Chapter 2 Newtonian Mechanics I

Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2 2-1

Chapter 2 Newtonian Mechanics I

! Why do seatbelts and airbags save lives?

! If you stand on a bathroom scale in a moving elevator, does its reading change?

! Can a parachutist survive a fall if the parachute does not open?

Make sure you know how to:

1. Draw a motion diagram for a moving object.

2. Determine the direction of acceleration using a motion diagram.

3. Add vectors graphically and by components for one-dimensional motion.

CO: The idea for safety belts followed very soon after the invention of the automobile. On

February 10, 1885, Edward J. Claghorn of New York City was granted United States Patent

#312,085 for a safety-belt for tourists. The patent said the invention was “designed to be applied

to the person, and provided with hooks and other attachments for securing the person to a fixed

object.” In the early 1950s, Volvo became the first car company to feature safety belts for all of

its cars. In 1968 Allen Breed submitted the first patent for airbags, and in 1973 General Motors

included air bags in some of their Chevrolet models. The main purpose of the bags was to prevent

seated passengers from flying forward and hitting the steering wheel or dashboard during a

vehicle crash. Airbags combined with seat belts significantly reduce the risk of injury (belts alone

by 40% and belts with airbags even more). They save about 250,000 lives worldwide every year.

How do seat belts and airbags provide this protection?

Lead: In the last chapter, we learned to describe motion—for example, to determine a car’s

acceleration when stopped abruptly during a collision. However, we did not discuss the causes of

the acceleration. In this chapter, we will learn why an object has a particular acceleration. This

knowledge will help us explain the motion of different objects: cars, car passengers, elevators,

skydivers, and even rockets!

2.1 Describe and represent interactions

In Chapter 1 we learned to represent an object’s motion using motion diagrams,

kinematics graphs, and equations. We learned to describe the motion of objects moving at

constant velocity, and objects speeding up and slowing down at a constant rate (constant

acceleration). Why do these different types of motion occur? A simple experiment allows us to

have a preliminary answer to this question. Imagine standing on rollerblades on a horizontal

floor. No matter how hard you swing your arms or legs you cannot start moving by yourself. In

order to start moving, you need to either push off the floor or have someone push or pull you.

Physicists say that the floor or this other person interact with you. It looks like interactions affect


motion. Objects can interact directly when they touch each other, or at a distance – you probably

observed a magnet attract or repel another magnet without directly touching it.

Sketching a process

When physicists want to analyze a process, they start by sketching it. For example,

consider a process in which a car skids to avoid a collision with a van. Where will the car stop

relative to the van? To analyze the process, we start with a sketch of the process and focus on an

object of interest in the sketch. In this case the object of interest is the car. We include the initial

and final states. (Fig. 2.1) The initial state is either when the car’s driver sees the van for the first

time or when the driver applies the brakes (the difference may be important and often depends on

the wording of the problem). The final state involves the location of the car when it stops and its

position relative to the van. We do not include in the sketch any other parts of the world.

Figure 2.1Sketch of process

Choosing a system

After sketching the initial and final states of the process, we choose an object of interest.

For the process involving the car and the van, we choose the car since our concern is the distance

it travels relative to the van before stopping. We call this object of interest the system. All other

objects that are not part of the system can interact with it (touch it, pull it, and push it) and are

said to be in the system’s environment. For example, the surface of the road is an external object

in the car’s environment that touches the car tires. Interactions between objects in the

environment and the system object are called external interactions.

System A system is the object of interest that we choose to analyze. Everything outside that

system is called the environment and consists of objects that might interact with the system

(touch, push, or pull it) and affect its motion. These are external interactions.

To show the system of interest you choose for analysis on the sketch, make a light

boundary (a closed dashed line) around the system object to emphasize the system choice (see the

example in Fig. 2.1). In this chapter we work with the systems that consist of only one object. In

later chapters we choose systems with more than one object. Sometimes, a single system object

has parts—like the wheels on a car and its axles. The parts interact with each other. Since both

parts are in the system, these are called internal interactions. In this chapter we will model an

object like a car as a point-like object and ignore such internal interactions.


Representing interactions

To understand how external interactions affect the motion of a system object, we need to

represent these interactions. When you hold a basketball in one hand, and a bowling ball in the

other (Fig. 2.2a), you feel a difference. To represent and analyze this difference, we choose one

ball as the system object and study it; then we do the same for the other ball. What are the objects

interacting with each ball? Our hand has to push up hard to keep the bowling ball steady (Fig.

2.2b) and pushes up much less to keep the basketball steady (Fig. 2.2c). We represent this upward

push exerted by each hand on a ball with an arrow. Notice that the arrow is longer for the

interaction of the Hand on the Bowling Ball than for the Hand on the Basketball. In physics we

say that the arrow represents “a force” that is exerted on the ball. A force is a physical quantity

that characterizes how hard and in what direction an external object pushes or pulls on the system

object. The symbol for the force is F!

. It has two subscripts indicating what external object exerts

the force and on what system object the force is exerted. In the case of your hand (external object)

pushing on the bowling ball (the system object), the full symbol is H on BBF!

. The force that the

hand exerts on the basketball is labeled H on BF!

. The arrow above the symbol indicates that force

is a vector quantity with a magnitude and a direction. The unit of a force is called 1 Newton (1N).

When you hold a 100-g ball steady, you push up on it exerting a force that is a little less than 1.0

N. We do not yet have a formal definition for force as we do not yet have a method for measuring

forces.

Figure 2.2(a)(b)(c) Representing interactions

Do any other objects interact with the balls and exert forces on them? Intuitively you

might think that there should be something pulling down to balance the upward force your hands

exert on the balls. What is this something? You might be thinking “Gravity”. On Earth, the word

gravity represents the interaction of Earth as a whole (the planet) with the ball (on other planets

“gravity” means the pull of that other planet). Earth pulls downward on an object toward Earth’s

center. Because of this interaction we need to include a second arrow to represent the force that

Earth exerts on the ball ( E on BF!

). How long should this arrow be? Intuitively you probably feel

that if there were no other objects interacting with the balls, then the two arrows for each ball


should be of the same length, as the ball is not moving anywhere (see Figs. 2.2d and e). Do other

objects besides the hand and Earth interact with the ball?

Figure 2.2 (d)(e)

Air surrounds everything close to the earth. Does it push down or up on the balls? Let’s

hypothesize that the air does push down. We can represent this push with a downward arrow

( A on BBF!

or A on BF!

). If air does push down, then our hands have to push up harder to balance the

combined effect of the downward push of the air and the downward pull of Earth on the balls

(Fig. 2.3). How can we test the hypothesis that air pushes down on the balls?

Figure 2.3 Possible effect of air on balls

Testing a hypothesis

To test a hypothesis in science means to first accept it as a true statement (even if we

disagree with it); then to design an experiment whose outcome we can predict using this

hypothesis (a testing experiment); then compare the outcome of the experiment and the

prediction; and finally, make a preliminary judgment about the hypothesis. If the outcome

matches the prediction, we can say that the hypothesis has not been disproved by the experiment.

When this happens, our confidence in the hypothesis increases. If the outcome does not match the


prediction, we could conclude that the hypothesis has been disproved. However, we should not

jump to that conclusion too quickly—there can be many reasons why an outcome does not match

the prediction. Perhaps there was something wrong with the way the experiment was performed;

or perhaps our prediction involved an unreasonable assumption. In any case, when the outcome

and prediction are inconsistent, there is a possibility that we need to reconsider the hypothesis and

perhaps reject it.

To test the hypothesis that the air exerts a downward force on objects, we use the

following experiment. Attach a ball to a spring and let it hang; the spring stretches. Place the ball

and spring inside a large jar that is connected to a vacuum pump (Fig. 2.4a). If the air inside the

jar does indeed push down on the ball, the force that the spring exerts on the ball has to balance

both the downward gravitational force that Earth exerts on the ball and the downward force that

the air exerts on the ball (as in Fig. 2.3). When we pump the air out of the jar it should be easier to

support the ball—the spring should stretch less. To summarize – if the air pushes down on the

ball, then the spring should stretch less when some of the air is pumped out. In the previous

sentence the statement after the word if is our hypothesis; the statement after the word then is the

prediction of the outcome of the experiment based on the hypothesis. When we do the

experiment, the spring actually stretches slightly more when the air is pumped out of the jar (Fig.

2.4b). Evidently the air was helping support the ball—exerting an upward force on the ball. This

contradicts our hypothesis. This outcome is surprising. In Chapter 10, you will learn the

mechanism by which air pushes up on objects.

Figure 2.4

It is important to reflect on what we have done here. We formulated an initial

hypothesis—air pushes down on objects. We tested that idea by using it to make a prediction

about what would happen in an experiment. We then performed the experiment and found that

something completely different happened. We revised our hypothesis—air pushes up slightly on

objects. We also learned that air’s upward push on an object (at least in our experiment) is very

small; thus, for many situations, this particular effect of air on objects can be ignored.


Force diagrams

A force diagram (sometimes called a free-body diagram) is a diagrammatic

representation of a system object and the forces that objects in its environment exert on it (see

Figs. 2.2d and e). The object is represented by a dot to show that we model it as a point-like

object. Arrows represent the forces. Unlike the motion diagram the force diagram does not show

us how the situation changes with time; thus we need to be aware of the instant for which the

diagram is constructed. For situations when no motion occurs, this makes no difference. But

when motion does occur, we need think if the force diagram is changing as the object moves.

Here’s how to construct a force diagram. The situation is a rock while sinking into sand

(making a small crater) after being dropped from above. We construct the diagram for one instant

of the rock’s motion. This instant is after the rock touched the sand but before it completely

stopped.

Reasoning Skill: Constructing a Force Diagram A falling rock sinking into sand.

Notice that on the force diagram, the arrow representing the force exerted by the sand on the rock

is longer than the arrow representing the force exerted by Earth on the rock. The difference in

lengths reflects the difference in the magnitudes of the forces. At this point in our study, we

cannot justify this choice and need simply to include in the diagram arrows for all of the forces

exerted on the system object.

Tip! Remember that on the force diagram, you only draw forces exerted ON the system object.

Do not draw forces that the system object exerts on other objects! For example, the rock exerts a

force on the sand, but we do not include this force in the force diagram since the sand is not part

of the system.

Conceptual Exercise 2.1 Force diagram for a book Book A sits on a table with book B on top

of it. Construct a force diagram for book A.


Sketch and Translate The situation is sketched in (Fig.

2.5a). We choose book A as the system object of interest.

Notice that the dashed line around book A passes between

the table and book A, and between book B and book A. It’s

important to be precise in the way you draw this line so that

the separation between the system and the environment is

clear. In this example, Earth, the table and book B are

external environmental objects that exert forces on book A. Figure 2.5(a)

Simplify and Diagram Represent book A by a dot and draw a

force arrow for each force being exerted on book A by an

object in the environment (Fig. 2.5b). Two objects touch book

A. The table pushes up on the bottom surface of the book

exerting a force T on AF!

and book B pushes down on the top

surface of book A exerting force B on AF!

. In addition, Earth

exerts a downward force on book A E on AF!

. We assume that

the air exerts a zero force on book A. Figure 2.5(b)

Try It Yourself: Construct a force diagram for book A if another book C is placed on top of book

B.

Answer: The same three objects interact with book A. Earth exerts the same downward force on

book A ( E on AF!

). C does not directly touch A and exerts no force on A. However, C does push

down on B, so B exerts a greater force on A ( B on AF!

). Because the downward force of B on A is

greater, the table has to exert a greater upward force on book A ( T on AF

!

), See Fig. 2.6.

Figure 2.6 Force diagram for A with books B and C on top


How can a table that is not an animate object exert a force on book A? Recall the forces

exerted on a bowling ball when you held it in your hand. You obviously felt that you had to push

up on it to prevent it from falling. A table also “feels” that it has to push up on the book to

prevent it from falling through the table. An object does not have to be alive to exert a force on

another object.

