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Goodman and Zavorotniy - 78

Dynamics

Introduction

While kinematics allows us to describe motion, dynamics lets us determine how that

motion will change. Dynamics is grounded in our intuition of how the world works, but

reaches beyond that intuition. It represents an approach that turns out to be so valuable

that it can be described as the very foundation of physics. It is the first step in our

journey to describe nature in a way that reaches beyond our common sense.

We all developed our common sense about how the world works based on the instincts

and understandings that have allowed humans to survive on this planet for so many

years. Those understandings have served us well throughout that time and they are

correct within their context. If they were not, we could not have relied on them to

survive. However, as we strive to extend our understanding beyond what is necessary

to survive, we will find that we need to question some of the very ideas that we have

relied upon.

Concepts that seem very familiar, such as force, mass, etc. will be revisited and defined

in very precise ways that allow them to be used to understand our universe in a new

way. While doing this, you will still need to maintain your intuition about how things

work. You will not be asked to discard those intuitions. In fact, we will rely on your

making use of them. But you will need to flexible enough to reach beyond them to

grasp their deeper meaning.

© Goodman and Zavorotiny 2007

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Inertial Reference Frames

The framework that will be developed in this chapter is built on Newton’s three laws of

motion. Those laws were developed to work in inertial reference frames; the same

reference frames that were discussed in the last chapter. An inertial reference frame is

anyplace that is not accelerating. For instance, if you are sitting in a room reading this

book, you can consider yourself in an inertial reference frame. Similarly, if you are in a

car driving at a constant velocity on a smooth road, you are also in an inertial reference

frame. The laws of motion that we will be developing in this chapter apply to you. On

the other hand, if you are in a car that is accelerating onto the highway, you are not in

an inertial reference frame. The system of dynamics being developed in this chapter

has to be modified to explain the motion of objects in that car.

A good test to see if you are in an inertial reference frame will turn out to be whether the

laws that we are developing in this chapter work for you. If you do an experiment and

find that Newton’s laws of motion are obeyed, then you are in an inertial reference

frame. If they are not obeyed, then you are not in one. That’s a bit circular, but it works

within the self-consistent system that we will be developing.

It turns out that there are no completely valid inertial reference frames. The room that

you are sitting in is spinning once around the axis of the earth every 24 hours. And the

earth is spinning around the sun once every 365 days. And the sun is spinning around

the center of the galaxy once every __ million years. And even our galaxy is moving

with respect to the Local Group of galaxies. As we will learn later, in our study of


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circular motion, all of these rotations involve accelerations, since the direction of our

motion keeps changing. As a result, there are no pure inertial reference frames easily

available to us. However, these effects turn out to be very small compared to the

problems that we will be solving so we can treat the room that you are sitting in an

inertial reference frame for most purposes. However, it is worth taking a moment to

consider the wild path that you and your room are taking through space as you read

this.

Force

To a great extent your intuitive understanding of force will work in dynamics. Whenever

you push on something you are exerting a force on it. Or when something pushes on

you, it is exerting a force on you. When you drop an object, you can think of the earth

as pulling it down, exerting a force on that object. These are a few examples of forces.

You’ll be learning about a number of forces in this book. Forces like friction, electricity,

magnetism, gravity, etc. However, there are some common features of all forces that

we will introduce here and develop further as they become necessary.

A very important property of a force is that it always involves two interacting

objects. You can exert a force by pushing on something, but it wouldn’t mean anything

to say that you are exerting a force by pushing on nothing. Similarly, something can

push on you, but it’s meaningless to say that you are experiencing a force because

“nothing” is pushing on you. For a force to exist there must be something pushing and

something being pushed. A force always involves the interaction between two different


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things. So when we describe a force we will always need to identify the two objects that

are involved in that force.

As problems get more complicated we will have to keep track of numerous forces. We

will use subscripts to do that. The first subscript will represent the object being pushed

while the second will indicate the object doing the pushing. So, for example, if I am

pushing an object with my hand, I would describe the force on the object by: Fobject hand or

Foh. Or if I want to describe the force on a book due to earth pulling down on it I would

write Fbook earth or Fbe. By reversing the subscripts I can describe the force on my hand

due to the object as Fho or the force on the earth due to the book as Feb. The subscript

for the object experiencing the force is always written first while the object

exerting the force is written second. The relationship between these pairs of forces,

for instance Fho and Foh, will be discussed when you are introduced to Newton’s Third

Law a bit later in this chapter.

The SI unit of force is the Newton (N). When a force is being specified its size, or

magnitude, is given in Newtons. However, a critical property of forces is that they

are vectors. As in the case of the vectors discussed in the prior chapter, a force must

be described by giving both its magnitude and its direction. So it is not sufficient to say,

for example, that Fbe (the force on a book due to the earth) is 20 N. You need to say

that Fbe is 20 N towards the center of the earth. You need to specify both the size of the

force (20N) and its direction (towards the center of the earth).


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When more than one force acts on an object those forces must be added together as

vectors. One implication of this is that forces that act in the same direction will add

together, while forces that act in opposite directions will reduce each other. The critical

determiner of how an object moves will not be any one force that acts on it, but rather,

the sum of all the forces that act on it. The vector sum of all the forces that act on

an object is called the net force on the object. This will be discussed further as part

of Newton’s Second Law.

Newton’s First Law

Galileo was actually the developer of the principal of inertia that was to become

Newton’s first law. Newton used that principal as the first of the three pillars that was to

support his theory of dynamics, which is how it came to be known as Newton’s First

Law. However, the thinking that developed the underlying principal came from Galileo.

His thinking represents an excellent example of how physics both uses human intuition

and moves beyond it.

Our common sense tells us that an object will naturally stay at rest unless something, or

someone, is pushing it. If a book were seen to be sliding along the floor, we would

wonder why. Who pushed it? However, if a book were to remain still on the floor, we

would not wonder why. It would seem perfectly reasonable that it remain in one place.

In fact, if we came back to the room, and it wasn’t there, we would reasonably wonder

who moved it.


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Our intuition tells us that the natural state of an object is at rest. Even if something is

moving, we expect that it will eventually come to a stop. If you give that book, which

had been sitting on the floor, a quick shove it will move for some distance and then

come to a stop. If you want to keep it moving you know that you’ll have to push it again.

Lift the book up off the floor and then let go of it. It’ll fall to ground as our intuition tells

us is natural for unsupported objects. But upon reaching the floor it will once again

come to a stop. Moving objects just naturally seem to come to rest and, once at rest,

they don’t start moving again on their own. Galileo was the first person that we know of

to question this idea and demonstrate that it is limited to the context of our everyday

experience. He recognized that moving beyond that intuition would lead to a much

deeper understanding of how the world works. Let’s follow his general approach to this

question.

First, imagine a device that consists of two ramps that are connected together (Figure

1). If we were to release a ball on the first ramp it would roll down that ramp and then

up the second ramp to a height that is less than the original height from which it was

released. If we make the ball and the ramp smoother and increase the density of the

ball, making it heavier for its size, we will find that it gets closer and closer to its original

height. The more ideal we make the ramps and the ball - the closer the ball gets to

reaching its original height on the second ramp. We can now imagine the ideal case; if

we were to make the ball and the ramp perfectly smooth and eliminate all air resistance


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that the ball would make it back to its original height. This step of doing practical

experiments but then extending the results by imagining a more ideal set of

circumstances was a critical element of Galileo’s genius. The critical step here is to

recognize that in the ideal case the ball will keep moving until it reaches its original

height.

