FORCE
Newton’s Laws
Three Laws of Motion
Aristotle’s Motion
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Natural Motion is up or down
 Down for falling objects
 Up for smoke
 Circular for heavenly bodies since without end
Violent Motion
 Due to imposed forces such as wind pushing a
ship or someone pulling a cart
Natural state of motion is rest
 A force is needed to keep something moving
Aristotle’s Basic Error
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Friction not understood as a force
Galileo’s Motion
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Force is a push or a pull
Friction is a force that
occurs when objects move
past each other
Friction due to tiny
irregularities
Only when friction is
present is a force required
to keep something moving
Galileo’s Inclined Planes
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Ball rolling downhill
speeds up
Ball rolling uphill
slows down
He asked about ball on
smooth level surface
Concluded it would
roll forever in absence
of friction
Inertia
Resistance to change in state of motion
 Galileo concluded all objects have inertia
 Contradicted Aristotle’s theory of motion
 No force required to keep Earth in motion
around sun because no friction
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Newton
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Born 1665
Built on Galileo’s
ideas
Proposed three laws of
motion at age of 23
Newton’s First Law
Ourtesy www.lakeheadu.ca/~alumni/ hockey.gif
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Every object continues in its state of rest, or of
motion in a straight line at constant speed, unless
compelled to change that state by forces exerted
on it.
Also called Law of Inertia: things move according
to their own inertia
Things keep on doing what they are doing
Examples: Hockey puck on ice, rolling ball, ball
in space, person sitting on couch
Mass
Amount of inertia depends on amount of
mass…or amount of material (number and
kind of atoms)
 Measured in kilograms
 Question: Which has more mass, a
kilogram of lead or a kilogram of feathers?
 Mass vs. Volume: volume is how much
space something occupies
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Experiencing Inertia
Inertia is resistance to shaking
 Which is easier to shake, a pen or a person?
 Why is it so hard to stop a heavy boat?
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Inertia in a Car
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Discuss three examples of inertia in a car
•Car hitting a wall
•Car hit from behind by a truck
•Car going around a corner
Newton’s Second Law
Law of Acceleration
 The acceleration produced by a net force on
an object is directly proportional to the
magnitude of the net force, and is inversely
proportional to the mass of the body.
 Acceleration = net force ÷mass
 F =ma
 Acceleration is in direction of net force
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Units
F
= ma
Unit of force is the Newton (N)
 1 N = 1 kg m/s2
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Net Force
Net Force means sum of all forces acting
 Sum is Vector sum
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F2
F1
Resultant force
Understanding the Second Law
The cause of acceleration is… Force
 _________
resists acceleration
Mass or inertia
 The greater the force, the ________
greater the
acceleration
______________
less
 The greater the mass, the _________
the
acceleration.
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F = ma is Three Equations
F and a are vectors
 So F = ma equation is really three
SFx = max SFy = may SFz = maz
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Examples
What force is required to accelerate a 1000
kg car at 2.0 m/s2 ?
Answer: F = ma = 1000 kg x 2.0 m/s2 =
2000 N.
 What is the acceleration of a 145 g baseball
thrown with a force of 20.0 N?
a = F/m = 20.0 N/0.145kg = 138 m/s2
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F = ma Example; m unknown
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An astronaut puts a 500.0 N force on an
object of unknown mass producing an
accelerations of 0.462 m/s2 . What was the
mass?
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M = F/a = 500.0N/0.462 m/s2 = 1082 Kg =
1.08 x 103 Kg
Net force example
If four teams are playing tug of war (imagine a rope
that looks like a cross, with the flag tied in the
middle). Each team is 90⁰ from each other.
Team A pulls with an overall force of 350 N to the
North, Team B pulls with an overall force of 270
N to the South, Team C pulls with an overall force
of 150 N to the East and Team D pulls with an
overall force of 250 N to the West. If the flag in
the middle has a mass of .25 kg, what is the
magnitude and direction of its acceleration?
Putting it all together…….
Calculate the change in force of a car that has
a mass of 2500 kg if it goes from 45 m/s to
rest in 7 seconds at a stop sign, then
accelerates up to 65 m/s in 5 seconds.
a= vf-vi/t
or
a = F/m
a1 = 0-45/7 = -6.42 m/s2
a2 = 65-0/5 = 13 m/s2
The difference between them is 19.42 m/s2.
