Chapter 4- Forces and Motion

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Chapter 4- Forces and Motion
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Think about the following questionsWhat is this
object? Where is it? Why does it look like that?
IO is a moon of Jupiter Competing forces between
Jupiter and the other Galilean moons cause the
center of Io compress and melt. Consequently
Io is the most volcanically active body in the
solar system.
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Other examples of forces
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What is a force?

• IPC definition A push or a pull exerted on some
  object
• Better definition Force represents the
  interaction of an object with its environment
• The Unit for Force is a Newton

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Two major types of forces

• Contact Forces Result from physical contact
  between two objects
• Examples Pushing a cart, Pulling suitcase
• Field Forces Forces that do not involve physical
  contact
• Examples Gravity, Electric/Magnetic Force

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Force is a vector! (yay more vectors ?)

• The effect of a force depends on magnitude and
  direction

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Force Diagrams (p. 126)

• Force Diagram A diagram that shows all the
  forces acting in a situation

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Free Body Diagrams p.127

• Free Body Diagrams (FBDs) isolate an object and
  show only the forces acting on it
• FBDs are essential! They are not optional! You
  need to draw them to get most problems correct!

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How to draw a free body diagram

• Situation A tow truck is pulling a car
• (p. 127)
• We want to draw a FBD for the car only.

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Steps for drawing your FBD

• Step 1 Draw a shape representing the car (keep
  it simple)
• Step 2 Starting at the center of the object,
  Draw and label all the external forces acting on
  the object

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Add force of gravity
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Add force of the road on the car(Called the
Normal Force)
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Finally add the force of friction acting on the
car
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A Free Body Diagram of a Football Being Kicked
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A person is pushed forward with a force of 185 N.
The weight of the person is 500 N, the floor
exerts a force of 500 N up. The friction force is
30 N.
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Forces you will need
Symbol of Force Description
Fg Gravitational Force is the Weight of the Object (equal to mass x g mg)
FN Normal Force Force acting perpendicular to surface of contact
Ff Frictional Force- Opposes applied force acts in direction opposite of motion
Fapp Applied Force
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Sample Problem p. 128 3

• Draw a free body diagram of a football being
  kicked. Assume that the only forces acting on the
  ball are the force of gravity and the force
  exerted by the kicker.

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Newtons 1st Law of Motion

• The Law of Inertia
• An object at rest remains at rest, and an object
  in motion continues in motion with constant
  velocity (constant speed in straight line) unless
  the object experiences a net external force
• The tendency of an object not to accelerate is
  called inertia

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Acceleration

• The net external force (Fnet) is the vector sum
  of all the forces acting on an object
• If an object accelerates (changes speed or
  direction) then a net external force must be
  acting upon it

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Equilibrium

• If an object is at rest (v0) or moving at
  constant velocity, then according to Newtons
  First Law, Fnet 0
• When Fnet 0, the object is said to be in
  equilibrium

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How do we use this information?Sample Problem p.
133 2

• A crate is pulled to the right with a force of
  82.0 N, to the left with a force of 115 N, upward
  with a force of 565 N and downward with a force
  of 236 N.
• A. Find the net external force in the x direction
• B. Find the net external force in the y direction
• C. Find the magnitude and direction of the net
  external force on the crate.

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Step 1 Draw a FBD
Fup 565 N
Fright 82 N
Fleft 115 N
Fdown 236 N
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Find the vector sum of forces

• A. 82 N (-115 N ) -33 N
• B. 565 N (-236 N) 329 N
• C. Find the resultant of the two vectors from
  part a and b.

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Newtons 1st Law

• Review Newtons 1st Law
• When Fnet0, an object is in equilibrium and will
  stay at rest or stay in motion
• In other words, if the net external force acting
  on an object is zero, then the acceleration of
  that object is zero

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Newtons 2nd Law (p.137)

• The acceleration of an object is directly
  proportional to the net external force acting on
  the object and inversely proportional to the
  objects mass

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Example p. 138 4

• A 2.0 kg otter starts from rest at the top of a
  muddy incline 85 cm long and slides down to the
  bottom in 0.50 s. What net external force acts on
  the otter along the incline?