Now examine the Devices “a” and “b” in Figure 2. While the “launching” ramp has the

same slope, the “target” ramp has less of a slope in each device. If the ball is to reach

the same height, it must move further to the right in each succeeding case. Now,

consider Device “c” in Figure 2, where the slope of the second ramp is zero. Instead of

a ramp, it is just a continuation of the straight part of the track. The ball could never get

back to its original height, so it would keep traveling to the right. If we imagine the ideal

case, it would keep moving forever, without anything pushing it. This directly opposes

the idea that an object must be pushed by something to keep moving. In fact, this

would indicate that if nothing is pushing an object its velocity will not change. Galileo

was perhaps the first to recognize that constant velocity is the natural state of an

object.

How do we reconcile this with our common sense perspective that something comes to

a stop unless it is being pushed? In our everyday experience there are a number of

forces that affect an object that we don’t think about because they’re always there. One

of these is friction. When you gave the book a shove across the floor, it came to a stop

because the friction between the floor and the book pushed on the book. Imagine now


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that instead of the floor, we were to give an object a shove on a sheet of ice. It will

move much farther than it did on the floor. But it will eventually come to stop. Now use

your imagination to picture a surface with no friction. Can you imagine that once you

shove the book on that surface, it will simply keep moving without coming to a stop…

ever. This turns out to be a core concept to physics, so if you can keep that image in

your mind it will prove very helpful to you.

Newton adopted and formalized this principal of Galileo’s and it is now known as

Newton’s First Law, or the Law of Inertia. An object will maintain a constant

velocity, both speed and direction, unless another object exerts a net force on it.

Please note that only a non-zero net force will result in a change of velocity. That

means that if the vector sum of all the forces acting on an object is zero, it will maintain

a constant velocity. That could be the case because there are no forces acting on an

object or because there a number of forces that are adding up to zero, canceling each

other out. In this circumstance the object is said to be in equilibrium.

How does this allow us to reconnect back to our intuition? Let’s look at a different case,

one where I push a book across the floor with a constant velocity. If the book is moving

with a constant velocity that means that there must be zero net force acting on it. That

means that if I add up all the forces acting on the book, they must add up to zero. I

know that I’m pushing the book in one direction. There must be an equal and opposite

force, friction, pointed in the opposite direction. The book maintains a constant velocity

not because there are no forces acting on it, but rather because the sum of all the


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forces acting on it is equal to zero. Our intuition says that I have to keep pushing

something to keep it moving. But it turns out that that’s only true because friction is

pushing the other way. If there were no friction, I wouldn’t have to keep pushing. But

we live in a world where friction is everywhere around us. So our intuition works, but

only in this context.

Newton’s Second Law

We now know that an object maintains a constant velocity unless a net force acts on it.

But how does it behave if a net force does act on it? The answer is that an object

subject to a net force accelerates. If I push on an object I can make it start moving,

thus changing its velocity from nothing to something. Acceleration is just the change in

velocity over a period of time. So if pushing something can make it start moving then a

net force must result in acceleration. Note, however, that only a net force results in

acceleration. If I push on an object in one direction and something else pushes that

object with an equal force in the opposite direction, no acceleration will result. So the

first part of Newton’s second law is that the acceleration of an object is

proportional to the net force acting on it.

But not all objects will accelerate the same amount when subject to the same net force.

An object’s acceleration also depends on its mass. Mass is a fundamental property of

matter. Understanding what mass is and where it comes from is a very complex issue

and would involve understanding Einstein’s General Theory of Relativity (his Special


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Theory of Relativity won’t do it!). That is well beyond the bounds of this book.

However, there are some properties of mass that you will need to understand.

1. Mass is intrinsic to an object. It does not depend on where the object is located.

No matter whether it is; on earth, in outer space or deep under the ocean, the

mass of an object will not change.

2. An object’s acceleration is inversely proportional to its mass. The more mass an

object has, the less it will accelerate when subject to a given net force. If you

have a certain amount of net force, you can compare the masses of different

objects by observing how they accelerate. The more massive an object is, the

less it will accelerate. If you double the mass of an object, it will accelerate half

as much when the same net force acts on it.

3. Mass is a scalar, it is not a vector. There is no direction associated with a mass,

only magnitude.

So Newton’s Second Law says that an object’s acceleration is proportional to the net

force acting on it and is inversely proportional to its mass.

a = Fnet / m

This is often written as:

Fnet = ma or ΣF = ma

In the second expression the Greek symbol “Σ”, sigma, acts as an instruction to “add up

all” the forces acting on the object. That’s the same process as is used to find the net

force, so Fnet is the same thing as ΣF. There are two reasons that we will often use ΣF

instead of Fnet. The first reason is that the “Σ” will serve to remind us that we have to

find all the forces and then add them together. The second reason is that that leaves us


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room for subscripts. Remember that there can be multiple forces acting on a single

object. We need to add all the forces together that are acting on that object to get ΣF.

But when we do that we have to remember which object those forces are acting on. So

for example, let’s say that there are three objects that are interacting. For simplicity let’s

name them by giving them each a number. Then to evaluate the motion of object #1 we

would use the second law as follows.

ΣF1 = m1a1

which in this case, with a total of three objects, means that

F12 + F13 = m1a1

While this would allow us to determine the acceleration of the first object, we could

equally well find the acceleration of the other two objects with the analogous equations.

(In general, the acceleration of each if the three object will not be the same.)

ΣF2 = m2a2 or F21 + F23 = m2a2

ΣF3 = m3a3 or F31 + F32 = m3a3

Once again, please remember that forces must be added as vectors, not simply as

numbers.

Units of Mass and Force

The SI unit of mass is the kilogram (kg). Since it is a fundamental unit of measure its value has been defined by international agreement. As is true for all fundamental units, such as the meter or the second, its value cannot be derived.

The unit of force is the Newton (N). The Newton is not a fundamental unit as it can be derived, using Newton’s Second Law, from the fundamental units of meters, kilograms and seconds. The SI units of mass are kilograms (kg) and of acceleration are meters per second squared (m/s2). Out of respect for Isaac Newton (1642-1727), the unit of force is named the Newton.


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Since ΣF = ma

Then the units of force must be:

N = (kg)(m / s2)

N = kg∙m/s2

One other important aspect of Newton’s second law is that the acceleration of an object

will always be in the same direction as the net force acting on it. That must be true

since in the expression Fnet = ma, only “Fnet” and “a” are vectors while “m” is not. That

means that “m” will change the size of the acceleration, but not its direction. The net

force and acceleration will always have the same direction.

Example 1: Determine the acceleration of a ball whose mass is 20 kg and which is

subject to a net force of 40N to the right due to a collision with a second object.

First, let us define the +x-direction as being to the right. That will allow us to interpret

the phrase “40N to the right” as +40N. Then we can apply Newton’s Second Law

ΣFb = mbab Start with the initial equation.

ab = ΣFb / mb

Because we are looking for the acceleration, we need to solve the equation for ab. To isolate ab, divide both sides by mb.

ab = +40 N / 20kg

With the found equation, plug in the given values for ΣFb and mb.

ab = (+40 kg∙m/s2) / 20 kg

ab = +2.0 m/s2


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ab = 2.0 m/s2 to the right Because we defined “to the right” as the positive direction.