F = m x a = 2500 kg x 19.42 m/s2
= 48550 N difference between the
two accelerations
Newton’s Third Law
Forces always come in pairs
 Two forces on different objects
 Every action has an equal and opposite
reaction
 Whenever one object exerts a force on a
second object, the second exerts an equal
and opposite force on the first
 Example: hammer hits nail
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Example: pushing on wall
What are the forces when you push on a
wall?
 You exert force on wall
 You accelerate in the opposite direction
 Wall must have exerted a force on you in
the direction you accelerated (by 2nd Law)
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Example: person walking
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Foot exerts force
backward on ground
Ground exerts force
forward on foot
Example: Throwing ball
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Pitcher exerts force on
ball
Ball exerts equal and
opposite force on
pitcher
Why doesn’t pitcher
move?
Example: Rocket
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Rocket engine exerts
rearward force on gas
molecules
Molecules exert
forward force on
rocket.
Book on Table
The mass of the book is one kg. What is the
force (magnitude and direction) on the
book?
 9.8 N upward
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Really putting it all together……
Calculate the Force necessary to launch a
cannonball with a mass of 15 kg if it is fired
at an angle of 43⁰ if it hits a target 210 m
away in 6.3 seconds?
What can we solve in this problem?
What equations do we need to solve this
problem?
What we need to solve the force
Vx = dx/t = 210/6.3 = 33.3 m/s
Vf2 = Vi2 + 2 a(d) Vi = 0 for this problem
a = Vf2/2d = 33.32 / 2(210) = 2.64 m/s2
Force of the cannon: F = m(a)
F = 15 kg (2.64 m/s2) = 39.6 N
The Horse and the Cart Problem
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If there is always an equal an opposite
reaction, how does anything move? For
example, if you have a horse and a cart,
how does the horse pull the cart?
The Horse and Cart Problem.
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These appear to be the equalizing forces.
A= - B
B= - C
C= -D A=B=C=D no acc!
The Horse and Cart Problem.
Because it is accelerating, the force the horse exerts on the cart has
increased. By Newton's third law, the force of the cart on the horse
has increased by the same amount. But the horse is also accelerating,
so the friction of the ground on its hooves must be larger than the
force the cart exerts on the horse. The friction between hooves and
ground is static (not sliding or rolling) friction, and can increase as
necessary (up to a limit, when slipping might occur, as on a slippery
mud surface or loose gravel).
So, when accelerating, we still have B = -C, by Newton's third law,
but D>C and B>A, so D>A.
More Examples
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Can you think of some more examples of
Newton’s Third Law in Action?
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Imagine an astronaut floating in deep space,
with only his spacesuit. Is there any way for
him to move himself back to earth?
Mass vs. Weight
Mass is intrinsic property of any object
 Weight measures gravitational force on an
object, usually due to a planet
 Weight depends on location of object
 Question 1: How does mass of a rock
compare when on Earth and on moon?
 Question 2: How does its weight compare?
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Review Mass vs. Weight
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What is mass?
Answer: quantity of
matter in something or
a measure of its inertia
What is weight?
Answer: Force on a
body due to gravity
Weight of 1 Kilogram
9.8 Newtons
 About 2.2 pounds
 Compare the weight of 1 kg nails with 1 kg
styrofoam
 Answer: Same
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Weight Examples
What does a 70 kg person weigh?
Weight = mass x g(acceleration due to gravity)
W = mg = 70 kg x 9.80 N/m2 = 686 N
 An object weighs 9800 N on Earth. What is its
mass?
 m = W/g = 9800 / 9.8 m/s2 = 1000 kg
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Atwoods Lab
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You have 25 washers on your lab setup, if you have a unbalanced
force, you will have acceleration. You will be using the stopwatch
function of your data collector.
Make a chart to record mass, time, acceleration and force.
Put all washers on one side, raise that side to the top, then release it
timing how long it takes to reach the bottom. Record this time.
The mass of one washer is 16 g. It is the difference in mass that
causes the acceleration. Calculate the difference in mass and
record in table. 1st mass is 25 x 16, 2nd mass is 23 x 16, 3rd mass is
21 x 16 etc.