Normal force

Another important point here is that the force that the table exerted on book A in this

example and the force that book B exerted on book A were both perpendicular to the surfaces of

the objects that exerted the forces – the surface of the table and the surface of book B. Those

forces are sometimes called “normal” forces, which in geometry means “perpendicular” and in

the future will be labeled using the letter N instead of F . Normal does not mean vertical,

although in this example they were vertical forces. Normal forces are contact forces as opposed to

the force exerted by Earth on books. In this chapter we will encounter many different contact

forces but only one non-contact force – gravitational force exerted by Earth as a planet.

Review Question 2.1

You slide toward the right at decreasing speed on a horizontal wooden floor. Choose yourself as

the system and list external objects that interact with and exert forces on you.

2.2 Adding and measuring forces

Most often, more than one environmental object exerts a force on a system object. How

can we add them to find the total or net force exerted on the system object? Suppose, for example,

that you are lifting a suitcase straight up. Earth pulls down on the suitcase exerting a force of

E on S 100 NF " and you exert an upward force of Y on S 150 NF " (Fig. 2.7a). What is the

magnitude and direction of the total force exerted on the suitcase? Intuitively you probably feel

that the result of those two forces exerted on the suitcase is a force of 50 N pointed up. Why?

Remember, that force is a vector. To add two vectors to find the sum E on S Y on SF F F# " $! ! !

, we

need to add them head to tail (see Fig.2.7b). As we see, the resultant vector is upward and its

length is 50 N. If we use the upward y direction as positive, then the sum of the forces has a y

component 50 Ny

F# " $ . If the positive direction is down, the sum of the forces has a y

component 50 Ny

F# " % . The skill box below shows you how to add forces graphically.


Figure 2.7 Sum of force exerted on suitcase

Skill box: Adding forces graphically 1 on O 2 on OF F F# " $! ! !

When several external objects in the environment exert forces on the system object and

when the forces point in different directions, we can still use the above vector addition method to

find the sum of the forces exerted on the object:

on O 1 on O 2 on O on O...n n

F F F F" $ $ $&! ! ! !

(2.1)

Suppose, for example, that three tug boats have cables attached to a barge carrying coal. Fig. 2.8a

A top view force diagram for the coal barge is shown in Fig. 2.8b. The tail-to-head diagrammatic

method for adding the forces to find the sum is illustrated in Fig. 2.8c. Later in this chapter and in

the next, we learn a mathematical way to add forces.


Figure 2.8 Adding force vectors

Tip! The sum of the force vectors is not a new force being exerted. Rather, it is the combined

effect of all the forces being exerted on the object. Because of this, the sum vector should never

be included in the force diagram for that object.

Measuring force magnitudes

Force is a vector physical quantity with a magnitude (how big is the force) and a

direction. In the next conceptual exercise, we develop a method for determining the magnitude of

a force.

Conceptual Exercise 2.2 Measuring forces A light hanging spring with a light plastic bag on the

end is not stretched. We place one golf ball in the bag and observe that the spring starts changing

length until it settles on a new length. We add a second ball and observe that the spring stretches

twice as far before it stops stretching. We add a third ball and observe that the spring stretches

three times as far. Explain how this experiment can be used to develop a method to measure the

magnitude of a force.

Sketch and Translate First, draw sketches to represent the three situations (Fig. 2.9a). On each

sketch, carefully show the change in the length of the spring. Choose the bag with the golf balls

as a system and analyze the forces exerted in each case. We focus on the instant when the spring

stops stretching in each case.

Figure 2.9(a) Calibrating a spring to measure force magnitude


Simplify and Diagram Assume that Earth exerts the same force on each ball

E on 1BallF!

independently of the presence of other balls. Thus, the total force exerted by Earth on the

three balls is three times greater than the force exerted on one ball. Draw a force diagram for the

system for each case (Fig. 2.9b). Assume that in each case the spring exerts a force on the balls

S on BallsF!

that is equal in magnitude and opposite in direction to the force that Earth exerts on the

balls E on BallsF!

so that the sum of the forces exerted on the balls is zero.

Figure 2.9(b) Calibrating a spring to measure force magnitude

Therefore, one method to measure an unknown force that some object exerts on a system

of interest is to calibrate a spring in terms of some standard force, such as the force that Earth

exerts on one golf ball. Then if some unknown force is exerted on a system object, we can use the

spring to exert a balancing force on that object. Then, the unknown force is equal in magnitude to

the force exerted by the spring and opposite in direction. In this case, we would be measuring

force in units equal to Earth’s pull on one golf ball. We could use some other standard object to

calibrate our spring scale.

Try It Yourself: Represent the relation between the force that the spring exerts on the bag with

golf balls system ( S on BallsF ) and spring stretch ( y ) with a -vs-F y graph. Draw a trend line.

Answer See Fig. 2.9c. Based on the graph’s trend line we can say that the spring elongates till the

force it exerts on the object balances the force that Earth exerts on it.

Figure 2.9(c)


In Conceptual Exercise 2.2 we devised a method to measure the force that

Earth exerts on a system by balancing it with the force exerted by a spring. In the Do It

Yourself part, we found that the force exerted by the spring increases in direct

proportion to the stretch. Thus, we can use any spring to balance a known standard

force (1 N or approximately the force that Earth exerts on a 100-gram object) and then

calibrate this spring in newtons by placing marks at equal stretch distances as you pull

on its end with increasing force. We thus build a spring scale – the simplest instrument

to measure forces (Fig. 2.10).

Figure 2.10

A spring scale

Force and language

Force is a word that is used in everyday life in many different ways. In physics it means

something very specific—a physical quantity that characterizes an interaction between two

objects—its direction and intensity. For a force to exist, there should be two objects that interact.

In real life a hug is a good analogy for a force—it takes two people for a hug. This analogy

highlights the most important attribute of force – two participating objects.

It took physicists a long time to agree on what to call a force. Even 300 years ago the

meaning of this word was different. It stood for something that a moving object carried with it –

the faster or heavier object carried a bigger force. Later, physicists agreed to call this property of

moving objects a different name (momentum) and did NOT call it a force. However, in everyday

language the idea that force resides in an object remains very strong; people say: “may the force

be with you”.

The word weight is often used in physics to denote the force that Earth as a planet exerts

on an object (or that the Moon or Mars exerts on the object if the object is on the Moon or on

Mars). In everyday speech we use the word very differently. For example, many people worry

about their weight. We think little about Earth as a planet when we say “I lost weight during this

hiking trip”. Expressions like these make weight sound like it belongs to an object and is not the

force that one object exerts on another object.

The word tension is another example. “There’s a lot of tension in my neck.” “This

climbing rope can withstand 2000 lb of tension.” Again, in everyday speech, tension often refers

to something an object has. In physics tension means the pulling force that a string (or rope,

cable, etc.) exerts on an object.

Since all interactions involve two objects, we will be careful in this book to always

identify those objects when speaking about any force. For an interaction to occur in a mechanical

system, the objects should be touching each other (with the exception of Earth or some other

planet, which can interact gravitationally at a distance). Remember, if you are thinking about a


force that is exerted on a moving object and you cannot find another object that interacts with it to

produce this force, then you are thinking of something else, not force.

Review Question 2.2

A book bag is partially supported while hanging from a spring scale and also by sitting on a

platform scale. The platform scale reads about 36 units of force and the spring scale reads about

28 units of force. What is the magnitude of the force that Earth exerts on the bag?

2.3 Conceptual relationship between force and motion

When we drew a force diagrams for a bowling ball held by a person, we intuitively drew

the forces exerted on the ball as being equal in magnitude. What if the person had to throw a ball

upward? Or slowly lift it upward? Or if she had to catch the ball falling from above? Would she

still need to exert a force on the ball of magnitude equal to that Earth exerts on the ball? In other

words, is there a relationship between the forces that are exerted on an object and the way the

object moves? Let’s do three simple experiments with a bowling ball to see if there is a pattern

between its motion and the forces exerted on it. In all cases, we assume that the friction force

exerted by the surface and by the surrounding air on the ball is very small. Since both the surface

and the bowling ball are hard and smooth and the ball moves relatively slowly, these are

reasonable assumptions. Consider the observations, analysis, and pattern in Observational

Experiment Table 2.2.

ALG 1.3.5


Observational Experiment Table 2.1 How are motion and forces related?

Observation experiment

Analysis

1. A bowling ball rolls on

a very hard, smooth surface; it does not slow

down.

Motion diagram

Force diagrams for first and third positions

2. A ruler lightly pushes the rolling bowling ball opposite the ball’s direction of motion.

Motion diagram


3. A ruler lightly pushes the rolling bowling ball in the direction of its motion.

Motion diagram


Pattern

! In all the experiments, the vertical forces add to zero and cancel each other. The sum of the forces is the horizontal direction.

! In the first experiment, the sum of the forces force exerted on the ball is zero; the ball’s velocity remains constant.

! In the second and third experiments, when the ruler pushes the ball, the v'!

arrow points in the same

direction as the sum of the forces.

! Summary: The change in velocity v'!

in all experiments is in the same direction as the sum of the

forces. Notice that there is no pattern relating the direction of the velocity v!

to the direction of the sum

of the forces. In experiment 2 they are in opposite directions, but in experiment 3 they were in the same direction.

Using the motion and force diagrams in the experiments in Table 2.1, we found that the v'!

arrow on the motion diagram for a system object and the sum of the forces F#!

that external

objects exert on it are in the same direction. If the sum of the forces is zero, the object’s velocity

does not change; its acceleration is zero.

This idea (hypothesis) contradicts our common sense beliefs. It seems that objects we

push or pull (grocery carts, cars, wagons or sleds) always move in the direction in which we push

or pull them, in other words, that it is the velocity of objects that is always in the same direction as


the sum of the forces exerted on them. Although experiment 2 in the table above contradicts this

idea, it seems like a plausible hypothesis worth additional testing.

Testing hypotheses about force and motion

We have two hypotheses concerning the relationship between motion and force:

Hypothesis 1: An object’s velocity v!

always points in the direction of the sum of the forces

F#!

that other objects exert on it.

Hypothesis 2: An object’s velocity change v'!

always points in the direction of the sum of

the forces F#!

that other objects exert on it.

To test these hypotheses, we use each hypothesis to predict the outcome of one or more

experiments. Then, we perform the experiments and compare the outcomes with the predictions

we have already made. From this comparison, we decide if we can reject one or both of the

hypotheses being tested. We’ll do this for the three experiments in Testing Experiment Table 2.2.

Testing Experiment Table 2.2 Testing ideas of how velocity and the sum of the forces F#!

are

related

Testing Experiments Prediction based on hypothesis

1 that an object’s velocity is in

the direction of the sum of the

forces.

Prediction based on

hypothesis 2 that an

object’s change in

velocity is in the

direction of the sum

of the forces.

Outcome

An air hockey puck moves freely on a friction-less surface. What happens to it if the table is very long?

The sum of forces is zero; thus the velocity should be zero and the puck should stop immediately.

The sum of forces is zero so the puck’s velocity should not change; it should continue moving the same way.

The puck continues to move at constant velocity.

Taisha on rollerblades coasts to the right. Her friend, who is behind her, pulls back lightly on a rope attached to Taisha. Predict what happens to Taisha’s velocity.

The sum of forces exerted on Taisha points to the left. Thus Taisha’s velocity should immediately change from right to left.

As the sum of forces points toward the left,

the v'!

arrow on

Taisha’s motion diagram should point left. Taisha should continue moving to the right slowing down until she stops.

Taisha slows down and eventually stops.

You throw a ball up. What happens to the ball after it leaves your hand?

The only force exerted on the ball after it leaves your hand points down.

