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Chapter 4 Lecture Notes
Chapter 4
Newton’s 1st Law of Motion
Aristotle was the first to attempt to classify different types of motion (at
least the first that we know about). He stated that there are two types
of motion, natural and violent. Natural motion would be dirt falling to
the ground (earth returns to the Earth) or a puff of smoke rising into the
air. He believed that everything has a natural state ant that it wants to
return to that state. Violent motion is imposed on an object by
applying a pushing or pulling force.
Aristotle believed that all objects wanted to be at rest and this meant
that the Earth must be at rest. This was agreed upon for nearly 2000
years.
Along comes Copernicus
Nicolaus Copernicus was the first to suggest that the Earth is not at rest.
He theorized that the reason that the Sun, moon, and planets move in
the sky is because the Earth moves around the Sun. Copernicus was
unsure of his theory but was finally convinced to publish De
Revolutionibus shortly before his death.
Galileo Galilei
Copernicus had contradicted Aristotle’s theory of motion. While we
now know that Copernicus was correct, no one believed him at the time.
In the early 17th century, Galileo also started to point out flaws in
Aristotle’s theories on motion. For nearly 2000 years people believed
that heavier objects fell faster than light ones. Galileo demolished this
theory by dropping objects from the Leaning Tower of Pisa. Galileo
dropped two stones from the top of the tower and found (as expected)
that they landed at basically the same time. He discovered that the pull
from the Earth is the same on all objects.
Now that Galileo had shown that object fall at the same rate, he started
looking at Aristotle’s theory of natural motion. Aristotle looked at
nature and stated that all objects would find a way to return to rest.
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Chapter 4 Lecture Notes
While Galileo agreed that this is what we see in nature, he wondered
why he should be limited to motion we see in nature.
When Galileo looked at inclined planes, he noticed that the ball would
speed up as it went down the plane and would slow down as it went up
one. He also noticed that if he used a very smooth set of inclined planes
(Why did they need to be very smooth?) then the ball would return to the
same height as when it was released. When the ball reached that same
height (on the other side) the ball would come to rest (for an instant)
and then start back down the incline.
• What if I have a very long horizontal plane that never comes
back up to the original height?
Galileo said that obviously the ball will continue forever because it will
never return to the same height (remember that the ball only stopped
when it reached the starting height). The property that the ball will keep
moving to the right is called inertia. Obviously, inertia contradicts
Aristotle’s theory of natural motion, but it would be until after Galileo’s
death that Aristotle’s theories would finally be discarded.
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Law of Inertia
Isaac Newton refined Galileo’s definition of inertia:
Every object continues in its state of rest, or uniform motion in a
straight line, unless it is compelled to change that state by forces
impressed upon it.
The is known as Newton’s 1st Law of Motion and was published in
Newton’s Principia.
Newton’s 2nd Law of Motion
A force can be thought of as a push or a pull. Forces can involve contact
(pushing a block on the floor) or not (gravity). Force is a vector
quantity. This means that we not only care about how big the force is
but also in which direction the force is applied. For example:
5N
We would say that the force on the block is 5 N î (Note: 1 N = 1 Newton
kg ∗ m
= 1 2 ).
s
What is the force on this block?
5N
10 N

Fnet = 10 N ˆi + 5 N ˆi = 15 N ˆi
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8N
12 N
The net force on the block is:

Fnet = 12 N ˆi + −8 N ˆi = 4 N ˆi
(
)
Therefore, our diagram above is equivalent to:
4N
Example 1: Find the net force in each of the following cases:
7N
A.
3N
5N
5N
B
10 N
5N
C.
8N
7N
3N
6N
5N
Solution:
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Equilibrium
An object is in static equilibrium if:

1. v = 0

2. ∑ F = 0
Example 2: All of the objects shown are in equilibrium. Find the
unknown force in each case.
A.
12 N
10 N
8N
F
B.
6N
F
15 N
7N
C.
2N
1N
2N
4N
2N
F
3N
Solution:
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Normal Forces

The force due to gravity on objects is called weight ( w ) .


w = mg

m
g = −9.8 2 ˆj
s
where
So when you are standing on a scale:
(1)

w

Your body pushes down on the scale with a force w . Newton’s 1st law
states that any object with a non-zero net force should be moving.
Is the person moving? NO!!!!!
Therefore the net force must be zero!!!!
This means that there must be an upward force acting on the person.
This upward force of the scale pushing on the person is called a normal

force ( n ) .
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
n

w

∑F = 0 ⇒ n − w = 0
n=w
In general, the net force on an object is:


