Forces and Newton’s Laws
The study of causing and changing motion
True or False?
1. When an object is at rest, it has no force
acting on it.
FALSE
When an object is at rest, it has no NET force acting on it.
It can have an infinite number of forces acting on it as
long as all the x-components add to zero and all the ycomponents add to zero. (the forces are balanced)
True or False?
2. When an object is moving it must have a
force acting on it.
FALSE
If an object is moving at a constant velocity, it will have
no NET force acting on it. This means that there could
either be equal and opposite forces or no forces at all.
True or False?
3. To make an object slow down, you must
remove all the forces acting on it.
FALSE
If an object has no force acting on it at all it will either be
at rest or moving at a constant velocity. To make an
object slow down, you must apply a NET force in the
direction opposite its motion.
True or False?
4. An object moving at constant speed will have
no net force acting on it.
FALSE
An object moving in a circle can be moving at a constant
speed, but its velocity is changing direction which means
its accelerating towards the center of the circle which is
caused by a centripetal force
True or False?
5. When an object is in outer space, there is
absolutely no force acting on it.
FALS E
The force of gravity on an object will never be zero. It
can get infinitely small, but objects can never truly be
‘weightless’
Newton’s Three laws
• 1st law: The Law of Inertia
• An object at rest will stay at rest, and an object in
constant velocity motion will stay in constant velocity
motion unless acted on by a net unbalanced force.
• INERTIA is an object’s “want” to resist a change in
motion
• INERTIA depends ONLY on the object’s mass
• The more massive, the more inertia regardless of the speed of
the object.
Newton’s Three laws
• 2nd Law: The law of acceleration
• Fnet = ma
• Ex 1: what is the acceleration of a 3kg mass being acted on by
a unbalanced force of 6N?
Newton’s Three laws
• 3rd Law: The law of action-reaction forces
• ANY force acting on an object, has an equal and
opposite reaction force.
• This will follow a sentence structure that looks like… “the force
object A exerts on object B has a magnitude of F, what is the
magnitude of the force object B exerts on object A?
Force Defined
• A force is any push or pull imparted on an object.
• Some have specific names and directions, others
are general and can point in any direction chosen.
• Force is a vector which means it has…
– A magnitude: how strong it is
– A direction: which way it points
• Some forces are calculated using specific formulas,
others can be found using logic and reasoning
based on the object’s state of motion.
Forces Represented
• Since force is a vector, it is represented by an arrow.
– The length of the arrow should denote its relative
strength.
– The arrow should point in the direction of its action
starting at the object and points AWAY from it.
• One object may have multiple forces acting on it
• The NET FORCE is the resultant of all the forces
acting on an object.
– Forces in the same direction, add their magnitudes
– Forces in opposite directions, subtract their magnitudes
– Forces at angles, use components and trig.
Examples of Forces we will use in this Unit
Name
Gravity
(Weight)
Symbol
Fg
Direction
Formula(s)
Towards the ground
𝐺𝑚1 𝑚1
𝐹𝑔 =
𝑟2
𝐹𝑔 = 𝑚𝑔
Friction (air
resistance)
Tension
Ff
T
Opposite the motion
along the surface
𝐹𝑓 = 𝜇𝐹𝑁
Along the string/chain
Derived based on
situation
Examples of Forces we will use in this Unit
Name
Normal Force
Push/Pull
Centripetal
Symbol
Direction
Formula(s)
FN
Perpendicular to the
Surface
Derived based on
situation
In their given direction
Derived based on
situation
Towards the center of
a circle
Derived based on
situation
𝑚𝑣 2
Fapp
Fc
𝐹𝑐 =
𝑟
Free Body Diagrams
•
•
A free body diagram is a graphical way to show all
the forces acting on an object.
To draw a free body diagram;
1. Represent the object in question with a dot
2. Represent each force acting on that object with an
arrow
-
-
The arrow should start at the dot and point in the direction
of the force
The length of the arrow should represent the relative
strength of the force.
The arrow should be labeled with the force it represents
Two Possible States of Motion
Equilibrium
• Equilibrium is a term used to describe an object that is…
– At rest
– Moving at a constant VELOCITY
• Constant speed AND direction
• Objects in equilibrium
– Have forces that are balanced by other equal and opposite forces.
– Remember, these can be ‘full’ forces or components of forces.
