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Physics - Newton's Laws
This is where the real physics begins. Physics is more than equations and math problems -- it is
the laws of the universe and, most importantly, understanding these laws. The laws, of course,
determine how everything works.
The first of these laws we will study were developed by Sir Isaac Newton while he camped out
on a farm having fled the London plague of 1665. An interesting thing about all of it is that he
didn’t publish them until 1687. Wonder why? Anyway, twenty-two years later in 1687 he
finally got around to publishing them in his book, Philosophiaie Naturalis Principia
Mathematica (Mathematical Principles of Natural Philosophy) which is usually known as the
Principia. It was written in Latin and wasn’t translated into English until 1729. Other trivia bits
on the thing? Okay. The Principia is perhaps the greatest scientific work ever written. In it
Newton set out how the universe operates. He explained how the planets orbit the sun, how the
moon orbits the earth, and how objects behave on earth. It basically founded the science of
physics.
So let us get into the physics.
Inertia:
Inertia is an important property of matter.
Inertia  property of matter that resists changes in its motion.
Basically, because of inertia, objects want to maintain whatever motion they have. This was
described initially by Galileo, later Sir Isaac Newton formulated it into one of his basic laws of
motion.
Inertia is proportional to mass. The more mass something has, the more inertia it has.
Mass  measurement of inertia
The unit for mass is the kilogram (kg). Mass is also defined as the amount of matter something
has. Mass is different than weight, which is the gravitational force of attraction between the
earth and an object.
More important definitions:
Force  push or pull
Contact Force  physical contact exists between the object and source of the force
Field Forces  No contact exists between the source of the force and the body being
acted upon: gravity, magnetic force, &tc.
Friction  A force that resists the motion between two objects in contact with one
another
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The First Law:
Newton’s First Law: An object at rest remains at rest, and an object in motion remains
in motion with constant velocity unless it is acted upon by an outside force.
This law really deals with inertia. It is because of its inertia that matter behaves according to this
law. The idea that something would keep moving at a constant velocity for like forever is
something that we don’t see happen very often on the earth, because when something is moving,
there is almost always an outside force acting on it – usually friction. This is why a ball rolling
along a straight section of road will come to a stop all on its own. Friction slows it down and
makes it stop.
When the net external force acting on an object is zero, the acceleration on the object is zero and
it moves with a constant velocity. Of course if it is at rest, it will remain at rest. (Unless and
outside force blah blah blah….)
If
F  0
Then
a0
And
v is constant
The Second Law:
Second Law  The acceleration of an object is directly proportional to
the net force acting on it and inversely proportional to its mass.
This is usually written as a simple formula,
F  ma
More properly, however, it should be thought of as:
 F  ma
This means that the acceleration of an object is a function of the sum of all forces acting on it.
The sum of these forces is known as the net force.
Force is a vector!
The unit for force is the Newton (N)
1 N 1
kg  m
s2
In the USA, the unit for force is the pound (lb).
1 N = 0.225 lb
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Newton’s second law is responsible for weight. Weight is a force, the force of gravity acting on
something. Using the second law, we see that the weight of an object is:
w  mg
Here, w is the weight.
Weight  force exerted by gravity on an object’s with mass.
The weight of an object depends on the acceleration of gravity. If the acceleration brought about
by gravity changes, then the weight can change. This does not happen with mass - the mass of
an object is a constant and has the same value everywhere. If you were to travel to the moon,
your weight would be only 1/6 of its value on earth, but your mass would be the same. This is
because the gravity on the moon is much smaller than the earth’s gravity.

An object has a mass of 10.0 kg, find its weight.
w  mg
m

 10.0 kg  9.8 2  
s 

98.0 N
Recall that accelerations change velocities. Therefore, the net force is the thing that causes
accelerations.

A 450 kg mass is accelerated at 2.5 m/s2. What is the net force causing this acceleration?
m

F  450 kg  2.5 2  
s 

 F = ma

A 2500 kg car is pushed with a net force of 250 N force, what is the acceleration acting on
the car?
 F = ma
a

1100 N
250 kg m 
1

 2500.0 kg
s2

a
F
m

 

a
0.10
250 N
2500.0 kg
m
s2
Now same problem, same 250 N force and all, but the mass of the car is twice as big, 5000.0
kg. Let's find the acceleration once again.
a
250 kg m 
1

 5 000.0 kg
s2


 

0.050
m
s2
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The acceleration is only one half the value of the first problem.

An artillery shell has a mass of 55 kg. The projectile is fired from the piece and has a
velocity of 770 m/s when it leaves the barrel. The gun barrel is 1.5 m long. Assuming the
force and therefore the acceleration is constant while the projectile is in the barrel, what is the
force that acted on the projectile?
v  vo  2ax
2
Find a:
2
v  0  2ax
2
2
m
1

