Download Chapter 2 Newtonian Mechanics I

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
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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.
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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
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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)
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pushing on the bowling ball (the system object), the full symbol is FH on BB . The force that the
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hand exerts on the basketball is labeled FH on B . 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
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Earth exerts on the ball ( FE on B ). 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
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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
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!
( FA on BB or FA on B ). 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
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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.
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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.
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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
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exerting a force FT on A and book B pushes down on the top
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surface of book A exerting force FB on A . In addition, Earth
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exerts a downward force on book A FE on A . 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
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book A ( FE on A ). C does not directly touch A and exerts no force on A. However, C does push
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down on B, so B exerts a greater force on A ( FB on A ). Because the downward force of B on A is
greater, the table has to exert a greater upward force on book A (
!
F
T on A
), See Fig. 2.6.
Figure 2.6 Force diagram for A with books B and C on top
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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
FE on S " 100 N and you exert an upward force of FY on S " 150 N (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?
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Remember, that force is a vector. To add two vectors to find the sum #F " FE on S $ FY on S , 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 #F y " $50 N . If the positive direction is down, the sum of the forces has a y
component #F y " %50 N . The skill box below shows you how to add forces graphically.
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Figure 2.7 Sum of force exerted on suitcase
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Skill box: Adding forces graphically #F " F1 on O $ F2 on O
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:
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&F
n on O
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" F1 on O $ F2 on O $ ... $ Fn on O
(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.
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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
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Simplify and Diagram Assume that Earth exerts the same force on each ball
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FE on 1Ball 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
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FS on Balls that is equal in magnitude and opposite in direction to the force that Earth exerts on the
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balls FE on Balls 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 ( FS on Balls ) and spring stretch ( y ) with a F -vs-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)
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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
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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.
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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
Force diagrams for first and third positions
3. A ruler lightly pushes
the rolling bowling ball in
the direction of its
motion.
Motion diagram
Force diagrams for first and third positions
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
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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
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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:
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Hypothesis 1: An object’s velocity v always points in the direction of the sum of the forces
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#F that other objects exert on it.
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Hypothesis 2: An object’s velocity change 'v always points in the direction of the sum of
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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.
The sum of forces is
zero so the puck’s
velocity should not
change; it should
continue moving the
same way.
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.
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,
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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
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The puck
continues to
move at
constant
velocity.
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'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.
Thus the ball should
immediately begin
moving downward
after you release it.
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.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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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
ALG
lands in a deep pile of leaves. Is the force that the leaves exert on the man greater than, equal to,
1.4.1or less than the gravitational force that Earth exerts on him while he is sinking 0.5 m into the
1.4.7
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.
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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 FE on M and the
!
leaves exert an upward force FL on M 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 ( FL on M ( FE on M ), 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 ( FL on M " FE on M ). How might you convince your friend that your
answer ( FL on M ( FE on M ) 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 ( #F " 0 ). Thus, on the motion diagram the velocity change arrow must also be zero
!
( 'v " 0 ). 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.
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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 ( #F " 0 ). Thus, a motion diagram for the woman must have a zero velocity change
!
( 'v " 0 ).
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
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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ALG
1.4.11.4.7
!
point down. This means the magnitude of the upward force FFloor on W that the elevator floor
!
exerts on her must now be less than the magnitude of the downward force FE on W that Earth
exerts on her ( FFloor on W ) FE on W ). Notice that the magnitudes of forces do not have arrows over
them (the magnitudes are positive numbers and have no direction).
a)
b)
c)
d)
e)
Let us summarize the skills you should have learned so far:
How to construct a motion diagram or how to interpret an already constructed motion diagram.
How to draw a force diagram or how to interpret an already constructed force diagram.
How to check the consistency of the motion diagram and the force diagram.
How to construct a motion diagram if you have a force diagram, and vice versa.
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 FC on El compare to the magnitude of the
!
downward force exerted by Earth on the elevator FE on El ? 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.
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Observational Experiment Table 2.3 Two observers watch the same mug.
Observational experiment
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.
Analysis done by each observer
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.
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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 noninertial 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
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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
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.
Analysis
Shown below are velocity and
acceleration-versus-time graphs for two
of the experiments when different
constant magnitude forces were exerted
on the cart.
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.
We see that the greater the force, the
greater the acceleration - in direct
proportion to the force.