• If an object is in equilibrium, it will stay in that state unless
and unbalanced force acts on it (Newton’s 1st Law)
Two Possible States of Motion
Accelerating (not Equilibrium)
• An object is not in equilibrium when…
– It is changing its velocity
• Changing how fast it is moving
OR
• Changing its direction of motion.
• If an object is experiencing either one of the two
listed above, there must be a net (unbalanced)
force acting on the object in the same direction as
the acceleration.
Using what we just discussed…
1. Decide if the object is in equilibrium or not.
2. Draw a free body diagram of the object. (remember to draw all
forces in their direction and label each arrow with the force it
represents)
A
C
B
At rest
Constant velocity
while pedaling
D
E
Constant speed
around a curve
Moon in orbit
around Earth
Free Falling with small air
resistance
F
Terminal velocity
Applying Newton’s Laws
Do Now: Recall Newton’s Laws and use them to
explain you state of motion right now. (try to
address all three)
Aim: How do we use Newton’s laws to
describe/determine the motion of an object?
Calculating the Force of Gravity
• The force of gravity acting on an object is also
known as the object’s weight
weight is not mass!
• Mass is a quantity that is determine by how much
physical matter you are made up of. To lose mass,
you must actually get rid of the matter you body is
made of.
Calculating the Force of Gravity on the
Surface of a Planet
• Weight is the force of gravity (Fg) in Newtons, acting
on an object. It depends on…
– the object’s mass (m) in kilograms
– the acceleration due to gravity (g) in m/s2 that object is
experiencing when it is near the surface of a planet.
• The formula is Fg = mg
• You can lose weight by either losing mass OR
moving to a place with a lower acceleration due to
gravity (ex. the moon)
Calculating the Force of Gravity
anywhere in the Universe
Gm1m2
Fg 
2
r
Fg
Force of gravity (weight) measured in Newtons
G
m
r
Universal Gravitational Constant 6.67x10-11 Nm2/kg2
Mass of the two objects (Kg)
Distance between the centers of the two objects (m)
The Newton
• The Newton is the unit of any force.
• It is a derived unit, which means it is a combination
of other fundamental units.
• To determine the fundamental units a Newton is
made up of
F  mg
g
1. Find a formula for force
2. Plug in the unit for each variable
3. Combine the units
You must know this!!
m
N  (kg)  2 
s 
 kgm 
N 2 
 s 
Newton’s 1st law- objects in equilibrium
• When an object is in equilibrium, the sum of all the
forces acting on the object is zero.
• We can make this statement more specific by saying
– The sum of all the y-components of the forces are equal
to zero
– The sum of all the x-components of the forces are equal
to zero.
1st law: The Law of Inertia
Using Newton’s first law, explain why the
table settings do not go flying
What does it mean to be in
Equilibrium?
• Objects that are AT REST or moving at a
CONSTANT VELOCITY are in equilibrium.
– The sum of all the horizontal components of force
acting on the object is zero.
– The sum of all the vertical components of force
acting on the object is zero.
Calculating the Normal Force
• In equilibrium on a horizontal surface with no
angles involved.
𝐹𝑦 = 𝑚𝑎𝑦
𝐹𝑁 − 𝐹𝑔 = 𝑚𝑎𝑦
𝐹𝑁 − 𝑚𝑔 = 𝑚𝑎𝑦
𝐹𝑁 − (28𝑘𝑔)(9.81𝑠𝑚2 ) = (28𝑘𝑔)(0)
𝐹𝑁 − 274.7 = 0
𝑭𝑵 = 274.7N
FN=?
28Kg
Fg=mg
Calculating the Normal Force
• In VERTICAL equilibrium
on a horizontal surface
with an angled pull.
𝐹𝑦 = 𝑚𝑎𝑦
FN=?
28Kg
Fy
=30o
Fx
Fg=mg
𝐹𝑁 + 𝐹𝑎𝑝𝑝𝑦 − 𝐹𝑔 = 𝑚𝑎𝑦
𝐹𝑁 + 𝐹𝑠𝑖𝑛𝜃 − 𝑚𝑔 = 𝑚𝑎𝑦
𝐹𝑁 + 200𝑁𝑠𝑖𝑛30 − (28𝑘𝑔)(9.81𝑠𝑚2 ) = (28𝑘𝑔)(0)
𝐹𝑁 + 100 − 274.7 = 0
𝑭𝑵 = 174.7N
Calculating the Normal Force
• In VERTICAL equilibrium
on a horizontal surface
with an angled push.