The only force exerted on the ball after it leaves your hand points down. The

The ball moves up at decreasing speed; then


Thus the ball should immediately begin moving downward after you release it.

v'!

arrow on the

motion diagram should point down, and the ball should slow down until it stops and then start moving back down at increasing speed.

reverses direction and starts moving downward.

Conclusion

! All outcomes contradict the prediction based on hypothesis 1—we can reject it.

! All outcomes are consistent with the prediction based on hypothesis 2. This does not mean it is necessarily true, but does mean our confidence in the idea increases.

Recall from Chapter 1 that the direction of the v'!

arrow in a motion diagram is in the same

direction as the object’s acceleration a!

. Thus, based on this idea and these testing experiments,

we can now assert hypothesis 2 with greater confidence.

Relating forces and motion The velocity change arrow v'!

in an object’s motion diagram (and

its acceleration a!

) point in the same direction as the sum of the forces that other objects exert on

it. If the sum of the forces points in the same direction as the system object’s velocity, the object

speeds up; if it is in the opposite direction, it slows down. If the sum of the forces is zero, the

object continues with no change in velocity.

Reflect

We have devised a rule that allows us to explain why objects slow down or speed up or

continue at constant velocity. The steps we used were as follows.

! Observe: We observed moving objects.

! Simplify: The objects were modeled as point-like objects and the motion was considered to

occur along a line.

! Describe: The motion was described using kinematics quantities and relationships between

them. These descriptions did not help us understand why the motion occurred.

! Explain: We developed the idea of force as an interaction between two objects and used it to

decide why the motion occurred as it did. We had two hypotheses relating the sum of the forces

exerted on an object and quantities describing its motion.

! Test: We used these hypotheses to make predictions about the outcomes of new experiments

and then performed the experiments to see if the predictions were correct. We were able to

reject one hypothesis and built confidence in the other.

The process resembles in some ways the processes that physicists use to acquire

knowledge. Their work is often more complicated and involves sidetracks. When making

simplifying assumptions about objects or interactions, physicists sometimes neglect features that

are later found to be important. Sometimes testing experiments fail because one or more

simplifying assumptions used in making the prediction were not reasonable.


It took physicists two thousand years to figure out things that you will learn in one year!

We will try to help you to avoid major sidetracks. However, we will try to preserve the spirit of

the process so that you can learn physics by using the processes important in the practice of

physics and of all types of science.

Before you proceed, think again why it is common to relate the sum of the forces exerted

on an object to its velocity and not to the change in its velocity (its acceleration). For example,

people often think that a moving object “has” a force, like the force of your hand continuing to

push up on a ball after it has left your hand so that the sum of the forces points up in the direction

of the velocity. Another reason for believing the sum of the forces is in the direction of the

velocity could be our everyday focus on instances when objects start moving from rest. In this

case they do indeed move in the direction of the sum of the forces. Thus, there can be at least two

reasons why your intuition conflicts with contemporary physics beliefs: the language that is used

(force as an entity) and the generalization of one example to other situations. Make sure that each

time you learn a new idea, you try to reconcile your intuition with that idea.

Review Question 2.3

Why do you need to keep pushing a grocery cart in a store in order to keep it moving?

2.4 Reasoning without mathematical equations

In this section you will learn how to use motion and force diagrams to reason qualitatively

about physical processes—for example, to decide the relative magnitudes of forces if information

about motion is provided or to decide about possible velocity changes if information about forces

is provided. The key here is to make sure that different representations of the process are

consistent with each other. This reasoning method is illustrated in the following conceptual

exercises. In the next exercise, we know enough about the motion of a person to make a motion

diagram and can then use it to answer a question about the relative magnitudes of forces that

external objects exert on the person.

Conceptual Exercise 2.3 Man lands in a pile of leaves A man slips off the roof of a house and

lands in a deep pile of leaves. Is the force that the leaves exert on the man greater than, equal to,

or less than the gravitational force that Earth exerts on him while he is sinking 0.5 m into the

leaves?

Sketch and Translate The process is represented by a sketch in 2.11a. When the man reached the

top of the leaves, he was moving medium fast. After sinking 0.5 m into the leaves, his motion had

stopped. Thus his speed was decreasing while he was sinking deeper into the leaves. Now that we

know roughly how to describe his motion, we can use that information to analyze the forces

exerted on him while he was sinking into the leaves.

ALG 1.4.1-1.4.7


Figure 2.11(a)

Simplify and Diagram We can now make a motion diagram that is a rough description of his

motion while sinking into the leaves (2.11b). The velocity change v'!

arrow points up since his

downward speed is decreasing. This means that the sum of the forces must also point up.

Choosing the man as the object of interest, we see that two external objects interact with him

while he is sinking into the leaves. Earth exerts a downward gravitational force E on MF!

and the

leaves exert an upward force L on MF!

that opposes his downward motion (Fig. 2.11c). Since the

velocity change v'!

arrow in the motion diagram points up, the sum of the forces exerted on him

as he is slowing down must also point up. Thus, independently of what instant of the motion we

choose to draw the force diagram, the magnitude of the force that the leaves exert on the man

during the sinking motion must be greater than the magnitude of the force that Earth exerts on

him ( L on M E on MF F( ), as shown in Fig. 2.11c.

Figure 2.11(b)(c) Compare forces that objects exert on man

Try It Yourself: Suppose that your friend when answering this question said that the magnitudes

of the forces must be the same ( L on M E on MF F" ). How might you convince your friend that your

answer ( L on M E on MF F( ) is correct?

Answer: You could agree with your friend and assume that he/she is correct. Then ask him or her

to construct a motion diagram that is consistent with this equal magnitude force idea and see if

this type of motion actually occurs. With equal magnitude forces, the sum of the forces must be

zero ( 0F# "!

). Thus, on the motion diagram the velocity change arrow must also be zero

( 0v' "!

). If the falling person is already moving when he encounters the leaves, he will continue

moving at constant velocity and will not stop—inconsistent with what happens.


In the next conceptual exercise, we are given information about the forces that external

objects exert on a woman and asked to answer a question about her motion. Try hard to answer

the question yourself before looking at the solution.

Conceptual Exercise 2.4 Force diagram jeopardy The force diagram in Fig. 2.12 describes the

forces that external objects are exerting on a woman (the force diagram does not change with

time). Invent three contextually different types of motion that are consistent with the force

diagram.

Figure 2.12 What might the woman be doing?

Sketch and Translate Two equal-magnitude oppositely directed forces are being exerted on the

woman ( 0F# "!

). Thus, a motion diagram for the woman must have a zero velocity change

( 0v' "!

).

Simplify and Diagram Three possible motions consistent with this idea are:

(1) She stands at rest on a level horizontal surface (Fig. 2.13a).

(2) She on roller blades glides at constant velocity on a smooth horizontal surface (Fig.

2.13b).

(3) She stands on the floor of an elevator that moves up at constant velocity (Fig. 2.13c).

Note that in all three of the above, the velocity change arrow is zero. This is consistent with the

sum of the forces being zero.

Figure 2.13 Woman’s motion consistent with motion diagram

Try It Yourself: Suppose the elevator in (Fig. 2.13c) above was moving up at decreasing speed

instead of at constant speed. How then would the force diagram in Fig. 2.12 be different?

Answer: A velocity change v'!

arrow for her motion would now point down opposite the

direction of her velocity. Thus, the sum of the forces F#!

that other objects exert on her must also

ALG 1.4.1-1.4.7


point down. This means the magnitude of the upward force Floor on WF!

that the elevator floor

exerts on her must now be less than the magnitude of the downward force E on WF!

that Earth

exerts on her ( Floor on W E on WF F) ). Notice that the magnitudes of forces do not have arrows over

them (the magnitudes are positive numbers and have no direction).

Let us summarize the skills you should have learned so far:

a) How to construct a motion diagram or how to interpret an already constructed motion diagram.

b) How to draw a force diagram or how to interpret an already constructed force diagram.

c) How to check the consistency of the motion diagram and the force diagram.

d) How to construct a motion diagram if you have a force diagram, and vice versa.

e) You want then to be able to use the diagrams to reason qualitatively about physical processes

and later to use the diagrams to evaluate quantitative answers to problems to see if the answers

are reasonable.

Review Question 2.4

An elevator in a tall office building moves downward at constant speed. How does the magnitude

of the upward force exerted by the cable on the elevator C on ElF!

compare to the magnitude of the

downward force exerted by Earth on the elevator E on ElF!

? Explain your reasoning.

2.5 Inertial reference frames and Newton’s first law

In Chapter 1 we learned that our description of the motion of an object depends on the

observer’s frame of reference. However, when we analyzed experiments in this chapter, we

tacitly assumed that all observers were standing on the earth’s surface. For example, in Section

2.3, we analyzed several experiments and concluded that if the forces exerted on one object by

other objects add to zero, then the chosen object moves at constant velocity. Are there any

observers who will see a chosen object moving with changing velocity even though the sum of

the forces exerted on the object appears to be zero?

Inertial reference frames

In Observational Experiment Table 2.3, we consider two different observers analyzing

the same situation.


Observational Experiment Table 2.3 Two observers watch the same mug.

Observational experiment Analysis done by each observer

Observer #1 is slouched down in the passenger seat of a car. All of a sudden he observes a coffee mug that was previously at rest on the dashboard starting to slide toward him.

Observer #1: A motion diagram and a force diagram for mug as created by Observer 1 in the car are shown below. On the motion diagram

increasing v!

arrows indicate that the mug’s speed

changes from zero to non-zero as seen by Observer 1.

Observer #2 stands on the ground beside the car. She observes that the car starts moving forward at increasing speed and that the mug remains stationary with respect to her.

Observer #2: A motion diagram and force diagram for the mug as created by Observer 2 on the ground are shown below.

Pattern

Observer 1: The forces exerted on the mug by Earth and by the dashboard surface add to zero. But the velocity of the mug increases as it slides off the dashboard. This is inconsistent with the rule relating the sum of the forces and the change in velocity. Observer 2: The forces exerted on the mug by Earth and by the dashboard surface add to zero. Thus the velocity of the mug should not change and it does not. This is consistent with the rule relating the sum of the forces and the change in velocity.

Observer 2 in Observational Experiment 2.3 can account for what is happening using the rule

developed earlier; but observer 1 cannot. For observer 1, the mug’s velocity changes for no

apparent reason.

We can observe other similar experiments; for example, a computer on the lap of a

passenger on a subway train all of a sudden starts to slide forward off the lap. Simultaneously,

another passenger standing in the train not holding onto anything all of a sudden moves forward,

bumping into the passenger standing in front of him. A person on the platform waiting for the

train to stop can explain these two events using the rule we developed. The train’s velocity started

decreasing as it approached the station, but the computer and standing person continued forward

at constant velocity.

It appears that the applicability of the rule depends on the reference frame of the

observer. Observers who can explain the behavior of the mug, the computer, and the standing

person by using the rule relating sum of the forces and changing velocity are said to be observers

in inertial reference frames. Those who cannot explain the behavior of the mug, the computer,

and the standing person (the velocity of these objects changes but the sum of the forces exerted

on them is zero) are said to be observers in non-inertial reference frames.


Inertial reference frame An inertial reference frame is one in which an observer sees

that the velocity of the system object does not change if no other objects exert forces on it

or if the sum of all forces exerted on the system object is zero. For observers in non-

inertial reference frames, the velocity of the system object can change even though the

sum of forces exerted on it is zero.

An observer in a car or train that is speeding up or slowing down with respect to the earth

is an example of a non-inertial reference frame. When you are in a car that abruptly stops, your

body jerks forward—yet nothing is pushing you forward. When you are in an airplane taking off,

you feel pushed back into your seat, even though nothing is pushing you in that direction. In these

examples, you are an observer in a non-inertial reference frame. Observers in inertial reference

frames can explain the changes in velocity of objects by considering the forces exerted on them

by other objects. Observers in non-inertial reference frames cannot. From now on, we will always

analyze phenomena from the point of view of observers in inertial reference frames. We can now

reformulate the rule invented earlier in this chapter.

