.

**Exam Tip from an AP Physics Veteran**

Do *not* put force components on the same diagram as the force itself. You won”t earn full credit. First, draw all forces at whatever angle is appropriate. Then, on a *separate diagram* , redraw the forces with the angled forces broken into components.

**FACT:** The acceleration of an object is *F* _{net} /*m* . This is the same thing as saying

*Whoa* . I get that the net force is the horizontal *T* cos68°, and that I can write that *T* cos68° = (0.1 kg)(*a* ). But the problem didn”t give me an acceleration, it didn”t give the tension—I”m stuck to solve for anything. The College Board screwed this problem up, right?

It”s vanishingly unlikely that the problem is unsolvable as posed. You obviously can”t ask questions of the College Board during the AP Exam. So if you”re absolutely sure the exam is screwed up. you can just state where you think the problem is unclear, make up the information you need, and do your best. Chances are, though, that you need to find a creative alternate way to solve the problem.

Look at the problem statement: It said that the dice have a mass of 0.10 kg. This means that the weight of the dice is 1.0 N.^{ }^{7}^{ }Since the dice are moving only in a horizontal direction, vertical acceleration (and the vertical net force on the dice) must be zero. Forces in opposite directions subtract to determine the net force. Here, that means that the up force must equal the down force of 1.0 N. The up force is *T* sin 68°, which equals 1.0 N. Plug in from your calculator that the sine of 68 degrees is 0.93, and then solve to find the tension is 1.1 N.

Now deal with the horizontal direction. The horizontal net force is *T* cos 68°, which is (1.1 N)(0.37) = 0.41 N. Since there”s no vertical net force, 0.41 N to the right is the entire net force. And then *a* = *F* _{net} /*m* , so acceleration is 0.41 N/0.10 kg = 4.1 m/s per second. That”s pretty much everything there is to calculate.

**The Mistake**

It”s tempting to use this equation in all sorts of circumstances. For example, I”m sitting in a chair. Since I”m near the Earth, the force of the Earth on me is equal to my weight of 930 N. I know my mass is 93 kg. Use *a* = *F* _{net} /*m* . My acceleration must be 930 N/93 kg, or 10 m/s per second.

Um, no. I”m sitting in a chair. My speed isn”t changing, so my acceleration is zero. The value of 10 m/s per second of acceleration means I”m in free fall. What went wrong?

I experience more forces than just the force of the Earth, of course. The chair is pushing up on me. Since I know that my acceleration is zero, the chair pushes up on me with 930 N of force. Now the up force and the down force on me subtract to zero. Phew.

**Only the net force equals mass times acceleration.** Never set a force equal to

*ma*unless it”s the net force.

**What Else Could You Be Asked?**

The AP Exam doesn”t like to ask for calculations. So what else could be asked relating to Example 2?

Here”s one thought: If the dice were to instead hang from a bigger angle than 22° from the vertical, would the tension go up, go down, or stay the same?

The best way to answer this type of question is to make the calculation, and then explain what part of the calculation leads to the correct answer. That”s the whole method behind answering questions that involve qualitative-quantitative translation, as discussed in __Chapter 8__ .

Make up a bigger angle: call it 60° from the vertical. (Or choose any number; just make a significant difference in the new situation. Don”t choose 23°.) Start back from the beginning: The dice still have a 0.10-kg mass, and the weight of the dice is still 1.0 N. The vertical acceleration is still zero, which means we can set the up force equal to the down force. But now the up force has changed, from *T* sin 68° to *T* sin 30°.^{ }^{8}^{ }Now we set *T* sin 30° equal to 1.0 N, giving a tension of 2.0 N.

The answer, then, is that the tension increases. The weight remains the same and the vertical component of tension must stay the same, but since to calculate tension we end up dividing the 1.0-N weight by the sine of the angle from the horizontal, a smaller angle from the horizontal gives a bigger tension.

**Exam Tip from an AP Physics Veteran**

If you are asked whether something increases, decreases, or stays the same, you might want to start by making a calculation to see numerically what happens to the answer. Be sure to explain *why* the calculation came out the way it did.

**Inclined Planes**

Treat objects on inclines the same as any other objects. Draw a free-body diagram, break angled forces into components, and use *a* = *F* _{net}_{ }*/m* in each direction. The only major difference is that you don”t use horizontal and vertical components for the forces. Instead, you look separately at the forces parallel to the incline and at the forces perpendicular to the incline.