a   770 
s  2 1.5 m 

 197 600
m
s2
v2
a
2x
 2.0 x 105
m
s2
Find the force:
m

F  ma  55 kg  2.0 x 105 2   110 x 105 N
s 


1.1 x 107 N
The Third Law:
Third Law  If two objects interact, the force exerted on object 1 by
object 2 is equal in magnitude but opposite in direction to the force
exerted on object 2 by object 1.
The classic way of saying this is, “For every action there is an equal and opposite reaction”.
Newton’s third law simply says that forces come in pairs. You push on a wall and the wall
pushes on you. We call these action/reaction force pairs.
One of the skills most people master is walking. We rarely think about the act of walking – you
don’t have to concentrate on it, it’s just something that you can do. It turns out, however, that
walking is a fairly sophisticated application of Newton’s laws of motion.
The key to walking is the third law. You push against the ground and the ground pushes you.
All there is to it. But how does that make you move? Okay, you takes your basic second law. It
goes like this; you exert a force on the ground and it exerts a force on you. The two forces are
equal and opposite. The force exerted on you by the earth causes you to accelerate. The size of
the acceleration depends on the force and your mass. F = ma. Because your mass isn’t very big
compared to the earth, you end up with a pretty good acceleration – enough to make you move.
What about the earth? It also had a force exerted on it – the one from your foot pushing on it.
Does it also get accelerated? Well, yes, it has to. The same magnitude force is acting on it. Why
then don’t you notice the earth moving away from you? The second law is involved here. F =
ma, but this time the mass of the earth is huge (6.0 x 1024 kg ) compared to your mass. If your
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mass is 60 kg, the earth is 1023 times more massive than you. That means that the acceleration
acting on the earth is 1023 times smaller than your acceleration. It is so small that it cannot be
measured. But it is real.
m
Because the forces are the same, the mass times the acceleration has to equal the same thing. So
we get:
earth
a
earth
=
m
Rockets work because of the third law. A common
misconception is that a rocket works by pushing against air.
We know this is not true because rockets work in outer space
where there is no air to push against. The way a rocket
works is like this: hot gases, the products of combustion, are
blasted out of the end of the rocket. They are pushed away
from the rocket, but according to the third law if the rocket
pushes the exhaust gases away, the exhaust gases must push
the rocket in the opposite direction. So the rocket goes.
you
a
you
Rocket pushes
gas
Here is the classic third law conundrum. An acquaintance
tells you, “There’s no way a horse can pull a wagon.”
You say, “No way that’s true because horses pull wagons all
the time!” (You’ve been around, you see, and no one can
fool with you.)
gas pushes
rocket
“No, I’m right. The horse can’t pull a wagon. See, if the
horse pulls on a wagon, then according to the third law the
wagon has to pull on the horse with an equal and opposite
force. Since the two forces are equal and opposite, they
cancel out, so the horse can’t pull the cart!”
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The argument sort of makes sense, but you know that it has to be flawed because it is clearly not
true. So what’s the deal?
Action/reaction force pairs never cancel each other out – they can’t because they are partners in
the same action force. Can you touch your mother without being touched by her? Does your
mother touching your finger when your finger touches her cancel out the touch? Can a singlehand clap?
Pretty heavy stuff.
In order to have forces cancel out; they have to be different forces. An example of this would be
if we had two horses hitched to the same wagon pulling in opposite directions. The forces the
horses exert would then cancel out and the wagon wouldn’t move. These would be two separate
forces that do cancel out.
The horse does indeed pull on the wagon and the wagon pulls on the horse in accordance with
the third law. The wagon pulling on the horse results in keeping the horse from running off really
fast – the horse is slowed down by the extra mass of the cart, so its acceleration is much smaller
than if it were not hitched to the wagon.
The horse moves, dragging the wagon with it, by pushing the earth away from it. The earth
pushes back on the horse, so the horse and wagon move. The earth moves away as well, but,
because its mass is so huge (as previously discussed), we can’t measure its acceleration.
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The horse and wagon move because the horse pushes against the earth. What would happen if
you tried to make you car move by sitting in it and pushing on the dashboard? Would the car
move?
What does this picture show about Newton’s laws?
Newton's Second Thought Laws:
1) A body at rest tends to watch television.
2) A body acted upon by a force usually breaks.
3) For every human action there is an overreaction.
Applying Forces
This section of your text will be very tedious, very tedious indeed. (The Physics Kahuna is just as
sorry as he can be.) It’s mostly just a bunch of complicated problems and how to solve them.
Don’t be discouraged by how dry it is – sometimes things that are useful don’t come in exciting
packages. Life cannot always be MTV and video game quality. This section is actually one of
the most important parts of the course.
Key Concept:
Of enormous importance in solving kinematic problems is this concept.
The sum of the forces acting on objects at rest or moving with constant
velocity is always zero.
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F =0
This is a special case of Newton’s second law; the special case where the net force acting on the
system is zero.
We can further simplify the situation! We can analyze the forces in both the x and y directions.
For an object in equilibrium (at rest or moving with constant velocity) the sum of the forces in
the x and y directions must also equal zero.
 FX = 0 and
 FY = 0
Next Key Concept:
Yet another key concept is that when a system is not in equilibrium,
the sum of all forces acting must equal the mass times the acceleration that is acting on it, i.e.,
good old Newton’s second law.
 F = ma
Free Body Diagrams:
When analyzing forces acting on an object, a most useful thing to
do is to draw a free body diagram or FBD. You draw all the force vectors acting on the system
as if they were acting on a single point within the body.
You do not draw the reaction forces.
A ball hangs suspended from a string. Let’s draw a FBD of the thing. First, draw the ball.
B
a
ll
What are the forces acting on it? In this simple case, there are only two forces, the weight of
the ball, mg, and the upward force, t, exerted by the string. We call forces that act along
strings and chains and such things tensions.
So there are two forces. The weight is directed downward and the tension is directed upward.
Draw the vectors from the center of the ball and label them. You have now made your first
free body diagram.
Tension  a pull tangent to a string or rope.
What can we say about the tension and the weight?
t
Well, is the ball moving?
No, it’s just hanging. So no motion; that means it is at rest.
What do we know about the sum of the forces acting on it?
m 60g
If a body is at rest, then the sum of the forces is zero. There are only two forces, the tension
and the weight.
F  0
t  mg  0
Therefore:
so
t  mg
Thus, behold! The two forces are equal in magnitude, but opposite in direction
(one is up and the other is down).
Problems that involve objects at rest (so the sum of the forces is zero) are called static
problems.
Let’s look at a typical static problem. We have a crate resting on a frictionless horizontal
surface. A force T is applied to it in the horizontal direction by pulling on a rope - another
tension. Let’s draw a free body diagram of the system.
There are three forces acting on the crate: the tension from the
rope (T), the normal force exerted by the surface (n), and the
weight of the crate (mg).
A normal force is a force exerted perpendicular to a surface
onto an object that is on the surface.
n
T
mg
It is important to realize that the normal force is not the reaction
force to the object’s weight. The reaction force to the object’s
weight is the force that the object exerts on the earth – recall that the object pulls the earth up just
as the earth pulls the object down. The normal force is the table pushing the object up, the
reaction force is the object pushing the table down. These action reaction pairs are separate
things.
Again, we do not draw the reaction forces on the FBD.
Useful Problem Solving Strategy:
Here is a handy set of steps to follow when
solving static problems.
1.
2.
3.
4.
5.
6.
Make a sketch.
Draw a FBD for each object in the system - label all the forces.
Resolve forces into x and y components.
Use  FX = 0 and  FY = 0
Keep track of the force directions and decide on a coordinate system so you can
determine the sign (neg or pos) of the forces.
Develop equations using the second law for the x and y directions.
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7.

Solve the equations.
A crate rests on very low friction wheels. The crate and the wheels and stuff have a weight of
785 N. You pull horizontally on a rope attached to the crate with a force of 135 N. (a) What
is the acceleration of the system? (b) How far will it move in 2.00 s?
(a) The forces on the system are: T, a tension (the pull on the rope), Fg, the weight of the cart,
and n, the normal force. Let’s draw the FBD.
Y direction: There is no motion in the y direction so the sum of the
forces is zero.
n
 Fy = 0. This means that the normal force magnitude equals the
weight. We can therefore ignore the y direction.
X direction: The motion in the x direction is very different. Since
there is only one force, the system will undergo an acceleration.
 Fx  ma
T
Fg
Writing out the sum of the forces (only the one), we get:
T  ma
a
We need to find the mass;
a  135
T
m
w  mg
kg  m  1

s 2  80.1 kg

kg  m
s2
m
9.8 2
s

 