Newton’s first law of motion For an observer in an inertial reference frame when no

other objects exert forces on an object of interest or if the forces exerted on it add to zero,

the object continues moving at constant velocity (including remaining at rest). Inertia is

the phenomenon when an object continues to move at constant velocity when the sum of

the forces exerted on it by other objects is zero.

Physicists have analyzed the motion of thousands of objects from the point of view of

observers in inertial reference frames and found no contradictions to the rule. They call the rule

Newton’s first law of motion. The first law limits the reference frames with respect to which the

other laws that you will learn in this chapter are valid—these other laws work only for the

observers in inertial reference frames.

Review Question 2.5

What is the main difference between inertial and non-inertial reference frames? Give an example.

2.6 Force and mass and their effects on an object’s acceleration

Our conceptual analyses in Sections 2.3 and 2.4 indicated that

an object’s acceleration is in the same direction as the sum of the forces

that other objects exert on it. We also learned that an object does not

change velocity and does not accelerate when the sum of all forces

exerted on it is zero. But we do not know how to predict the magnitude

ALG

3.1.1

ALG

3.3.1a


of the acceleration of an object if we know the forces exerted on it. Do other physical quantities

affect the acceleration? Note for example that the strong man shown in Fig. 2.14 is causing a tiny

acceleration while pulling the 40-ton train but could easily cause a large acceleration when he

pulls a small wagon. The amount of matter being pulled must affect the acceleration. We next

examine these ideas quantitatively using a computer probe (a device that can exert known forces)

and a motion detector (a device that uses sound similar to a bat’s echolocation to record an

object’s velocity and acceleration.

The experiments in Observational Experiment Table 2.4 will help us construct a

quantitative relationship between the sum of the forces exerted on a system object and its

acceleration.

Observational Experiment Table 2.4 Forces and resulting acceleration

Observational experiments Analysis

A cart starts at rest on a low-friction horizontal track. A force probe continuously exerts 1 unit of force in the positive direction. The same experiment is repeated five times and each time, the force probe exerts one more unit of force on the cart (up to 5 units at step 5). A computer records the value of the force, and a motion detector on the track records the cart’s speed and acceleration.

Shown below are velocity and acceleration-versus-time graphs for two of the experiments when different constant magnitude forces were exerted on the cart.

We see that the greater the force, the greater the acceleration - in direct proportion to the force.

The same five experiments are repeated, only this time, the cart is moving in the positive direction, and the probe pulls back on the cart in the negative direction so that the cart slows down.

Shown below are velocity and acceleration-versus-time graphs for the cart when two different magnitude forces opposing motion were continuously exerted on the cart.


Pattern

! When the sum of the forces exerted on the cart is constant, its acceleration is constant – the cart’s speed increases at a constant rate.

! Plotting the acceleration-versus-force using the five positive and five negative values of the force, we obtain the graph at the right.

! The acceleration is directly proportional to the force exerted by the force probe (in this case it is the sum of all forces) and points in the direction of the force.

The outcome of these experiments expressed mathematically is as follows:

a F* #!!

(2.1)

where F#!

is the sum of all the forces that other objects exert on the object (not an additional

force), and a!

is the object’s acceleration. The symbol * means, “is proportional to”. In other

words, if the sum of the forces doubles, then the acceleration doubles. When the sum of the forces

is zero, the acceleration is zero. When the sum of the forces exerted on an object is constant, the

object’s resulting acceleration (not velocity) is constant.

Mass, another physical quantity

Let’s perform another experiment to find the quantitative effect of the amount of material

being pulled. We use the force probe to pull one cart, then two carts stacked on top of each other,

and then three and four carts on top of each other. In each case, the force probe exerts the same

force on the carts, regardless of how many carts are being pulled. The experiment is summarized

in Observation Experiment Table 2.5.

Observational Experiment Table 2.5 Amount of material and acceleration.

Observational experiment Analysis

Pull the indicated number of carts (stacked on top of each other) using an identical pulling force. Measure the acceleration with a motion detector.

# n of carts Acceleration m/s2

1 1.00

2 0.49

3 0.34

4 0.25

A graph of acceleration-versus-number of carts for constant pulling force.

From the graph we see that increasing the number of carts decreases the acceleration. To check whether this relationship is inversely

proportional, we plot 1

vs a

n

.

ALG

3.1.2


From the pattern observed in Observational Experiment Table 2.5, we conclude that the

greater the amount of material being pulled, the smaller the object’s acceleration when the same

force is exerted on it. This property of an object that affects its acceleration is called mass.

To measure the mass of an object quantitatively, you first define a standard unit of mass.

A commonly used standard is a 1.0-kilogram (1.0-kg) object. The choice for the unit is arbitrary,

but after the unit has been chosen, the masses of all other objects can be determined. Suppose for

example that you exert a constant pulling force on a 1.0-kg object (all other forces exerted on this

object are balanced), and you measure its acceleration. Suppose you exert the same pulling force

on another object of unknown mass. Your measurement indicates that it has half the acceleration

of the standard 1.0-kg object. Then its mass is twice the standard mass of 2.0 kg. This method is

too complicated for everyday life. Later we will learn another method to measure the mass of an

object, a method that is simple enough to use in everyday life.

The SI metric system standard of mass, the kilogram, is a cylinder made of a platinum-

iridium alloy. It is kept in a museum of measurements near Paris. Copies of this cylinder are

available in most countries. You can make fractional kilograms by cutting this cylinder in pieces

and multiple kilograms by putting together identical cylinders. A quart of milk has a mass of

about 1 kilogram. Our experiments indicate that when the same force is exerted on two objects,

the one with the greater mass will experience a smaller acceleration. Specifically, if object A has

twice the mass of object B, the acceleration of object A will be half the acceleration of object B.

Mathematically:

1a

m* (2.2)

Mass m characterizes the amount of material comprising an object. When the same

unbalanced force is exerted on two objects, the object with greater mass has a smaller

acceleration. The unit of mass is called the kilogram (kg). Mass is a scalar quantity and

masses add as scalars.

Pattern

As the graph 1

vs a

n

is a straight line, we conclude that 1

a

n

* .


Newton’s second law

We found that the acceleration a!

of a system is proportional to the sum of the forces F#!

exerted on it by other objects [Eq. (2.1)] and inversely proportional to the mass m of the system

[Eq. (2.2)]. We can combine these two proportionalities into a single equation.

on system

system

system

Fa

m

#*

!

! (2.3)

Rearrange the above to get system system on system m a F* #!!

. We can turn this into an equation if we

choose the unit of force to be kg•m/s2. Because force is such a ubiquitous quantity, physicists

have given the force unit a special name called a newton (N):

21 N = 1 kg•m/s . (2.4)

The above important relationship, rewritten with the equality sign on system

system

system

Fa

m

#"

!

! is called

Newton’s second law. It is written using a new symbol # (the Greek letter sigma), which means

that you must add everything that follows the # .

Newton’s second law The acceleration systema!

of a system object is proportional to the sum

of all forces being exerted on the object and inversely proportional to the mass m of the

object:

Other objects on System object

system

system

Fa

m

#"

!

! (2.5)

The sum of all the forces being exerted on the system by other objects is:

Other objects on System object 1 on System 2 on System n on System...F F F F# " $ $ $! ! !

The acceleration of the system object points in the same direction as the sum of the forces.

Tip! Notice that the “vector sum of the forces” mentioned in the definition above does not mean

the algebraic sum. Vectors are not added as numbers; their directions affect the magnitude of the

vector sum. See the example in Fig. 2.7. Two forces are exerted on the suitcase: you exert a 150-

N force up and Earth exerts a 100-N force down. The vector sum of the forces is the vector of 50

N in magnitude pointing up.

Now that we have a new equation, think if it makes sense. One way to do it is to consider

extreme cases. Imagine an object with an infinitely large mass – according to the law it will have

zero acceleration no matter how large the sum of the forces exerted on it:

Other objects on System object

system 0F

a#

" "+

!

! . It means that if the object is already moving, it will

continue moving at the same velocity independently of other objects, if it is at rest- it will remain

at rest. On the other hand, an object with a zero mass will have an infinitely large acceleration


even if the tiniest force is exerted on it:Other objects on System object

system0

Fa

#" " +

!

! . Both extreme cases

make sense.

Another important issue here is if the sum of the forces exerted on an object is constant,

not its velocity but its acceleration is constant – the object changes its velocity at constant rate.

Newton’s second law represents a new kind of relation, a cause–effect relation. In a

cause–effect relation, one side of the equation (the right side) contains the cause (in this case the

sum of the forces being exerted on the system and the system’s mass) and the other side (the left

side) contains the effect (in this case, the acceleration of the system).

on system

system

system

Fa

m

#"

!

!

Compare this relation to the definition of acceleration v

at

'"'

!!

, which is called an

operational definition. An operational definition tells us how to determine the quantity but does

not tell us why it has a particular value. For example, you measure an elevator’s speed, which

changes from 2 m/s to 5 m/s in 3 s as it moves vertically along a straight line in the positive

direction of the chosen axis. You can use these data to operationally determine the elevator’s

acceleration (using the definition of acceleration):

25 m/s - 2 m/s1 m/s

3 sx

a " " $

But you do not know the reason for the acceleration. If you know that the mass of the elevator is

500 kg and that Earth exerts a 5000-N downward force on the accelerating elevator, then the

cable must exert a 5500-N upward force on it since according to Newton’s second law (using the

cause-effect relationship):

25500 N + (-5000 N)1 m/s

500 kg" $ .

You obtain the same number using two different methods – one from kinematics and the other

one from dynamics.

Another important idea here is that Newton’s second law allows us to explain a unit of

force – 1 newton. We can think if a force of 1 N as a force that causes an object of the mass of 1

kg to accelerate at 1 m/s2.

Force components used for forces along one axis

When the vector sum of the forces exerted on a system of interest point along one

direction (which could be called an x direction or a y direction), you can use the component form

of the Newton’s second law equation for the x direction (or the y direction) instead of the vector

equation [Eq. (2.5)]:

Cause Effect


Other on System

system

system

x

x

Fa

m

#"

(2.6)

A similar equation applies for the y-direction, if the forces are all along the y-axis. To use the

above equation (or the y-version of the equation), you first need to identify the positive direction

of the x-axis. Then find the components of all the forces being exerted on the system. Forces that

point in the positive direction have a positive x- component, and forces that point in the negative

direction have a negative x-component.

In this chapter, we will analyze situations in which the forces that external objects exert

on the system are either: (a) all along one axis, or (b) cases that can be analyzed along only one

axis because the forces pointing along the other axis balance and don’t contribute to the

acceleration of the system.

All forces along one axis

Consider first a situation where all forces are along one axis.

Example 2.5 Lifting a suitcase Earth exerts a downward 100-N force on a 10-kg mass suitcase.

To lift it very slowly, you need to exert about the same magnitude upward force (it will take a

long time interval to raise it). Suppose instead that you exert an upward 150-N force on it. If the

suitcase starts at rest, what time interval is needed to raise it 0.50 m?

Sketch and Translate We make an initial-final sketch of the process (Fig. 2.15a). This important

step helps us visualize the process and also provides all the known information thus allowing our

brains to focus on other aspects of solving the problem. One common aspect of such problems is

the use of a two-step strategy. Here, we use Newton’s second law to determine the acceleration of

the suitcase and then use kinematics to determine the time interval needed to lift it 0.50 m.