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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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
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
Analysis
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
1
proportional, we plot a vs .
n
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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ALG
3.1.2
Pattern
As the graph a vs
1
n
is a straight line, we conclude that a *
1
n
.
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 platinumiridium 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:
a*
1
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.
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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.
!
Rearrange the above to get msystem asystem
!
#F on system
!
asystem *
(2.3)
msystem
!
* #F on system . 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):
1 N = 1 kg•m/s 2 .
(2.4)
!
The above important relationship, rewritten with the equality sign asystem "
!
#F on system
msystem
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 asystem 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:
!
#FOther objects on System object
!
asystem "
msystem
(2.5)
The sum of all the forces being exerted on the system by other objects is:
!
!
!
#FOther objects on System object " F1 on System $ F2 on System $ ... $ Fn on System
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 150N 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:
!
#FOther objects on System object
!
asystem "
" 0 . 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
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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!
#FOther objects on System object
!
even if the tiniest force is exerted on it: asystem "
" + . Both extreme cases
0
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).
!
#Fon system
!
asystem "
msystem
Cause
Effect
!
Compare this relation to the definition of acceleration a "
!
'v
, which is called an
't
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):
ax "
5 m/s - 2 m/s
" $1 m/s 2
3s
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):
5500 N + (-5000 N)
" $1 m/s 2 .
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)]:
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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asystem x "
#FOther on System x
msystem
(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 FY on S " 150 N . Earth pulls down exerting a force of magnitude FE on S " 100 N . The
!
upward force is larger thus the suitcase will have an upward acceleration a .
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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Figure 2.15(b)
Represent Mathematically The vector form of Newton’s second law applied to suitcase and this
process, is given below:
!
! !
! #F FY on S $ FE on S
aS "
"
mS
mS
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):
aS y "
#Fon S y
mS
"
FY on S y $ FE on S y
mS
"
, $ FY on S - $ (% FE on S ) " FY on S
mS
% FE on S
mS
The y components of the forces exerted on the suitcase are Your upward pull on the Suitcase
FY on S y " $ FY on S " $150 N and Earth’s downward pull on the Suitcase
FE on S y " – FE on S " –100 N .
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:
1
y – y0 " v0 y t $ a y t 2 .
2
Substituting the initial velocity v0 y " 0 into the above and rearranging to find the time t when
the suitcase is at y " 0.50 m (its initial position is y0 " 0 ), we get:
t"
2( y – y0 )
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:
aS y "
FY on S % FE on S 150 N % 100 N
"
" $5 m/s 2
mS
10 kg
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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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:
t"
2( y – y0 )
2(0.5 m – 0)
"t "
" 0.45 s . 0.5 s .
a
5 m/s 2
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 $5 m/s 2 = (+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
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!
surface, Earth, and you. You exert a horizontal force on the Mower FY on M ; Earth exerts a
!
downward gravitational force on the Mower FE on M . The Grassy surface exerts a force on the
!
Mower that we will represent with two force arrows —a normal force N G on M perpendicular to
!
the surface (in this case it points upward) and a horizontal friction force f G on M opposite the
!
!
direction of motion. The downward force FE on M and the upward normal force N G on M 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:
aM x "
#Fx FY on M x $ f G on M x
"
.
mM
mM
The x-component of your force on the mower is positive since that force points in the positive xdirection ( FY on M x " $ FY on M ). 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 ( f G on M x " % f G on M ).
Therefore the acceleration of the mower is:
aM x "
FY on M $ (– f G on M ) FY on M – f G on M
"
mM
mM
Solve and Evaluate We can now determine the mower’s acceleration by substituting the known
information into the above equation:
aM x "
FY on M – f G on M 100 N – 80 N
"
" $0.7 N kg " $0.7 m s 2
mM
30 kg
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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 9.8 m s 2 . A force diagram for an object falling inside the tube has only
!
the force exerted by Earth on the Object FE on O 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:
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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aO y "
F
1
1
FE on O y "
, $ FE on O - " $ E on O
mO
mO
mO
Every object dropped in our experiment had the same free fall acceleration g = 9.8 m s 2 , 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:
aOy "
FE on O
mO
"
FE on O
mO
"g
This reasoning leaves just one possibility for the magnitude of the force exerted by Earth on the
object:
FE on O " mO g " mO , 9.8 m/s 2 - .
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:
FE on O " mO g
(2.7)
where g " 9.8 m/s 2 " 9.8 N kg 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 9.8 m/s 2 when calculating
the gravitational force.