Example 2: Determine the acceleration of a package whose mass is 20 kg and which is

subject to two forces. The first force is 40N to the right while the second force is 20N to

the left.

First, let us define the +x-direction as being to the right. That will allow us to interpret

the phrase “40N to the right” as +40N and “20N to the left” as -20N. We’ll use “p” as the

subscript to denote the package and “1” and “2” to label the two forces. Then we can

apply Newton’s Second Law keeping in mind that we need to add the forces to find ΣFp.

ΣFp = mpap

ap = ΣFp / mp

ap = (Fp1+ Fp2) / 20kg

ap = (+40 N - 20N) / 20kg

ap = (+20 kg∙m/s2) / 20 kg

ap = +1.0 m/s2

ap = 1.0 m/s2 to the right

Newton’s Second Law in Two Dimensions

When working with vectors, (e.g. displacement, velocity, acceleration and force) it is

important to first define a set of axes. We will define the +x axis to the right and the +y

axis as up, away from the center of the earth, unless otherwise noted. If the axes are


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chosen in this way the vector equation ΣF = ma can be broken into two independent

equations, one along the horizontal, x-axis, and the other along the vertical, y-axis.

The vector equation ΣF = ma becomes ΣFx = max and ΣFy = may. The motion and the

forces along the x-axis and the y-axis can then be treated separately. The dynamics

and kinematics in the horizontal and vertical directions are independent and unaffected

by each other so Newton’s second law must be true in both directions.

Example 3: Determine the acceleration of an object whose mass is 20 kg and which is

subject to three forces. The first force, Fo1, is 80N to the right, the second force, Fo2, is

20N upwards and the third force, Fo3, is 20N downwards.

With our convention of defining the +x-direction as being to the right and the +y-

direction as up, we can interpret the phrase “80N to the right” as +80N along the x-axis,

“20N upwards” as +20N along the y-axis and “20N downwards” as -20N along the y-axis

and. Then we can apply Newton’s Second Law independently along each axis keeping

in mind that we need to add the forces to find Fnet.


ΣF = ma

x-direction y-direction

ΣFo = moao ΣFo = moao

ao = ΣFo / mo ao = ΣFo / mo

ao = (Fo1 / mo) ao = (Fo2 + Fo3) / mo

ao = (+80N/20kg) ao = (+20N-20N)/(20kg)

ao = +1.0 m/s2 ao = 0

a = 1.0 m/s2 to the right

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In this example, all the vectors (forces and acceleration) were directed either parallel or

perpendicular to the x-axis. In nature, forces and accelerations can be directed in any

arbitrary direction. That does not represent any new physics, but would require the use

of trigonometry. Since this book does not presume a prior background in trigonometry,

we will only work with vectors that are either parallel or perpendicular to the x-axis.

Dealing with vectors at arbitrary angles will not represent much additional difficulty once

you have mastered basic trigonometry.

Newton’s Third Law

We indicated earlier that forces can only exist between two objects. A single isolated

object cannot experience a force. Newton’s third law indicates the very simple

relationship between the force that each object exerts on the other; they will always be

equal and act in opposite directions. So if I push on a box with a force of 20 N to the

right, the box will be pushing on me with a force of 20 N to the left. Using our subscripts

this simply says that Fmb = -Fbm, with the subscript “m” standing for me and “b” standing

for box.

Another way of saying this is that forces always occur in equal and opposite pairs.

Some people get confused by this law in that they think that these two forces will cancel

out and no acceleration will ever result. However, that is not the case since each force


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is acting on a different object. Only one force acts on each object so they cannot cancel

out. This can most easily be made clear by working through an example.

Example 3: I push on a box whose mass is 30 kg with a force of 210 N to the right.

Both the box and I are on a frictionless surface so there are no other forces acting on

either the box or me. My mass is 90 kg. Determine the acceleration of both the box

and me? Are they the same?

There are no forces in the y-direction so I only need to use one axis. First, let’s solve

for the acceleration of the box.

ΣFb = mbab

ab = ΣFb / mb

ab = Fmb / mb

ab = +210 N/ 30kg

ab = (+210 kg∙m/s2) / 30 kg

ab = +7.0 m/s2

ab = 7.0 m/s2 to the right

Now let’s solve for my acceleration.

ΣFm = mmam

am = ΣFm / mm

am = Fmb / mm but Fmb = - Fbm

am = -Fbm / mm (Note the change in the order of the subscripts)

am = -210 N/ 90kg


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am = -210 kg∙m/s2 / 30 kg

am = -2.3 m/s2

am = 2.3 m/s2 to the left

So as I push the box to the right, the box pushes me to the left. Since my mass is

greater, my acceleration is less. The accelerations are not the same. Nor is either of

them equal to zero.

Sketches and Free Body Diagrams

I. The first step in solving a problem is to make a quick sketch, unless one is provided.

Without having a rough sketch, it’s easy to go in the wrong direction or not know how to

get started. Often, a sketch is provided, but if isn’t make this your first step. After

you’ve done that, you’ll need to draw a free body diagram for each of the objects in

your sketch.

II. Before we get started with that, it’s important to define what we mean by an object.

Obviously any single entity, like a wooden block, ball, book, car, person, etc. represents

an object. But also, if two blocks of wood are connected together so that if one moves,

the other must move, those two blocks together can be treated as an object. In this

case, you can generate three free body diagrams, one for each block of wood

separately and a third for the combination of the two blocks. Anytime two objects are

locked together so that they have the same acceleration, they can be treated as a single

object. This technique may give you the crucial bit of information that you need to solve

a problem.


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III. Now let’s go through how to draw a free body diagram.

1. Draw and label a dot to represent the first object.

2. Draw an arrow from the dot pointing in the direction of one of the forces that is

acting on that object. Clearly label that arrow with the name of the force, for

instance Fc1.

3. Repeat for every force that is acting on the object. Try to draw each of the

arrows to roughly the same scale, bigger forces getting bigger arrows.

4. Once you have finished your free body diagram, recheck it to make sure that

you have drawn and labeled an arrow for every force. This is no time to forget a

force.

5. Draw a separate arrow next to your free body diagram indicating the likely

direction of the acceleration of the object. This will help you use your free body

diagram effectively.

6. Repeat this process for every object in your sketch.

IV. Now let’s use your free body diagram to apply Newton’s Second Law. One of the

wonderful things about his law is that it will work for each object independently. That

means that you can to any of your free body diagrams and proceed to analyze it.

1. Write the law in its general vector form using a subscript to indicate the object

being analyzed, ΣFo = moao.

2. If there are forces acting along more than one axis, break this general equation

into its x and y components.


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3. Now read your free body diagram to write down all the forces on the right side of

the equation. Using whatever convention you wish, as long as you are

consistent, indicate the direction of each force by a + or – sign. For instance if

the forces were in the vertical direction, you might use + for up and – for down.

4. Now’s when you can make use of the acceleration vector that you drew next to

your free body diagram. Be consistent. If you used a negative sign for a

downward or leftward directed force and the acceleration vector is also pointing

down or to the left, you need to put a negative sign in front of the “ma” on the

right side of the equation. On the other hand, if the acceleration vector is pointed

upwards or to the right, it should have a positive sign.

