History
 Aristotle : 384-322 BCE – Force required to keep
objects “in” motion
 Galileo : 1564-1642 - Objects stay in motion given that
we neglect friction
 Isaac Newton : 1642-1727 - Net Forces acting on the
object is zero
 Principia - 1687
What causes a change in the state of
motion?
Types of Forces (Variables)
Applied Force
Tension
Weight
Normal Force
Friction
Newton’s
st
1
Law
 Objects at rest stay at rest
 Objects in motion stay in motion
 Until a net force acts on it
 “Inertia” – Tendency of an object to
resist change
 “An object in motion tends to stay in motion, and an
object at rest tends to stay at rest, unless the object is
acted upon by an outside force”
What causes motion?
Newton’s First Law of Motion
(Inertia)
“If a force acts on a body, it produces a change in the
motion of the body that is in the direction of the
applied force”
Force = mass x acceleration
F=ma
Force is measured in Newtons
1N = 1kg x 1m/s2
What are the units of Force?
Newton’s Second Law of Motion
EXAMPLE PROBLEM
A 0.02 kg PIGEON FLIES INTO A PANE OF GLASS
WITH A FORCE OF 2N. AT WHAT RATE WAS THE
PIGEON ACCELERATED WHEN IT HIT THE PANE
OF GLASS?
EXAMPLE #2
A .3 kg ARROW IS SHOT THROUGH AN APPLE. IT
TAKES .01 seconds TO TRAVEL THROUGH THE
APPLE. IF THE ARROW ENTERS THE APPLE WITH
A VELOCITY OF 30 m/s AND LEAVES THE APPLE
WITH A VELOCITY OF 25 m/s IN THE SAME
DIRECTION, WITH WHAT FORCE HAS THE APPLE
RESISTED THE ARROW?
Kinematics Force Example
What average force is needed to accelerate a 7.0 g pellet from rest to
175 m/s over a distance of 0.700 m along the barrel of a rifle?
Newton’s Third Law of Motion
(Reaction)

When you hit something, it hits you
back!
Action - Reaction
Action - Reaction
Action: Rocket
Pushes Gas
Reaction: Gas
Pushes Rocket
Newton’s 3rd Law
 For every force there is an equal and
opposite force
action = reaction
Fc = Fcb
(ma)c = (ma)cb
(
ma) = (ma)
c
cb
So for a canon vs canon ball, why does
the cannon ball get launched from the
cannon if the force is equal?
The cannon has a greater mass than
the cannon ball, therefore the cannon
ball gets a larger acceleration than the
cannon.
Isaac Newton’s three laws of motion give scientists a
clear and thorough understanding of the way force and
motion were related
b Inertia
b Force
b Reaction
Mass vs. Weight
Mass – the quantity of matter
in an object (determines state
of motion) – Not Volume!
• Example: equal bags of nails and
cotton
Mass vs. Weight
Weight – The force of
gravity acting on an
object’s mass
• Example: Fweight = mass x gravity
Force Example One
What is the weight of a 66 kg
astronaut on the Earth? The
astronaut weighs 224.2 N on Mars.
What is the acceleration due to
gravity on Mars?
Pressure
What is Pressure?
How is Pressure related to force?
What are some examples?
Pressure
 Pressure is the force per area
 P = F/A
 As area increases pressure decreases and vice versa.
 Area = l x w
 Area(circle) = pr2
Pressure Example
Edgar cuts his knee in a fall while chasing a soccer ball.
If a 6 N force is exerted on Edgar’s knee during the fall,
applying a pressure of 1000 N/m2 on an area of his
skin, what is the area( in cm2) of the cut that results
from the impact?
Free body diagrams
FN
Ff
FA = applied force
Ff = friction force
W = weight
FN = normal force
FA
W = mg
Normal Force = Forces that act
perpendicular to the common surface of
contact (Bathroom Scales) – Support Force
Sum of Forces Statements
 ΣF = ???
 Accounts for the forces on an object
 Direction of motion is positive
 Opposite directions = opposite sign
 Always = “ma”
Summation of forces
FN
Ff
FA
Only forces in
exactly the same
direction can be
added
W = mg
Fx = FA – Ff = ma
Fy = FN – mg = ma
The x direction notes
parallel to the surface and
the y direction is
perpendicular to the
surface
Normal Forces & Candy Boxes
A 10 kg box of candy is resting on a smooth horizontal surface of a table.
a.) Determine the weight of the box and the normal force acting on it. b.)
Determine the weight and Normal force acting on it if you apply 40 N of
force down on the box c.) Determine the weight and Normal force acting
on it if you apply 40 N of force up on the box
Did you remember your free body diagram?
Tension
T
Fy = T – W = may
Fy = T – mg = may
W = mg
A pull on an object, typically a wire or string
Free Body Diagrams & F 1
A crate, which has a mass of 55.0 kg. is being accelerated
straight up by a rope at a rate of 3.80 m/sec2. What will be the
tension in the rope?
Did you remember your free body diagram?
An Elevator Problem…
How much tension must a rope
withstand if it is used to accelerate a
1200 kg elevator car vertically
downward at 0.80 m/s2?
Free Body Diagrams & F 2
A tall building has elevators equipped with
bathroom scales (naturally). While
descending in this elevator, the scale reads
a weight of 680 N. If you have a mass of 65
kg, what is the magnitude and direction of
the elevator’s acceleration?
Did you remember your free body diagram?
Atwood’s Machine
m1 > m2
Direction in an Atwood’s
Machine should remain
constant with the system
T1 = T2
T
T
m2
m1
m1
W1
m2
Fm1 = m1g – T = m1a
W2
Fm2 = T – m2g = m2a
Tension is the same: Thus,
Fy = g(m1 - m2)/ (m1 + m2)= ay
Forces at angles
Ty = Tsin
Tx = Tcos
Fx = Tcos – Ff = ma
Fy = FN + Tsin - mg = ma
Forces with Angles
A 30 kg sled is pulled across the top of a mountain. A force of 50
N is applied to a rope at an angle of 20º to the horizontal from
the sled. Neglecting any frictional forces, calculate the
acceleration of the sled. Calculate the upward force the snow
covered ground exerts on the sled as it pulled.
Did you remember your free body diagram?
Forces with Angles - Tension
A 100 kg chandelier hangs by two cables from a ceiling. The
cables form an angle of 45º with respect to the horizontal. What
is the Tension in each of the cables.
Did you remember your free body diagram?
Forces at angles
Fx = mgsin – Ff = ma

