Forces and Newton’s Laws of
Motion
The most important chapter of
physics EVAR
What is a Force?
• A force is a push or a pull
• Contact forces:
– Boat pulling a water skier
– Pushing a car with a dead battery
• Noncontact forces:
– Gravitational attraction of moon to earth
– Magnetic attraction/repulsion
– Electrostatic attraction/repulsion
• The SI unit of force is the Newton (N)
Newton’s First Law
• “An object at rest will stay at rest. An object in
motion will stay in motion in a straight line
unless acted upon by an external net force.”
• Take a moment to contemplate this truth of
the universe.
• Now, let’s look at the individual pieces of that
statement….
Deconstructing Newton’s First Law
• “…An object at rest will stay at rest…”
– Things don’t just move spontaneously. If an object is
just sitting there, it will continue to sit there unless
something else pushes or pulls it.
• “…An object in motion will tend to stay in motion
in a straight line…”
– The reason you are used to seeing things slow down in
the real world is because they are usually acted upon
by friction (or air resistance). In the absence of
friction, however, objects would continue in a straight
line forever until they hit something.
Deconstructing Newton’s First Law
• “…by some external net force…”
– ‘external’ means it comes from ‘outside’. Forces
from within do not count as external forces
– A ‘net force’ is the resultant force if you have
several different forces acting on an object.
– If there is only one force, then the net force is that
force.
– If there are several, you add them together as
vectors to find the one final resultant force.
– In order to show that, we need to know that….
What else can we say about forces?
• Forces are vectors
– That means they have magnitude and direction
• A group of forces acting on the same object
will add together graphically to produce a “net
force”
• This is why you learned to add vectors
together. It wasn’t just for your health.
Example of finding net force – Tug of
war
Freshman Class
Senior Class
800 Newtons
1000 Newtons
Add individual
vectors together
resultant
=
Net force = 200 N @ 00
Example of finding net force – Pushing
a car
Bubba
Earl
800 Newtons
Add individual
vectors together
resultant
=
1000 Newtons
Net Force =
1800 Newtons
@ 00
But Dr. Mason, those examples were
far too easy
Let’s look at a contrived example:
Three kids pulling on a tire.
ropes
Witness the artistry
Top down view
tire
Kids pulling on tire problem
Let’s look at the forces from each individual rope
10N
10N
Θ2 = 1350
Θ3 = 2700
14.14N
Θ1 = 450
Kids pulling on tire problem
10N
Θ1 = 450
FY = 7.07N
What are the horizontal and vertical components
of each force?
F = 10N
FX = 7.07N
Find the horizontal and vertical components of
force the same way you do for other vectors.
Kids pulling on tire problem
F = 10N
10N
Θ2 = 1350
FX = -7.07N
FY = 7.07N
What are the horizontal and vertical components
of each force?
Now let’s look at all the forces again
• Forces acting horizontally:
– X-component of force 1 = 7.07N
– X-component of force 2 = -7.07N
– X-component of force 3 = 0 N
• Forces acting vertically:
– Y-component of force 1 = 7.07N
– Y-component of force 2 = 7.07N
– Y-component of force 3 = -14.14 N
• So, all forces in the X-direction add to Zero
and all forces in Y-direction add to Zero
• So net force on tire is Zero Newtons
Inertia and Mass
• Inertia: the tendency for an object to remain
at rest or in motion in a straight line.
• Mass: a quantitative measure of inertia
– SI unit for inertia/mass: kilogram (kg)
• We will discuss mass at length shortly
Newton’s Second Law
• This is the most important physics law you will
learn.
• It is way more important than that E=MC2
business.
• You ready for it?
• Here it is:
• F = MA
– Force = mass X acceleration
Investigating Newton’s Second Law
• F = ma
• Or, more specifically, ΣF = ma
– Where ‘Σ’ = the greek letter “sigma” and means
“sum of”
– So the sum of all the forces on an object are equal
to that objects mass times its acceleration
– As before, the forces on an object add together
vectorally. This is how you find the resultant.
Examples of Newton’s 2nd law
• You are pushing a 200 kg sled over a
completely frictionless frozen lake. Your
applied force is 150 N. What is the sled’s
acceleration?
• F = ma -> 150 N = (200 kg)(a)
• a = (150 N)/(200 kg) = 0.75 m/s2
More Examples of Newton’s 2nd law
• Now, you are pushing a 200 kg sled over a
frozen lake with some friction. Your applied
force is still 150 N, but now it is counteracted
by a 100 N friction force. What is the sled’s
acceleration?