5. You can now repeat this process for as many of your other free body diagrams

as is necessary to solve the problem.

Let’s do a couple of examples.

Example 4: Two objects are connected together by a string. Object 1 has a mass of 20

kg while object 2 has a mass of 10 kg. Object 2 is being pulled to the right by a force of

60 N. Determine the acceleration of the objects and the tension in the string that

connects them. .


1 2

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First, note that there are three possible objects, 1, 2 and 1+2. The combined object 1+2

exists because the string connecting objects 1 and 2 insures that they will always move

together and have the same acceleration.

Here are the three free body diagrams.

Note that in the diagrams, F12=- F21 so the arrows are the same size but pointed in

opposite directions. Fapplied is the same size for both objects 2 and (1+2) so those arrows

are the same size. Since the acceleration of all three objects is the same and must be

to the right, all three of the arrows labeled “a” are the same size and directed to the

right.

First, let’s solve for the acceleration of the combined object (1+2). That’s the easiest to

solve since I know the combined mass and the one force that acts on it.

ΣF(1+2) = m(1+2) a(1+2)

a(1+2) = ΣF(1+2) / m(1+2)

a(1+2) = Fapplied / m(1+2)

a(1+2) = +60 N/ (20kg + 10k)

a(1+2) = (+60 kg∙m/s2) / 30 kg


Object (1+2)Object 2Object 1

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a(1+2) = +2.0 m/s2

a(1+2) = 2.0 m/s2 to the right

Now let’s solve for tension in the string, F12, by using the free body diagram for object 1.

This will be simplest since I now know the acceleration and the mass of object 1 and F12

is the only force acting on it.

ΣF1 = m1a1

F12 = +m1a1

F12 = (20 kg)(+ 2.0 m/s2)

F12= + 40 kg∙m/s2

F12 = 40 N to the right

Just to show that all three free body diagrams would work we can also solve it using the

free body diagram for Object 2. This can be used to check our work but will not give us

any extra information.

ΣF2 = m2a2

Fapplied+ F21 = m1a1

Fapplied - F21 = +m1a1

F21 = - m1a1 + Fapplied

F21 = -(10 kg)(+ 2.0 m/s2) + 60N

F21= + 40 kg∙m/s2

F21 = 40 N to the left

Some Examples of Forces


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There are a wide variety of forces that we experience every day. One of the goals of

physics is to show the source of these forces and how they are connected to one

another. That theme will present itself as you continue your study of physics. However,

in order to reach that stage of understanding you will first need to master the skills that

are based on the everyday, not the fundamental, forces of nature. In this section of the

book, we will present four of those everyday forces. Their connection to more

fundamental forces will be touched upon, but a full exploration of those connections will

have to wait until later in this book. A complete understanding is not yet available in

physics. There is much that remains to be understood.

Weight

Everyone has seen that if you drop an object, like a book or a pen, it falls downwards.

Since the object wasn’t moving when you dropped it (that’s the meaning of the word

“dropped”) and is moving after you let go of it that must mean the object accelerated

upon being released. In general, unsupported objects near the surface of the earth

accelerate downwards. (I say “in general” since this won’t work if you “drop” a helium

balloon, but we’ll deal with that later in the book when we talk about buoyancy.) This

attraction of objects towards the center of the earth is a result of an important

fundamental force, gravity. A fuller discussion of gravity in general requires its own

chapter. However, we will now explore its consequences near the surface of the earth.

From Newton’s Second Law we know that if an object accelerates there must be a net

force acting on it. In this case, the force being exerted on the object is due to the earth


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so we can call it, Foe. However, this force is so universally felt by all of us on the planet

that it has received its own name and is often referred to as the weight (W) of the object.

It turns out that the weight, W or Foe, of an object is proportional to its mass. The more

massive the object - the greater its weight. The constant of proportionality between an

objects weight and its mass has been given its own symbol, “g”, and is equal to about

9.8 m/s2. This is expressed by the following formula.

W = Foe = mg

A surprising result of the fact that the weight of an object is proportional to its mass is

that all unsupported objects near the surface of the earth fall with the

same acceleration, 9.8 m/s2, as long as no other force acts on them. We can see how

this would be true if we use the idea that W = mg and combine that with Newton’s

Second Law. If an object only has the force of its weight acting on it, it’s free body

diagram would show only one force, W, pointing downwards. Its acceleration would

also point downwards. So…

ΣFo = moao

W = m0ao

mg = ma

a = g

a = 9.8 m/s2


wa

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On the one hand this seems like a reasonable result. A more massive object is heavier,

so it has more force pulling it towards the earth. However, it’s also more massive in

exactly the same proportion, and the more mass an object has the more force it takes to

give it the same acceleration. These two effect, more weight but also more mass,

exactly cancel out and all objects fall with the same acceleration.

On the one hand, this is a surprising result since it seems like heavy objects should fall

faster than light objects. And that is often the case. If you drop a feather and a book at

the same time its clear that the feather falls slower than the book. However, we have to

recall the condition we put in our statement about falling objects having the same

acceleration. We said that this would be true if “no other forces are acting on the

object”. This is much like the case we discussed earlier where the force of friction made

it hard for us to see that an object in motion will continue to move at a constant velocity

if no net force acts on it. In this case, it’s the resistance of the air to the motion of the

feather that serves as a sort of “friction” which opposes the weight of the feather. If you

remove the air and repeat the experiment, the feather and book will fall with the same

acceleration. Since we not only live on the earth, we also live in the earth’s

atmosphere, it’s reasonable that our intuition would tell us that light objects don’t fall to

the ground as quickly as heavy objects. But it’s important that we go beyond our

intuition to try to separate out the two different effects, gravity and air resistance, and

study each on its own.


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One way you can prove this to yourself is to first drop a piece of 8 ½ x 11 paper at the

same time as a pen. You’ll see that the pen falls much faster than the paper. However,

if you then tightly crumple the paper into a ball and repeat that experiment, you’ll see

that they fall at about the same rate. Now if you imagine, as Galileo did in his

experiments, that you could make that ball of paper small enough, then it would fall at

the exact same rate as the pen.

Surface Forces

Whenever the surfaces of two objects come in contact there are two types of forces that

can result.

The Normal Force – which always acts perpendicularly to the surfaces

Friction - which always acts parallel to the surfaces

Both of these forces are due to microscopic effects within the objects. They are both

very complicated to understand at that level and require a foundation in electrostatics to

get any reasonable idea of how they operate. Understanding these forces in detail

requires a grasp of very advanced physics. However, as we will often find to the be the

case, the characteristics of how these forces work can be understood without fully

understanding the underlying mechanism that makes them work.

The Normal Force is a result of the atoms in an object trying to avoid being pushed

closer together. When they are pushed closer together by another object, they resist

that by pushing back on that object. The harder they are pushed together the harder

they push back.


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Friction is a result of two surfaces rubbing together as they slide, or try to slide, by each

other. Each of the surfaces has a texture as well as surface atoms that have

electrostatic properties. The interaction of the textures and atoms on the two surfaces

determines how slippery they feel to one another.

