Chapter 2
Newton’s Laws
Force
 Force is a push or pull on a body.
 It usually causes:



a distortion in the body,
a change in the body’s velocity, or
both.
 Force is the agent for a change in motion.
2
Force, cont’d
 The units of force include:


SI: newton (N), dyne or
metric ton (2000 N)
English: pound (lb), ounce (oz) or
ton (2000 lb).
 To convert between SI and English:
1 N  0.225 lb
or 1 lb  4.45 N
3
Weight
 Weight is the force of gravity acting on a
body.

We typically use the symbol W.
 Your weight depends on which planet you are
on.

A 200 lb person on Earth weighs about 33 lb
on the Moon.
4
Weight, cont’d
 Although weight always acts toward the
Earth’s center, it might not always appear that
way.
 Consider a box on a ramp.
5
Weight, cont’d
 The weight is always straight down.
 We can break the weight into “components”
parallel and perpendicular to the ramp.
6
Weight, cont’d
 The perpendicular component only acts to
hold the box on the ramp.
 The parallel component acts to accelerate the
box down the ramp.
7
Weight, cont’d
 As the ramp becomes more steep, the
parallel component increases:

The box tends to accelerate more quickly.
8
Friction
 Friction is a resistive force to relative motion
between two bodies or substances in contact.


Between your tire and the road;
Between your car and the air.
9
Friction, cont’d
 Static friction is the resistance you
experience when you try to start sliding an
object.

It is static because there is no relative motion
between the surfaces.
 Kinetic friction is the resistance you
experience while sliding an object.

It is kinetic since there is relative motion.
10
Newton’s first law of motion
 Newton’s First Law of Motion states: An
object will remain at rest or in motion with
constant velocity unless acted on by a net
external force.
 An external force is a force applied to the
object from some other object.

force from an impact, gravity, air resistance,
etc.
11
Newton’s first law of motion, cont’d
 The net force is the vector sum of all
external forces.


If you and a friend push on opposite ends of a truck
with the same force, the net force is zero — the
forces cancel.
If you push in the same direction, there is a non-zero
net force.
12
Newton’s first law of motion, cont’d
 If the two forces are
in different
directions, the net
force is an
“in-between”
direction.
13
Newton’s first law of motion, cont’d
 Remember centripetal acceleration?

The acceleration experienced by an object
moving in a circular path.
 There must be a force pulling the object
toward the center.

Otherwise the object would fly off.
14
Newton’s first law of motion, cont’d
 You can create an “artificial gravity” in a
space station by rotating it.
15
Mass
 Mass is a measure of an object’s resistance
to acceleration, i.e., changes in the object’s
motion.
 It also indicates the amount of matter in an
object.


SI units: kilogram (kg) or gram (g);
English unit: slug.
16
Mass, cont’d
 An object of little mass requires little force to
accelerate it.
 A massive object requires a much larger force
to give it the same acceleration.
17
Mass, cont’d
 The weight of an object depends on which
planet you measure the weight.
 The mass of the object is independent of the
planet.
18
Newton’s 2nd law of motion
 Newton’s 2nd Law of Motion states that an
object is accelerated when a net external
force acts on it.
 The net force equals the object’s mass times
its acceleration:
F  ma
19
Newton’s 2nd law of motion, cont’d
 Think about why you might add more dogs to
a sled…
20
Newton’s 2nd law of motion, cont’d
 The object will accelerate according to
aF
m
 If the object experiences a centripetal
acceleration, the centripetal force is
F  ma  m
v
2
r
21
Newton’s 2nd law of motion, cont’d
 The SI unit of force, the newton, is defined
according to Newton’s 2nd law:
F  ma
kg m
 m
1 N  1 kg   1 2   1
2
s
 s 
22
Newton’s 2nd law of motion, cont’d
 A falling object accelerates due to the Earth’s
gravity at
g  9.8 m/s
2
 So the force the object feels from Earth, i.e.,
its weight, is:
F  ma  W  mg
23
Example
Example 2.2
An automobile manufacturer decides to build a
car that can accelerate uniformly from 0 to 60
mph (27 m/s) in 10 s. The car’s mass is to be
about 1,000 kilograms. What is the force
required?
24
Example
Example 2.2
vi  0 m/s
ANSWER:
The problem gives us:
v f  27 m/s
The force is:
m  1000 kg
t  10 s
F  ma
 1000 kg 
27 m/s  0 m/s
10 s
 2700 N
25
Example
Example 2.2
DISCUSSION:
What produces this force?