Calculate the Acceleration = 2d/t2 (d = 1 m for the fall) so a = 2/ t2
Calculate the Net force of the fall and record. (F= ma)
Move one washer at a time over to the other side and repeat.
Continue until the machine no longer turns (12 or 13 trials)
FRICTION
Sliding (motion) & Static (stationary)
Sliding Friction
Often called kinetic friction
 A force opposite to direction of motion
 Due to bumps in surfaces and electric forces
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Surface under
microscope
Ff
Kinetic Friction is…
Dependent on nature of the two surfaces
 Directly proportional to the normal force
between the surfaces
 Normal Force is perpendicular to the
surface. If it is on a flat surface, it is equal
to the weight of the object.
 Independent of velocity
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Reducing Friction
In order to reduce friction we can:
 A. Reduce surface area
 B. Reduce weight of object
 C. Change type of friction
 - sliding(the greatest amount)
 - rolling (use wheels to ease friction)
 - fluid ( Eliminate contact by using
liquids or gases)
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Coefficient of friction mk
Generally between zero and one
 Based on comparing Friction Force to
Normal Force
 Normal Force is always perpendicular to
surface
 Calculate from Ff / FN = µk
 Can be more than one for special rubber
 Very low for ice, Teflon, lubricated
surfaces, ball bearings
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Friction: Good or Bad
Mostly undesirable since reduces useful
force and wastes energy
 Friction produces heat
 Necessary for walking!
 Necessary for braking
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Static Friction
Force to start something moving
 Usually larger than kinetic friction for same
surfaces
 Requires force to be exerted
 Before sliding begins, is equal and opposite
to applied force
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Where are all the forces?
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Block on an inclined plane
Free Body Diagram Example 1
If the box below accelerates to the right at
1 m/s2 Solve all of the following:
Solution 1
Fgrav = m x g = 5 x 9.8 = 49 N
 Using the angle and the F applied, we can
calculate the X and Y component of that
force.
 Fx= 15 sin 45 Fy = 15 cos 45
 Fx = 10.6 N
Fy = 10.6 N
 If the force of gravity is 49 N down and the
applied force is 10.6 N up, then the normal
force applied is the difference between the
two. F norm= 49-10.6 = 38.4 N
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Solution 1 cont.
If the object has an a of 1 m/s2 and a mass
of 5 kg, then it has a net force of 5 N in the
X direction.
 If the applied force in the X is 10.6 and the
net is 5, then the force of friction is the
difference between the two.
 Ffric= 10.6-5 = 5.6 N
 To solve the coefficient of friction we use
this equation: Ff = mkFN
mk= Ff/FN = 5.6/ 38.4 = .145
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Flat pull
If you pull a 2505 g box with a force of 15
N at an angle of 53⁰ to the horizon and the
box accelerates at 2.0 m/s2 to the right,
calculate the following:
 Fn, Fg, Ff, Fnet, Fapp, Fx, Fy and µ
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Friction Lab
Put a ramp flat in your lab space. Place two
photogates relatively close together.
 If the mass of the sled is .040 kg calculate
the Fnormal (Fn=Fg if on flat surface)
 Now, using your sled car (no wheels)
launch the car with your rubber band. Make
sure that it goes through both photo gates
(you may have to adjust photo gates). Use
our acceleration procedure from lab and
calculate the rate of deceleration.
 Calculate Ffric= mass of sled x deceleration
 Calculate µ = Ff/Fn
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Free body diagram example 2
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Say a box is sitting on 30⁰ slope and is
frictionless, so the only forces are the
normal force and gravity. What is the
block's acceleration down the slope if the
mass is 3.0 kg? What is the normal force?
Free Body Diagram example 3
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A box is sitting on a 35⁰ inclined plane. It is
being pulled up the ramp by you with an
acceleration of 2.5 m/s2. If the box has a
mass of 25 kg and the force of friction is 3.5
N, solve all of the following: Fnet, Fnormal,
Fgravity, Fapplied, and µ.
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A 50-N applied force (30 degrees to the
horizontal) accelerates a box across a horizontal
sheet of ice (see diagram). Glen Brook, Olive N.