Normal Force

If you put a book on a flat table and let go of it you will see that it does not move, its

acceleration is zero. We know that there is at least one force acting on the book, its

weight. We also know that if there were a net force acting on the book it would

accelerate, and in this case it is not accelerating. The only way that the total force

acting on the book could be zero, if we know that there is a least one force acting on the

book, is if a second force is acting on the book in the opposite direction (there could be

more than one extra force but let’s keep this simple for now).

We also know that a force requires the interaction of two objects. In this case, one of

the objects is the book. The other object must be the table. This becomes clear if you

imagine what would happen if the table suddenly disappeared (or if you try to put the

book down next to, but not on, the table). The book will accelerate without the table to

support it. We can conclude from this that the table must be furnishing the force that is

canceling the effect of the earth on the book.


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Using our convention for naming the force between two objects, let’s call that force Fbt,

the force on the book due to the table. This force must be equal and opposite to the

force of the earth on the book, in order to cancel out the pull of the earth. Since this

direction is perpendicular, or normal, to the surface of the table this force is also called

the Normal Force, Fn. You can use either Fbt or Fn to name this force.

Let’s now analyze our book sitting on a table using a free body diagram and Newton’s

Second Law to determine the normal force exerted on a 10 kg book by a table.

ΣFo = moao

W – Fbt = 0

W = Fbt

Fbt = mg

Fbt = (10kg) (9.8 m/s^2)

Fbt = 98 N

While we have only shown two objects, and one pair of forces, in this picture, there are

really three objects and two pairs of forces. Newton’s Third Law tells that forces only


N

W

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exist between objects and in pairs. The first pair of forces is between the earth and the

object, Foe and Feo. The second pair of forces is between the object and the table, Fot

and Fto. In order to keep the diagram simple, we have not shown the earth. But none of

the above would occur if the earth were not present. If you were to put a book on a

table in deep space, far from the earth, there would no weight to pull the book against

the table; therefore there would be no resulting normal force.

A more comprehensive diagram would show all four forces, as below. However,

in many cases this added complication is not needed and may make drawings harder to

read – which would defeat their purpose. This does show, however, that the table is

experiencing a normal force downwards, due to the book, that is equal and opposite to

the normal force that it is supplying to the book upwards. If that downwards force, were

to become too great, the table would break.

While we derived the normal force as a force that opposes gravity in the case of a book

resting on a table, it has a much wider range of applications.


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Feo Fot


When you stand in an elevator, the floor of the elevator exerts a normal force

upwards on your feet. We’ll find that you can be accelerated upwards, or

downwards, when that normal force does not equal your weight.

When you lean against a wall, you exert a normal force on that wall. By

Newton’s Third Law, you know that the wall will exert an equal and opposite

normal force on you. Since the wall is vertical, and the normal force is always

perpendicular to the surfaces, the provided normal force is in the horizontal

direction.

When a baseball is hit by a bat, the normal force exerted by the bat on the ball is

in a direction perpendicular to the surfaces of the bat and ball when they collide.

Since the ball is a sphere and the bat is a cylinder, there’s a wide range of

directions in which that normal force can be directed depending on where they

come in contact. That’s why the ball can accelerate in ways that leads to line

drives, grounders or pop-ups.

Friction

There are two types of friction that can be generated between a pair of surfaces. The

difference between the two types depends on the relative velocity of the surfaces to

each other. Please note that that means that if both surfaces are moving along

together, their relative velocity is zero.

Kinetic friction is the result of the surfaces of two objects rubbing together as they

slide by each other with some relative velocity. In this case, the force of friction will act


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on each object in a direction opposite to its velocity relative to the other. Kinetic friction

acts in a way so as to bring those objects to rest relative to each other.

Static friction results when an object experiences a net force parallel to its contact

surface with a second object. The force of static friction is directed opposite to the

direction of that net force in order to keep the objects at rest relative to each other.

There are two factors that determine the maximum amount of friction force that can be

generated; the slipperiness of the surfaces and the amount of force pushing the

surfaces together. This makes sense in that if two surfaces are perfectly slippery, they’ll

never generate any friction between them no matter how hard you squeeze them

together. On the other hand, if two surfaces are not very slippery but are also not

actually being pushed together, they also will not generate any friction. They’ll just slide

by each other, barely touching.

The amount of slipperiness between two surfaces is indicated by the coefficient of

friction between those surfaces. This is given the symbol “μ” which is a Greek letter

and is pronounced “mu”. Values of μ range from zero to one. Two surfaces that have a

coefficient of friction of zero will generate no friction regardless of how hard they are

squeezed together; they are perfectly slippery. On the other extreme, two surfaces with

a μ of 1 generate a very high level of friction if they are pressed together. Note that

values of μ can only be given for pairs of surfaces. There is no way to compute a value


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for each surface and then combine them together to figure out how the surfaces will

behave when placed together.

If you’ve ever tried to slide something very heavy across the floor you’ve probably

noticed that it’s harder to get it started then to keep it going. Once it starts moving, you

don’t have to push as hard to keep it going. That’s a result of the surfaces becoming

more slippery once they start sliding by each other. On a microscopic level, the atoms

on the surfaces just don’t have time to make bonds with each other as they slide by. As

a result, there are two different coefficients of friction for each pair of surfaces. The

static coefficient of friction, μs, gives the measure of slipperiness between the surfaces

while they are not moving relative to each other, while the kinetic coefficient of friction,

μk, gives a measure of that slipperiness when they are. Since higher values of μ result

in more friction that means that μs is always greater than μk. Please note that

determining which coefficient to use is based on whether the surfaces are moving

relative to one another, sliding past each other, or not. Surfaces that are moving

together are treated the same as if they were standing still and you would use μs, not μk.

The other factor, aside from slipperiness, that determines the force of friction is how

hard the surfaces are being squeezed together. A force that squeezes two surfaces

together must be perpendicular to them. This is just another way of describing the

normal force that we discussed earlier. So we now have enough information to write a

formula for the maximum amount of friction that can be generated by two surfaces.

If the two surfaces are gliding by each other than:


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Ffr max = μk Fn

If the two surfaces are not moving relative to each other than

Ffr max = μs Fn

We now have just one last step to take and we’ll have rounded out our picture of friction.

We have to determine when the amount of friction will be at the maximum values given

by the above formulas and when that will not be the case. In the latter case, we need to

figure out what it will be if it is not at a maximum.

The force of kinetic friction is the simplest because it’s always at its maximum value.

Once the surfaces are sliding by each other, the force of kinetic friction will be at its

maximum value and nothing will affect this value as long as that relative motion

continues. So we can replace our maximum formula with:

Ffr = μk Fn for the case of kinetic friction

We indicated above that the force of static friction is directed opposite to the direction of

an applied force in order to keep the two objects at rest relative to each other. That

means that if there is no net force applied to an object parallel to its contact surface with

a second object there will not be any static friction. As a result, the minimum static

friction must be zero. The maximum amount is given by Ffr max = μs Fn. The formula for

static friction becomes.

Ffr ≤ μs Fn

There is static friction only if there is a net applied force that would, in the absence of

static friction, accelerate the object parallel its contact surface. As that net applied force

increases, so does the amount of static friction, always in the direction opposite to the


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net applied force. However, this static friction cannot increase without limit. Once the

maximum is reached, Ffr max = μs Fn, the object will begin to slide. At that point the static

friction formula becomes inapplicable and the kinetic friction formula become relevant,

Ffr = μk Fn. This will remain the case until such time as the surfaces once more become

at rest with respect to each other.