ALG 3.3.7
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.
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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
1 ft/s 2 or 1 lb = (1 slug)(1 ft/s2 ) . Note that 1 slug = 14.6 kg and 1 foot = 0.31 m . Thus,
1.0 lb = (1.0 slug)(1.0 ft/s2 ) " [(1.0 slug)(
14.6 kg
0.31 m
)][(1.0 ft/s2 )(
)]
slug
1.0 ft
" 4.45 kg!m/s 2 " 4.45 N
.
For example in the British units, suppose that the gravitational force that Earth exerts on a person
/ 4.45 N 0
2 = 710 N.
3 1.0 lb 4
is 160 lb. This force in newtons is (160 lb) 1
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
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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
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 ydirection) while stopping using kinematics:
!
!
NS-G on H on Holmes. Earth exerts a downward gravitational force FE on H .
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 NS-G on H must be
longer to match the motion diagram, which shows the velocity change arrow
pointing up.
v y % v0, y
2
ay "
2
.
2( y % y0 )
From the motion and force diagrams we know that the y-component of
Newton’s second law with the positive y-direction up is:
#Fon H y
ay "
m ,
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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.
y
The component of the force exerted by shrubbery-ground is
N S-G on H y " $ N S-G on H
y
and the component of the force exerted by Earth
F
" % FE on H " – mH g
. Therefore
force is E on H y
ay "
N S-G on H y $ FE on H y
mH
"
( $ N S-G on H ) $ ( % FE on H )
mH
5 N S-G on H " mH a y $ mH g
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.
"
$ N S-G on H % mH g
mH
.
Holmes’ average acceleration is
0 % , %36 m/s 2
ay "
2 , 0 % 2.0 m -
2
" $324 m/s
2
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:
N S-G on H " mH a y $ mH g " (70 kg)(324 m/s ) + (70 kg)(9.8 N/kg)
2
= 22,680 kg 6 m/s + 686 N = 23,366 N = 23,000 N
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.
2
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
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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 FE on Y = myou g = 490 N , and your acceleration is: (a) a y = +1.0 m/s 2 (the upward velocity
is increasing); (b) a y = 0 (v is a constant 4.0 m s upward); and (c) a y = –1.0 m/s 2 (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
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!
!
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:
aY y "
#Fy
mY
"
FE on Y y $ N S on Y y
mY
The y-component of the force that Earth exerts on you FE on Yy " – mg is negative and the ycomponent of the normal force that the scale exerts on you NS on Yy " $ NS on Y is positive. Thus
we can rewrite the above equation as:
aY y "
– mg $ NS on Y
mY
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Multiplying both sides by mY we get aY y mY " – mg $ NS on Y . We can now move % mg to the
left side: mY aY y $ mg " N S on Y or
NS on Y " mY aY y $ 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:
NS on Y " mY aY y $ mg " mY aY y $ 490 N.
(a) Early in the trip, the elevator is speeding up and its acceleration is aY y " +1.0 m/s 2 . Then,
the force exerted by the scale on you should be:
NS on Y = mY aY y $ 490 N = (50 kg)(+1.0 m/s 2 ) + 490 N = 540 N.
(b) In the middle of the trip when the elevator moves at constant velocity, your acceleration is
zero and the scale should read:
NS on Y = mY aY y $ 490 N = (50 kg)(0 m/s 2 ) + 490 N = 490 N .
(c) When the elevator is slowing down near the end of the trip, its acceleration points downward
and is a y " –1.0 m/s 2 . Then, the force exerted by the scale on you should be:
NS on Y = mY aY y $ 490 N = (50 kg)(–1.0 m/s 2 ) + 490 N = 440 N .
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 –1.0 m s 2 , then
moves down at constant velocity, and finally slows its downward trip with an acceleration of
$1.0 m s 2 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
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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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 ( a y " –9.8 m s 2 , 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:
ax "
– fS on O
(50 kg)
NS on O – (50 kg)(9.8 N/kg) " 0
0 – (20 m/s)2 = 2ax (25 m)
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.
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Solve We can solve the second equation for the normal force:
NS on O = (50 kg)(9.8 N/kg) " 490 N .
We can solve the third equation for the acceleration ax:
ax "
–(20 m/s) 2
= –8.0 m/s 2
2(25 m)
.
Now substitute this result in the first equation to find the magnitude of the force fS on O:
– fS on O = (50 kg)(–8.0 m/s2 ) " –400 N .