Glenveau, and Warren Peace are discussing the
problem. Glen suggests that the normal force is 50
N; Olive suggests that the normal force in the
diagram is 75 N; and Warren suggests that the
normal force is 100 N. While all three answers
may seem reasonable, only one is correct. Indicate
which two answers are wrong and explain why
they are wrong.
Review: Newton’s Laws of Motion
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Newton’s First Law:
Every object continues in its state of rest, or of motion in a
straight line at constant speed, unless compelled to change
that state by forces exerted on it.
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Newton’s Second Law:
The acceleration produced by a net force on an object is
directly proportional to the magnitude of the net force, and
is inversely proportional to the mass of the body.
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Newton’s Third Law:
Whenever one object exerts a force on a second object, the
second exerts an equal & opposite force on the first
Action- Reaction Lab
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Adjust the smart track (or lab table) to be as level as
possible(may have to put lab book under) put rubber
band around one car.
Squeeze two cars together and attach with the car link.
Position car in middle of track, making sure all wheels
are on track.
With a quick upward motion, pull the link straight up
and out from the cars.
Describe how the cars move in a data table.
Start adding marbles to cars and repeat procedures
above
Make all these combinations of marbles in cars
0,0 0,1 0,2 0,3 1,1 1,2 1,3 2,2 2,3 3,3
Sum up the action reaction effect on cars and marbles.
Draw the free body diagram
Draw the free body diagram, if a = .1 m/s2
and the force you push on the lawnmower is
25 N, solve for every force you know.
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Say a box is sitting on 40⁰ slope ramp. If
the mass is 3.0 kg? What are all the forces
acting on the box and what is µ?
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Renee is on Spring Break and pulling her
21-kg suitcase through the airport at a
constant speed of 0.47 m/s. She pulls on the
strap with 120 N of force at an angle of 38°
above the horizontal. Determine the normal
force and the total resistance force (friction
and air resistance) experienced by the
suitcase.
For each collection of listed forces,
determine the vector sum or the net force.
 Set A
58 N, right
42 N, left
98 N, up
98 N, down
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Hector is walking his dog (Fido) around the
neighborhood. Upon arriving at Fidella's
house (a friend of Fido's), Fido turns part
mule and refuses to continue on the walk.
Hector yanks on the chain with a 67.0 N
force at an angle of 30.0° above the
horizontal. Determine the horizontal and
vertical components of the tension force.
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Helen is parasailing. She sits in a seat
harness which is attached by a tow rope to a
speedboat. The rope makes an angle of 51°
with the horizontal and has a tension of 350
N. Determine the horizontal and vertical
components of the tension force.
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Jerome and Michael, linebackers for South’s varsity
football team, delivered a big hit to the halfback in last
weekend’s game. Striking the halfback simultaneously
from different directions with the following forces:
FJerome = 1230 N at 53°
FMichael = 1450 at 107°
Determine the resultant force applied by Jerome and
Michael to the halfback. (The directions of the two forces
are stated as counter-clockwise angles of rotation with
East.)
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2. A box is pulled at a constant speed of
0.40 m/s across a frictional surface. Perform
an extensive analysis of the diagram below
to determine the values for the blanks.
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Use your understanding of force
relationships and vector components to fill
in the blanks in the following diagram and
to determine the net force and acceleration
of the object. (Fnet = m•a; Ffrict = μ•Fnorm;
Fgrav = m•g)
Friday Problem 1
A 5-kg mass below is moving with a an
acceleration of 4 m/s2 to the right. The
coefficient of friction for this surface is .2.
Use your understanding of force
relationships and vector components to
determine all your forces.
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Friday Problem 2
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You are pushing a 200 kg block up a 20 ⁰
hill with a force of 200 N. If the box moves
up the hill with a constant speed of 2 m/s,
calculate all the forces involved and
calculate µ.