1.69
m
s2

3.38 m
m
w
g
785
 80.1 kg
(b) How far does it travel in 2.00 s?
1
x  at 2
2
1
m 
2
 1.69 2   2.00 s 
2
s 
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Adding Forces:
When adding two or more vectors, you find the components of the
vectors, then add the components. So you would add the x components together which gives you
the resultant x component. Then add the y components obtaining the resultant y component.
Then you can find the magnitude and direction of the resultant vector.
You want to add two forces, a and b. They are shown in the drawing. The resultant force, r, is
also shown. To the right you see the component vectors for a and b.
b
ay
r
a
b
ax
a
ay
by
bx
ry
r
by
ax bx
rx
We add the component vectors – it looks like this:
See how you end up with the resultant vector after you’ve added up the components?
Now let’s do a problem where we have to add two forces.
Here is a drawing showing the two vectors:
a
15 N
75
b
13 N
a
We do a quick sketch showing how the forces add up:
38
b
r 63
Okay, here’s how to add them up.
1. Resolve each vector into its x and y components.
2. Add all the x components to each other and the y components to each other. This
gives you the x and y components of the resultant vector.
3. Use the Pythagorean theorem to find the magnitude of the resultant vector.
4. Use the tangent function to find the direction of the resultant.
1. Find the x and y components for force a:
ax  a cos
 15 N cos 75o
 3.88 N
a y  a sin 
 15 N sin 75o
 14.5 N
15 N
a
ay 75
ax
2. Find x and y components for force b
bx  a cos
 13 N cos 38
 10.2 N
by  a sin 
 13 N sin 38o
  8.00 N
o
bx13N
by b38
2. Add the components:
rx  ax  bx
 3.88 N  10.2 N
 14.08 N
ry  a y  by
 14.5 N   8.00 N   6.50 N
3. Find the magnitude of the resultant vector (which we shall call
r 2  rx 2  ry 2
r
rx 2  ry 2

r ):
14.08 N 2   6.50 N 2
4. Find the direction of the resultant force:
 rY 

 rX 
  tan 1 
ry

15.5 N
r

rx
 6.50 N 
 
14.08
N


  tan 1 
24.8o
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
A 85.0 kg traffic light is supported as shown. Find the tension in each
cable.
There are three forces acting on the traffic light, T1, T2, and mg. T1 is the
tension from the left cable, T2 is the tension from the right cable, and mg is
the weight of the light.
35.0
55.0
T2
mg
T1
To solve the problem, we must resolve the two tensions into their x and y
components and then add up the forces in the x and y directions. We know
that the sum of these forces must equal zero.
Here is the FBD for the problem: We can identify the two angles by using a
little geometry. 1 is 35.0 and 2 is 55.0.
1
Let’s look at the forces acting in the x direction.
 FX  T2 cos 2  T1 cos 1  0
T2
T1
2
mg
The only forces acting in the x direction are the two x components from the
tension. The weight has no x component since its direction is straight down.
The two x component forces are in opposite directions.
Now we can write out an equation for the sum of the forces in the y direction.
 Fy  T2 sin 2  T1 sin 1  mg  0
We’ve let down be negative and up be positive.
We now have two equations with two unknowns, so we can solve the equations simultaneously.
Solve for T1 in the first equation:
 FX  T2 cos 2  T1 cos 1  0
T1  T2
cos  2
cos 1
T2 cos  2  T1 cos 1  0
 cos 55.0o 
 T2 
 0.7002 T2
o
cos
35.0


Plug this value into the second equation:
T2 sin 2   0.7002 T2  sin 1  mg  0 T2  sin 2  0.7002 sin 1   mg
mg
T2 
 sin 2  0.7002 sin 1 
65
m

85.0 kg  9.8 2 
s 

T2 
sin55.0  0.7002 sin 35.0



682 N
Now we can find T1 by plugging the value of T2 into the first equation, which we already solved
for T1.
T1  0.7002 T2
 0.7002  682 N  
478 N
Lovely Ramp Problems:
A common type of kinematic problem involves an object at
rest upon or moving along the surface of an elevated ramp.
Here’s a simple problem. A frictionless ramp is elevated at a 28.0 angle. A block rests on the
surface and is kept from sliding down by a rope tied to a secure block as shown
n
28.0
T
If the block has a weight of 225 N, what is the force on the rope holding it up?
First, let’s draw a FBD:
Fg

Next we have to choose x and y coordinates:
The positive x direction is up the surface of the ramp – parallel to the
surface.
y
x
The positive y direction is perpendicular to the surface of the ramp.
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There is a component of the block’s weight, Fg , that is directed
Fg cos 
down the surface of the ramp, which would be along the x axis.
This force is Fg sin . The normal force will be Fg cos .
T
 Fx = 0 and  Fy = 0
x direction: T is balanced by a force down the ramp
T  Fg sin  0
T  Fg sin
T   225 N  sin 28o

in
Fg s



106 N
A 5.00 kg ball slides down a 18.0 ramp. (a) What is the acceleration of the ball? Ignore
friction. (b) If the ramp is 2.00 m long, how much time to reach the bottom?
18.0
n
Here’s the FBD:
We’ll let the direction down the ramp be the positive x direction; the y
direction will be perpendicular to the surface of the ramp.
(a) First we look at the sum of the forces in the x direction (up and down
the ramp).
mg sin 

mg
 Fx = mg sin  = ma
There is only one force acting in this direction, the component of the weight that is down the
slope:
mg sin   ma a  g sin 
m

  9.8 2  sin 18o
s 


3.03
m
s2
67
(b) Since we know the acceleration and the distance it goes down the ramp, it’s a simple matter
to calculate the time it takes to do this.
1
x  at 2
2
t
2x
a


 1
2  2.00 m  
 3.03 m
s2







1.15 s
Notice that we pretty much ignored the y direction. This was because there was no motion in that
direction.
Two Body Problems:
So far we’ve dealt with only one body. Let’s
expand the use of Newton’s laws to deal with multiple body situations.
To solve these problems, each body is treated separately. You draw a
FBD for each object an then analyze the forces that are acting. This will
give you several equations that can be used to solve the problem.
 Two masses, 4.00 kg and 5.25 kg are connected by a light string to a
frictionless pulley as shown. Find the tension in the string, and the
acceleration on the system.
4.00 kg
5.25 kg
A single pulley as we have here simply changes the direction of the forces. With the weights
arranged as they are, we can see that the heavy weight will move downward and the lighter mass
will move up. We will treat the as if they are in one dimension, however.
As we have two bodies, we must draw a FBD for each of them.
Each body experiences two forces; the tension in the string (T) which has the same magnitude for
each of them (although it is directed in opposite directions), and their weight (m1g and m2g).
Here are the FBD’s for each:
T
4.00kg
mg
1
T
5.25kg
m2g
For the forces on the rising mass, we use up as the positive direction:
 Fy  m1a
T  m1g  m1a
For the falling mass, down is positive
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 Fy  m2a
m2 g  T  m2a
Note that the acceleration on both masses is the same.
Add the 2 equations:
m1a  T  m1g

m2a  m2 g  T
m1a  m2a  T  m1g  m2 g  T
a  m1  m2   m2 g  m1g
ag
m   5.25 kg  4.00 kg 

a   9.80 2 
s   5.25 kg  4.00 kg 

 m2  m1 
 m1  m2 
m  1.25 
m

  9.80 2 
  1.32 2
s  9.25 
s

We’ve solved for the acceleration, so we can use that to find the tension:
m1a  T  m1g
T  m1a  m1g
m
m