Any normal force will be perpendicular to the incline, and so won”t have to be broken into components as long as the object is moving up or down the incline. Any friction force will be parallel to the incline and so won”t have to be broken into components. It”s the weight—the force of the Earth—that will be broken into components.

**FACT:** On an incline of angle *θ* (measured from the horizontal), break the weight into components:

- The component of the weight that is parallel to the incline is equal to the weight times sin
*θ*. - The component of the weight that is perpendicular to the incline is equal to the weight times cos
*θ*.

Example problems and extra drills on this frequently tested topic are available in __Chapter 18__ .

**Multiple Objects**

When two masses are connected over a pulley, it”s often easiest to start by considering both objects as a single system. Draw the free-body diagram for the entire system, and use *a* = *F* _{net} /*m* to find the acceleration of the system. Then, if you need to find the tension in the connecting rope, or if you need to talk about just one of the two connected objects, draw a new free-body diagram just for that object.

**FACT:** One rope has just one tension.^{ }^{9}

An alternative approach is to start by drawing two separate free-body diagrams, one for each object. Write *F* _{net} = *ma* for each object separately. Then, recognizing that the tension is the same in each equation, solve algebraically for the acceleration and tension.

** Practice Problems**

** Note: **Extra drills on problems including ropes and inclined planes can be found in

__Chapter 18__.

** 1 .** A 7.0-N block sits on a rough surface. It is being pulled by a force

**F**at an angle

_{1}*θ*= 30° above the horizontal, as shown above. The block is initially moving to the right with speed 5 m/s. The coefficient of friction between the block and the surface is

*μ*= 0.20. Justify all answers.

(a) Is it possible for the block to be slowing down? If so, give a possible value of the magnitude of *F* _{1} that would allow the block to slow down. If not, explain why not with reference to Newton”s second law.

(b) In order to double the block”s initial speed to 10 m/s, how must the magnitude of the force *F* _{1} change?

(A) It must double.

(B) It must quadruple.

(C) It does not have to change.

** 2 .** A drag-racing car speeds up from rest to 22 m/s in 2 s. The car has mass 800 kg; the driver has mass 80 kg.

(a) Calculate the acceleration of the drag racer.

(b) Calculate the net force on the drag racer.

(c) Which experiences a greater net force?

(A) The driver

(B) The car

(C) Both the driver and the car experience the same net force.

** 3 .** A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?

(A) The normal force must be equal to the car”s weight.

(B) The normal force must be less than the car”s weight.

(C) The normal force must be greater than the car”s weight.

(D) The normal force must be zero.

** 4 .** Bert, Ernie, and Oscar are discussing the gas mileage of cars. Specifically, they are wondering whether a car gets better mileage on a city street or on a freeway. All agree (correctly) that the gas mileage of a car depends on the force that is produced by the car”s engine—the car gets fewer miles per gallon if the engine must produce more force. Whose explanation is completely correct?

**Bert says:** Gas mileage is better on the freeway. In town the car is always speeding up and slowing down because of the traffic lights, so because *F* _{net} = *ma* and acceleration is large, the engine must produce a lot of force. However, on the freeway, the car moves with constant velocity, and acceleration is zero. So the engine produces no force, allowing for better gas mileage.

**Ernie says:** Gas mileage is better in town. In town, the speed of the car is slower than the speed on the freeway. Acceleration is velocity divided by time, so the acceleration in town is smaller. Because *F* _{net} = *ma* , then, the force of the engine is smaller in town giving better gas mileage.

**Oscar says:** Gas mileage is better on the freeway. The force of the engine only has to be enough to equal the force of air resistance—the engine doesn”t have to accelerate the car because the car maintains a constant speed. Whereas in town, the force of the engine must often be greater than the force of friction and air resistance in order to let the car speed up.

** Solutions to Practice Problems**

** 1 .** (a) When an object slows down, its acceleration (and therefore the net force it experiences) is opposite the direction of its motion. Here, the motion is to the right, so if the net force is left, it will slow down. To make the net force to the left, choose a value of

*F*

_{1}such that the rightward component

*F*

_{1}cos(30°) is less than the friction force acting left.