Figure 2.15(a)

Simplify and Diagram The suitcase is chosen as the system object of interest. A force diagram for

the suitcase while being lifted is shown in Fig. 2.15b. You pull upward exerting a force of

magnitude Y on S 150 NF " . Earth pulls down exerting a force of magnitude E on S 100 NF " . The

upward force is larger thus the suitcase will have an upward acceleration a!

.


Figure 2.15(b)

Represent Mathematically The vector form of Newton’s second law applied to suitcase and this

process, is given below:

Y on S E on SS

S S

F FFa

m m

$#" "

! !!!

The above is not very useful, as we can’t simply add vector quantities. Since all the forces are

along the y axis, we instead use the y -component form of the law (notice how the subscripts in

the equation below change from step to step):

, -on S Y on S E on S Y on S E on S Y on S E on SS

S S S S

( )y y y

y

F F F F F F Fa

m m m m

# $ $ $ % %" " " "

The y components of the forces exerted on the suitcase are Your upward pull on the Suitcase

Y on S Y on S 150 Ny

F F" $ " $ and Earth’s downward pull on the Suitcase

E on S E on S– –100 Ny

F F" " .

After determining the acceleration of the suitcase using the above application of

Newton’s second law, we can then use kinematics to determine the time interval needed to raise

the suitcase 0.50 m:

2

0 0

1–

2y y

y y v t a t" $ .

Substituting the initial velocity 0 0y

v " into the above and rearranging to find the time t when

the suitcase is at 0.50 my " (its initial position is 0 0y " ), we get:

02( – )y yt

a"

Solve and Evaluate We now substitute the known information in the Newton’s second law

component equation above to find the acceleration of the suitcase:

2Y on S E on SS

S

150 N 100 N5 m/s

10 kgy

F Fa

m

% %" " " $


Then insert this and other known information into the kinematics equation above to find the time

when the suitcase is 0.50 m above its starting position:

0

2

2( – ) 2(0.5 m – 0)0.45 s 0.5 s

5 m/s

y yt t

a" " " " . .

The unit for time is correct. The 0.5 s time interval is short. Our body might be better served by

lifting it slower exerting a smaller force.

Try It Yourself: Use the acceleration 25 m/s = (+5 m/s)/s$ and the lifting time interval 0.5 s to

estimate in your head the velocity of the suitcase at the end of the 0.5 s.

Answer: If lifted for 1 s, its velocity would be + 5 m/s. Since you lifted only for 0.5 s, its velocity

would be about +2.5 m/s.

Tip! Notice that if you were lowering the suitcase when it is already moving down and exerted

the same force, the suitcase would continue moving down at decreasing speed. The sum of the

forces determines the direction and magnitude of the acceleration of the object, not its velocity.

Horizontal accelerated motion with balanced forces perpendicular

In this chapter we focus only on situations where we consider the forces exerted on

objects along the line of motion. In Chapter 3, we analyze more complex situations involving

forces in different directions. However, there are situations when forces of the same magnitude in

the opposite directions are exerted on an object perpendicular to the line of motion. They add to

zero and can be ignored. Consider the next example.

Example 2.6 Pulling (or pushing) a lawn mower You pull horizontally on the handle of a lawn

mower that is moving across a horizontal grassy surface. The lawn mower mass is 30 kg. You

exert a force of magnitude 100 N on the mower. The grassy surface exerts an 80-N resistive

friction-like force on the mower. Earth exerts a downward force on the mower of magnitude 290

N, and the grassy surface exerts an upward normal force of magnitude 290 N. What is the

acceleration of the mower?

Sketch and Translate A sketch of the situation is

shown in Fig. 2.16a. We choose the mower as the

system.

Figure 2.16(a)

Simplify and Diagram We model the mower system as a point-like object and draw a force

diagram for it (Fig. 2.16b). As the forces do not change during motion, the diagram can represent

any instant during the motion. Three external objects interact with the system—the grassy


surface, Earth, and you. You exert a horizontal force on the Mower Y on MF!

; Earth exerts a

downward gravitational force on the Mower E on MF!

. The Grassy surface exerts a force on the

Mower that we will represent with two force arrows —a normal force G on MN!

perpendicular to

the surface (in this case it points upward) and a horizontal friction force G on Mf!

opposite the

direction of motion. The downward force E on MF!

and the upward normal force G on MN!

have the

same magnitudes and point in opposite directions. This means that these two forces cancel and do

not contribute to the horizontal acceleration of the mower. We can ignore them.

Figure 2.16(b)

Represent Mathematically Since the acceleration of the system is along the x-axis and the forces

perpendicular to the grassy surface cancel, we can use the x-component form of Newton’s second

law to determine the acceleration of the mower:

Y on M G on M M

M M

x x x

x

F F fa

m m

# $" " .

The x-component of your force on the mower is positive since that force points in the positive x-

direction ( Y on M Y on MxF F" $ ). The x-component of the friction force exerted by the grass on the

mower is negative since that force points in the negative x-direction ( G on M G on M xf f" % ).

Therefore the acceleration of the mower is:

Y on M G on M Y on M G on MM

M M

(– ) –x

F f F fa

m m

$" "

Solve and Evaluate We can now determine the mower’s acceleration by substituting the known

information into the above equation:

2Y on M G on M

M

M

– 100 N – 80 N0.7 N kg 0.7 m s

30 kgx

F fa

m" " " $ " $


The acceleration is in the positive x-direction. If the mower started at rest, it would be moving at

speed of about 0.7 m/s in the positive x-direction after 1 s, at speed 1.4 m/s after 2 s, and so on.

The units for acceleration are correct, and the magnitude is reasonable.

Notice the number of significant digits in the answer. If you were doing the analysis with

a calculator, you would get 0.666666667 m/s2 for the answer. Since the measurements we had to

work with had just one significant digit, we report the result as 0.7 m/s2.

Try It Yourself: Imagine that you pulled the mower as described in the above example. After the

mower had accelerated to a velocity with which you are comfortable, what is the magnitude of

the force you should exert on the mower so that it now moves at constant velocity?

Answer: 80 N.

Isaac Newton

Newton began his work on the second law of motion in the summer of 1665. The Great

Plague (bubonic) had devastated England that year. One tenth of London’s population perished.

Isaac Newton, then a 22-year-old student at Trinity College in Cambridge, returned to his

mother’s farm in Woolsthorpe, England. During the next eighteen months, Newton initiated one

of the most productive periods of achievement by an individual in the history of science. The

period culminated about 22 years later with the publication of Newton’s major work, a book

entitled Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural

Philosophy). During these two decades, Newton invented differential and integral calculus,

formulated the law of universal gravitation, developed a new theory of light, and put together the

ideas for his three laws of motion (we have learned only two so far).

Review Question 2.6

Jim says that ma!

is a special force exerted on an object and it should be represented on the force

diagram. Do you agree or disagree with Jim? Explain your answer.

2.7 Gravitation force law

In the last example, we were given the mass of the mower (30 kg) and the magnitude of

the force that Earth exerts on the mower (290 N). Is it possible to determine the magnitude of this

force by just knowing this mass of the mower? In fact it is.

Imagine that we evacuate all the air from a 3.0-m long Plexiglas tube, place a motion

sensor at the top, and drop objects of various sizes, shapes, and compositions through the tube.

The measurements taken by the motion sensor reveal that all objects fall freely straight down with

the same acceleration 29.8 m s . A force diagram for an object falling inside the tube has only

the force exerted by Earth on the Object E on OF!

during the whole flight. If we choose the positive

y-axis pointing down and apply the y-component form of Newton’s second law, we get:


, - E on OO E on O E on O

O O O

1 1y y

Fa F F

m m m" " $ " $

Every object dropped in our experiment had the same free fall acceleration g = 29.8 m s , even

those with very different masses (such as a ping-pong ball and a lead ball). Thus, the gravitational

force that Earth exerts on each object must be proportional to its mass so that the mass cancels

when we calculate the acceleration. This means that Earth exerts a 10 times larger force on a 10

kg object than on a 1-kg object:

E on O E on OO

O O

y

F

m

a gF

m" " "

This reasoning leaves just one possibility for the magnitude of the force exerted by Earth on the

object:

, -2

E on O O O 9.8 m/s .F m g m" "

The magnitude of the gravitational force that Earth exerts on the object is proportional to the mass

of the object. The ratio of the force and the mass is a constant for all objects—the so-called

gravitational constant g, already familiar to us as free-fall acceleration.

Gravitational force The magnitude of the gravitational force that Earth exerts on any object FE on

O when on or near its surface equals the product of the object’s mass m and the constant g:

E on O OF m g" (2.7)

where 29.8 m/s 9.8 N kgg " " on or near Earth’s surface. This force points toward the center

of Earth.

The value of the free-fall acceleration g in the above gravitational force law [Eq. (2.7)]

does not mean that the object is actually falling. The g is just used to determine the magnitude of

the gravitational force exerted on the object by Earth as it is not the real acceleration of the object

(it is the acceleration that the object would have if Earth were the only object exerting a force on

it). To avoid confusion, we will use g = 9.8 N/kg rather than 29.8 m/s when calculating

the gravitational force.

We learn in Chapter 4 that the value of the gravitational constant g at a particular point

depends on the mass of Earth, and on how far this point is from the center of Earth. If we lived on

Mars or the Moon, the gravitational constant there depends on the mass of Mars or the Moon and

not on the mass of Earth. The gravitational constant on the Moon is 1.6 N/kg and on Mars is 3.7

N/kg. You could hit a golf ball farther on the Moon, as it has a smaller g and would exert a

smaller gravitational force on the ball.

ALG 3.3.7


Tip! In some physics books, the force that Earth exerts on an object is called the object’s weight.

Weight of the object on another planet is the force that that planet exerts on the object. As we said

earlier, we will not use the term “weight of an object” as it implies that weight is a property of the

object rather than an interaction between two objects.

British Units In everyday life in the U.S. and Great Britain, the commonly used force unit is the

pound (lb). The unit of mass in this English system is the slug, and the unit of acceleration is

foot/second2. One pound equals the force that causes a 1-slug mass to have an acceleration of

21 ft/s or 21 lb = (1 slug)(1 ft/s ) . Note that 1 slug = 14.6 kg and 1 foot = 0.31 m . Thus,

2 214.6 kg 0.31 m1.0 lb = (1.0 slug)(1.0 ft/s ) [(1.0 slug)( )][(1.0 ft/s )( )]

slug 1.0 ft"

24.45 kg m/s 4.45 N" "!

.

For example in the British units, suppose that the gravitational force that Earth exerts on a person

is 160 lb. This force in newtons is (160 lb)4.45 N

1.0 lb

/ 01 23 4

= 710 N.

Review Question 2.7

Earth pulls on an apple exerting a force of 1.0 N. What is the mass of the apple in standard units

and in British units?

2.8 Skills for applying Newton’s second law for one-dimensional

processes

When astronomers announce that a comet will pass within one million miles of Earth on

a certain month, how do they know? They are using Newton’s second law to predict the location

of the comet. That’s a more complicated situation than we will be able to handle, but we will use

the same ideas and the same physics in our applications and problems. We start by developing a

strategy that can be used whenever a process involves force and motion. The name for this area of

physics, the study of motion and forces, is called dynamics. The general strategy we will use is

outlined on the left side of the Example 2.7 and illustrated for an interesting process described

briefly next.

In 2007, sky diving champion Michael Holmes jumped from an airplane 12,000 ft above

Lake Taupo in New Zealand. He had already made more than 7000 sky diving jumps.

Unfortunately, his main parachute failed to open and his backup chute became tangled in its

cords. Traveling at about 80 mph (slowed from the normal 120 mph terminal speed by his

partially opened backup parachute), he reached a 2-meter-high thicket of wild shrubbery. Holmes


plunged through the shrubbery, which decreased his speed, before he thudded on the ground. His

right lung collapsed, and his left ankle broke, but he survived!