Tension Force

Another way to support an object is to hang it from a string. For instance, imagine

hanging a plant from a string that is then connected to the ceiling. The downward force

due to the weight of plant will cause the string to stretch a bit and get taut. The tautness

in the string results in a tension force that pulls the plant upward and the ceiling

downward.

For a tension force to act the string must be connected to objects at both ends. In this

example the objects are the plant and the ceiling. Tension always acts on both objects

with equal magnitude but in opposite directions. It has the effect of pulling the objects

towards one another.

Tension force is not just present in string. It can be found in rope, cable, and metal

rods, etc. In fact, any material that resists being stretched can exhibit a tension force

when connected between two objects. It can be labeled either as F t or by the using the

subscript system using the object under study first and the material that provides the


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tension force second. For instance, in the above example the tension force could be

named Ft or Fps, where “p” refers to the plant and “s” the string.

Let’s work through an example where a 5 kg plant is hung from a string connected to

the ceiling. First, let’s do a free body diagram for the plant

Then use Newton’s Second Law, keeping mind that the acceleration is zero since the

plant is not falling.

ΣFp = mpap

W – Ft = 0

Ft = W

Ft = mg

Ft = (5kg) (9.8 m/s^2)

Ft = 49 N

So far we have discussed tension as a vertical force that offsets the weight of an object

when it is hanging. However, tension does not have to act vertically. When an object is


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Ft

W


pulled across a horizontal surface by a string, the tension force is acting horizontally.

The things that are always true about the tension force are:

Tension force always acts along the direction of the string, or whatever material

is under tension. It can never act at an angle to the string.

Tension force requires that there be two connected objects.

Tension force acts in opposite directions on those two objects so as to bring

them together.

Elastic Force – Hooke’s LawThe elastic force, Fe, is the force exerted by a stretched or compressed material.

Robert Hooke originally explored the force in the context of springs, but we will see that

it has wider applications. It will also take us a step towards better understanding both

the tension force and the normal force.

Hooke observed that it takes very little force to stretch a spring a very small amount.

However, the further the spring is stretched, the harder it is to stretch it further. The

force needed increases in proportion to the amount that it has already been stretched.

The same was observed to be true when compressing a spring. This can be stated

mathematically as

Fspring = - k x

In this equation, k represents the spring constant (a characteristic of the individual

spring), and x represents the magnitude of the distance the spring is stretched or

compressed from its natural length. The negative sign tell us that the force that the

spring exerts is back towards its equilibrium length, its length when it is not being


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stretched or compressed. Note that x is always either zero or positive since it is a

magnitude. The negative sign accounts for the direction of the force.

Therefore, if the spring constant for a particular spring were 100 N/m, I would need to

exert a force of 100 N to stretch, or compress, it by a length of 1m. If I were to exert a

force of 50N, it would stretch 1/2 m. A force of 10 N would stretch, or compress, it by a

distance of 1/10 m. This is shown graphically below.

The elastic force also underlies the tension and normal forces. In the case of the

tension force, the rope or string that is exerting that force actually stretches a bit as that

force is exerted, just like a spring. The difference is that in the case of the tension force,

the amount of stretch is relatively small and is neglected. We assume that the string

does not stretch for the purpose of our calculations. But the string will exert its force so

as to go back to its equilibrium length, so it does act just like a spring in this respect.

Another key difference between a string and a spring is that you can’t compress a

string. Strings, ropes, etc. only exhibit an elastic force when they are stretched, not

when they are compressed. Also, they follow Hooke’s law for only a small amount of


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stretch. After that they get a lot more complicated. So while the source of the force is

the same as the elastic force, tension is treated differently.

The same holds for the normal force, however, in this case a force is only exhibited

upon compression. When you put a book on a table, the table bends just the slightest

bit as it compresses in the middle. The bend is so small that it’s hard to see. But if you

put a lot of books on a bookshelf, you can see that bending take place. Once again, the

table or bookshelf wants to go back to its equilibrium position. To do this, it exerts a

force in a direction opposite to that which causes it to compress. This is the normal

force. It’s just like the elastic force due to compression but, as in the case of tension, it

follows Hooke’s law for only a small distance before getting too complicated. Also, the

amount of compression is small enough to be neglected in our calculations. However,

that compression is the source of the normal force. Without compression, there is no

normal force.

Chapter Questions

1. What is inertia? Give some examples.

2. What is the difference between inertial and non-inertial reference frames?

3. State the relationship between mass and inertia.

4. A boy seems to fall backward in an accelerating bus. What property does this

illustrate?

5. A fisherman stands in a boat that is moving forwards towards a beach. What

happens to him when the boat hits the beach?


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6. A passenger sits in a stationary train. There are some objects on a table: an

apple, a box of candy, and a can of soda. What happens to all these objects with

respect to the passenger when the train accelerates forward?

7. An object is in equilibrium. Does that mean that no forces act on it?

8. A rock is thrown vertically upward and stops for an instant at its highest point. Is

the rock in equilibrium at this point? Are there forces acting on it?

9. Is it possible for an object to have zero acceleration and zero velocity when only

one force acts on it?

10. Is it possible for a car to move at a constant speed when its engine is off?

11.Two boys are pulling a spring scale in opposite directions. What is the reading of

the spring scale if each boy applies a force of 50 N?

12.Much more damage is done to a car than a truck when the two collide. Is that in

agreement with Newton’s Third Law?

Chapter Problems

Newton’s First Law, Force1. How many forces are applied in each of the following situations and what can

you say about their directions and magnitudes?

a. An air balloon hovers in midair;

b. A submarine is on the bottom of ocean;

c. A car travels at constant speed in a straight line;

d. A coin lies at rest on a rotating table, and the table rotates at constant

speed.


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2. Is it possible for a car to travel at constant speed on straight road when its engine

is off?

3. Compare the accelerations of two balls with the same radii if one is made of steel

and the second one is made of lead.

4. A pair of fuzzy dice is hanging by a string from the rearview mirror of a car. What

happens to the dice when the car accelerates from a stoplight?

Newton’s Second and Third Laws

5. A 0.40 kg toy car moves at constant acceleration of 2.3 m/s2. Determine the net

applied force that is responsible for that acceleration.

6. If a net horizontal force of 175 N is applied to a bike whose mass is 43 kg what

acceleration is produced?

7. A car travels at constant acceleration of 2.2 m/s2. Find the mass of the car if a 5.3

kN net force is required to produce this acceleration.

8. A wooden block is pulled at a constant acceleration of 1.4 m/s2. Find the net

applied force on the block if its mass is 0.6 kg.

9. A 95 N net force is applied to an ice block with a mass of 24 kg. Find the

acceleration of the block if it moves on a smooth horizontal surface.

10.A net force of 345 N accelerates a boy on a sled at 3.2 m/s2. What is the

combined mass of the sled and boy?

11.What average net force is required to stop an 8500 kg truck in 10 s if it’s initially

traveling at 72 km/h?


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12.What average net force is required to accelerate a 9.5 g bullet from rest to 650

m/s over a distance of 0.85 m along the barrel of a rifle?

13.A 7.5 kg cannon ball leaves a cannon with a speed of 185 m/s. Find the average

net force applied to the ball if the cannon muzzle is 3.6 m long.

