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## Chapter 4- Forces and Motion

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### Chapter 4- Forces and Motion 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. – PowerPoint PPT presentation

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Title: Chapter 4- Forces and Motion

1
Chapter 4- Forces and Motion
2
Think about the following questions What 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.
3
Other examples of forces
4
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

5
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

6
Force is a vector! (yay more vectors ?)
• The effect of a force depends on magnitude and
direction

7
Force Diagrams (p. 126)
• Force Diagram A diagram that shows all the
forces acting in a situation

8
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!

9
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.

10
• 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

11
12
Normal Force)
13
Finally add the force of friction acting on the
car
14
A Free Body Diagram of a Football Being Kicked
15
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.
16
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
17
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.

18
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

19
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

20
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

21
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.

22
Step 1 Draw a FBD
Fup 565 N
Fright 82 N
Fleft 115 N
Fdown 236 N
23
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.

24
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

25
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

26
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?

27
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! ?

28
(No Transcript)
29
Newtons 3rd Law
• Forces always exist in pairs
• For every action there is an equal and opposite
reaction

30
Action- Reaction Pairs
Some action-reaction pairs
31
Although the forces are the same, the
accelerations will not be unless the objects have
the same mass.
32
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

33
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

34
Kinetic Friction
• Kinetic Friction (Fk) is the frictional force on
an object in motion

35
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

36
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

37
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)

38
Coefficient of Kinetic Friction
• The problem states that the chair is moving with
constant velocity, which means Fnet0. Therefore,
Fapp must equal -Fk.

39
Solve for Coefficient of Kinetic Friction
40
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.

41
FBD
42
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

43
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.

44
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.

45
FBD
46
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

47
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

48
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

49
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?

50
DRAW FBD
FN
Fapp,x
Ff
Fapp,y
Fg319 N
Fapp 485 N
51
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

52
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
53
• 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

54
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

55
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.

56
Draw a Free Body Diagram
?
57
Closer look at gravity triangle.
• Solve for Fg,y and Fg,x

58
Coordinate system for inclined planes
Y axis
X axis
59
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.

60
Forces In the y-direction
• So what are the forces acting in the y direction?
• We have normal force and Fg,y
• Since they have to cancel out
• FN mgcos(?)

61
Forces in the x direction
• What is the force that makes the object slide
down the inclined plane?
• Gravitybut only in the x direction

62
Remember that Vectors can be moved parallel to
themselves!!
?
Fg,y
?
Fg,x
63
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

64
What if there is an additional applied force?
• Example a box is being pushed up an inclined
plane

Fapp
Fg,x
Ff
Fg,y
?
65
In that case
• FN mgcos?
• Fnet Fapp- Fg,x Ff
• If the object is in equilibrium then
• Fapp Fg,x Ff

66
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.

67
Draw a FBD
Ff
Fg,x
Fg,y
Fg
?
68
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?

69
Find Force of Friction
• Fnet 0
• Fg,x Ff mgsin?5.4(9.81)sin(15)
• Ff 13.7N

70
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?

71
FBD
Ff
Fg,x
Fg,y
Fg
?
72
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

73
Solve for Ff
• Fnet Fg,x Ffmgsin? Ff
• Ff mgsin? Fnet
• Ff 75kg(9.8)sin(25) 270 N
• Ff 40.62 N

74
Finish the Problem
• We are trying to solve for µk