The friction between the car’s tires and the
road.
If the road is too slick, the tires cannot get
enough “grip” and you “peel-out.”
26
Example
Example 2.3
In Example 1.5, we computed the centripetal
acceleration of a car going 10 m/s around a
curve with a radius of 20 meters. If the car’s
mass is 1,000 kilograms, what
is the centripetal
force that acts
on it?
27
Example
Example 2.3
v  10 m/s
ANSWER:
The problem gives us:
r  20 m
m  1000 kg
The force is:
F  ma  m
v
2
r
 1000 kg 
10 m/s 
20 m
2
 5000 N
28
Example
Example 2.3
DISCUSSION:
What causes this force?


It is the friction between the tire’s and the
road.
If the road is too slick, the road could not
generate this force and you slide into the
ditch.
29
Types of motion
free-fall
 Free-fall is a type of motion in which the only
force acting on the object is the object’s
weight.

There is no friction, no air resistance, etc.
 Gravity always acts toward the Earth’s center.
30
Types of motion
free-fall, cont’d
 A ball thrown upward will slow-down
because gravity is “pulling” down on the ball.
31
Types of motion
free-fall, cont’d


Eventually it will stop.
It will then fall back to the Earth due to
gravity.

The time it
takes to go
up is the
same amount
of time for it
to fall.
32
Types of motion
free-fall, cont’d
 Why does a
football follow
a curved
trajectory?
33
Types of motion
free-fall, cont’d
 Why does a football follow a curved
trajectory?
ANSWER:
 Gravity acts in the vertical.

So the vertical speed changes.
 Ideally, there is no force in the horizontal
direction.

The horizontal speed does not change.
34
Types of motion
free-fall, cont’d
 Why does a football follow a curved
trajectory?
ANSWER:
 So the ball goes up and down just like if you
had thrown it straight-up.
 But the horizontal speed carries it downfield.
35
Types of motion
simple harmonic motion
 Imagine a block
attached to a spring.
 If we stretch the
spring and release it,
the block moves in a
regular, periodic
fashion.
36
Types of motion
simple harmonic motion
 This type of motion
is called simple
harmonic motion.

Notice that the speed
is a maximum when
the block is farthest
from its rest position.
37
Types of motion
falling with air resistance
 What happens when you “really” drop an
ball?
 Think about the forces acting on the ball.


Gravity is accelerating it downward.
The air offers resistance, trying to prevent the
ball’s descent.
38
Types of motion
falling with air resistance, cont’d
 The ball’s weight pulls it down.
 The air has to be forced out of the ball’s way
— it
exerts an
upward
trying to
slow the
ball down.
39
Types of motion
falling with air resistance, cont’d
 As the ball’s speed increases, the air
resistance force increases.
 Eventually the ball no longer accelerates.
 At this point,
you’ve reached
the terminal
speed.
40
Newton’s third law of motion
 Newton’s 3rd Law of Motion
states that forces always occur
in pairs:

When one object exerts a
force on a second object, the
second exerts an equal and
opposite force on the first.
41
Newton’s third law of motion,
cont’d
 If object A exerts a force on object B, then
object B exerts an equal force in the opposite
direction on A:
FB on A   FA on B

When you fall down, you feel the Earth exerting a
force on you but you also exert that same force
on the Earth.
42
Newton’s third law of motion,
cont’d
 Think about pushing off against a wall.




You push against the wall.
The wall pushes back.
If the wall is
weak, it might
fall down.
If not, you move
away.
43
Newton’s third law of motion,
cont’d
 Consider an airplane’s wing.



Due to the angle of attack, the air impacts the
bottom of the wing.
The wing pushes the air out of the way.
The air pushes
back and
provides some
lift.
44
The law of universal gravitation
 Newton’s Law of Universal Gravitation
states that every object exerts a gravitational
force on every other object.


The force increases as either object’s mass
increases.
The force decreases as the objects move
farther apart.
45
The law of universal gravitation,
cont’d
 Mathematically, we can write
F
m1m2
d



2
m1 is the mass of one object
m2 is the mass of the other object
d is the center-to-center distance separating the
objects.
46
The law of universal gravitation,
cont’d
 Note that Newton’s 3rd law means:


the Earth pulls down on you with a force equal
to your weight, and
you pull “up” on the Earth with you’re a force
equal to your weight.

You move more than the Earth because the Earth
is much more massive than you.
47
The law of universal gravitation,
cont’d
 If you could stand on a tower 4,000 miles
high, you would weigh one-fourth your usual
weight.