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Solving the problem

• To calculate Fnet, we need m and a
• M2.0 kg
• What is a?
• Vi 0 m/s, t0.50 s,
• displacement85 cm.85 m
• Welcome back kinematic equations! ?

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Newtons 3rd Law

• Forces always exist in pairs
• For every action there is an equal and opposite
  reaction

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Action- Reaction Pairs
Some action-reaction pairs
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Although the forces are the same, the
accelerations will not be unless the objects have
the same mass.
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Everyday Forces

• Weight Fg mg
• Normal Force FN Is always perpendicular to the
  surface.
• Friction Force Ff
• Opposes applied force
• There are two types of friction static and
  kinetic

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Static Friction

• Force of Static Friction (Fs) is a resistive
  force that keeps objects stationary
• As long as an object is at rest
• Fs -Fapp

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Kinetic Friction

• Kinetic Friction (Fk) is the frictional force on
  an object in motion

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Coefficients of Friction

• The coefficient of friction (µ) is the ratio of
  the frictional force to the normal force
• Coefficient of kinetic Friction
• Coefficient of Static Friction

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Sample Problem p. 145 2

• A 25 kg chair initially at rest on a horizontal
  floor requires a 365 N horizontal force to set it
  in motion. Once the char is in motion, a 327 N
  horizontal force keeps it moving at a constant
  velocity.
• A. Find coefficient of static friction
• B. Find coefficient of kinetic friction

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Coefficient of Static Friction

• In order to get the chair moving, it was
  necessary to apply 365 N of force to overcome
  static friction. Therefore Fs 365 N.
• The normal force is equal to the weight of the
  chair (9.81 x 25 245 N)

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Coefficient of Kinetic Friction

• The problem states that the chair is moving with
  constant velocity, which means Fnet0. Therefore,
  Fapp must equal -Fk.

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Solve for Coefficient of Kinetic Friction
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Forces at an angle

• A woman is pulling a box to the right at an
  angle of 30 above the horizontal. The box is
  moving at a constant velocity. Draw a free body
  diagram for the situation.

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FBD
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What is Fnet?

• Since the suitcase is moving with constant
  velocity, Fnet0.
• That means the forces in the x direction have to
  cancel out and the forces in y direction have to
  cancel out
• Fk Fapp,x
• FN Fapp,y Fg
• NOTICE THAT NORMAL FORCE DOES NOT EQUAL WEIGHT IN
  THIS SITUATION

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Lets do an example. P. 154 42

• A 925 N crate is being pushed across a level
  floor by a force F of 325 N at an angle of 25
  above the horizontal. The coefficient of kinetic
  friction is 0.25. Find the magnitude of the
  acceleration of the crate.

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What do we need to know?

• So we need mass and Fnet.
• We have weight (925 N). So what is mass?
• How to find Fnet?
• Find vector sum of forces acting on crate.

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FBD
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Finding Fnet,y

• Is box accelerating in y direction?
• No. Therefore Fnet in y direction is 0
• So FN Fapp,y Fg
• So FN Fg- Fapp,y 925 N- 325sin(25)
• FN 787.65 N

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Finding Fnet,x

• Is box accelerating in x direction?
• Yes. Therefore Fnet,x is not 0
• Fnet,x Fapp,x Ff
• Fapp,x Fappcos(25)294.6 N
• Use coefficient of friction to find Ff
• FfµFN(0.25)(787N)197 N

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Finish the Problem

• Fnet,x 294 N 197 N 97 N
• So now we know that the Fnet on the box is 97 N
  since Fnet,y is 0

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Another example. P. 154 54 part a

• A box of books weighing 319 N is shoved across
  the floor by a force of 485 N exerted downward at
  an angle of 35 below the horizontal.
• If µk between the floor and the box is 0.57, how
  long does it take to move the box 4.00 m starting
  from rest?