F
∑ = ma
(2)
This is known as Newton’s 2nd Law of Motion. Imagine that a 1 kg
block rests on a frictionless table. We now push the block to the right
with a force of 5 N for 3 seconds. How fast is the block moving after this
period of time? This problem can be solved as follows:
1. Draw a force diagram for the block (also called a Free-Body
Diagram).
n
5N
1 kg
w
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Chapter 4 Lecture Notes
2. Find the net force on the block (in each direction but the net
vertical force is zero in this case):
∑F
y
∑F
x
= n − w =0 ⇒ n = w
F 5N
m
=ma x ⇒ a x = =
=
5 2
m 1 kg
s
3. Now compute the final speed of the block:
m
 m  m
v f = v i + at =  0  +  5 2  (3 s ) = 15
s
 s  s 
Example 3: Your car has a mass of 2000 kg. What net force must be
km
(from rest) in 6
applied to the car to achieve a final speed of 100
hr
seconds? How far will the car travel during this time?
Solution:
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Chapter 4 Lecture Notes
Imagine that a girl is sliding down a frictionless inclined plane. What is
her acceleration?
1. Draw the free body diagram:
2. Find the sum of forces in x and y:
∑F =
y
n − w y = n − mg cos θ= 0 ⇒ n= mg cos θ
∑ F=
x
θ ma x ⇒ a=
w=
mg sin=
g sin θ
x
x
Now that we have the acceleration, let’s find the time that it takes for
the girl to reach the bottom. If the length of the hill is d, then the time is:
1
1
x f = x i + v ix t + a x t 2 ⇒ d = ( g sin θ ) t 2
2
2
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t=
2d
gsin θ
And the speed of sled at the bottom of the hill is:
2
2
+ 2a x d
v=
v 0x
x
=
vx
=
2a x d
2dg sin θ
Example 4: A lamp of mass m = 25.0 kg is hanging from the wires as
shown below. Find the tensions in each of the wires.
Solution:
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Chapter 4 Lecture Notes
Example 5:
The system above is in equilibrium. What is the value of m? If the mass
m is doubled, what is the acceleration of the system?
Solution:
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Chapter 4 Lecture Notes
Example 6: Two 100 kg boxes are dragged along a frictionless surface
m
with a constant acceleration of 1.0 2 , as shown below. Each rope has a
s
mass of 1.0 kg. Find the force F and the tension in the ropes at points A,
B, and C.
Solution:
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Chapter 4 Lecture Notes
Example 7: For the diagram below, find the acceleration of either mass
and the tension in the string (this is called an Atwood’s Machine).
Solution:
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Chapter 4 Lecture Notes
Apparent Weight
Have you ever noticed that you feel heavier when an elevator starts
moving upward? In order to explain this we first need to note that

scales read the normal force ( n ) . This is because the normal force is
the reaction of the scale to your weight pushing on it. Now imagine that
we have the following situation:
Elevator
Scale
If we draw a free-body diagram for the scale:
n
w
Now we apply Newton’s 2nd law to the scale:
∑F
y
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=n − w =ma y
n − mg =
ma y
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Chapter 4 Lecture Notes
(
=
n m ay + g
)
If a y = 0 (constant velocity or at rest):
(
(3)
)
n= m a y + g = m (0 + g )= mg
If a y > 0 then n > w (elevator is accelerating upward)
If a y < 0 then n < w (elevator is accelerating downward)
Example 8: A 50.0 kg woman stands in an elevator which accelerates
m
upward at 2.00 2 . What is her apparent weight?
s
Solution:
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Example 9: A 70.0 kg man is in an elevator that is accelerating
m
downward at 3.00 2 . He decides that he needs to go back up to his
s
m
office and rides back up. The elevator accelerates upward at 2.0 2 .
s
What is the difference in the man’s apparent weight during the two
parts of his trip?
Solution:
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Chapter 4 Lecture Notes
Action and Reaction
So far, we have thought of forces as either pushes or pulls. Now we
want to look at them in a different way. Imagine that a book is sitting on
a table. We know that the table exerts a normal force on the book
because the weight of the book is pushing on it. In other words, the
normal force is in reaction to the weight of the object pushing down on
the table. We say that the weight is an action force and that the
normal force is a reaction force. Some examples of action-reaction
pairs are:
Newton’s 3rd Law of Motion
Newton’s 3rd law states:
Whenever one object exerts a force on a second object, the second
object exerts an equal and opposite force on the first.
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Chapter 4 Lecture Notes
This is more commonly stated as: Every action has an equal and
opposite reaction.
Example 10: A farmer wants the mule shown above to pull the items in
the cart from one side of the farm to another. The mule states that he
can’t do it because Newton’s 3rd law states that every action has an
equal and opposite reaction so the cart won’t move. Where did the mule
go wrong?
Solution:
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Chapter 4 Lecture Notes
We have stated that every action has an equal and opposite reaction.
Now let’s look at the case of the rock falling toward the Earth:
m
W=mg
Earth (mass = M)
Our action force is:
( )
w= mg − ˆj
By Newton’s 3rd law, our reaction force is:
Freaction
= mg ˆj
Does the Earth actually move because of the rock? YES! It just does not
move very far. To see why the Earth barely moves, let’s apply Newton’s
2nd law to the Earth:
Freaction
= mg
= Ma
Solving for the acceleration of the Earth:
a=
m
g
M
m
 1 . This means that the acceleration of the Earth is
M
so small that it barely moves at all!
Since m  M,
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Chapter 4 Lecture Notes
Example 11: A 10 gram bullet is fired from a 5 kg gun. As the bullet is
m
fired the gun accelerates backward at 6 2 . What is the acceleration of
s
the bullet? If the force on the bullet is applied for 0.1 s, how fast will it
be moving when it exits the gun?
Solution:
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