Choosing a value is a bit tricky: the value of the friction force itself depends on *F* _{1} , because the normal force on the block is 7 N minus the vertical component of *F* _{1} . Try choosing an *F* _{1} much smaller than the block”s weight, like 1 N. Then the normal force on the block is (7 N) – (1 N)sin 30° = 6.5 N. The friction force becomes *μF* _{n} = 1.3 N. The component of *F* _{1} pulling right is 0.9 N, so the net force will be to the left as required.

(Any value of *F* _{1} that”s less than a bit over 1.4 N will work here. Try it.)

(b) While the net force is related to acceleration, the net force has no effect on an object”s speed. Beyond that, no one has said anything about what happens before the problem, about how that initial speed came about. The forces can all be as indicated, and the object can have any initial speed.

** 2 .** (a) The car”s speed changes by 22 m/s in 2 s. So the car changes its speed by 11 m/s in 1 s, which is what is meant by an acceleration of 11 m/s per second.

(b) Newton”s second law says that the net force on the racer is the drag racer”s mass of 800 kg times the 11 m/s per second acceleration. That gives a net force of 8,800 N.

(c) The driver and the car must experience the same acceleration because they move together; when the car changes its speed by 11 m/s in one second, so does the driver.^{ }^{10}^{ }To calculate the net force on the driver, the driver”s 80-kg mass must be used in Newton”s second law, *F* _{net} = *ma* . With the same *a* and a smaller mass, the driver experiences a smaller net force (and the car experiences a greater net force).

** 3 .** (B) The normal force exerted on an object on an inclined plane equals

*mg*(cos

*θ*), where

*θ*is the angle of the incline. If

*θ*is greater than 0, then cos

*θ*is less than 1, so the normal force is less than the object”s weight.

** 4 .** Although Bert is right that acceleration is zero on the freeway, this means that the

*net*force is zero; the engine still must produce a force to counteract air resistance. This is what Oscar says, so his answer is correct. Ernie”s answer is way off—acceleration is not velocity/time, acceleration is a

*change*in velocity over time.

** Rapid Review**

- Only gravitational and electrical forces can act on an object without contact (in AP Physics 1).
- When an object moves along a surface, the acceleration in a direction perpendicular to that surface must be zero. Therefore, the net force perpendicular to the surface is also zero.
- The friction force is equal to the coefficient of friction times the normal force,
*F*_{f}=*μF*_{n}. - The force of Object A on Object B is equal in amount and opposite in direction to the force of Object B on Object A. These two forces, which act on different objects, are called Newton”s third law companion forces.
- If the net force has both a vertical and a horizontal component, use the Pythagorean theorem to determine the magnitude of the net force, and use the tangent function to determine the direction of the net force.
- When a force acts at an angle
*θ*measured from the horizontal: - The vertical component of that force is equal to the amount of the force itself times sin
*θ*. - The horizontal component of that force is equal to the amount of the force itself times cos
*θ*. - The acceleration of an object is
*F*_{net}/*m*(which is the same thing as saying*F*_{net}=*ma*). - On an incline of angle
*θ*(measured from the horizontal), break the weight into components: - The component of the weight that is parallel to the incline is equal to the weight times sin
*θ*. - The component of the weight that is perpendicular to the incline is equal to the weight times cos
*θ*. - One rope has just one tension.

__ ^{1}__ In AP Physics 1, anyway.

^{2}__Chapter 15__ , on gravitation, explains more about how to find the weight of an object in a gravitational field.

__ ^{3}__ Why not call it “force of gravity on the car?” Well, because all forces must be exerted by an

*object*on another object. Since when is “gravity” an object?

__ ^{4}__ It makes no difference what you label the arrow, as long as you define the label in a list. You want to call it

*N*instead of

*F*

_{n}? Be my guest.

__ ^{5}__ Yes, both the friction force and the normal force are properly listed as the force of the ground on the car.

__ ^{6}__ Or, sometimes, this is called a Newton”s third law force pair.

__ ^{7}__ On the Earth, 1 kg of mass weighs 10 N. This fact is discussed in more detail in

__Chapter 15__, Gravitation.

__ ^{8}__ Remember we had to measure from the horizontal according to the fact on the previous page.

__ ^{9}__ This is true unless the rope is tied to or connected over a mass. For example, if the pulley itself had mass, then the rope can have different tensions on each side of the pulley. But that”s a rare happening, and that certainly shouldn”t require any calculation.

__ ^{10}__ Otherwise, the driver would fall out of the car.