Wy = mgcos
Fy = FN – mgcos = ma

Wx = mgsin
Forces with Angles – Inclined Planes
You and your friends are playing with the slip and slide. You
slide down a hill that is 10 meters long with an angle of 25º.
What is the acceleration of that friend #2 (100kg) if he starts
with a speed of 6 m/s? What is his speed at the end of the slide?
What is the force of the ground acting up on slider #2 – Joey?
Did you remember your free body diagram?
What moves a car?
 Tires or road?
 Why doesn’t the road move instead of the car?
Friction
 Friction - a force which acts to oppose the motion of an
object.
Types of Friction
 Static Friction - force that opposes the
start of motion. These forces have
maximum values.
 Sliding Friction (Kinetic) - force between
surfaces already in motion. This type of
friction is less than static friction.
 Rolling Friction – force between a surface
and a rolling object
 Fluid Friction – force of a gas or a liquid as
an object passes through (i.e. Air
Resistance
The coefficient of friction
 The coefficient of friction(µ) is a number that
describes the type of surface. The greater the
coefficient for a surface, the greater the friction(Ff )
 µ can be determined by using the following equation:
 Ff = µ x N
 N (normal force) is the force pushing surfaces
together. It is a force that acts perpendicular to the
surface.
Pairs of Forces
 Forces act in pairs.
 *When studying friction - there are two pairs
of forces acting on an object.
 1st - parallel force moving an object and an
opposing frictional force to the surfaces
touching.
 2nd - acts perpendicular to the 2 surfaces,
weight of object pushes downward and an
upward push or normal force.
Summation of forces
FN
Ff
FA
Only forces in
exactly the same
direction can be
added
W = mg
Fx = FA – Ff = ma
Fy = FN – mg = ma
The x direction notes
parallel to the surface and
the y direction is
perpendicular to the
surface
Friction Example One
A 10.0 kg box rests on a surface. The
coefficient of static friction is 0.3. The
coefficient of kinetic friction is 0.1. What
force is needed to start the box moving?
What force is needed to accelerate the
box at 1 m/s2?
Forces at angles
Ty = Tsin
Tx = Tcos
Fx = Tcos – Ff = ma
Fy = FN + Tsin - mg = ma
Atwood’s Machine
m1 > m2
Direction in an Atwood’s
Machine should remain
constant with the system
T1 = T2
T
T
m2
m1
m1
W1
m2
Fm1 = W1 – T = m1a
Fy = W1 – T + T – W2 = mTa
W2
Fm2 = T – W2 = m2a
Fy = g(m1 - m2)/ (m1 + m2)= ay
Atwood’s Machine
m1 > m2
Tension in the Atwood
Machine
T1 = T2
T
T
m2
m1
m1
W1
m2
Fm1 = T = W1- m1a
Fy = W1 – T + T – W2 = mTa
W2
Fm2 = T = W2 +m2a
Fy = g(m1 - m2)/ (m1 + m2)= ay