T   4.00 kg  1.32 2    4.00 kg   9.80 2  
s 
s 



44.5 N
2 blocks hang in an elevator as shown. The elevator accelerates
upward at 3.00 m/s2. Find the tension in each rope.
It is important to realize that both blocks will experience the same
accelerations as the elevator, 3.00 m/s2. We will also treat each object
separately.
2 0 .0 k g
3 .0 0 m / s
2
2 0 .0 k g
Look at the forces on the upper block in a skillfully drawn FBD:
There is:
T1
T1 up
T2 down
m g
m1g down
We sum these forces:
T2
1
T1  T2  m1g  m1a
We can’t solve anything here because we have too many unknowns – the two tensions to be
specific.
69
Let us now look upon the lower block:
T2
T2  m2 g  m2a
m
This we can solve as there is only one
unknown.
T2  m2 g  m2 a
2
g
m
m
T2  m2  g  a   20.0 kg (9.8 2  3.0 2 ) 
s
s
256 N
Now we can find the tension on upper block:
T1  T2  m1 g  m1 a
T1  T2  m1 g  m1 a
m
m
T1  256 N  20.0 kg (9.8 2  3.0 2 ) 
s
s

 T2  m1  g  a 
512 N
A 20.0 kg cart with very low friction wheels sits on a table. A light string is attached to it and
runs over a low friction pulley to a 0.0150 kg mass. What is the acceleration experienced by
the cart?
20.0kg
0.0150kg
FBD’s:
n
T
T
m1 g
m2 g
We shall choose the direction of motion to be positive. So for the cart, positive is to the right and
for the weight positive will be down. Okay?
Sum the forces on each object:
70
T  m1a
Cart:
Note that there is only one force acting on the cart in the
horizontal direction.
m2 g  T  m2a
Hanging mass:
Two forces act on the hanging mass.
Both bodies experience the same acceleration.
We can add the two equations together.
T  m1a  m2 g  T  m2a
T  m2 g  T  m1a  m2a
m2 g  m1a  m2a
Note how the unknown tension canceled out, leaving us with a single equation with only one
unknown, a thing we can now solve.
a  m1  m2   m2 g
a
m2 g
m1  m2
m 
1

a  0.0150 kg  9.8 2  
s   20.0 kg  0.0150 kg



 

0.00734
m
s2
3 masses hang as shown, they are connected by light strings and your basic frictionless
pulley. (a) Find the acceleration of each mass and (b) the tensions in the 2 strings.
T1
F
BD’s:
m
3
m g
1
5.00 kg
4.00 kg
3.00 kg
m
m
2
1
T2
T1
T2
m2 g
m3 g
Key point: the magnitude for the acceleration for each mass is the
same.
For falling masses (left side) down is positive. Up is positive on the
upper mass.
m1 :
m1g – T1 = m1a
m2 :
m2g + T1 – T2 = m2a
m3 :
T2 – m3g = m3a
71
Add the equations and solve for the acceleration:
m1g + m2g – m3g = m1a + m2 a + m3a
g  m1  m2  m3   a  m1  m2  m3 
a
g  m1  m2  m3 
 m1  m2  m3 
m   3.00 kg  4.00 kg  5.00 kg 

a   9.8 2 
s   3.00 kg  4.00 kg  5.00 kg 

Find the tensions:
T1  m1  g  a 
T2  m3 g  m3a
m1g  T1  m1a

m
1.63 2
s
m1g  m1a  T1
m
m

T1  3.00 kg  9.80 2  1.63 2  
s
s 

T2  m3  g  a 
m
m

T2  5.00 kg  9.80 2  1.63 2  
s
s 

24.5 N
57.2 N
Friction
Friction! You’ve heard that word, heck, maybe you’ve even used it. So what in the world is
friction? We've all heard about it and blame it for all sorts of things. It is the cause of many
problems, but we couldn't get along without it.
Press your hand down on your desktop and then pull it sideways. The force that tugs on your
hand trying to stop it is friction. Friction is experienced by cars rolling down a road, a horse
walking in a field, a sailboat gliding through the water, or an airplane cruising at 45,000 feet.
Friction is a force that resists the motion between two objects in contact
with one another.
For a fancier definition, try this:
Friction  force caused by interaction of a body with its surroundings.
Cause of Friction:
Friction is brought about by many things. One major cause of friction
has to do with the electrons of the two surfaces in contact with one another. When the objects
are close, electrons from one object will form weak bonds with the atoms of the other (and vice
versa). These bonds, individually weak, are present in great numbers and add up to a significant
72
force. Another cause of friction is the surfaces themselves. Even the smoothest surface is, when
examined under high magnification, rough as a cob. When an object lies on a surface, its little
hills and dales presses down and gets stuck in the rocky mountains of the other surface. To make
them move you have to dislodge all the little catches and hooks that have developed. Once the
thing is moving, the surface irregularities still try to catch and lodge with each other.
Another cause of friction is deformation of the surface. An object sitting on a surface causes the
surface to sag a little. So the object is sort of sitting in a little depression. To make it move, you
have to dislodge it from the depression it’s in. (This is similar to what the Physics Kahuna
experiences with a class on Monday morning, to get the class moving, he must dislodge it from
the depression each student is in.)
Value of Friction:
Is friction bad? Yes! Friction is a force that slows things down,
makes airplanes go slow, uses up energy, makes you tired, costs a lot of money, wears thing out,
&tc.
If friction good? Yes! You couldn't walk without friction. Sans friction, cars wouldn't go and
they couldn't be stopped. Your pants would fall off. &tc.
Forces Involved in Friction:
Friction opposes motion. If you pull on an object to
make it slide across the floor, the frictional force opposes that pull and will be in the opposite
direction. The frictional force is always parallel to the surface and in a direction that opposes the
motion of the object. There are several forces acting on an object that is experiencing friction.




The weight of the object which pushes it down onto the surface it sits upon. This is Fg (or
mg), the weight vector.
The force exerted on the object by the surface, pushing it perpendicular to the surface. This
is, of course, our old friend, the normal force, n.
The force that is pulling the object (or pushing it). This is called the applied force, F.
The frictional force, parallel to the surface and opposing the applied force, f.
n
f
F
Fg
73