Example 2.7 Holmes Sky Dive Michael Holmes (70 kg) was moving downward at 36 m/s (80

mph) and was stopped by 2.0-m-high shrubbery and the ground. Estimate the average force

exerted by the shrubbery and ground on his body while stopping his fall.

Sketch and Translate

! Make a sketch of the process.

! Choose the system object.

! Choose a coordinate system.

! Label the sketch with everything you know about that situation.

A labeled initial–final sketch of the process is shown. We choose Holmes as the system of interest. We want to know the average force that the shrubbery and ground exert on him from when he first touches the shrubbery to the instant when he stops. We choose the y-axis pointing up and the origin at the ground where Holmes comes to rest. We use kinematics to find his acceleration while stopping, and Newton’s second law to find the average force that the shrubbery and ground exerted on him while stopping him.

Simplify and Diagram

! Make appropriate simplifying assumptions about the process. For example, can you neglect the size of the system object or neglect frictional forces? Can you assume that forces or acceleration are constant.

! Then, represent the process diagrammatically with a motion diagram and/or a force diagram(s). Make sure the diagrams are consistent with each other.

We model Holmes as a point-like object and assume that the forces being exerted on him are constant so that they lead to a constant acceleration. A motion diagram for his motion while stopping is shown. along with the corresponding force diagram. To draw the force diagram we first identify the objects interacting with Homes. There are two objects interacting with him as he slows down: the shrubbery and the ground (combined as one interaction) and Earth. The shrubbery and ground exert an upward normal force

S-G on H N!

on Holmes. Earth exerts a downward gravitational force E on H F!

.

The force diagram is the same for all points of the motion diagram as the

acceleration is constant. On the force diagram the arrow for S-G on H N!

must be

longer to match the motion diagram, which shows the velocity change arrow pointing up.

Represent Mathematically

! Convert these qualitative representations into quantitative mathematical descriptions of the situation using kinematics equations and Newton’s second law for motion along the axis. After you make the decision about

We can determine Holmes’ average acceleration (its component in the y-direction) while stopping using kinematics:

2 2

0,

02( )

y y

y

v va

y y

%"

%.

From the motion and force diagrams we know that the y-component of Newton’s second law with the positive y-direction up is:

on H

H

y

y

F

a

m

#"

,


the positive and negative directions, you can determine the signs for the force components in the equations. You need to add the force components (with either positive or negative signs) to find the sum of the forces.

The y

component of the force exerted by shrubbery-ground is

S-G on H S-G on H yN N" $

and the y

component of the force exerted by Earth

force is E on H E on H H– y

m gF F "" %. Therefore

S-G on H E on H S-G on H E on H S-G on H H

H H H

( ) ( )y y

y

N F N F N m g

m m m

a$ % $ %$ $

" " "

S-G on H H HyN m a m g5 " $

.

Solve and Evaluate

! Substitute the known values into the mathematical expressions and solve for the unknowns.

! Finally, evaluate your work to see if it is reasonable (check units, limiting cases, and whether the answer has a reasonable magnitude). Check if all representations – mathematical, pictorial and graphical are consistent with each other.

Holmes’ average acceleration is

, -, -

22

20 36 m/s

324 m/s2 0 2.0 m

ya

% %" " $

%

Holmes’ initial velocity is negative, since he is moving in the negative direction. His initial position is + 2.0 m (when he first hits the shrubbery), and

his final position is at the origin. The diver’s velocity in the negative direction is decreasing, which means the velocity change and the acceleration point in the opposite direction (positive). The average magnitude of the force exerted by the shrubbery and ground on Holmes is:

2

S-G on H H H

2

(70 kg)(324 m/s ) + (70 kg)(9.8 N/kg)

= 22,680 kg m/s + 686 N = 23,366 N = 23,000 N

yN m a m g" $ "

6

The force has a magnitude greater than the force exerted by Earth – thus the results is consistent with the force diagram and the motion diagram. The magnitude is huge, the units are correct. The limiting case – for zero acceleration for example, gives us a correct prediction – the force exerted on Holmes by shrubbery is equal to the force exerted by Earth.

Try It Yourself: Use the strategy discussed above to estimate the average force that the ground

would have exerted on Holmes if he had stopped in 0.20 m with no help from the shrubbery.

Answer: 230,000 N (over 50,000 lb).

Think about the result in the above example - the force has a magnitude greater than 5000

lb! You can see why Holmes’ survival is so miraculous. One thing that helped was that the force

was exerted over a significant area of his body, which occurred thanks to the shrubbery. Most of

all, the shrubbery increased the stopping distance. If Holmes had landed directly on dirt or

something harder, his acceleration would have been much greater and so would the force exerted

on him.

Tip! Look at the last equation in the above example. On the left side, an N in italics is used to

indicate the magnitude of the normal force exerted by the shrubbery and ground on Holmes. On

the right side, an N in Roman is used to indicate the unit of force. Be careful not to confuse these

two similar looking notations.

Holmes was not unique in surviving a parachute jump in which the chute did not open. In

other cases, skydivers whose parachutes did not open landed in snow banks or on the slope of a


snow-covered hill. The long stopping distance led to a reduced force being exerted on the

skydivers and survival in most cases. In the next example, we consider a much less dangerous

process, one you could carry out the next time you ride an elevator.

An elevator ride standing on a bathroom scale

You will stand on a bathroom scale in an elevator as it makes an upward trip from the

first floor of a hotel to the tenth floor. When you are standing on the scale in the stationary

elevator before the trip starts, the scale reading indicates how hard the scale is pushing up on you.

This normal force that it exerts on you balances the downward force that Earth exerts on you (in

every day language called your weight) resulting in your zero acceleration. Thus, the scale

reading when the elevator is at rest is equal in magnitude to the downward gravitational force that

Earth exerts on you. What will the scale read during your upward trip?

Example 2.8 Elevator ride You stand on a bathroom scale in an elevator as it makes a trip from

the first floor to the tenth floor of a fancy hotel. Your mass is 50 kg. When you stand on the scale

in the stationary elevator, it reads 490 N (110 lb). (a) What will the scale read early in the trip

while the elevator’s upward acceleration is 1.0 m/s2, (b) for the middle of the trip while the

elevator moves up at a constant speed of 4.0 m/s, and (c) near the end of the trip when the

elevator slows to a stop changing its speed by 1 m/s every second, a downward acceleration of

1.0 m/s2 magnitude.

Sketch and Translate A sketch of the situation is shown in Fig. 2.17. We choose you as the

system object of interest, and the coordinate axis pointing upward with its origin at the first floor

of the elevator shaft. Your mass is m = 50 kg, the magnitude of the force that Earth exerts on you

is E on Y you = = 490 NF m g , and your acceleration is: (a) 2= +1.0 m/s

ya (the upward velocity

is increasing); (b) = 0y

a (v is a constant 4.0 m s upward); and (c) 2= –1.0 m/s

ya (the

upward velocity is decreasing, so the acceleration points in the opposite negative direction).

Figure 2.17 Standing on scale in elevator

Simplify and Diagram We model you as a point-like object and represent you as a dot in both the

motion and force diagrams, shown for each part of the trip in Fig. 2.18a, b, and c. The downward


force that Earth exerts does not change magnitude (equal to mg!

and neither m nor g!

changes).

Notice that the force diagrams and motion diagrams are consistent with each other for each part

of the trip: the length of the normal force arrows representing the force that the scale exerts on

you changes from one case to the next so that the sum of the forces points in the same direction as

your velocity change arrow. On the diagram E represents Earth, Y is you, and S is the scale.

Figure 2.18 Elevator upward trip

Represent Mathematically The motion and the forces are entirely along the vertical y -axis. Thus,

we can use the vertical y -component form of Newton’s second law [Eq. (2.6)] to analyze the

process. There are two forces exerted on you (the system object) so there will be two forces on

the right side of the equation:

E on Y S on Y

Y

Y Y

y y y

y

F F Na

m m

# $" "

The y-component of the force that Earth exerts on you E on Y –y

F mg" is negative and the y-

component of the normal force that the scale exerts on you S on Y S on YyN N" $ is positive. Thus

we can rewrite the above equation as:

S on Y Y

Y

–y

mg Na

m

$"


Multiplying both sides by Ym we get Y Y S on Y –y

a m mg N" $ . We can now move mg% to the

left side: Y Y S on Yym a mg N$ " or

S on Y Y Y yN m a mg" $

This is the expression for the magnitude of the force that the scale exerts on you. Remember that

mg = 490 N is the magnitude of the force that Earth exerts on you. Thus we see that the scale and

Earth exert the same magnitude forces on you ONLY when your acceleration is zero.

Solve and Evaluate We can now use the last equation to predict the scale reading for the three

parts of the trip:

S on Y Y Y Y Y 490 N.y y

N m a mg m a" $ " $

(a) Early in the trip, the elevator is speeding up and its acceleration is 2

Y +1.0 m/sy

a " . Then,

the force exerted by the scale on you should be:

2

S on Y Y Y = 490 N = (50 kg)(+1.0 m/s ) + 490 N = 540 N.y

N m a $

(b) In the middle of the trip when the elevator moves at constant velocity, your acceleration is

zero and the scale should read:

2

S on Y Y Y = 490 N = (50 kg)(0 m/s ) + 490 N = 490 Ny

N m a $ .

(c) When the elevator is slowing down near the end of the trip, its acceleration points downward

and is 2–1.0 m/s

ya " . Then, the force exerted by the scale on you should be:

2

S on Y Y Y = 490 N = (50 kg)(–1.0 m/s ) + 490 N = 440 Ny

N m a $ .

When the elevator is at rest or moving at constant speed, the scale reading equals the

same as the magnitude of the force that Earth exerts on you. When the elevator accelerates

upward, the scale reads more. When it accelerates downward, the scale reads less. If you ride an

elevator standing on a scale, you will find qualitatively that these predictions are correct; so the

scale indeed measures the force it exerts on the person and not the force that Earth exerts on the

person. What is also important is that the motion and force diagrams in Fig. 2.18a, b, and c are

consistent with each other and the force diagrams are consistent with the predicted scale readings.

This is an important part of evaluating your results—a consistency check of the motion diagrams,

force diagrams, math, and whatever else you use to analyze the process.

Try It Yourself: What will the scale read when the elevator starts from rest on the tenth floor and

moves downward with increasing speed and a downward acceleration of 2–1.0 m s , then

moves down at constant velocity, and finally slows its downward trip with an acceleration of

21.0 m s$ until it stops at the first floor?

Answer: 440 N, 490 N, and 540 N.

Tip! The example above shows that the reading of the bathroom scale is not always equal in

magnitude to the force that Earth exerts on a person (the person’s weight). The readings change


depending on the acceleration of the person standing on the scale. Think of what the scale will

read if you stand on it in an elevator that is freely falling (2–9.8 m s

ya " , assuming the upward

direction is positive).

Why should we worry about the scale reading in an elevator; people rarely weigh

themselves in moving elevators. However, the changing reading of the scale means that the cable

supporting the elevator cabin exerts a variable force on it. The force is larger when the elevator

accelerates upward (either speeding up when moving up or slowing when moving down). Thus

when choosing a cable we must remember that it should be able to handle the force more than the

weight of the maximum number of passengers plus the cabin. Also, the analysis of this problem

illustrates important general dynamics problem-solving strategies such as:

! sketching the process and choosing a system,

! identifying all the external forces that other objects exert on the system object,

! choosing the directions of coordinate axes,

! using the sketch and diagrams to help apply important general principles (kinematics

equation and the component form of Newton’s second law), and

! evaluating work.

Equation Jeopardy Problems

Hopefully, you are getting used to representing the processes in worked examples and

problems in multiple ways: the words in the problem statement, a sketch, one or more diagrams,

and a mathematical description. Another extremely important skill for success in physics is being

able to read the equations and make sense of them. This is what we do in Equation Jeopardy

Problems—you had a chance to try them in Chapter 1. Here is an example of an Equation

Jeopardy Problem involving Newton’s second law and a kinematics equation.