Represent Mathematically The third equation in the problem statement looks like the application
of the following kinematics equation to the problem process:
vx2 – v02 x " 2ax ( x – x0 )
By comparing the above general version of the kinematics equation to the third provided
equation, we see that: the final velocity vx " 0 , the initial velocity v0 x " $20 m/s , and the
displacement of the object was (x – x0 ) = 25 m . 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)
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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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
3.3.4 –
3.36
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ALG
3.1.3;
3.1.4
Observational Experiment Table 2.6 Analyze the force that two carts exert on each other.
Observational experiment
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
Analysis
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 F1 on 2 had the same magnitude but
!
pointed in the opposite direction from the force that cart 2 exerted on cart 1 F2 on 1 .
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.
!
!
Fobject 1 on object 2 " % Fobject 2 on object 1
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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:
!
!
Fobject 1 on object 2 " % Fobject 2 on object 1
(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:
aPaddle
"
FBall on Paddle
m
and
a
Ball
"
FPaddle on Ball
Paddle
mBall
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
Etkina/Gentile/Van Heuvelen Process physics 1/e Chapter 2
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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 FE on B , and the table exerts an upward normal (contact) force on
!
the book N T on B . 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 ( FE on B ). According to Newton’s third law, the book
!
!
must exert an equal magnitude upward gravitational force on Earth ( FB on E " % FE on B ), as shown
!
in Fig. 2.24c. The table exerts a contact upward force on the book ( N T on B ), so the book must
!
!
exert an equal magnitude contact downward force on the table ( – N B on T " N T on B ).
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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.
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Example 2.11 Froghopper jump Professor Burrows found that the froghopper’s speed changed
from zero to 4 m/s in about 1 ms = 1 x 10 –3 s —an acceleration of 4,000 m/s 2 . 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 y -axis 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 ( F " mfroghopper 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 N S on F on the froghopper and
!
Earth exerts a downward gravitational force FE on F . 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 y -components 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)
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Note in the force diagram that on the force diagram shows the normal force exerted by
!
the Surface on the Froghopper N S on F . But we need to determine the force that the Froghopper
!
exerts on the Surface N F on S . 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 NS on F , than the magnitude of N F on S is the same.
Represent Mathematically The process occurs along the vertical direction, so we apply the
vertical y -component form of Newton’s second law:
ay "
#Fy
mF
"
FE on F y $ NS on F y
mF
"
– mF g $ NS on F
.
mF
Rearranging the above, we get an expression for the force of the surface on the froghopper while
pushing off:
NS on F " mF aF y $ mF g .
Solve and Evaluate We can now substitute the know information in the above noting that the
/ 1 g 0 / 1 kg 0
–6
2 1 3 2 = 12 x 10 kg :
3
10
mg
10
g
3
43
4
froghopper mass 12 mg = (12 mg) 1
NS on F " (12 x 10 –6 kg)(4000 m/s 2 ) $ (12 x 10–6 kg)(9.8 m/s2 )
= 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
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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,
10 , 9.8 m/s 2 - " 98 m s 2 . 100 m s 2 ] 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
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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 (30mph) 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.
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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 FA on D , shown in the force diagram in Fig.
!
2.25b. We ignore the downward gravitational force that Earth exerts on the dummy FE on D and
!
the upward normal force that the car seat exerts on the dummy NS on D 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:
ax "
vx2 % v0,2 x
2 , x % x0 -
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.
ax "
FA on D x % FA on D
F
"
" % A on D
mD
mD
mD
5 FA on D " %mD ax
Solve and Evaluate We know that v0 x = +13.4 m/s and vx = 0 (the dummy has stopped). The
initial position of the dummy is x0 " 0 and the final position s x = 0.65 m . The acceleration of
the dummy while in contact with the air bag and seat belts is:
ax "
02 % (13.4 m/s) 2
" % 138 m/s 2
2(0 m % 0.65 m)
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ALG
3.4.73.4.14
Thus, the magnitude of the average force exerted by the air bag and seat belts on the dummy is:
FA on D " %(60 kg)( % 138 m/s2 ) = 8300 N .
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.
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Summary
Words
Pictorial and physical
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).
!
Gravitational force FE on O 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)
Mathematical
representations
!
FE on O
Interacting objects
Magnitude
FE on O " mg
where
2
g " 9.8 m/s = 9.8 N/kg
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 & Fx " 0 , then
vx = 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)
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