If the surface is horizontal and there are no other vertical forces n = Fg.
The normal force, n, is always perpendicular to the surface.
If F is greater than f, the object must accelerate with a net force of F – f.
If f is greater than F, the object won't move at all.
If f = F then the object is moving at a constant speed (or is at rest) as the net force on the
object is zero.
Types of Friction:
There are several types of friction. The one we'll be dealing with is
called sliding friction. There is also rolling friction (which is much smaller). This is the friction
of wheels, rollers, ball bearings, &tc. There is also friction caused by passing through fluids water resistance and air resistance for example.
Figuring Friction: Friction exists between two things.
No one, single thing has friction.
Friction is the result of the interaction of two surfaces. A rocket traveling through the
atmosphere experiences friction with the air but a rocket traveling in the vacuum of space
experiences no friction at all. (Well, maybe a teeny amount because the vacuum of space does
have a slight smattering of loose atoms and things, but they are so far apart that we can virtually
ignore them.)
The magnitude of the frictional force depends on the normal force and the material of the two
objects in contact. Wood on wood would have a different frictional force than steel on wood,
and so on. Heavy objects have more friction than very light ones, &tc.
When an object is at rest, it takes a larger force to make it move than it takes to keep it moving
once it’s got going. Resting objects have to have their static friction overcome before they can
move.
Fs  static force of friction
If an object has a force applied to it but remains at rest, then fs = F. The static frictional force
can have a max value of F.
Moving objects require an applied force to keep them moving that overcomes the kinetic force of
friction. Kinetic is based on a Greek word and means “moving”.
fk  kinetic force of friction.
If the object moves, then the applied force has to be greater than the static frictional force (to get
it started) and must be as big or bigger than the kinetic frictional force.
74
Kinetic friction is always
smaller than static friction.
This is because an object at
rest on a surface has its
microscopic surfaces
embedded in the surface.
Also, the electrons form
their little bonds (as
mentioned before), and you have the depression deal. When an object is in motion, you just
have the rough surfaces bouncing off each other. The electrons don’t have time to set up bonds,
so that force is near zero, and the depression thing gets eliminated or rendered pretty
insignificant. Therefore the kinetic frictional force is always less than the static frictional force.
Finding the frictional force is pretty simple. The frictional force is proportional to the normal
force. Since there are two types of friction, static and kinetic, we have two equations:
fs
 s n
fk
 k n
s  coefficient of static friction
k  coefficient of kinetic friction
The value of the coefficients depends on the two surfaces in contact with one another. These
values are found by experiment. Useful tables can sometimes be found that have the different
coefficient values for common materials worked out and ready for use by the enterprising
physicist.
Let’s do a simple friction problem.

A 25.0 N wood block is pulled across a wooden table at constant speed. What is the force
needed to do this? For the coefficient of kinetic friction use 0.35.
The net force must be zero in the x and y direction. This is because the block is not moving up
or down, so the sum of the vertical forces must be zero. It is moving at a constant speed
horizontally, so , again, the sum of the forces in the x direction must be zero as well.
 Fx = fk – F = 0
so F = fk
n
Behold! The frictional force is equal to the applied force!
 Fy = n – Fg = 0
therefore n = Fg
fk
F
Fg
75
Since everything is flat and there are no extra vertical forces, the normal force equals the weight
of the block. Knowing the normal force, we can calculate the value of the frictional force, which
is, of course, equal to the applied force.
fk
 k n
fk
 0.35  25 N   F 
8.8 N
Here is a table of values for some example coefficients of friction.
Materials
Steel on steel
Aluminum on steel
Wood on brick
Copper on steel
Rubber on concrete
Wood on wood
Glass on glass
Waxed wood on wet snow
Waxed wood on dry snow
Metal on metal (lubricated)
Ice on ice
Teflon on teflon
Synovial Joints in humans
Coefficients of Friction
Static Friction
0.74
0.61
0.60
0.53
1.0
0.25 – 0.50
0.94
0.14
-0.15
0.10
0.040
0.010
Kinetic Friction
0.57
0.47
0.45
0.36
0.80
0.20
0.40
0.10
0.040
0.060
0.030
0.040
0.0030
Let’s look at a slightly more complicated problem where the applied force is at an angle. The
force that friction will oppose will be the horizontal component of the applied force.

You pull a 65.0 kg crate of mass m across the
floor with a force F at an angle of 34.0. The
coefficient of friction between the surfaces is
0.235. If the crate moves at a constant speed,
what is the applied force?
F
O
m
The complication in this problem is that F has a horizontal and vertical component. We draw the
FBD. The components are F sin  and F cos .
Fx  F cos  f  0
Fy  n  F sin   mg  0
n
Fsin 
F

Fcos 
Normally we assume that the normal force which is
responsible for the frictional force is equal to the
mg
76
weight of the object, but since there is an upward component from the applied force, this is not
true any longer. So we write out an equation for the sum of the forces in the vertical direction
and use this equation to find the normal force.
n  F sin   mg  0
We also know that:
n  mg  F sin
f  k n
We now substitute the value for n found in the first equation into the friction equation:
f  k  mg  F sin  
f  k mg  F k sin 
We substitute in the value for the frictional force into the equation for the sum of the horizontal
forces and solve the thing for the applied force:
Fx  F cos  f  0
F cos   k mg  F k sin    0
so
F cos  k mg  F k sin   0
F  cos  k sin    k mg
F cos  F k sin   k mg
F
k mg
 cos  k sin  
We know what all this stuff is, so we can now plug and chug.
m
0.235  65.0 kg  9.8 2
s
F
cos34.0  0.235sin 34.0



149.7 N
0.9604

156 N
Friction on a Ramp:
Frictional ramp problems are extremely popular with physics affectionados. The basic setup is
that a block is on a ramp that is elevated at some angle. The block is at rest, it slides down at a
constant speed, it accelerates down the slope, etc. The lucky physics
student gets to solve the beast for various things. Let’s look at a simple
n
system. An object rests on the ramp at some angle .
fs
mg sin 

77
mg
Let us draw a FBD and examine the forces.
The weight of the block is straight down, mg. The normal force, n, is perpendicular to the
slope’s surface. There is a component of the weight force directed down the slope, mg sin . fs
is the frictional force which is directed up the ramp.
If  is increased till the block just slides, then fs must equal the force component directed down
the ramp.
fs = mg sin 
If block slides down ramp at constant speed, then fk must equal the force component down the
ramp
fk = mg sin 
If the block accelerates down the ramp, then the net force down the ramp is:
Fx = mg sin  - fk = ma
Now let’s do a real problem.

A 15.0 kg block slides down a ramp at constant speed. The elevation angle is 42.5. What is
the coefficient of kinetic friction?
n
fk
mg cos 
mg sin 

 
42.5
mg
If the block moves at constant velocity, then the net force down the ramp is zero. This means
that the kinetic force of friction must equal the force component down the surface of the ramp.
First let us find the kinetic force of friction:
 Fx = fk – mg sin  = 0
fk = mg sin 
We know that the force of friction is also:
fk
 k n
k 
fk
n
78
n = mg cos 
Also
k 
mg sin 
mg cos 

Interesting Point:
sin 
cos 
so

sin 42.5
cos 42.5

0.916
Look at the solution we got for the coefficient of friction:
k 
From trigonometry we know that:
sin 
cos 
k 
sin 
cos 
 tan 
Thus for constant motion or when the friction force = the force down the slope, the coefficient
of friction is simply the tangent of the angle.
  tan
Now let’s look at a wonderfully complicated problem. This one is a friction problem crossed
with a second law problem like we’ve just mastered.