Example 2.9 Equation Jeopardy Problem The first and third equations below are the horizontal

x-component form of Newton’s second law and a kinematics equation representing a process. The

second equation is for the vertical y-component form of Newton’s second law for that same

process:

S on O–

(50 kg)x

fa "

S on O – (50 kg)(9.8 N/kg) 0N "

20 – (20 m/s) = 2 (25 m)x

a

First, determine the values of all unknown quantities in the equations. Then, work backward and

construct a force diagram and a motion diagram for a system object and invent a process that is

consistent with the equations (there are many possibilities). The problem solving strategy is

reversed.


Solve We can solve the second equation for the normal force:

S on O = (50 kg)(9.8 N/kg) 490 NN ".

We can solve the third equation for the acceleration ax:

22–(20 m/s)

= –8.0 m/s2(25 m)

xa "

.

Now substitute this result in the first equation to find the magnitude of the force fS on O:

2

S on O– = (50 kg)(–8.0 m/s ) –400 Nf ".

Represent Mathematically The third equation in the problem statement looks like the application

of the following kinematics equation to the problem process:

2 2

0 0– 2 ( – )x x x

v v a x x"

By comparing the above general version of the kinematics equation to the third provided

equation, we see that: the final velocity 0x

v " , the initial velocity 0 20 m/sx

v " $ , and the

displacement of the object was 0( – ) = 25 mx x . The first equation indicates that there is only

one force exerted on the system object in the horizontal x-direction. It has a magnitude of 400 N

and points in the direction of the object’s acceleration, which is opposite the direction of the

initial velocity. The second equation indicates that the object of interest has a 50-kg mass. It could

be that some kind of a friction force causes an 50-kg object to slow down and stop while it is

traveling 25 m horizontally.

Simplify and Diagram We can now construct a motion diagram and a force diagram for the object

(see Fig. 2.19a and b).

Figure 2.19(a)(b)

Sketch and Translate This could have been a skier skidding to a stop on a horizontal surface after

coming down a steep hill (see Fig. 2.19c).

Figure 2.19(c)


One Possible Problem Statement A 50-kg skier is moving at 20 m/s after traveling down a steep

hill. At the bottom the skier skids to a stop in 25 m on a horizontal surface. What was the average

friction force that the snow exerted on the skier while he was slowing down?

Review Question 2.8

Three friends are arguing about the type of information a bathroom scale reports: Eugenia says

that it reads the weight of a person, Alan says that it reads the sum of the forces exerted on the

person by Earth and the scale, and Mike says that the scale reads the force that the scale exerts on

the person. Who do you think is correct? Why?

2.9 Forces come in pairs

So far, we have been interested in the acceleration of a system object due to the forces

exerted on it by external objects. But, what effect does the system have on these other objects? To

answer this question, we need to observe the interaction of two objects and analyze what happens

to each of them. Suppose you wear rollerblades and push abruptly on a wheeled cart that is loaded

with a heavy box. If you and the cart are on a hard smooth floor, the cart starts moving away (thus

it accelerates), and you also start moving (you accelerate too) in the opposite direction. To

explain both accelerations we need to assume that you exerted a force on the cart, and the cart

exerted a force on you. Since you and the cart accelerate in opposite directions, the forces must

point in opposite directions. Imagine that you are on an ice skating rink wearing ice skates

holding a bowling ball. You throw the ball horizontally forward as hard as you can. You will find

that you start gliding backwards! This means that during the moment of the throw the ball

accelerated in one direction; but you accelerated in the opposite direction. It could only happen if

while you exerted a force on the ball the ball exerted an oppositely directed force on you.

Magnitudes of the forces that two objects exert on each other

Are the forces that two interacting objects exert on each other the same in magnitude or

different in magnitude? Consider two dynamics carts (small carts that have very low-friction

wheels so they can roll freely) and add some extra mass to one of them (see Observational

Experiment Table 2.6). Mount force probes on each cart so that we can measure the force that one

cart exerts on the other while they are colliding, and a motion sensor on each end of the track to

record the velocity of each cart. Place the carts on a low friction, horizontal track and send them

on a collision course toward each other.

ALG 3.1.3;

3.1.4

ALG 3.3.4 – 3.36


Observational Experiment Table 2.6 Analyze the force that two carts exert on each other.

Observational experiment Analysis

Two carts of different mass on a level track move toward each other. The motion detector indicates their speed before the collision and the force probes attached to each cart record the forces exerted by each cart on the other. Before the collision: m1 = 2 kg, v1 x = –2 m/s; m2 = 1 kg, v2 x = +2 m/s

As both carts change their velocities due to the collision, they must have exerted forces on each other. The computer recording the signals from the force probes shows that the forces that the carts exert on each other vary with time, and at each time, the forces have the same magnitude and point in the opposite direction Cart 1 exerts a force on cart 2 toward the right and the cart 2 exerts a force on 1 toward the left.

Cart masses and velocities before collision: m1 = 1 kg, v1 x = 0 m/s (at rest); m2 = 1 kg, v2 x = +1 m/s

Although the forces that the carts exert on each other are smaller than in the first experiment, the magnitudes of the forces at each time are still the same.

Cart masses and velocities before collision: m1 = 2 kg, v1 x = –2 m/s; m2 = 1 kg, v2 x = +1 m/s

The same analysis applies.

Pattern

In each experiment, independent of the masses and velocities of the carts before the collisions, at every

instant during the collision the force which cart 1 exerted on cart 2 1 on 2F!

had the same magnitude but

pointed in the opposite direction from the force that cart 2 exerted on cart 1 2 on 1F!

.

The pattern that emerges from the data is that when two carts collide, the force that one

cart exerts on the other is equal in magnitude and opposite in direction to the force that the other

cart exerts on the first.

object 1 on object 2 object 2 on object 1F F" %! !


Will the pattern that we found allow us to correctly predict the results of some new experiments?

Testing the idea

Imagine that you have two spring scales. You connect the scales

together and attach one scale to a hook on the wall (Fig. 2.20). What

happens if you pull on the other scale? If the pattern described by the

equation above is correct, the scale on which you pull should have the

same reading as the scale fixed to the wall as they exert the forces on each

other that are the same in magnitude and opposite in direction. If you reverse the scales and

repeat the experiment, they always have the same readings.

The equation relating the forces that two interacting objects exert on each other gave us

predictions for this experiment that matched the outcome. To be convinced that it works, we need

many more testing experiments. So far, physicists have found no experiments involving the

dynamics of everyday processes that violate this equation. This rule is called Newton’s third law.

Newton’s third law When two objects interact, object 1 exerts a force on object 2. Object

2 in turn exerts an equal-magnitude, oppositely directed force on object 1:

object 1 on object 2 object 2 on object 1F F" %! !

(2.8)

Two important issues to note: These forces are exerted on different objects and can’t be

added to find the sum of the forces exerted on one object. If one force is a contact force – the

Newton’s third law pair force to it is the contact force; if one force is gravitational – the

Newton’s third law pair force is gravitational.

It seems counterintuitive that two interacting objects always exert forces of the same

magnitude on each other. Imagine playing table tennis (ping-pong). A paddle hits the ball and the

ball flies rapidly toward the other side of the table. However, the paddle seems to move forward

with little change in motion. How is it possible that the light ball exerted a force of the same

magnitude on the paddle as the paddle exerted on the ball?

To resolve this apparent contradiction, think about the masses of the interacting objects

and their corresponding accelerations. If the forces are the same, the object with larger mass has

smaller magnitude acceleration than the object with smaller mass:

Paddle

Ball on Paddle Paddle on BallBall

Paddle Ball

and a

m

F F

ma" "

There is a much greater change in velocity of the ball than of the paddle. Just because the ball’s

velocity changed by a great deal does not mean a larger force was exerted on it. It’s actually

because the mass of the ball is so small. The paddle’s mass on the other hand is much larger.

Thus, the same magnitude force leads to an almost zero velocity change.

Another example is a head-on collision of a small car with a large truck. The forces that

they exert on each other are the same. But the acceleration of the small car will be much greater


than that of the large truck (see the above equation). The head and internal body parts of a driver

in the small car have a much higher acceleration during the collision and the chance that there

will be internal hazards to their body compared to that of the driver of the large truck with smaller

acceleration.

Remember that the forces in Newton’s third law are exerted on two different objects.

This means that the two forces will never show up on the same force diagram, and they

should not be added together to find the sum of the forces. You have to choose the system

and consider only the forces exerted on it! Another important feature of the forces in

Newton’s second law is that the forces in the pair should be of the same type – if one object

exerts a force on the other through contact, the other object cannot exert a gravitational force

on the first one – both forces should be contact forces.

Conceptual Exercise 2.10 A book on the table A book sits on a tabletop. Identify the forces

exerted on the book by other objects. Then, for each of these forces, identify the force that the

book exerts on another object. Explain why the book is not accelerating.

Sketch and Translate First, draw a sketch of the situation (Fig. 2.21a) and choose the book as the

system.

Figure 2.21(a)

Simplify and Diagram Assume that the tabletop is perfectly horizontal and model the book as a

point-like object. A force diagram for the book is shown in Fig. 2.21b. Earth exerts a downward

gravitational force on the book E on BF!

, and the table exerts an upward normal (contact) force on

the book T on BN!

. Newton’s second law explains why the book is not accelerating; the forces

exerted on it by other objects are balanced and add to zero. These two forces are exerted on the

same object and are of different types, so they cannot be a Newton’s third law pair of forces.

The two-letter subscripts on each force identify the two objects involved in the interaction. The

Newton’s third law pair force will have its subscripts reversed. For example, Earth exerts a

downward gravitational force on the book ( E on BF!

). According to Newton’s third law, the book

must exert an equal magnitude upward gravitational force on Earth ( B on E E on BF F" %! !

), as shown

in Fig. 2.24c. The table exerts a contact upward force on the book ( T on BN!

), so the book must

exert an equal magnitude contact downward force on the table ( B on T T on B–N N"! !

).


Figure 2.21 Forces on book and equal and opposite forces of book on other objects

Try It Yourself: A horse pulls on a carriage that is stuck in mud and not moving. Your friend

Chris says this happens because the horse exerts on the carriage the same magnitude force that the

carriage exerts on the horse. Since the sum of the forces is zero, there is no acceleration. What is

wrong with Chris’ reasoning?

Answer: He added the forces exerted on two different objects and did not consider all forces

exerted on the carriage. If you choose the carriage as the system object, then the horse pulls

forward on the carriage, and the mud exerts an equal magnitude backward, resistive force—a

different type of force (friction). These two horizontal forces add to zero, and the carriage does

not accelerate horizontally. If on the other hand we choose the horse as the system, the ground

exerts a forward force on the horse’s hooves (since the horse is exerting a force backward on the

ground) and the carriage pulls back on the horse with equal magnitude forces. The net horizontal

force is again zero, and the horse does not accelerate.

Tip! Notice that the forces in third law pairs in the example 2.10 are both either normal forces

(contact), or both are gravitational forces. The forces exerted on the same book (second law

forces) are different – one is normal (contact) and the other one gravitational.

Record animal jumper

According to Professor Malcolm Burrows, Head of the

Zoology Department at Cambridge University in the United Kingdom,

“the best jumper in the animal world is the froghopper.” (Fig. 2.22)

The froghopper (Philaenus spumarius) can jump 0.7 m vertically, over

100 times higher than its 6-mm length.


