Download Chapter 4- Forces and Motion

Chapter 4- Forces and Motion
Think about the following questions:
What is this object? Where is it? Why does it look
like that?
Erupting Volcano!!
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.
Other examples of forces
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
kgm
1N  1 2
s
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
Force is a vector! (yay more vectors )

The effect of a force depends on
magnitude and direction
Force Diagrams (p. 126)

Force Diagram: A diagram that
shows all the forces acting in a
situation
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!
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.
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
Force of Tow
Truck on Car=
5800 N
Add force of gravity
Force of Tow
Truck on Car=
5800 N
Gravitational
force (Weight
of car)=
14700 N
Add force of the road on the car
(Called the Normal Force)
Normal Force =
13690 N
Force of Tow
Truck on Car=
5800 N
Gravitational
force (Weight
of car)=
14700 N
Finally add the force of friction acting
on the car
Normal Force =
13690 N
Force of Tow
Truck on Car=
5800 N
Gravitational
force (Weight
of car)=
14700 N
Force of Friction= 775 N
A Free Body Diagram of a Football
Being Kicked
Fkick
Fg
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.
FN= 500 N
Ff= 30 N
Fapp= 185 N
Fg= 500 N
Forces you will need
Symbol of Description
Force
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
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.
Newton’s 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
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
Equilibrium


If an object is at rest (v=0) or
moving at constant velocity, then
according to Newton’s First Law,
Fnet =0
When Fnet =0, the object is said to
be in equilibrium
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.
Step 1: Draw a FBD
Fup = 565 N
Fleft = 115 N
Fright = 82 N
Fdown = 236 N
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.
R = 331 N at 84.3 North of West
329 N
33 N
Newton’s 1st Law

Review Newton’s 1st Law:


When Fnet=0, 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
Newton’s 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
object’s mass
Fnet
a
m
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?
Solving the problem

To calculate Fnet, we need m and a

M=2.0 kg
What is a?
Vi= 0 m/s, t=0.50 s,
displacement=85 cm=.85 m

Welcome back kinematic equations! 



1 2
x  vi t  at
2
x  vi t 0.85m  00.50
m
a

 6.8 2
1 2
1
s
0.502
t
2
2
Fnet
m

 ma  2kg  6.8 2   14 N
s 

Newtons’ 3rd Law


Forces always exist in pairs
For every action there is an equal
and opposite reaction
Action- Reaction Pairs
Some action-reaction pairs:
Although the forces are the same, the
accelerations will not be unless the objects
have the same mass.
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
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
Kinetic Friction

Kinetic Friction (Fk) is the frictional
force on an object in motion
Coefficients of Friction


The coefficient of friction (μ) is the ratio
of the frictional force to the normal force
Coefficient of kinetic Friction
Fk Kinetic Friction Force
Coefficien t of Kinetic Friction   k 

FN
Normal Force

Coefficient of Static Friction
Fs Static Friction Force
Coefficien t of Static Friction   s 

FN
Normal Force
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
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)
F s 365 N
s 

 1.5
FN 245 N
Coefficient of Kinetic Friction

The problem states that the chair is
moving with constant velocity,
which means Fnet=0. Therefore, Fapp
must equal -Fk.
Fk= 327 N
Fapplied= 327 N
Solve for Coefficient of Kinetic Friction
Fk 327 N
k 

 1.3
FN 245 N
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.
FBD
FN= Normal Force
Fapp= Applied Force
F app,y
F app,x
Ff= Friction Force
Fg=Weight
What is Fnet?


Since the suitcase is moving with
constant velocity, Fnet=0.
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
Let’s 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.
What do we need to know?
So we need mass and Fnet.
Fnet
a
m
We have weight (925 N). So what is mass?
weight
925 N
mass 

 94.3 kg
m
g
9.81 2
s
How to find Fnet?
Find vector sum of forces acting on crate.
FBD
FN= Normal Force
Fapp= 325 N
F app,y
Ff= Friction Force
F app,x
Fg=Weight=925 N
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
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

Ff=μFN=(0.25)(787N)=197 N
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
Fnet
97 N
m
a

 1.03 2
m 94.3kg
s
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?
DRAW FBD
FN
Ff
Fapp,x
Fapp,y
Fg=319 N
Fapp= 485 N
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
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
Ff=μFN=(0.57)(598)=341 N
Fnet, x = 397.29- 341= 57.29 N
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
Fnet 57.29 N
m
a

 1.76 2
m 32.52kg
s
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
x  4m
vf 
vi  0
m
s
a  1.76
m
s2
m
v  2ax  3.8
s
2
i
m
m
v f  vi 3.8 s  0 s
t

 2.13s
m
a
1.76 2
s
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.
Draw a Free Body Diagram
FF
FN
Fg,y
θ
Fg
Fg,x
θ
Closer look at gravity triangle.

Solve for Fg,y and
Fg,x
Fg , y
a
Fgcos

mg
cos(

)



,y
h Fg
F
o sin(
Fgsin(

mg

)
,x  )  
g ,x
h
Fg
Fg,y
θ
Fg
Fg,x
Coordinate system for inclined planes
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.
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(θ)
Forces in the x direction

What is the force that makes the
object slide down the inclined
plane?

Gravity…but only in the x direction
Remember that Vectors can be
moved parallel to themselves!!
FF
FN
Fg,y
θ
Fg
Fg,x
θ
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
What if there is an additional applied force?

Example: a box is being pushed up
an inclined plane…
Fapp
FN
Fg,x
Fg,y
Ff
θ
In that case…

FN= mgcosθ

Fnet = Fapp- Fg,x – Ff

If the object is in equilibrium then

Fapp= Fg,x + Ff
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.
Draw a FBD
Ff
FN
Fg,x
Fg,y
Fg
θ
Solve the Problem


The block is in equilibrium so…

Fnet=0

Fg,y= FN=mgcosθ=(5.4kg)(9.81)cos(15)

FN=51 N
Additionally, what is the force of
friction acting on the block?
Find Force of Friction

Fnet= 0

Fg,x= Ff= mgsinθ=5.4(9.81)sin(15)
Ff= 13.7N
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?
FBD
Ff
FN
Fg,x
Fg,y
Fg
θ
What is Fnet?

They give mass and acceleration

So Fnet= ma= 75kg x 3.60 m/s2

Fnet= 270 N

FN= mgcosθ

Fnet= Fg,x – Ff=mgsinθ - Ff
Solve for Ff


Fnet= Fg,x – Ff=mgsinθ – Ff
Ff= mgsinθ – Fnet
Ff = 75kg(9.8)sin(25) – 270 N
Ff = 40.62 N
Finish the Problem

We are trying to solve for μk
Ff
40.26 N 40.26 N
 k


 .061
FN mg cos 
667 N