A 0.500 kg block is on a ramp attached to a mass hanging over a low friction pulley via a
light string. The ramp is elevated at 25.0. The block is accelerated up the ramp at 0.256
m/s2. What is the hanging mass? For k use 0.285.
m2
0.2
56 m
s2
m1
0.50
0 kg
25.0
First, we draw the appropriate FBD’s. Then we can
analyze the forces acting on each block.
m2 g  T  m2a
T  m1g sin   f k  m1a
We have two equations and two unknowns, T and m2.
+
T
T
m2
+
m2 g
fk
m 1 g sin O
79
We can eliminate T by adding the two equations:
 m2 g  T
 m2a  
 T  m1g sin  fk  m1a 
m2 g  m1g sin   f k  m1a  m2a
We can now solve this equation for m2:
m2 g  m2a  m1a  m1g sin   f k
m2  g  a   m1a  m1g sin   f k
m2 
m1a  m1g sin   f k
 g  a
But what is fk?
Well, we know that it has to be:
fk  k n  k m1g cos
So we plug that into the equation we set up for m2:
m2 
m2 
m1a  m1g sin   k m1g cos
 g  a
m1  a  g sin   k g cos 
 g  a


m 
m
m

0.500 kg  0.256 2   9.8 2  sin 25.0   0.285   9.8 2  cos 25.0 
s
s 
s 




m2 
m
m

 9.8 2  0.256 2 
s
s 

m2 
3.464 kg
9.544

0.363 kg
One could say that this was a complicated problem. To be true, the Physics Kahuna supposes
that it probably is. It is not a difficult problem, however. One just has to be methodical –
account for all the forces, write out the appropriate equations, and solve for what you need to
find. Take your time, be careful, and enjoy the process.
Now we are ready to look at an AP Test problem.
80

Blocks 1 and 2 of masses m1 and m2, respectively, are connected by a light string, as shown
above. These blocks are further connected to a block of mass M by another light string that
passes over a pulley of negligible mass and friction. Blocks 1 and 2 move with a constant
velocity v down the inclined plane, which makes an angle  with the horizontal. The kinetic
frictional force on block 1 is f and that on block 2 is 2f.
v
m2
m1
M
O
(a) On the figure below, draw and label all the forces on block m1. Express your answers to each
of the following in terms of m1, m2, g, , and f.
O
(b) Determine the coefficient of kinetic friction between the inclined plane and block 1.
(c) Determine the value of the suspended mass M that allows blocks 1 and 2 to move with
constant velocity down the plane.
(d) The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it
is on the inclined plane.
Okay, let’s solve the problem. The first task is simply to draw a FBD. Piece of cake.
(a) On the figure below, draw and label all the forces on block m1.
81
Express your answers to each of the following in terms of m1, m2, g, , and f.
n
T1
f
O
m g
1
We did not draw in the force down the ramp as it is a component of the weight vector, mg which
we did draw. It would not have been wrong to draw it in, however. The forces are: the weight,
mg; the normal force, n; the tension in the string T1; and the frictional force, f. The frictional
force is directed up the ramp because the block is sliding down the ramp and the force of friction
is always opposite to the direction the block is moving.
(b) Determine the coefficient of kinetic friction between the inclined plane and block 1.
The coefficient of friction is given by:
f k  k n
m1g cos
The normal force n is:
Plug that into the friction equation and we can solve for the coefficient of kinetic friction.
f  k m1g cos 
k 
f
m1g cos 
(c) Determine the value of the suspended mass M that allows blocks 1 and 2 to move with
constant velocity down the plane.
We draw the FBD’s for each of the blocks. We’ve chosen the positive direction to be down the
ramp.
+
82
F=0
For block 1:
m1g sin   T1  f  0
For block 2:
m2 g sin   T1  T2  2 f  0
For block M:
T2  Mg  0
m1g sin   T1  f  0

m2 g sin   T1  T2  2 f  0
Add the three equations:

T2  Mg  0
m1g sin 1  m2 g sin 1  f  2 f  Mg  0 We do a bunch of algebra stuff:
g sin   m1  m2   3 f  Mg  0
Mg  g sin   m1  m2   3 f
Solve for M:
M
g sin   m1  m2   3 f
g
M  sin   m1  m2  
3f
g
(d) The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it
is on the inclined plane.
f
This is really a simple problem. With the string cut, the tension,
T1, is gone and the only forces left are the frictional force, f, and
the weight component down the ramp.
 F = m1a
a
m1g sin   f
m1
m1g sin   f  m1a
a  g sin  
f
m1
Free-Body Exercises: Linear Motion
In each case the rock is acted on by one or more forces. All drawings are in a vertical plane, and
friction is negligible except where noted. Draw accurate free-body diagrams showing all forces
acting on the rock. LM-1 is done as an example, using the "parallelogram" method. For
83
convenience, you may draw all forces acting at the center of mass, even though friction and
normal reaction force act at the point of contact with the surface. Please use a ruler, and do it in
pencil so you can correct mistakes. Label forces using the following symbols: w = weight of
rock, T= tension, n = normal reaction force,/= friction.
3. Draw lines
parallel to the two
strings to complete
the parallelogram.
LM - 2 Equilibrium
LM - 3 Friction prevents sliding
2. Draw a line
the same length
as the weight
vector.
1. Draw arrow
representing
the weight.
4. Draw the
tension vectors along
the strings.
mg
The free body
diagram is made up
only of the three
vectors.
LM - 4 Equilibrium
LM - 7 Equilibrium
LM - 5 Equilibrium
LM - 8 Equilibrium
LM - 6 Equilibrium
LM - 8 Equilibrium
84
LM - 10 Rock is falling.
No friction.
LM - 11 Rock sliding at constant
speed on frictionless surface.
LM - 12 Rock is falling at
constant (terminal) velocity.
LM - 13 Rock decelerating
because of kinetic friction.
LM - 14 Rock rising along
free fall trajectory.
LM - 15 Rock at top of
parabolic trajectory
LM - 16 Rock tied to string and
pulled straight up, at 9.8 m/s2 .
LM - 17 Rock pulled by rope
so that it moves at a constant
horizontal velocity. Friction.
LM - 18 Rock pulled by rope
with constant horizontal
acceleration of 2 g.
No friction.
85
LM - 2 Equilibrium
3. Draw lines
parallel to the two
strings to complete
the parallelogram.
2. Draw a line
the same length
as the weight
vector.
1. Draw arrow
representing
the weight.
LM - 3 Friction prevents sliding
t
f
t
4. Draw the
tension vectors along
the strings.
mg
The free body
diagram is made up
only of the three
vectors.
LM - 4 Equilibrium
mg
mg
LM - 6 Equilibrium
LM - 5 Equilibrium
n
t1
t1
t2
t2
mg
mg
LM - 8 Equilibrium
LM - 8 Equilibrium
LM - 7 Equilibrium
t
n
mg
n
n1
n2
mg
mg
86
LM - 11 Rock sliding at constant
speed on frictionless surface.
LM - 10 Rock is falling.
No friction.
LM - 12 Rock is falling at
constant (terminal) velocity.
n
Fdrag
mg
mg
LM - 13 Rock decelerating
because of kinetic friction.
LM - 14 Rock rising along
free fall trajectory.
mg
LM - 15 Rock at top of
parabolic trajectory
n
f
mg
mg
mg
LM - 16 Rock tied to string and
pulled straight up, at 9.8 m/s2 .
LM - 17 Rock pulled by rope
so that it moves at a constant
horizontal velocity. Friction.
LM - 17 Rock pulled by rope
with constant acceleration of
2 g. Moves horizontally.
No friction.
t
F (2mg)
f
mg
mg
mg
Newton’s Laws Wrapup
There aren’t a lot of equations in this section of the old curriculum that you’ll have on
your equation sheet. Just two.
Here they are:
 F  FNet  ma
F fric   N
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The first one is simply the second law. The second equation is a general frictional force thing works for any sort of sliding friction, either kinetic or static.
Remember that there are only a few applications of the first equation that you will have to worry
about.
1. The first one is for when a system is at rest or moving at a constant velocity. If this is
true, then you know that the sum of the forces must be zero. Look for this condition –
it is used over and over on the AP Test in different problems.
This is really important. So important that the Physics Kahuna will highlight it with
bold type, italics, and make it big.
If a body is at rest or moving with a constant velocity the sum of the
forces acting on it is zero.
2. Another type of problem is when the system is being accelerated. When this is true,
the net force is equal to the mass of the system times the acceleration. This is
Newton’s second lae. The key thing is that it is the sum of the forces that equal ma.
We’ve done a lot of problems of this sort.
Here’s what you’ll be expected to do:
1. Static Equilibrium (First Law)
Be able to analyze situations in which a particle remains at rest, or moves with
constant velocity, under the influence of several forces.
This is using F = ma. And, since the particle remains at rest or is
moving with a constant velocity, the ma part of it is zero, i.e., the
sum of the forces is zero. This makes it simple to solve the
problems. And didn’t we have a wonderful time doing them? You
betcha!
2. Dynamics of a Single Body (Second Law)
a. You should understand the relation between the force that acts on a body and the
resulting change in the body’s velocity so you can:
(1) Calculate, for a body moving in one direction, the velocity change that results
when a constant force F acts over a specified time interval.
88
Use F = ma to solve for acceleration. Then use the acceleration to find the velocity
using the kinematic equations. Piece of pie.
(2) Calculate, for a body moving in one dimension, the velocity change that results
when a force F acts over a specified time interval.
Pretty much the same thing as above.
(3) Determine, for a body moving in a plane whose velocity vector undergoes a
specified change over a specified time interval, the average force that acted on the
body.
Pretty much the same stuff as above except that you solve for the force using
acceleration and mass.
b. You should understand how Newton’s Second Law, F = ma applies to a body
subject to forces such as gravity, the pull of strings, or contact forces so you can:
(1) Draw a well-labeled diagram showing all real forces that act on the body.
This is the good old FBD, free body diagram deal. By now you should be one of the
best FBD drawers in the whole of the known universe! Always draw a FBD for force
problems. The graders will no doubt be lulled into the belief that you know what you
are doing if you make a nice force drawing. Could get you a couple of undeserved
extra points.
(2) Write down the vector equation that results from applying Newton’s Second Law
to the body, and take components of this equation along appropriate axes.
We did this with the whole vector component addition thing. You have been trained to
do this.
c. You should be able to analyze situations in which a body moves with specified
acceleration under the influence of one or more forces so you can calculate the
magnitude and direction of the net force, or of one of the forces that makes up the net
force, in situations such as the following:
(1) Motion up or down with constant acceleration (in an elevator, for example).
We did many of these. Draw a FBD!
d. You should understand the significance of the coefficient of friction so you can:
(1) Write down the relationship between the normal and frictional forces on a surface.
This is the F fric   N equation. If everything is horizontal, then the normal force is
just the weight. If, however, the applied forces have vertical components, you will have
to solve for the normal force. It won’t be the weight. If this doesn’t make sense, go
89
back to the handout and your carefully taken notes and do some practice problems
until it does make sense.
(2) Analyze situations in which a body slides down a rough inclined plane or is pulled
or pushed across a rough surface.
No doubt the friction on the ramp problems were your personal favorites. Here you’ve
got to deal with the components of the forces that are parallel to the ramp and
perpendicular to it. You are well trained in how to do this.
(3) Analyze static situations involving friction to determine under what circumstances
a body will start to slip, or to calculate the magnitude of the force of static friction.
This is simply the use of the F fric   N equation. The body will start to slip when the
static frictional force is exceeded. If it isn’t exceeded, then the object won’t move.
3. Systems of Two or More Bodies (Third Law)
a. You should understand Newton’s Third Law so that, for a given force, you can
identify the body on which the reaction force acts and state the magnitude and
direction of this reaction.
Well, you shouldn’t have any trouble doing this. You have been well prepared by the
Physics Kahuna himself. All those lovely homework, quiz, and test problems that you
got to struggle with. But now, can’t you see, it was worth it!
b. You should be able to apply Newton’s Third Law in analyzing the force of contact
between two bodies that accelerate together along a horizontal or vertical line, or
between two surfaces that slide across one another.
This means friction problems. You have survived a great many of these problems, so
they should now be pie.
c. You should know that the tension is constant in a light string that passes over a
massless pulley and should be able to use this fact in analyzing the motion of a system
of two bodies joined by a string.
Ditto on this.
You’ve solved bunches of problems like the ones described above, so you should be good to go
(as the Physics Kahuna’s compatriots in the Marine Corps were fond of saying) on this section.
None the less, the Physics Kahuna, being a pretty good guy (mostly) will be glad to give you
some example problems from old tests.
Here’s one we looked at after the last unit. There were some parts that we couldn’t do back then.
But we can do’em now! This one, you will recall, is from 1998.
90

Two small blocks, each of mass m, are connected by a string of constant length 4h and
negligible mass. Block A is placed on a smooth tabletop as shown below, and block B hangs
over the edge of the table. The tabletop is a distance 2h above the floor. Block B is then
released from rest at a distance h above the floor at
time t = 0.
a.
Determine the acceleration of block B as it
descends.
We analyze the forces on each block:
T
A
Note that the table is “smooth” so we assume
there is no sliding friction.
T
B
The FBD for each block is shown to the right.
 FA  ma  T
ma  T
 FB  mg  T  ma
mg
ma  mg  T
Two unknowns and two equations. Got to solve them simultaneously or something.
Hey, presto, why not add the two equations together:
ma  ma  T  mg  T The two tension terms annihilate one another.
ma  ma  mg
b.
2a g
g
2
a
Block B strikes the floor and does not bounce. Determine the time t1 at which block B
strikes the floor.
1
y  at 2
2
c.
2 ma  mg
t
2y
a

2h
g
 
2
t
4h
g

2
h
g
Describe the motion of block A from time t =0 to the time when block B strikes the
floor.
It will accelerate at the same rate as B,
a
g
2
, till B hits the floor.
91
d.
Describe the motion of block A from the time block B strikes the floor to the time block
A leaves the table.
No force is acting on it, so it moves at constant velocity due to Newton’s first law – its
inertia.
e.
Determine the distance between the landing points of the two blocks.
 g  h 
v  0     2

 2   g 
v  vo  at
v  hg
1
y  gt 2
2
4h
t
g
x  vo x t
2h 
x

1 2
gt
2
 h
hg  2

g



x
2h
Okay, one more problem.

Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected
by massless strings, one of which passes over a frictionless pulley
of negligible mass, as shown. Calculate each of the following.
a.
The acceleration of the 4 kilogram block.
We can treat the two masses on the opposite side of the pulley
from the 4.0 kg block as a single block of 3.0 kg, since we
don’t have to find out anything for them.
Here is the FBD
for the blocks:
We sum the forces:
 F2  m2a  m2 g  T
T
m1 g
T
m2 g
Add the two equations:
92
m2a  m2 g  T   m1a  T  m1g 
m2a  m1a  m2 g  m1g
 a  m2  m1   g  m2  m1   a  g
m   4 kg  3 kg 

a   9.8 2 
s   4 kg  3 kg 

b.

1.4
 m2  m1 
 m2  m1 
m
s2
The tension in the string supporting the 4 kilogram block.
m2a  m2 g  T
 T  m2 g  m2a  T  m2  g  a 
m
m

T  4 kg  9.8 2  1.4 2  
s
s 

c.
 m2a  m1a  m2 g  T  T  m1g
34 N
The tension in the string connected to the 1 kilogram block.
The string connecting the two weights holds 1.0 kg instead of 3 kg. So it is probably 1/3 the answer in part b, or 11 N.
Here’s the FBD for all three weights. The small
1.0 kg mass has been designated as ms by the
Physics Kahuna.
If we look at the sum of the forces on it, we get:
 Fs  ms a  T2  ms g  T2  ms a  ms g
m
m

T2  1.0 kg 1.4 2  9.8 2  
s 
 s
T
2
T
1
ms g m g
1
T
1
T
2
m2g
 ms  a  g 
11 N
93
1. A rock is thrown at an angle of 35.0 to the horizontal with a speed of 11.5 m/s. How far
does it travel?
2. A 450 kg mass is accelerated at 2.5 m/s2. (a) What is the force causing this acceleration? (b)
If the mass of the car is doubled, what happens to the acceleration?
3. How much does a 34.5 kg gymnast weigh?
4. A 2500 kg car experiences a net 250 N force, (a) what is the acceleration acting on the car?
(b) What is the car’s speed at the end of 35 seconds?
94
5. An artillery shell has a mass of 75 kg. The projectile is fired from the weapon and has a
velocity of 670 m/s when it leaves the barrel. The gun barrel is 2.7 m long. (a) Assuming the
force and therefore the acceleration is constant while the projectile is in the barrel, what is the
force that acted on the projectile? (b) If the elevation angle is 52, what is the horizontal
range of the projectile?
6. A ball is attached to a string and hangs from the ceiling. Draw a FBD of the system. Label
the vectors.
7. The space shuttle has a mass of 2.0 x 106 kg. At lift off the engines generate an upward
thrust of 1.3 x 108 N. (a) Draw a FBD of the space shuttle system.
(b) What is the weight of the space shuttle?
95
(c) What is the acceleration of the shuttle when it is launched?
(d) The average acceleration of the shuttle during its 7.5 minute run is 18 m/s2. What velocity
does it theoretically achieve at the end of that time?
8. A 3.45 g hockey puck rests on a flat, smooth table. A horizontal net force of 85.0 N acts on it
for 1.10 seconds. The puck slides across the table at the end of that time and then falls off the
table. The table’s top surface is 85.0 cm above the deck. Find (a) the acceleration of the
puck, (b) the speed of the puck after the 1.10 s, (c) the horizontal distance from the table edge
to where the puck impacts the deck.
96
9. A rock is thrown at an angle of 35.0 to the horizontal with a speed of 11.5 m/s. How far
does it travel?
m
m
m
m
v y  v sin   11.5 sin 35.0o  6.596
vx  v cos   11.5 cos 35.0o  9.42
s
s
s
s
m
m
6.596
 6.596
vy  vy 0
s
s
v y  v y 0  at t 

 1.346 s
m
a
9.8 2
s
m
x
vx 
x  vxt  9.42 1.346 s  
12.7 m
t
s
10. A 450 kg mass is accelerated at 2.5 m/s2. (a) What is the force causing this acceleration? (b)
how fast is the mass traveling at the end of 3.0 s?
m

(a) F  ma  450 kg  2.5 2  
(b)
1100 N
s 

m
m
v  vo  at  2.5 2  3.0 s  
7.5
s
s
11. How much does a 34.5 kg gymnast weigh?
m

w  mg  34.5 kg  9.8 2  
338 N
s 

12. A 2500 kg car is pushed with a 250 N force, (a) what is the acceleration acting on the car?
(b) what is the acceleration if the mass is doubled?

F
kg  m 
1
(a) F  ma a 
(b)
 250 2 
0.10 N
 
m
s  2500 kg 
0.10 m
m

0.050 2
2
2 s
s
13. An artillery shell has a mass of 75 kg. The projectile is fired from the weapon and has a
velocity of 670 m/s when it leaves the barrel. The gun barrel is 2.7 m long. (a) Assuming the
force and therefore the acceleration is constant while the projectile is in the barrel, what is the
force that acted on the projectile? (b) If the elevation angle is 52, what is the horizontal
range of the projectile?
a
v2  vo2  2ax
a
v2
2x
2
m
1

a   670 
s  2  2.7 m 

 83130
m
s2
m

F  75 kg  83130 2  
6.2 x 106 N
s 

14. A ball is attached to a string and hangs from the ceiling. Draw a FBD of the system. Label
the vectors.
F  ma
15. The space shuttle has a mass of 2.0 x 106 kg. At lift off the engines generate an upward
thrust of 1.3 x 108 N. (a) Draw a FBD of the space shuttle system.
(b) What is the weight of the shuttle?
m

w  mg  2.0 x 106 kg  9.80 2  
2.0 x 107 N
s


(c) What is the acceleration of the shuttle when it is launched?
97
F
Net
 FThrust  w  1.3 x 108 N  0.20 x 108 N
F  ma
a
 1.1 x 108 N
F
m

kg  m 
1
m
m
 0.55 x 102 2 
55 2

2 
6
s
s
s  2.0 x 10 kg 
(d) The average acceleration of the shuttle during its 7.5 minute run is 18 m/s2. What velocity
does it theoretically achieve at the end of that time?
 60 s 
m
m
v
a
v  18 2  7.5 min  
8100
v  at
 
t
s
s
 1 min 
16. A 345 g hockey puck rests on a flat, smooth table. A horizontal net force of 125 N acts on it
for 2.00 seconds. The puck slides across the table at the end of that time and then falls off the
table. The table’s top surface is 85.0 cm above the deck. Find (a) the acceleration of the
puck, (b) the speed of the puck after the 2.00 s, (c) the horizontal distance from the table edge
to where the puck impacts the deck.
a  1.1 x 108
98