You are 2×’s
farther from
the Earth’s
center.
48
The law of universal gravitation,
cont’d
 Using the gravitational constant,
G  6.67 10
11
N  m / kg
2
2
we can write the law of gravity as
F G
m1m2
d
2
49
Example
Problem 2.25
A space probe is launched from the Earth,
headed for deep space. At a distance of
10,000 miles from the Earth’s center, the
gravitational force on it is 600 lb. What is the
size of the force when it is 20,000 miles from
the Earth’s center?
50
Example
Problem 2.25
ANSWER:
The problem gives us:
d1  10, 000 mi
F1  600 lb
d 2  20, 000 mi
The force is:
F
GMm
d
2
51
Example
Problem 2.25
ANSWER:
From the given info, we know:
F1 
GMm
2
1
d
and F2 
GMm
d2
2
We can write these equations as
2
1 1
Fd
 GMm and F2 d 2  GMm
2
52
Example
Problem 2.25
ANSWER:
So we have,
2
1 1
Fd
 F2 d 2
2
53
Example
Problem 2.25
ANSWER:
The requested force is
2
F2  F1
d1
d2
2
  600 lb 
10, 000 mi 
2
 20, 000 mi 
2
  600 lb  
1
4
 150 lb
54
Example
Problem 2.25
DISCUSSION:
We could solve for the mass from the first
force and distance.
We’d have to convert a lot.
But this approach is less work and gives us
some insight into the nature of the
gravitational force.
55
The law of universal gravitation,
cont’d
 We can combine the law of gravity and
Newton’s 2nd law to calculate the gravitational
acceleration.
 Let’s take:


M to be the Earth’s mass,
R to be the Earth’s radius

distance from you to the Earth.
56
The law of universal gravitation,
cont’d
 Equating our usual formula for weight to the
actual gravitational force:
Weight  Fgravity
mg  G
mM
R
g
2
GM
R
2
57
The law of universal gravitation,
cont’d
 This gives us a method to calculate the
gravitational acceleration:
g
GM
R
2
58
Example
Let’s calculate the acceleration due to gravity at
the Earth’s surface:
M Earth  5.98 10
24
kg
REarth  6.4 10 m
6
59
Example
ANSWER:
g
GM
R


2
6.67  10
11

5.98  10
 6.4 10 
6
 9.74 m/s
24

2
2
60
Example
DISCUSSION:
The Earth is not a perfect sphere.
We used an average radius so we obtained an
average acceleration (at sea level).
Notice that we never had to worry about the
mass of the accelerating object.
61
Example
Let’s calculate the acceleration due to gravity at
the Moon’s surface:
M Moon  7.22 10
22
kg
RMoon  1.75 10 m
6
62
Example
ANSWER:
g
GM
R
2
 6.67 10  7.22 10 

1`.75 10 
11
22
6
 1.57 m/s

1
6
2
2
 g Earth
63
Example
DISCUSSION:
You would weigh one-sixth your usual weight
on the Moon.
Things fall at one-sixth the rate.
You could jump about six times higher.
64
Orbits
 Consider an object orbiting the Earth in a
circular orbit.

Let’s say it is a distance r from the Earth’s
center.
 There is one force acting on it:

Gravity.
 It’s acceleration is centripetal.
65
Orbits, cont’d
 Newton’s second law gives:
F  ma
GmM
r
2
GM
m
v
2
r
v
2
r
66
Orbits, cont’d
 So we can determine what speed a satellite
has if we know its distance from the Earth’s
center.
GM
v
r

The altitude (height above the surface) is just
the distance less the Earth’s radius.
67
Example
A communications satellite is at an altitude of 35,900
kilometers. At what speed is the satellite orbiting the
Earth?
68
Example
ANSWER:
The problem gives us:
h  35, 900 km
This is the distance above the Earth. We need
the distance from the center:
R  RE  h
 6.4 10  35.9 10 m
6
6
 42.3 10 m
6
69
Example
ANSWER:
v
 6.67 10
11
 5.98 10
42.3 10
24

6
 3100 m/s
This is almost 7000 mph.
70
Example
DISCUSSION:
This type of orbit is called “geosynchronous”
because the satellite remains above the
same location on the Earth’s surface.
That’s where you want a communications
satellite.
71
Orbits, cont’d
 Here is a graphic detailing the orbit of Halley’s
comet.
 There are some general results
from the law of
gravity.
72
Orbits, cont’d
 The orbits are ellipses.

Either circles or stretched-out circles.
 The Sun is at a focus of the ellipse.

Or at the center if
the orbit is a
circle.
73
Orbits, cont’d
 Halley’s comet moves faster when its nearer
the Sun.

If the distance from the Sun is smaller,
our orbital speed
formula says the
speed is higher:
v
GM
r
74
Orbits, cont’d
 You can draw an ellipse using two push-pins
and a piece of string.

Each push-pin corresponds to a focus of the
ellipse.
75