Title: FORCE
1FORCE
2Newtons Laws
3Aristotles Motion
- 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
4Aristotles Basic Error
- Friction not understood as a force
5Galileos Motion
- 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
6Galileos Inclined Planes
- 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
7Inertia
- Resistance to change in state of motion
- Galileo concluded all objects have inertia
- Contradicted Aristotles theory of motion
- No force required to keep Earth in motion around
sun because no friction
8Newton
- Born 1665
- Built on Galileos ideas
- Proposed three laws of motion at age of 23
9Newtons First Law
Ourtesy www.lakeheadu.ca/alumni/ hockey.gif
- 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
10Mass
- Amount of inertia depends on amount of massor
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
11Experiencing 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?
12Inertia in a Car
- Discuss three examples of inertia in a car
- Car hitting a wall
- Car hit from behind by a truck
- Car going around a corner
13Newtons 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
14Units
- F ma
- Unit of force is the Newton (N)
- 1 N 1 kg m/s2
15Net Force
- Net Force means sum of all forces acting
- Sum is Vector sum
F2
F1
Resultant force
16Understanding the Second Law
Force
- The cause of acceleration is
- _________ resists acceleration
- The greater the force, the ________ the
- ______________
- The greater the mass, the _________ the
acceleration.
Mass or inertia
greater
acceleration
less
17F ma is Three Equations
- F and a are vectors
- So F ma equation is really three
- SFx max SFy may SFz maz
-
18Examples
- 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
19F ma Example m unknown
- 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? - M F/a 500.0N/0.462 m/s2 1082 Kg 1.08 x
103 Kg
20Net 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 90 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?
21Putting 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.
22a 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
23Newtons 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
24Example 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)
25Example person walking
- Foot exerts force backward on ground
- Ground exerts force forward on foot
26Example Throwing ball
- Pitcher exerts force on ball
- Ball exerts equal and opposite force on pitcher
- Why doesnt pitcher move?
27Example Rocket
- Rocket engine exerts rearward force on gas
molecules - Molecules exert forward force on rocket.
28Book on Table
- The mass of the book is one kg. What is the
force (magnitude and direction) on the book? - 9.8 N upward
29Really 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 43 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?
30What 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
31The Horse and the Cart Problem
- 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?
32The Horse and Cart Problem.
- These appear to be the equalizing forces.
A - B B - C C -D ABCD no acc!
33The 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 DgtC and BgtA, so DgtA.
34More Examples
- Can you think of some more examples of Newtons
Third Law in Action? - Imagine an astronaut floating in deep space, with
only his spacesuit. Is there any way for him to
move himself back to earth?
35Mass 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?
36Review Mass vs. Weight
- 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
37Weight of 1 Kilogram
- 9.8 Newtons
- About 2.2 pounds
- Compare the weight of 1 kg nails with 1 kg
styrofoam - Answer Same
38Weight 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
39Atwoods Lab
- 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)
40FRICTION
- Sliding (motion) Static (stationary)
41Sliding Friction
- Often called kinetic friction
- A force opposite to direction of motion
- Due to bumps in surfaces and electric forces
Surface under microscope
Ff
42Kinetic 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
43Reducing 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)
44Coefficient 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
45Friction Good or Bad
- Mostly undesirable since reduces useful force and
wastes energy - Friction produces heat
- Necessary for walking!
- Necessary for braking
46Static 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
47Where are all the forces?
- Block on an inclined plane
48Free Body Diagram Example 1
If the box below accelerates to the right at 1
m/s2 Solve all of the following
49Solution 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
50Solution 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
51Flat pull
- If you pull a 2505 g box with a force of 15 N at
an angle of 53 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
52Friction 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 (FnFg 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
53Free body diagram example 2
- Say a box is sitting on 30 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?
54Free Body Diagram example 3
- A box is sitting on a 35 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.
55- 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.
56Review Newtons Laws of Motion
- Newtons 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. - Newtons 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. - Newtons Third Law
- Whenever one object exerts a force on a
second object, the second exerts an equal
opposite force on the first -
57 Action- Reaction Lab
- 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.
58Draw the free body diagram
59Draw 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.
60- Say a box is sitting on 40 slope ramp. If the
mass is 3.0 kg? What are all the forces acting on
the box and what is µ?
61- 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.
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64- For each collection of listed forces, determine
the vector sum or the net force. - Set A58 N, right42 N, left98 N, up98 N, down
65- 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.
66- 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.
67- Jerome and Michael, linebackers for Souths
varsity football team, delivered a big hit to the
halfback in last weekends game. Striking the
halfback simultaneously from different directions
with the following forces - FJerome 1230 N at 53FMichael 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.)
68- 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.
69- 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 ma
Ffrict µFnorm Fgrav mg)
70Friday Problem 1
The 5-kg mass below is moving with a constant
speed of 4 m/s to the right. 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 ma Ffrict
µFnorm Fgrav mg)
- 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. -
71Friday Problem 2
- You are pushing a 200 kg block up a 20 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.
72Tuesday Problem 1
- 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 ma Ffrict µFnorm Fgrav
mg)
73- 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, upFfrict
27.6 N, leftFgrav 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.
74- A box is sliding down a ramp at an angle of 47
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?
75Ramp Problem 1
- Say a box is sitting on 30 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?
76Rotation Centripetal Force
Coutesy Space.com
- How to Keep it Straight Without Getting Dizzy
77Rotation
- 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
78Speed on a Wheel
- Which horses on a carousel move the fastest,
inner or outer?
Outer v radius x angular speed v rw
79Mass at the End of a String
- What force must the
- string exert on the mass?
- What is the direction of
- this force?
A force toward the center of the circle
80Centripetal Force
- 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.
81How 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
82Is a mass moving at steady speed in a circle
accelerating?
- Yes. The direction is changing
- What is the direction of this acceleration?
Toward the center of the circle
83Car on a Curve
- When auto rounds corner, sideways acting friction
between tires and road provides centripetal force
that holds car on road
84Dont Confuse Inertia With Force
- Tubs 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
85How Can Water Stay In The Bucket?
- Bucket swung in a
- vertical circle
- What force pushes on the
- water?
Weight and normal force down
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
86Centrifugal Force
- 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
87Change 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
88Book on a Car Seat
- When a car goes around a curve to the left, a
book slides - Which way does it slide?
- Why doesnt 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?
89Roller Coaster Lab- Centripetal Force
- You are dropping the ball from 45, 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?