•
•
•
•
ΣF = ma
So, 150 N – 100 N = ma
50 N = (200 kg)(a)
a = (50 N)/(200 kg) = 0.25 m/s2
Mini-competition 1: Question 1
• Chuck Norris (who once successfully ordered a
Big Mac at Burger King) has a fist of mass 0.7kg.
Starting from rest, his fist achieves a velocity of
8m/s in 0.15 sec. What avg force does he
generate to achieve this level of performance?
• 37 N
• If paper beats rock, rock beats scissors, and
scissors beats paper, what beats all 3 at the same
time?
• Chuck Norris.
Mini-competition 1: Question 2
• Carla Wong serves a 58 gram tennis ball and
accelerates it from rest to 45m/s. The impact
gives it a constant acceleration over 44cm.
What is the magnitude of the force on the
ball?
• 133 N
Newton’s 3rd Law
• “Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal magnitude
on the first body.”
• Whenever you push on something, that
something pushes back on you with the same
force.
Newton’s Third law example
• From the book:
– An astronaut (92kg) pushes on a satellite
(11,000kg) with a force of 36N. What is the accel
of the astronaut? Of the Satellite?
• Fastronaut on satellite = - Fsatellite on astronaut
• Aastronaut = -F/mastronaut = -36N/92kg = -0.30m/s2
• Asatellite = F/msatellite = 36N/11,000kg =
0.0033m/s2
Mini-competition 2: Question 1
• Charles Steigerwald pushes on a wall to the
right with a force of 200 N. What is the force
that the wall exerts on Charles?
• -200N
Mini-competition 2: Question 2
• Kurt Stegmeier (mass 63 kg) and Dr. Mason
(mass 110 Kg) are sitting opposite one another
in rolling chairs. They push off one another in
opposite directions with a force of 100 N.
Assuming perfectly frictionless school chairs
and floors, what is the magnitude of the
acceleration each experiences?
• Dr. Mason: 0.909 m/s2
• Kurt: 1.58 m/s2
Gravitational Force
•
•
•
•
•
•
Gravitational force is always attractive
Fg = Gm1m2/r2
Fg = force of gravity
M1 = mass of first object
M2 = mass of second object
r = distance between them (or specifically,
between their centers of mass
• G = 6.67X10-11 Nm2/kg2
A note on the center of mass
• Every object has a center of mass
• For purposes of calculating forces like gravity,
you can assume that all the mass of the object
is at the center of mass.
• For a sphere (assuming uniform density),
center of mass is the actual center
r
m1
is
like
m2
r
m1
m2
Gravitational Force example 1
• What is magnitude of gravitational force
between two particles of masses m1 = 12kg
and m2 = 25 kg separated by distance r =
1.2m?
• F = Gm1m2/r2
• = 6.67X10-11 Nm2/kg2 (12kg)(25kg)/(1.2m)2
• = 1.4 X 10-8 N
• Fg is very small for normal sized objects, which
is why you don’t feel yourself being pulled
sideways every time a car passes by.
Gravitational Force example 2
• What is the magnitude of Fg between the
earth and the moon?
• Mearth = 5.9742 × 1024 kilograms
• Mmoon = 7.36 × 1022 kilograms
• According to the internet (which is never
wrong), distance between centers of mass =
384,403 Km, or 384,403,000m
• Fg = 1.98 X 1020 N
Super-neato explanation of g
• Recall that F = Gm1m2/r2
• If we set m1 = mass of the earth and r = radius
of the earth, let’s see what happens:
• rearth = 6 378.1 km = 6.3781X106m
• F = (6.67X10-11Nm2/kg2)(5.9742 × 1024
kg)(m2)/(6.3781X106m)2
• F = (9.795 m/s2)(m2)
• i.e. all those numbers other than m2 work out
to be 9.8 m/s2
Mini-competition 3: Question 1
• Saturn has an equatorial radius of 6 X 107 m
and a mass of 5.67 X 1026 kg. What is the
acceleration due to gravity at the equator of
Saturn? What is the ratio of Sami Fekadu’s
weight on Saturn to that on Earth?
• 10.5 m/s2
• Ratio = 1.07
Mini-competition 3: Question 2
• The mass of the earth is 5.9742 × 1024 kg and
the mass of the sun is 1.99 X 10 30 kg. They
are separated by a distance of 146 X 109 m.
What is the magnitude of the gravitational
attraction on the earth by the sun?
• 3.72 X 1022 N
Gravity and Weight
• Mass is the amount of material that makes up
an object
• Weight is the gravitational force on that mass
• W = mg
• W = weight, m = mass and g = -9.8m/s2
Weight example: Hubble Telescope
• Hubble telescope has a mass of 11,600kg
• What is it’s weight
– Resting on the surface of the earth
– In orbit at 598km above the surface of the earth?
