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The Laws of Motion
Unit 3 Presentation 1
What is a force?


A force is a push or a pull on an
object.
Two types of forces:


Contact Force: Two objects are in
contact with each other causing a force
(ex: springs, physically pulling
something, etc.)
Field Force: A force is induced by two
objects that are not touching (ex:
Electrostatic forces, magnetism,
gravity)
Units of Force
kg  m
N
2
s

SI Unit = Newton (N)

English Unit = Pound (lb)
Isaac Newton (1642 – 1727)

English Physicist and Mathematician

Credited with inventing Calculus

Formulated the basic laws of
mechanics, discovered the law of
gravitation, and numerous smaller
discoveries
Newton’s 1st Law of Motion




An object moves with a velocity that is
constant in magnitude and direction,
unless acted on by a nonzero net force.
The “Lazy Law” or the “Law of Inertia”
Inertia: The tendancy of an object to
continue in its original state of motion.
“An object stays in motion (or stays at
rest) unless acted upon by another net
force.
Newton’s 2nd Law

The acceleration of an object is directly
proportional to the net force acting on it
and inversely proportional to its mass.


F  ma
Newton’s 3rd Law

If object 1 and object 2 interact, the force

F12 exerted by object 1 on object 2 is
equal in magnitude but opposite in
direction to the force F21 exerted by
object 2 on object 1.
“Every action force has an equal but
oppositely-directed reaction force.”
Types of Forces




Gravity: The force of attraction
between two massive bodies. Often
called “weight” on Earth.
Normal Force: The force exerted
by an object on another object in
direct contact. Always
perpendicular to the surface.
Tension: Force that is exerted
along a rope
Friction: Force that always
opposes motion
Net Forces


Net Force: The vector sum of all
forces acting on an object.
Free Body Diagram: A pictoral
representation of all of the vector
forces acting on an object’s center
of mass.
Free Body Diagram Example
Bob is standing in the street. Draw a
FBD for Bob.
First, draw a dot representing Bob’s center of mass.
Next, consider every force that is acting on Bob and its
direction.
Forces:
Normal Force
1) Gravity
(downward)
2) Normal
Force (upward)
Gravity
There are no ropes present (no
tension) and Bob isn’t moving
(no friction).
Another Free Body Diagram Example

Bob is getting pulled down the
street (to the east) by his friend Joe
with a rope. Draw a FBD for Bob.
Forces:
Normal Force
Friction
Tension
Gravity
1) Normal force of
street onto Bob
2) Gravity
3) Tension
(towards the east)
4) Friction
(opposing the
tension, pointed to
the west)
Another Free Body Example

Bob is pulling a crate up an inclined
plane by a rope. Draw a FBD for
the crate.
Lets draw a picture.
Forces:
1) Gravity
(downward)
2) Normal Force
(perpendicular to
surface)
3) Tension (up
the incline)
4) Friction (down
the incline)
Normal Force
Friction
Tension
Gravity
Using Newton’s 2nd Law



Calculate the weight of an object
using Newton’s 2nd Law.
What is the appropriate
acceleration? Use the acceleration
due to gravity on the Earth’s
surface! (g = 9.81 m/s2)
Hence, weight = mass * g
Weight Example Problems

Calculate the weight of a 5 kg bowling ball.
m  5kg
g  9.81m / s
2
Fw  mg  5kg(9.81m / s 2 )  49 N
Fw  ?

Calculate the mass of a 150 N doll.
m?
g  9.81m / s 2
Fw  150 N
Fw  mg
Fw
150 N
m
 15.29kg
2
g
9.81m / s
Newton’s 2nd Law Example

An airboat of mass 3.50 x 102 kg, including
passengers, has an engine that produces a net
horizontal force of 7.70 x 102 N, after accounting for
forces of resistance.
(a) Find the acceleration of the airboat
(b) Starting from rest, how long does it take the airboat
to reach a speed of 12.0 m/s?
(c) After reaching this speed, the pilot turns off the
engine and drifts to a stop over a distance of 50.0 m.
Find the resistance force, assuming its constant.
Newton’s 2nd Law Example Cntd.

First, apply Newton’s 2nd Law to find the
acceleration.
F  7.70 x10 2 N
m  3.50 x10 2 kg
a?

F  ma
F
7.70 x10 2 N
2
a

2
.
2
m
/
s
m
3.50 x10 2 kg
Now, use this acceleration and a kinematics equation
to find the time. Remember, the boat starts from
rest.
vo  0m / s
v  v  at
v f  12.0m / s
a  2 .2 m / s
t ?
2
f
o
v f  vo
a
12m / s  0m / s
t 
 5.45 sec
2
2.2m / s
Newton’s 2nd Law Example Cntd.

Now, lets find the resistance force. First, lets find the
resistance acceleration (or deceleration) using kinematics.
d  50m
vo  12.0m / s
v f  0m / s
a?

v 2f  vo2
2d
0  (12m / s) 2
a
 1.44m / s 2
2(50m)
Now, lets use the deceleration and mass to find the
resistance force using Newton’s 2nd Law:
a  1.44m / s 2
2
2
2
F

ma

3
.
50
x
10
kg

(

1
.
44
m
/
s
)


5
.
04
x
10
kg
F ?
m  3.50 x10 2 kg