Tuesday Problem 1
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5. The following object is being pulled at a
constant speed of 2.5 m/s. Use your
understanding of force relationships and
vector components to fill in the blanks in
the following diagram and to determine the
net force and acceleration of the object. (Fnet
= m•a; Ffrict = μ•Fnorm; Fgrav = m•g)
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At one moment during a walk around the block,
there are four forces exerted upon Fido - a 10.0 kg
dog. The forces are:
Fapp = 67.0 N at 30.0° above the horizontal
(rightward and upward)
Fnorm = 64.5 N, up
Ffrict = 27.6 N, left
Fgrav = 98 N, down
Resolve the applied force (Fapp) into horizontal
and vertical components, then add the forces up as
vectors to determine the net force and calculate the
acceleration.
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A box is sliding down a ramp at an angle of
47⁰ to the horizontal. If it is accelerating at
2.5 m/s2 and has a mass of 150 kg, what is
the Fnormal, Fnet, Fgravity, Ffric and µ?
Ramp Problem #1
 Say
a box is sitting on 30⁰ slope and is
frictionless, so the only forces are the
normal force and gravity. What is the
block's acceleration down the slope if
the mass is 3.0 kg? What is the normal
force?
Coutesy Space.com
Rotation & Centripetal Force
How to Keep it Straight Without
Getting Dizzy
Rotation
In addition to side to side (linear) motion,
rotation plays an important role in physics,
engineering, and life.
 Name some common phenomena or devices
that show rotation

Tops, planets, bicycle, car wheels, gears, pulleys, fans
etc
Speed on a Wheel
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Which horses on a
carousel move the
fastest, inner or outer?
Outer
v = radius x angular
speed
v = rw
Mass at the End of a String
What force must the
string exert on the mass?
What is the direction of
this force?
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A force toward the center of the
circle
Centripetal Force
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Any force directed toward the center of a circle is
called centripetal.
Centripetal forces have clear causes such as
tension in a string, gravity, friction etc.
Some people call centripetal force a
“pseudoforce.” (not real)
They say “a real force such as friction provides
centripetal force.”
How Big is Centripetal Force?
Fc = mv2/r
 The faster the speed the more the force
 The tighter (smaller) the radius the more the
force
 v2/r is called centripetal acceleration
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Is a mass moving at steady speed
in a circle accelerating?
Yes. The direction is changing
 What is the direction of this acceleration?
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Toward the center of the circle
Car on a Curve
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When auto rounds corner, sideways acting
friction between tires and road provides
centripetal force that holds car on road
Don’t Confuse Inertia With Force
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Tub’s inner wall exerts
centripetal force on
clothes, forcing them into
circular path
Water escapes through
holes because it tends to
move by inertia in a
straight line path
Clothes Washer
Photo courtesy HowStuffWorks.com
How Can Water Stay In The
Bucket?
Bucket swung in a
vertical circle
 What force pushes on the
water?
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You have to swing the bucket fast enough
for the bucket to fall as fast as the water
There must be a “normal” force exerted by
the bottom of the bucket on the water, in
addition to gravity
Weight and
normal
force down
Centrifugal Force
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The force ON THE
PAIL is inward
(centripetal)
The force ON THE
STRING is outward
(centrifugal)
If the string broke,
which way would the
can go?
Tangent to the circle
Change Your Point of View

In rest frame of the
can there appears to
be a centrifugal force.
This pseudoforce(or
fictitious force) is a
result of rotation
Unlike real forces,
centrifugal force is not
part of an interaction
Book on a Car Seat
When a car goes around a curve to the left,
a book slides
 Which way does it slide?
 Why doesn’t it keep moving with the car?

There is not enough static friction force to keep it
going in a circle. This friction must provide the
necessary centripetal force.
The explanation in the rotating rest frame is different.
How?
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Roller Coaster Lab- Centripetal Force
You are dropping the ball from 45⁰, practice dropping the steel ball
and the plastic ball to observe when it gets around the track.
Attach the photogate and calculate the speed and centripetal force
of the marble at the top of the loop from various distances for both
marbles. Width of ball= .019 m
Complete the table for both marbles. (as many trials as necessary)
steel = .028 kg plastic = .004 kg Fc= mv2/r radius of loop = .05 m
Draw a free body force diagram when the ball is at the top of the
loop, label all forces. Do the following lab to solve for the minimum
force needed to keep the ball (steel and plastic) on the loop.
Mass(kg)
Weight(N)
Photogate
Time (sec)
Speed (m/s)
Centripetal
Force (N)
Did the
marble stay
on track?