• Wsurface = 1.14 X 105 N
• Worbit: need total distance from earth center =
6.38 X 106 m + 598 X 103 m = 6.98 X 106m
• Worbit = 0.95 X 105 N
• Weighs less in orbit, as expected
More about gravity
• According to F = Gm1m2/r2 , force of gravity
decreases as square of distance from center of
mass
www.rpdp.net/sciencetips_v2/P12B4.htm
Mini-competition 4: Question 1
• On the surface of the mysterious planet Jay-Z,
the acceleration due to gravity is 14.5 m/s2. If
an object weighing 45 N on earth were
transported to this other planet, how much
would it weigh?
• 66.58N
Mini-competition 4: Question 2
• In the World’s Strongest Man competition, the
past three different winners were Žydrūnas
Savickas (mass 175 kg), Mariusz Pudzianowski
(mass 142 kg) and Phil Pfister (mass 156 kg).
What is their combined weight in Newtons? If
one kg weighs 2.2 pounds, what was their
combined weight in pounds?
• 4635.4N
• 1040 pounds
Other forces: Normal force
• ‘Normal’ means ‘perpendicular’
• FN = Normal force = the force that surface
exerts on an object sitting on it.
• If a 10kg block is sitting on a table, it weighs
98N
• The table, then, exerts a 98N force upwards
on the block
Normal force examples
• A 5kg book is resting on a table. What is the
normal force exerted on the book by the table?
• A 15 N block is on a table. I push on the block
with an additional 11 N. What is FN on the
block?
• The same 15N block is on the table, but now I
pull upwards on it with 11N. What is FN on the
block?
Normal force and Apparent Weight
Explanation of apparent weight
• Remember, ΣF = ma
• When standing in an elevator, FN is the normal
force (force from floor on you).
• Your weight = W = mg
• ΣFy = FN – mg = ma, so FN = mg + ma
• FN is your apparent weight, mg = your true
weight
Example: moving elevators
• A 100 kg man is on an elevator that is
accelerating upwards at a rate of 5 m/s2. What
is the man’s apparent weight?
• FN = mg + ma (a is positive if going upwards)
• FN = (100kg)(9.8m/s2) + (100kg)(5m/s2)
• FN = (100kg)(14.8m/s2) = 1480 N
• Now, how about 5m/s2 downwards?
• FN = (100kg)(4.8m/s2) = 480 N
Friction
• Friction is a force that resists the motion of
two surfaces sliding along one another.
• Motion must be relative
• Recall the normal force, FN. The normal force
is perpendicular to the surface.
• By contrast, the frictional force is parallel to
the surface.
• Friction always opposes the motion. Always
always always.
Coefficient of Friction
• The coefficient of friction is a dimensionless
number (no units! The only one you’ll get all
year!) that is a proportionality constant
between the normal force and the frictional
force between two surfaces.
• Ff = μFN
• The symbol for the coefficient of friction is the
greek letter μ (written “mu” and pronounced
“myoo”)
• Fun Fact: there is no friction in New Zealand
But Dr. Mason, tell us more interesting
and awesome facts about friction
• There are two kinds of friction, static and
kinetic
– Static friction is friction between two objects that
are not moving relative to one another
• Example: a heavy crate on a carpet floor that when you
push on it, you can’t move it.
– Kinetic friction is friction between two objects
that are moving relative to one another
• Example: same heavy crate that Tim Tebow is pushing
on, who can move it.
Why the distinction between static
and kinetic friction?
• Because, in general, the two are different
magnitudes, even when you have the same two
surfaces in contact with one another.
• For example, to pull two random materials off the
internet, the coefficients of static and kinetic
friction between glass and nickel are
• μS = 0.78
• μK = 0.56
• In general, kinetic friction is less than static
friction.
• I.e. it’s harder to get a ‘stuck’ object to move than
to keep a sliding object moving
A freaky note on static friction
• Picture, if you will, a block resting on the
ground. Let us say that the block is very heavy
and is hard to get moving.
• The force of static friction can change
depending on how much sideways force you
apply to the block.
• If you apply a small force, the frictional force is
small to counteract your push. If you apply a
medium-sized force, the frictional force will be
medium-sized to counteract your push.
Freaky note continued
• In fact, the frictional force will vary up to a
maximum value of FFmax = μS FN
• After it starts moving, there is a constant force
of kinetic friction on the block = μK FN.
• What would the implication be if the static
frictional force did not vary with applied
horizontal force?