14.What average force is needed to stop a 15000 kg train in 5 s if it’s traveling at

120 km/h?

15.A hockey puck with a mass of 0.18 kg is at rest on the horizontal frictionless

surface of the rink. A player applies a horizontal force of 0.5 N to the puck. Find

the speed and the traveled distance 5 s later.

16.A sportsman is trying to push a weight upward. Assuming that the weight starts

from rest accelerates then moves at constant speed and decelerates to rest.

Explain what happened to the force applied by the sportsman on the floor during

this trial.

17.A boy who is inside a boat tries to move the boat by pushing on its side. Explain

why he failed in his attempt. What must be done to move the boat?

18.A beaker is ¾ filled with water and then placed on one plate of a lab balance. A

student uses lab weights to bring the balance to equilibrium. What happens to

the equilibrium when a small steel cylinder is submerged into the water without

touching the bottom of the beaker?

Mass and Weight19.A woman weighs 580N. What is her mass?

20.Find the weight of a 2000 kg elephant.


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21.An astronaut has a mass of 85 kg. Calculate his weight on Earth and on the

Earth’s moon. Does his mass change when he goes from Earth to its moon?

22.A car weighs 14500 N. What is its mass?

23.Calculate the weight of a 4.5 kg rabbit.

24.At the surface of Mars the acceleration due to gravity is 3.8 m/s2. A book weighs

34 N at the surface of the Earth. What is its mass on the earth’s surface? What

are its mass and weight on Mars’s surface?

25.Calculate the apparent weight of a 75 kg astronaut as she accelerates in a

spacecraft from the surface of the Earth at constant rate of 18 m/s2.

26.A 65 kg woman is inside an elevator. Calculate her apparent weight for the

following cases:

a) The elevator moves at constant speed upward;

b) The elevator moves at constant speed downward;

c) The elevator accelerates upward at a constant rate of 2.4 m/s2;

d) The elevator accelerates downward at a constant rate of 2.4 m/s2.

27.A 56 kg object is attached to a rope, which can be used to move the load

vertically.

a) What is the tension force in the rope when the object moves upward at a

constant velocity?

b) What is the tension force in the rope when the object accelerates upward

at a constant acceleration of 1.8 m/s2?

c) What is the tension force in the rope when the object accelerates

downward at a constant acceleration of 1.8 m/s2?


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28.An 88 kg worker stands on a bathroom scale in a motionless elevator. When the

elevator begins to move, the scale reads 650 N. Find the magnitude and

direction of the elevator’s acceleration.

29.A cable in which there is 12500 N of tension force supports an elevator. What is

the magnitude and direction of the acceleration of the elevator if its total mass is

1175 kg?

Free-Body Diagrams30.A box sits at rest on a tabletop. Draw and clearly label all the forces acting on the

box; compare their magnitudes and directions.

31.A wooden block moves at a constant speed on a rough horizontal surface. Draw

a free-body diagram clearly showing all the forces applied to the block; compare

their magnitudes and directions.

32.A boy pulls a sled horizontally at a constant speed by holding a rope that is

connected to the sled.

a) Show all the forces exerted on the sled (do not ignore friction);

b) Show all the forces exerted on the boy (do not ignore friction);

c) Show all the forces acting on the rope;

d) Use Newton’s Law to explain the directions and the magnitudes of all the

forces; compare “action” and “reaction”.

33.A crane lifts a load at a constant speed. Draw a free-body diagram for the load

and compare the magnitudes and directions of the all forces.

34.A crate is accelerated at a constant rate along a rough horizontal floor. Draw a

free-body diagram for the crate and compare all the forces exerted on the crate.


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35.A hokey puck slides on a rough horizontal surface. Draw a free-body diagram for

the puck and compare the magnitudes and the directions of all the forces exerted

on it.

Kinetic Friction

36.The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.25. The weight of the object is 20N. What is the force of friction?

37.The force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What is the coefficient of kinetic friction?

38.The force of friction between an object and the surface upon which it is sliding is 12N and the coefficient of friction between them is 0.70. What is the weight of the object?

39.The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40. The mass of the object is 3.2kg. What is the force of friction?

40.The force of friction between an object and the surface upon which it is sliding is 1.5N. The mass of the object is 20kg. What is the coefficient of kinetic friction?

41.The force of friction between an object and the surface upon which it is sliding is 250N and the coefficient of friction between them is 0.80. What is the mass of the object?

42.The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40. The weight of the object is 80N. What is the force of friction?

43.The force of friction between an object and the surface upon which it is sliding is 36N. The weight of the object is 85N. What is the coefficient of kinetic friction?

44.The force of friction between an object and the surface upon which it is sliding is 46N and the coefficient of friction between them is 0.30. What is the weight of the object?

45.The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object is 8.0 kg. What is the force of friction?


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46.The force of friction between an object and the surface upon which it is sliding is 360N. The mass of the object is 95kg. What is the coefficient of kinetic friction?

47.The force of friction between an object and the surface upon which it is sliding is 3.6N and the coefficient of friction between them is 0.30. What is the mass of the object?

A horizontal Applied Force is pushing an object along a horizontal surface. The only other force acting on the object is the Force of Friction. In words, compare the size and relative direction of those two forces if:

48.The object is slowing down.49.The object is moving at constant velocity.50.The object is speeding up.

A 23 kg object is on a horizontal table near the surface of the earth. The coefficient of friction between the object and table is 0.35. A 150 Newton force is pushing it across the table.

51.Draw a free body diagram for the object. Label all the forces.52.Determine the Force of Friction.53.Determine the object’s acceleration.


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Static FrictionA stationary 15 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.40 and of kinetic friction is 0.25.

54.A horizontal force of 20 N is applied to the object. a. Draw a free body diagram with the forces to scale.b. Determine the force of friction.c. Determine the acceleration of the object.

55.A horizontal force of 40 N is applied to the object. d. Draw a free body diagram with the forces to scale.e. Determine the force of friction.f. Determine the acceleration of the object.

56.A horizontal force of 60 N is applied to the object. g. Draw a free body diagram with the forces to scale.h. Determine the force of friction.i. Determine the acceleration of the object.

57.A horizontal force of 100 N is applied to the object. j. Draw a free body diagram with the forces to scale.k. Determine the force of friction.l. Determine the acceleration of the object.

A stationary 250 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.30 and of kinetic friction is 0.15.

58.A horizontal force of 300 N is applied to the object. m. Draw a free body diagram with the forces to scale.n. Determine the force of friction.o. Determine the acceleration of the object.

59.A horizontal force of 500 N is applied to the object. p. Draw a free body diagram with the forces to scale.q. Determine the force of friction.r. Determine the acceleration of the object.

60.A horizontal force of 750 N is applied to the object. s. Draw a free body diagram with the forces to scale.t. Determine the force of friction.u. Determine the acceleration of the object.

61.A horizontal force of 1500 N is applied to the object. v. Draw a free body diagram with the forces to scale.w. Determine the force of friction.x. Determine the acceleration of the object.


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A stationary 2.0 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.80 and of kinetic friction is 0.65.

62.A horizontal force of 5 N is applied to the object. y. Draw a free body diagram with the forces to scale.z. Determine the force of friction.aa.Determine the acceleration of the object.

63.A horizontal force of 10 N is applied to the object. bb.Draw a free body diagram with the forces to scale.cc. Determine the force of friction.dd.Determine the acceleration of the object.