𝐹𝑦 = 𝑚𝑎𝑦
FN=?
Fapp=200N
28Kg
Fy
Fg=mg
=30o
Fx
𝐹𝑁 − 𝐹𝑎𝑝𝑝𝑦 − 𝐹𝑔 = 𝑚𝑎𝑦
𝐹𝑁 − 𝐹𝑠𝑖𝑛𝜃 − 𝑚𝑔 = 𝑚𝑎𝑦
𝐹𝑁 − 200𝑁𝑠𝑖𝑛30 − (28𝑘𝑔)(9.81𝑠𝑚2 ) = (28𝑘𝑔)(0)
𝐹𝑁 − 100 − 274.7 = 0
𝑭𝑵 = 374.7N
Calculating the Normal Force
• In vertical equilibrium on
an inclined surface
FN=?
𝐹𝑦 = 𝑚𝑎𝑦
𝐹𝑁 − 𝐹𝑔 𝑝𝑒𝑟𝑝 = 𝑚𝑎𝑦
𝐹𝑁 − 𝑚𝑔𝑐𝑜𝑠𝜃 = 𝑚𝑎𝑦
𝐹𝑁 − 28𝑘𝑔 9.81𝑠𝑚2 cos 30 = (28𝑘𝑔)(0)
𝐹𝑁 − 237.9 = 0
𝑭𝑵 = 237.97N
=30o
Fg=mg
Solving any Force Question
1.
2.
3.
4.
Draw your free body diagram
Break any angled force into its components
Decide your positive and negative directions
Apply newton’s second law for horizontal and
vertical separately
𝐹 = 𝑚𝑎
5. Plug in formulas for each force
6. Then plug in numbers and solve.
6. A 30Kg child sits on top of a 10kg crate
a. What is the normal force acting on the child?
b. What is the normal force acting on the crate?
Day 3: Free Body Diagrams and Tension
Do Now: Draw a free body diagram of the sign
pictured below
40o
T1
T2
10kg
Aim: How do we use free body diagrams and
equilibrium to find unknown forces?
Hanging Signs and Tension in a string
• Draw a Free Body Diagram for all the signs
seen below.
40o
mA=5kg
mB=5kg
40o
mC=5kg
7. Finding the Tension Force
a. Draw a free body diagram
b. Break down any angled forces if necessary.
c. Examine all the x-components and set their sum equal to
zero
d. Examine all the y-components and set their sum equal to
zero
e. Resolve your resultant vector into magnitude and direction.
40o
mA=5kg
mB=5kg
40o
mC=5kg
8. A 23Kg girl is sitting at rest in a 8kg tire swing.
a. What is the normal force acting on the girl?
b. What is the tension in the rope?
9. Find T1 and T2
40o
T1
10kg
T2
10. Find T1 and T2
T2 55o
T1
10kg
30o
11. Based on the diagram below (frictionless surface)
a. Determine the tension in the string.
b. Find the mass needed to keep the block at rest.
M=?
30o
Pulleys are usually frictionless which
Allows them to redirect force without
Changing the magnitude of the force.
12. Challenge: Find all unknowns
Day 4: Accelerating Objects
Do Now: Draw the free body
diagram of a rocket taking
off
Aim: How do we use
Newton’s laws to
determine the motion of
an accelerating object?
Newton’s 2nd law- objects accelerating
• When an object is not in equilibrium, the sum of all
the forces acting on the object is NOT zero.
• There will be a net force pointing in the direction
the object is accelerating.
Newton’s Second Law
Example: A car accelerating to the right of a level
road…
• Since there are two dimensions, we must analyze
each separately….
– In the y-direction (up and down), the car is not
accelerating which means the sum of all the ycomponents of the forces are equal to zero
– In the x-direction (left and right) the car is accelerating
to the right, which means the sum of all the xcomponents of the forces IS NOT equal to zero and
the direction of the resultant must be to the right.
a. What is the normal force acting on the object?
b. What is the magnitude and direction of the net force
acting on the object?
c. What is the magnitude and direction of the object’s
acceleration?
11. A 6N force to the right acts concurrently with a 12N force
to the left on a 3kg object.
a. What is the normal force acting on the object?
b. What is the magnitude and direction of the net force
acting on the object?
c. What is the magnitude and direction of the object’s
acceleration?
