Physics 1A
Lecture 4C
"I believe there are
15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,9
14,527,116,709,366,231,425,076,185,631,031,296
protons in the universe and the same number of electrons.”
--Sir Arthur Eddington
Frictional Force
The static force will be in the direction
opposite the applied force that is attempting
the motion.
Before you can get an object to move you
must overcome the maximum static friction.
Once you have an object moving over a
surface, the friction will become kinetic
friction, fk.
Kinetic friction is less than the maximum
static friction for a given surface.
Frictional Force
To calculate kinetic friction use:
where μk is the coefficient of kinetic friction and FN
is the normal force.
For static friction, we only calculate the maximum
possible static friction via. We express this situation
as:
Newton’s laws
1st: An object continues in its state of motion at
a constant speed along a straight line, unless
compelled to change that state by a net force.
2nd:
3rd: “Whenever one body exerts a force on a
second body, the second body exerts an oppositely
directed force of equal magnitude on the first
body.”
Newton’s 3rd Law
Newton’s Third Law
“Whenever one body exerts a force on a second
body, the second body exerts an oppositely
directed force of equal magnitude on the first
body.”
This law is sometimes shortened to:
“For every action, there is an equal, but opposite
reaction.”
The third law is the one that is most often
misconstrued.
Let’s look at a game of Tug-Of-War.
In Class Question
Who wins a game of Tug-Of-War between a 80kg
person and a 60kg person?
A) A tie, Newton’s Third Law tells that neither can win,
because the force between them is equal and opposite.
B) The 80kg person, because the 80kg person will
always exert a greater force than the 60kg person.
C) The 60kg person, because the force exerted by this
much mass is greater than the 80kg person.
D) You can’t tell until you draw your force diagram for
this situation, and know more about the coefficients of
friction for both people and the surface they stand on.
Newton’s 3rd Law
Draw a force diagram for each person separately.
Fnormal, ground on 80kg
Ffriction, ground on 80kg
Fgravity, Earth on 80kg
80kg
Ftension, 60kg on 80kg
Fnormal, ground on 60kg
Ftension, 80kg on 60kg
60kg
Fgravity, Earth on 60kg
Ffriction, ground on 60kg
The winner of the tug-of-war contest is not who is
the strongest (FA on B = FB on A), but whoever had the
most friction with the ground.
Example
Example
A box lies on an inclined plane at an angle of
30.0o to the horizontal plane. The mass of
the box is 5.00kg. What is value of the
coefficient of static friction if it is noted that
the static friction is at a maximum?
Answer
Go through the guidelines.
First, you must define a coordinate system.
Example
Answer
We will make a clever choice of coordinate
systems:
The angled coordinate
system will allow us to only
break one force into
components.
Next we should draw a force diagram for the box:
Ffriction, ground on box
Fgravity, Earth on box
Fnormal, plane on box
box
Example
Answer
The normal force only points in the y-direction and
the friction force only points in the x-direction.
We do not need to break them into components.
But we do need to
break the
gravitational force
into x and y
components.
We get:
Fgx = mg sinθ
Fgy = mg cosθ
Example
Answer
Next we should apply the appropriate Newton’s Laws.
Since the box is at rest we shall apply Newton’s First
Law in both independent directions (x and y).
ΣFx = 0
and
ΣFy = 0
Now we can perform the math, let’s start with the xdirection:
Example
Answer
Next we can turn our attention to the y-direction:
Turning to the equation for maximum static friction:
<- unitless
A word of Caution !!
The action force and the associated reaction force
is usually called a Third Law Pair.
FA on B and FB on A are called Third Law Pairs.
But normal force and gravitational force on a chair
are not each other’s Third Law Pairs (Fground on chair
and FEarth on chair), because these are two forces
that act on the same body. The two forces are
only equal because the body is in equilibrium.
Equilibrium
An object either at rest or moving with a constant
velocity is said to be in equilibrium.
The net force acting on the object is zero (since
the acceleration is zero):
Though we usually work with components when
dealing with equilibrium.
ΣFx = 0
and
ΣFy = 0
Equilibrium
Example
Find the tension in the
two wires that support
the 100N light fixture in
the following diagram.
Answer
First, you must define a
coordinate system.
Let’s choose up as positive
y and to the right as
positive x.
Equilibrium
Answer
Next we should draw a force diagram:
Ftension, left wire on light
Ftension, right wire on light
Fgravity, Earth on light
light
Now we need to break the forces into components:
Tleft
Tleft,y
40o
Tleft,x
Tleft,x = Tleft cos40o
Tleft,y = Tleft sin40o
Equilibrium
Answer
Similarly we can break the right wire into:
Tright
Tright,y
Tright,x = Tright cos40o
Tright,y = Tright sin40o
40o
Tright,x
Since this light is in equilibrium, we can apply
Newton’s 1st Law to this situation:
ΣFx = 0
and
ΣFy = 0
Let’s look at the x-direction and sum the forces:
Equilibrium
Answer
ax = 0
ΣFx = 0
ΣFx = Tright,x - Tleft,x = 0
Tright,x = Tleft,x
Tright cos40o = Tleft cos40o
Tright = Tleft
Now, let’s look at the y-direction and sum the
forces:
ΣFy = Tright,y + Tleft,y - Fgravity = 0
ay = 0
Tright,y + Tleft,y = Fgravity
ΣFy = 0
Tright sin40o + Tleft sin40o = Fgravity
2Tright sin40o = 100N
Tright = (100N)/2sin40o = 78N
For Next Time (FNT)
Start Chapter 4 HW.
Finish Reading Chapter 4.