64.A horizontal force of 16 N is applied to the object. ee.Draw a free body diagram with the forces to scale.ff. Determine the force of friction.gg.Determine the acceleration of the object.

65.A horizontal force of 100 N is applied to the object. hh.Draw a free body diagram with the forces to scale.ii. Determine the force of friction.jj. Determine the acceleration of the object.


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General Problems66.A train with a mass of 25000 kg increases its speed from 36 km/h to 90 km/h in

20 s. Assume that the acceleration is constant and that you can neglect friction.

e) Find the acceleration of the train;

f) Find the distance traveled during this 20 s?

g) Draw a free- body diagram for the train;

h) Find the average net force supplied by the locomotive.

67.A 150 kg motorcycle starts from rest and accelerates at a constant rate along a

distance of 350m. The applied force is 250 N and the coefficient of kinetic friction

is 0.03.

e) Draw a free-body diagram for the motorcycle showing all applied forces to

scale. Next to that diagram show the direction of the acceleration;

f) Find the net force applied to the motorcycle;

g) Find the acceleration of the motorcycle;

h) What is its speed at the end of 350 m?

i) Find the elapsed time of this acceleration.


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m1 m2 F

68.Two blocks, with masses m1 = 400 g and m2 = 600 g, are connected by a string

and lie on a frictionless tabletop. A force F = 3.5 N is applied to block m2.

a. Draw a free-body diagram for each block showing all applied forces to

scale. Next to each diagram show the direction of the acceleration of that

object.

b. Find the acceleration of each object.

c. Find the tension force in the string between two objects.


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A B

F

69.Two boxes are placed on a horizontal frictionless surface, as shown above. Box

A has a mass of 10 kg and box B has a mass of 16 kg. A force of 54 N is pushing

box A.

a. Draw a free-body diagram for each block showing all applied forces to

scale. Next to each diagram show the direction of the acceleration of that

object.

b. Find the acceleration of the system of two boxes.

c. Find the force of contact that each box exerts on its neighbor.


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15 kg

12 kg

70.A 12 kg load hangs from one end of a rope that passes over a small frictionless

pulley. A 15 kg counterweight is suspended from the other end of the rope. The

system is released from rest.

a. Draw a free-body diagram for each object showing all applied forces in

relative scale. Next to each diagram show the direction of the acceleration

of that object.

b. Find the acceleration each mass.

c. What is the tension force in the rope?

d. What distance does the 12 kg load move in the first 3 s?

e. What is the velocity of 15 kg mass at the end of 5 s?


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A

B

71.The masses of blocks A and B are 4.5 kg and 3.7 kg respectively. The blocks are

initially at rest and are connected by a massless string passing over a massless,

frictionless pulley. The system is released from rest.

a. Draw a free-body diagram for each block showing all the applied forces in

relative scale. Next to each diagram show the expected direction of

acceleration.

b. What is the acceleration of blocks?

c. What is the tension force in the rope?

d. How high will the 3.7 kg block move in the first 2.5 s?

e. Find the speed of the 4.5 kg block at the end of 5th second.


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500 g

300 g

72.A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction

between the block and the surface is 0.25. The block is connected by a massless

string to the second block with a mass of 300 g. The string passes over a light

frictionless pulley as shown above. The system is released from rest.

a. Draw clearly labeled free-body diagrams for each of the 500 g and the

300g masses. Include all forces and draw them to relative scale. Draw

the expected direction of acceleration next to each free-body diagram.

b. Use Newton’s Second Law to write an equation for the 500 g mass.

c. Use Newton’s Second Law to write an equation for the 300 g mass.

d. Find the acceleration of the system by simultaneously solving the system

of two equations.

e. What is the tension force in the string?


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35 kg

F

45 kg

73.A crate with a mass of 45 kg is suspended from a massless rope that runs

vertically upward over a light pulley. The other end of the rope is connected to a

35 kg crate, which lies on a tabletop. The coefficients of the kinetic friction and

the static friction between the crate and the surface are 0.3 and 0.5 respectively.

An applied force, F, pulls the 35 kg crate to the right.

a. In the first case, the applied force is just sufficient to keep the crates from

sliding. Draw clearly labeled free-body diagrams for each crates including

all forces drawn to scale.

b. How much force would need to be applied in this first case?

c. In the second case, the 35 kg crate is sliding to the right with a constant

velocity. Draw clearly labeled free-body diagrams for each crate including

all forces drawn to scale.

d. How much force would need to be applied in this second case?

e. In the third case, the 35 kg crate moves to the right at a constant

acceleration of 0.5 m/s2. Draw clearly labeled free-body diagrams for


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each crates including all forces drawn to scale. In this instance, draw the

direction of acceleration next to each diagram.

f. How much force would need to be applied in the third case?


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A

F B

F

74.A block A with a mass of 0.3 kg sits on the top of block B, whose mass is 2.4 kg.

The coefficient of kinetic friction between all surfaces is 0.2.

a. In the first case, block B is being pulled to the left with constant velocity

Draw clearly labeled free-body diagrams for both block A and B. Include

all forces, drawn to scale.

b. Find the magnitude of the horizontal force F in this first case.

c. In the second case, block B is being pulled to the left with an acceleration

of 0.6m/s2. Draw clearly labeled free-body diagrams for both block A and

block B. Include all forces, drawn to scale. Indicate the direction of

acceleration for block B next to its diagram.

d. Find the magnitude of the horizontal force F in the second case.


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B

A C

75.Block B has a mass of 2.6 kg and sits on a horizontal frictionless table. Block A

has a mass of 1.7 kg and is suspended from a sting at the left side of the table.

Block C has a mass of 5.4 kg and suspended from a string at the right side of the

table. The coefficient of kinetic friction between the surface and block B is 0.05.

The system starts from rest.

a. Draw clearly labeled free-body diagrams for each of the three blocks.

Include all forces and drawn to scale. Draw the expected direction of

acceleration next to each free-body diagram.

b. Find the acceleration of the system of three blocks.

c. Find the tension force in the string between blocks A and B.

d. Find the tension force in the string between blocks B and C.


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40

Velocity (m/s)

50 Time (s)

76.A 2000 kg car travels in a straight line on a horizontal road. The relationship

between car’s velocity and the time are given by the above graph.

a. What is the car’s acceleration during first 20 s?

b. What is the net force applied by the engine during the first 20 s?

c. What is the car’s acceleration from 20 s to 40 s?

d. What is the net force applied by the engine during this time?

e. What is the car’s acceleration from 40 s to 50 s?

f. What is the net force applied by an engine during this time?

20


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Velocity (m/s)

25

Time (s)

77.A 180 kg motorcycle travels in a straight line on a horizontal road. The

relationship between motorcycle’s velocity and time are given by the above

graph.

a. What is the acceleration during the first 5 s?

b. What is the net force during first 5 s?

c. What is the acceleration from 5 s to 10s?

d. What is the net force from 5 s to 10 s?

e. What is the acceleration from 10 s to 15 s?

f. What is the net force from 10 s to 15 s?

g. What is the acceleration from 15 s to 25s?

h. What is the net force from 15 s to 25 s?


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dynamics - bergenusers.bergen.org/laumul/dynamics chapter.doc · web viewa 0.40 kg toy car moves at...

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