12. A 2Kg crate is accelerating to the right along a rough
surface at 4m/s2 when acted on by a 14N push force.
a. What is the normal force acting on the object?
b. What is the friction force acting on the object?
8. A 2000kg car is moving at a constant velocity down a high
way. If the engine provides a force of 300N,
a. what is the frictional force acting on the car?
b. What is the normal force acting on the car?
13. A 30Kg child decides to ride in an elevator while standing
on a scale that can read her weight.
a. What does the scale read while the elevator is at rest on the
bottom floor?
b. What does the scale read as the elevator accelerates upward at
2m/s2
c. What does the scale read as she moves upwards at a constant
speed?
d. What does the scale read as she slows down at a rate of 1.5m/s2
when reaching the top floor?
e. What does the scale read as she accelerates downward at 3m/s2
on her way back to the loby?
f. What does the scale read as she moved downwards at a
constant speed?
g. What does the scale read as she slows down at a rate of 2.5m/s2
when returning back to the ground floor?
14. Two children are fighting over a 2kg toy. If one child pulls
to the right with a force of 20N and the other child pulls to
the left with a force of 14N, what is the acceleration of the
toy?
15. While pulling a 30Kg sled across the ice, one person uses a
40N force to the North while the other person uses a 60N
force to the East?
a. What is the magnitude and direction of the resultant
force?
b. What is the magnitude and direction of the resulting
acceleration?
What is the acceleration of the crate?
16. A worker pushes a 20Kg crate across a horizontal surface at a
constant speed. If the force the worker applies is 55N and an angel
of 60o to the horizontal.
a. What is the vertical component of the push force?
b. What is the horizontal component of the push force?
c. What is the magnitude of the friction acting on the crate?
d. What is the normal force acting on the crate?
17. A mother pulls her 20Kg child in a wagon across a frictionless
surface by applying a 100N force at an angle of 55o above the
horizontal.
a. What is the normal force acting on the wagon?
b. What is the magnitude of the wagon’s acceleration?
Atwood Machines
Newton’s Second Law for a System
𝐹𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚𝑠𝑦𝑠𝑡𝑒𝑚 𝑎𝑠𝑦𝑠𝑡𝑒𝑚
Assuming m2 is greater than m1.
1. Derive an expression for the
acceleration of this system.
2. Derive an expression for the time it
would take for m2 to fall a distance
d
Newton’s Second Law for a System
𝐹𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚𝑠𝑦𝑠𝑡𝑒𝑚 𝑎𝑠𝑦𝑠𝑡𝑒𝑚
A 12kg mass, m2 as seen in the diagram is
attached over a massless frictionless pulley to
a 5kg mass, m1. If m2 is dropped from a
height of 1.3m above the ground,
a. What is the acceleration of m2 as it falls?
b. How long does it take m2 to hit the
ground?
c. What is the tension in the string as it
falls?
Newton’s Second Law for a System
𝐹𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚𝑠𝑦𝑠𝑡𝑒𝑚 𝑎𝑠𝑦𝑠𝑡𝑒𝑚
Assuming m2 is greater than m1.
1. Derive an expression for the
acceleration of this system.
2. Derive an expression for the time it
would take for m2 to fall a distance
d
Newton’s Second Law for a System
A 6kg mass, m2 as seen in the diagram
is attached over a massless
frictionless pulley to a 7kg mass, m1.
If m2 is dropped from a height of 0.8m
above the ground,
a. What is the acceleration of m2 as
it falls?
b. How long does it take m2 to hit the
ground?
c. What is the tension in the string as
it falls?
CHALLENGE: What is the acceleration of
the system?
(assume the surfaces are frictionless)
Day 5: Calculating the Friction Force
Do Now: What happens when an object
experiences a friction force?
Aim: What does friction depend on and how to
we calculate friction acting on a horizontal
surface?
Coefficient
of Friction
(no units)
F f  FN
Friction
Force (N)
Normal
Force (N)
The Coefficient of Friction µ
• The coefficient of friction
depends on
– The two surfaces in contact
with each other
– The type of friction
• Static- if the block is initially
at rest
• Kinetic- if the block is already
in motion
• Kinetic friction is always less
than the static friction for
the same object!
• µ can be found in a table in
your reference tables.
• You know to look for it when
the question is very specific
with the two surfaces
Ex: - Wooden block on wooden floor
- Waxed skis on snow