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DRAW FBD
FN
Fapp,x
Ff
Fapp,y
Fg319 N
Fapp 485 N
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Find Fnet

• Is box accelerating in y direction?
• No. Therefore Fnet in y direction is 0
• So FN Fapp,y Fg
• So FN 485sin(35) 319 N 598 N

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Fnet,x
Is box accelerating in x direction? Yes.
Therefore Fnet,x is not 0 Fnet,x Fapp,x
Ff Fapp,x 485cos(35)397.29 N Use coefficient
of friction to find Ff FfµFN(0.57)(598)341
N Fnet, x 397.29- 341 57.29 N
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• So now we know that the Fnet on the box is
• 57.29 N since Fnet,y is 0
• Weight of box is 319 N.
• Find mass by dividing by 9.81
• m 32.52 kg

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Finish the problem

• We want to know how long it takes for the box to
  move 4.00 m.
• Find vf so that you can solve for t
• Solve for t

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Forces on An Incline

• A block slides down a ramp that is inclined at
  30 to the horizontal. Write an expression for
  the normal force and the net force acting on the
  box.

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Draw a Free Body Diagram
?
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Closer look at gravity triangle.

• Solve for Fg,y and Fg,x

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Coordinate system for inclined planes
Y axis
X axis
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Fnet in the y direction

• When a mass is sliding down an inclined plane, it
  is not moving in the y direction.
• Therefore Fnet,y 0 and all the forces in the y
  direction cancel out.

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Forces In the y-direction

• So what are the forces acting in the y direction?
• Look at your FBD
• We have normal force and Fg,y
• Since they have to cancel out
• FN mgcos(?)

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Forces in the x direction

• What is the force that makes the object slide
  down the inclined plane?
• Gravitybut only in the x direction

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Remember that Vectors can be moved parallel to
themselves!!
?
Fg,y
?
Fg,x
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Forces in the x direction

• So what are the forces acting in the x direction?
• Friction Force (Ff) and Gravitational Force
  (Fg,x)
• If the box is in equlibrium
• Fg,x Ff
• If the box is accelerating
• Fnet Fg,x - Ff

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What if there is an additional applied force?

• Example a box is being pushed up an inclined
  plane

Fapp
Fg,x
Ff
Fg,y
?
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In that case

• FN mgcos?
• Fnet Fapp- Fg,x Ff
• If the object is in equilibrium then
• Fapp Fg,x Ff

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An Example p. 153 40

• A 5.4 kg bag of groceries is in equilibrium on an
  incline of angle. Find the magnitude of the
  normal force on the bag.

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Draw a FBD
Ff
Fg,x
Fg,y
Fg
?
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Solve the Problem

• The block is in equilibrium so
• Fnet0
• Fg,y FNmgcos?(5.4kg)(9.81)cos(15)
• FN51 N
• Additionally, what is the force of friction
  acting on the block?

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Find Force of Friction

• Fnet 0
• Fg,x Ff mgsin?5.4(9.81)sin(15)
• Ff 13.7N

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Example p. 147 3

• A 75 kg box slides down a 25.0 ramp with an
  acceleration of 3.60 m/s2.
• Find the µk between the box and the ramp
• What acceleration would a 175 kg box have on this
  ramp?

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FBD
Ff
Fg,x
Fg,y
Fg
?
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What is Fnet?

• They give mass and acceleration
• So Fnet ma 75kg x 3.60 m/s2
• Fnet 270 N
• FN mgcos?
• Fnet Fg,x Ffmgsin? - Ff

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Solve for Ff

• Fnet Fg,x Ffmgsin? Ff
• Ff mgsin? Fnet
• Ff 75kg(9.8)sin(25) 270 N
• Ff 40.62 N

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Finish the Problem

• We are trying to solve for µk