Example 2.11 Froghopper jump Professor Burrows found that the froghopper’s speed changed

from zero to 4 m/s in about –31 ms = 1 x 10 s —an acceleration of

24,000 m/s . The froghopper

achieves this huge acceleration using its leg muscles, which occupy 11 percent of its 12-mg body

mass. What is the average force that the froghopper exerts on the surface during the short time

interval while pushing off during its jump?

Sketch and Translate We start by constructing an initial-final sketch of the process (see Fig.

2.23a). Choose the froghopper as the system object of interest. This is an example of a case where

we cannot regard the froghopper as a point object. We need to consider the extension of its rear

legs as it pushes off the surface. We use a vertical -axisy with the positive direction pointing up

and the origin of the coordinate system at the surface. The froghopper’s acceleration while

pushing off has been given. Some students tend to determine the force that the bug exerts on the

surface by simply multiplying its mass times the known acceleration ( froghopperF m a" ). This is an

incorrect approach as two objects exert forces on the bug—we need to include both in the

analysis.

Figure 2.23(a)

Simplify and Diagram A motion diagram for the pushing off process is shown in Fig. 2.23b and a

force diagram in c. The surface exerts an upward normal force S on FN!

on the froghopper and

Earth exerts a downward gravitational force E on FF!

. Both forces need to be included in the

application of Newton’s second law to the problem. Note that the acceleration is up so the sum of

the -componentsy of the forces must also be up; the upward normal force exerted by the surface

is greater in magnitude than the downward gravitational force exerted by Earth.

Figure 2.23(b)(c)


Note in the force diagram that on the force diagram shows the normal force exerted by

the Surface on the Froghopper S on FN!

. But we need to determine the force that the Froghopper

exerts on the Surface F on SN!

. We can do it using Newton’s third law. These two forces are

Newton’s third law forces and thus they have the same magnitudes and point in opposite

directions. So if we determine S on FN , than the magnitude of F on SN is the same.

Represent Mathematically The process occurs along the vertical direction, so we apply the

vertical -componenty form of Newton’s second law:

E on F S on F F S on F

F F F

–y y y

y

F F N m g Na

m m m

# $ $" " " .

Rearranging the above, we get an expression for the force of the surface on the froghopper while

pushing off:

S on F F F FyN m a m g" $ .

Solve and Evaluate We can now substitute the know information in the above noting that the

froghopper mass –6

3 3

1 g 1 kg12 mg = (12 mg) = 12 x 10 kg

10 mg 10 g

/ 0/ 01 21 23 43 4

:

–6 2 –6 2

S on F (12 x 10 kg)(4000 m/s ) (12 x 10 kg)(9.8 m/s ) N " $

= 0.048 N + 0.00012 N = 0.048 N .

Note that this force that the surface exerts on the froghopper is 400 times greater than the force

that Earth exerts on the froghopper—so we could have ignored the force that Earth exerted on it

after all. Also, notice that the force magnitudes and both the motion and force diagrams are

consistent with each other – a nice way to check our work. See the Try It Yourself question to

consider what would happen if a human could do the same thing.

Try It Yourself: Suppose an 80-kg basketball player could push off a gym floor with a force (like

the froghopper) that is 400 times greater than the gravitational force that Earth exerts on him and

that the push off lasted 0.10 s (unrealistically short). How fast would he be moving when he left

contact with the floor?

Answer: About 40 m/s or 90 mph (faster if he pushed off longer exerting the same force). The

froghopper is a remarkable jumper.

Newton’s laws help us reconcile our intuition about common phenomena. For example,

people believe that when a person throws a ball upward, the ball continues to carry the force of

the thrower until it “runs out” at the top of the flight. Newton’s third law helps us understand that

the hand stops exerting a force on the ball right after the ball leaves contact with the hand. The

ball and the hand are two separate objects and exert Newton’s third law forces on each other. The

ball exerts a force on the hand as long as it touches the hand, and the hand exerts a force on the

ball only as long as it touches the ball. So why does the ball continue going up after it leaves


contact with the hand? It left the hand with considerable velocity. The downward gravitational

force caused a downward acceleration, which caused the ball to slow down and eventually stop (a

Newton’s second law application).

Review Question 2.9

Identify force pairs for the following interactions: A rollerblader and the floor; a volleyball player

and the volleyball; and a speeding up car’s tires and the road.

2.10 Putting it all together—seat belts and air bags

At the beginning of the chapter we posed a question about airbags. How do they save lives?

We now have all the physics needed to investigate this question. Consider Fig. 2.24. A driver’s

air bag is like a balloon with heavy-walled material that is packed in a small box in your steering

wheel. If your car has an acceleration of magnitude 10g [that is,

, -2 2 210 9.8 m/s 98 m s 100 m s" . ] or more due to a rapid decrease in speed during a

collision, the bag inflates with nitrogen gas in about 0.04 s and forms an air cushion for the

driver’s chest and head. The bag has two important effects:

1. It spreads the force that stops the person over a larger area of the body (very important for

improving the chance of survival); and

2. It increases the stopping distance and consequently the stopping time interval and thus

reduces the average force stopping the driver.

Figure 2.24 Airbag stops driver

Why is spreading the stopping force over the air bag an advantage? We were very

explicit earlier in the chapter about modeling objects as point-like objects. It is definitely not

reasonable to model the human body as a point-like object during a car collision. For example, if

a passenger uses only seat belts, his head is not belted to the seat and tends to continue moving

forward during a collision even though his chest and waist are restrained. To stop the head

without an air bag, the neck must exert considerable force on the head. This can cause a


dangerous stretching of the neck, a phenomenon known as “whiplash”. This can cause

considerable damage to neck muscles and possibly to the spinal cord. The air bag exerts a more

uniform force across the upper body and head and helps make all parts stop together. Actually, if

all parts of the body are moving together, it becomes more reasonable to model the body as a

point-like object.

How does the air bag increase the stopping distance? The answer depends on the type of

car, on the type of collision, and on the car’s initial speed. Imagine that we videotape collisions of

a test car with a crash test dummy and then analyze the videotape with a ruler that can measure

distances in centimeters. Imagine that the test car is traveling at a constant speed of 13.4 m/s (30-

mph) until it collides head-on into a concrete wall. The front of the car crumples about 0.70 m.

Since the crash test dummy is rigidly attached to the car’s seat and is further protected by the

rapidly inflating airbag, the dummy also travels about 0.70 m before coming to rest. Without an

air bag or a seat belt, the dummy would continue to move forward at the initial velocity of the car.

The dummy would then crash into the steering wheel or windshield and stop in a distance much

shorter than 0.70 m. If the stopping distance is smaller, the acceleration is greater, and therefore

the force that is exerted on the dummy is greater. This is why it is much safer to be in a car whose

front end collapses significantly during a collision and is equipped with seat belts and airbags. It

translates into a longer stopping distance. What are the stopping distance, the stopping time

interval, and the average stopping force exerted on the crash test dummy during a collision? Here

are the details for a collision

We observe that during the 0.04 s while the air bag is inflating, the front of the car

collapses and the body of the car moves forward about 0.43 m. The dummy flies forward at speed

13.4 m/s (30 mph, the initial speed of the car before the collision). After the dummy reaches the

air bag, the car collapses an additional 0.70 m – 0.43 m = 0.27 m. Auto makers often design the

steering wheel so that it sinks about 0.20 m into the steering column when the driver starts

pressing against the air bag. Thus, the dummy can now stop in 0.27 m + 0.20 m = 0.47 m. Finally,

after the initial inflation of the bag, it starts to leak air and consequently decreases in thickness

(from about 0.26 m to 0.08 m). Thus, the dummy gets to move forward an additional 0.18 m.

Combining everything, we find that the dummy stops in a distance of

0.27 m + 0.20 m + 0.18 m = 0.65 m . That’s a lot better than crashing into a hard surface!

Let’s estimate the average force exerted by the air bag on the body during such a stop.

Example 2.12 Force exerted by airbag on driver during collision A 60-kg crash test dummy

moving at 30 miles per hour (13.4 m/s) stops during a car collision in a distance of 0.65 m.

Estimate the average force that the air bag and seat belts exerts on the dummy.

Sketch and Translate The situation is sketched and labeled in Fig. 2.25a. We choose the crash test

dummy as the system object. The positive x direction will be in the direction of motion and the

origin will be at the position of the dummy at the start of the collision.


Figure 2.25(a)

Simplify and Diagram We model the dummy as a point-like object and assume that the force

exerted by the airbag on the dummy is constant. The primary force exerted on the dummy while

stopping is the force due to the air bag and seatbelts A on DF!

, shown in the force diagram in Fig.

2.25b. We ignore the downward gravitational force that Earth exerts on the dummy E on DF!

and

the upward normal force that the car seat exerts on the dummy S on DN!

since they balance and do

not contribute to the acceleration of the dummy.

Figure 2.25(b) Analysis of dummy during car collision

Represent Mathematically To determine the dummy’s acceleration, we use kinematics:

, -

2 2

0,

02

x x

x

v va

x x

%"

%

Once we have the dummy’s acceleration, we can find the force exerted by the air bag and seat

belt on the dummy using the x component of Newton’s second law.

A on D A on D A on Dx

D D D

A on D D x

xF F F

am m m

F m a

%" " " %

5 " %

Solve and Evaluate We know that 0 = +13.4 m/sx

v and = 0

xv (the dummy has stopped). The

initial position of the dummy is 0 0x " and the final position s = 0.65 mx . The acceleration of

the dummy while in contact with the air bag and seat belts is:

2 220 (13.4 m/s)

138 m/s2(0 m 0.65 m)

xa

%" " %

%

ALG 3.4.7-3.4.14


Thus, the magnitude of the average force exerted by the air bag and seat belts on the dummy is:

2

A on D (60 kg)( 138 m/s ) = 8300 NF " % % .

This is a huge force —8300 N(1lb/4.45 N) = 1900 lb , or almost one ton. Is this estimate

reasonable? The magnitude is large but experiments with car dummies in the real world are

consistent with a force this large in magnitude.

Try It Yourself: Find the acceleration of the dummy and the magnitude of the average force

needed to stop the dummy if the airbag does not leak.

Answer: –191 m/s2 and 11,400 N.

Review Question 2.10

Explain how an airbag reduces the stopping force exerted on the driver of a car.


Summary

Words Pictorial and physical

representations

Mathematical

representations

System and environment A system is the object of interest. We circle the system in a sketch of a process. Objects that are not part of the system are external and might interact with the system and affect its motion.(2.1)

Force The force that one object exerts on another characterizes an interaction between the two objects (a pull or a push) denoted by a symbol with an arrow above it and with two subscripts indicating the two interacting objects (2.1).

E on OF!

Interacting objects

Gravitational force E on OF!

The gravitational

force that Earth exerts on an object of mass m when on or near the earth’s surface

depends on the gravitational constant g of

Earth. If on or near the Moon or some other planet, the gravitational constant near those objects is different. (2.7)

Magnitude

E on OF mg"

where 2

9.8 m/s = 9.8 N/kgg "

when object is on or near the earth’s surface.

A Force Diagram (see page 9) indicates with arrows the forces that external objects exert on the system object. The arrows point in the directions of the forces. The arrow lengths indicate roughly the relative magnitudes of the forces. (2.1)

Relating forces and motion The v'!

arrow in

an object’s motion diagram is in the same direction as the sum of the forces that other objects exert on it. (2.3)

Inertial reference frame An inertial reference frame is one in which an object’s velocity does not change if no other objects interact with it or if the sum of all these interactions is zero. (2.5)

Newton’s first law of motion If no other objects interact with an object of interest or if the forces add to zero, then the object continues moving at constant velocity (as seen by observers in inertial reference frames). (2.5)

If 0x

F& " , then

xv = constant

(same for y direction)

Mass Mass m characterizes the amount of “material” in an object and its resistance to a change in motion. (2.6)

Chapter 2 Newtonian Mechanics I

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Transcript of Chapter 2 Newtonian Mechanics I