Example from the book
• A sled is resting on some snow with μS = 0.35
• Sled + rider have mass = 38 kg.
• What is the magnitude of horizontal force
needed to get the sled moving?
• Fneeded = μS FN = μS mg = (0.35)(38kg)(9.8m/s2)
= 130N
A different example from the book
• A sled is traveling with an initial velocity of 4m/s
on a horizontal stretch of snow. The coefficient of
kinetic friction μK = 0.05. How far does the sled go
before coming to a halt?
• Ff = μK FN = μKmg
• Omg, we don’t know m! Whatever shall we do??
• By Newton’s 2nd law, ax = -Ff/m = - μKmg/m = - μKg
• Now can use Vf2 = Vi2 + 2ax to find x
• X = 16.3m
Mini-competition 5: Question 1
• A rocket blasts off from rest and attains a
speed of 45 m/s in 15 sec. An astronaut on
board has a mass of 57 kg. What is his
apparent weight during takeoff?
• 730N
Mini-competition 5: Question 2
• A box of calculators that students perpetually
borrow weighs 45N and sits at rest on a
horizontal table. The coefficients of static and
kinetic friction are 0.65 and 0.42, respectively.
• A horizontal force of 36N is applied to the box.
Will it move and, if so, what will its
acceleration be?
• Yes, it will move @ 3.72m/s2.
Mini-competition 5: Question 3
• Edward is driving a car at constant speed and
Jacob is in the passenger seat. A cup of hot
coffee is on the dashboard and μS = 0.3
between it and the dash. What is the
maximum acceleration the car can have
without spilling the coffee on Jacob and
making him go all aggro ‘n’ stuff?
• A = F/m = Ff/m = μKmg/m = μKg = 2.94m/s2.
• You know vampires can’t go out during the
day, right? I mean, like, seriously?
Equilibrium
• Equilibrium means that the sum of all the
forces acting on an object = zero.
• If ΣF = ma and ΣF = 0, then what does a =?
• a=0
• For an object to be truly in equilibrium, we
must look at both horizontal and vertical
directions.
– ΣFX = 0
– ΣFY = 0
You must achieve….balance….
• How to analyze equilibrium situations:
• Identify the system you wish to analyze
– Is it a single object? Are two objects acting together?
Are there ropes or pulleys or other things?
• Draw a free body diagram
• Choose a set of x- and y-axes. Resolve all forces
along these axes.
• Apply ΣFX = 0 and ΣFY = 0
• Solve for any unknown quantities
Example from the book – replacing an
engine
100
T1
Yoon Jae Sung is replacing the engine to his car.
There is a rope with tension T1 that goes up to a
pulley and he pulls with tension T2 as shown in
the diagram. If the engine weighs 3150 N, find
the tensions T1 and T2.
100
Engine
W
T2
Example from the book – replacing an
engine
100
T1
Step 1: identify the system
-The system is the engine plus two ropes
Step 2: Draw the free body diagram
All forces are acting on the ring at the
center, so use that point as the center of
your free body diagram.
100
Engine
W
T2
Example from the book – replacing an
engine
100
T1
100
T1
T2
W
100
Engine
W
100
T2
Free Body Diagram
Example from the book – replacing an
engine
100
T1
Step 3: Choose a set of x- and y-axes.
-Done
Step 4: Resolve all forces along these
axes.
100
T2
W
Force X-comp
T1
T1cos(1000)
T2
T2cos(3500)
W
0
Y-comp
T1sin(1000)
T2sin(3500)
-W
Example from the book – replacing an
engine
Step 5: Apply ΣFX = 0 and ΣFY = 0
100
T1
ΣFX = 0
so T1cos(1000) + T2cos(3500) = 0
cos(1000) = -0.173
cos(3500) = 0.985
100
T2
W
so T1*(-0.173) + T2*(0.985) = 0
Example from the book – replacing an
engine
Step 5: Apply ΣFX = 0 and ΣFY = 0
100
T1
ΣFY = 0
so T1sin(1000) + T2sin(3500) – W = 0
sin(1000) = 0.985
sin(3500) = -0.173
100
T2
W
so T1*(0.985) + T2*(-0.173) – W = 0
Example from the book – replacing an
engine
Step 6: Solve for the unknowns
T1*(-0.173) + T2*(0.985) = 0
T1*(0.985) + T2*(-0.173) – W = 0
T1*(-0.173) + T2*(0.985) = 0, so T1 = T2[(0.985/0.173)]
Substitute this into second equation
T2[(0.985/0.173)]*(0.985) + T2*(-0.173) – 3150 = 0
T2 = 579 N, so then T1 = 3300 N