Download Friction Video Script

Physics 503 – Friction
Instructor
Today we tackle a new topic, friction. Let’s start with a physics challenge.
I have here a flask with a narrow neck and a cork stopper. Notice how difficult it is to push the stopper
into the flask. Now the challenge is to get the stopper out without breaking the flask. You can see that
shaking is not going to work. But, with just one simple item, I can get the cork out, without changing
the flask or stopper. Both the problem and the solution are based on friction.
See if you can figure out how I do it.
I’ll reveal the secret at the end of the program.
(Read objectives on screen.)
Instructor
During this course, we’re going to blame lots of things on friction, like this puck slowing down and
stopping after we give it a push.
We’ve already said that without friction, once an object like this air puck is in motion, it would
continue in uniform motion forever. If we could eliminate friction, we could put a car in motion, then
turn off the motor, and it would keep moving. That would solve lots of our energy problems. The only
hitch is that without friction, we wouldn’t be able to stop.
And sometimes, without friction, we wouldn’t be able to start moving because you need traction
between surfaces to get that action-reaction thing going. Have you ever tried to walk on an icy surface
when you’re wearing shoes with slick soles?
So friction can be desirable as well as undesirable. Before we can control it, we need to know what it
is, what causes it, and what factors affect it. Time to take some notes.
(green chalkboard on screen)
VO
Friction is a force that resists motion.
Friction always involves surfaces that are in contact with each other.
Friction is caused by the interlocking of irregularities on the two surfaces.
Instructor
To understand the causes of friction, think about what happens when a tire drops into a rut in the road.
It’s hard to get the car going, isn’t it? So what does this have to do with the cause of friction?
You must understand that all surfaces are rough on a microscopic level. So the irregular places on the
surfaces will interlock, no matter what. That’s the major cause of friction
Now let’s go back to the car and rut. You probably know from experience that it’s easier to get across
ruts in the road if you keep the car moving. Once it stops and the tire sinks down into the rut, the
resistance to motion is much worse. So there are two main types of friction.
(green chalkboard on screen)
VO
The first type is static friction. Static means stationary, so this type involves objects at rest. Static
friction is also called starting friction.
The second type is called kinetic friction. Since kinetic means moving, this type of friction involves
objects already in motion. Sometimes kinetic friction is called sliding friction.
Static friction between two surfaces will always be greater than kinetic friction between the same
surfaces.
Instructor
Now here’s another thing you need to know about static friction. It acts in a direction that is parallel to
the surfaces in contact and opposite the applied force. Before you write that down, let me show you
what I mean.
Notice that even though I’m pushing this container of orange juice to your left, it’s not moving. What
does this tell you about the net force on the container? Tell your teacher.
Did you say that the net force on the container is zero and that it’s in a state of equilibrium? You’re
right. Now I’m going to push a little harder. But the container still isn’t moving.
Finally, I push with enough force to start the container moving. Let’s draw some force diagrams to
show what’s happening in each case.
(pictures of OJ containers on screen)
VO
In the first picture, a small force is being applied by the hand but the container is not moving. Let’s
start by drawing a big dot to represent the object. Now let’s draw the applied force, which is a small
push to the left. Remember that pushes are drawn as pulls from the other side, so we’ll draw a short
vector to the left and label it “F sub a” for applied force. Now here’s what we mean when we say static
friction acts parallel to the surfaces and opposite the applied force. The force of friction will be drawn
to the right and parallel to the table. Since the object isn’t moving, the force of static friction must be
equal in magnitude to the applied force of the hand. So we’ll draw our vector the same length and label
it “Ff.”
Now in the second picture, the hand pushed harder, but the container still didn’t move. How would you
draw the force diagram for this one? Tell your teacher.
For this situation, we draw a longer vector for the applied force, and we draw a vector of equal length
for the force of static friction. The net force is still zero.
But there’s a limit to the force of static friction between two surfaces, and in the third picture, we’re
pushing harder than that limit. So the applied force is greater than the force of static friction, and the
container starts to move. It’s accelerating in the direction of the net force.
Instructor
Now let’s look at a situation involving kinetic or sliding friction. I’m applying a force to the container,
but this time the container is moving uniformly to your right. Draw a force diagram showing the
applied force and the force of friction in this case. Your teacher will pause the tape until you’ve
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finished.
(Pause Tape Now graphic)
(picture of OJ container on screen)
VO
In this situation, the container is moving uniformly to the right. That means the applied force is to the
right, and the force of kinetic or sliding friction is to the left. Because the container is in uniform
motion, the net force is still zero, so applied force and force of friction are equal in magnitude and in
the opposite direction.
Instructor
There’s another force we need to consider when we study friction. So we’ll need to include it in our
force diagrams. It’s called the normal force. You may have heard this term in math class. In math,
normal means “perpendicular to.”
(green chalkboard on screen)
VO
The normal force exerted on an object is perpendicular to the surface.
A normal force is a reaction to the action force of the object pushing down on the surface beneath it.
The object pushes down and the surface beneath it pushes back up with the same force.
Instructor
Let’s see how this looks in a force diagram.
(diagram on screen)
VO
This is a diagram of a box that is sliding uniformly to the right. You’ll want to draw it with me so that
it’s in your notes.
Because the box is sliding to the right, the applied force, whether it’s a push or a pull, must be to the
right.
Since the box is sliding uniformly, the force of friction must equal this applied force. We draw the
friction force vector to the left , parallel to the surfaces and in the opposite direction of the applied
force.
The next force vector we need to draw is the weight of the box. We always draw this vector straight
down and label it “F sub w.”
The last force acting on the box is the normal force exerted by the floor on the box. Since the box and
floor are horizontal, the normal force will equal the weight. The box pushes down on the floor and the
floor pushes up on the box.
Instructor
This is a complete free-body diagram because it shows all forces acting on this box. The weight shows
the earth pulling it down, the normal force shows the floor pushing it up, the applied force pushes or
pulls it to the right, and the force of friction resists that applied force.
There are other forces involved in this situation, but these four are the only forces acting on the box.
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That’s what a free-body diagram is all about.
Now we’re ready to investigate factors that could affect the force of friction between two surfaces. The
best way to do this in the lab. This is a friction board with four different surfaces: cardboard, cork,
sandpaper, and rubber. We’ll place the board flat on the table and place this wood block on top of one
of the surfaces. To measure the force of kinetic friction, we’ll use a spring scale to pull the bock
horizontally until the block slides uniformly across the surface. There will be some bumps, but we’ll
take an average reading.
To get ready to collect and analyze the data, you’ll need to copy this data table. Your teacher will put it
on the board or overhead projector and give you time to copy them.
VO
Local teachers, please pause the tape and give students the data table from the facilitator's guide.
(Pause Tape Now graphic)
VO
The first thing we need to do is measure the weight of the block by hanging it from a spring scale.
The weight is 2.0 N. Record this as “Fw” in data table A under trial one.
Next, we determine the applied force needed to make the block slide uniformly across the table.
The average applied force is 0.6 N. Record this as “Fa” in trial one.
Next, we’ll keep the block in the same position but change its weight by placing a 100-gram mass on
top of the larger side and pulling it as before.
This time the average applied force is 0.90 N. Record this as “Fa” under trial two.
We also need the weight of the block and mass together. It is 3.0 N. Record the weight in trial two.
Next we’ll go back to the original weight and determine the effect of the area of contact on friction by
placing the block on its smaller side and pulling as before.
This time the average applied force is 0.60 N. Record this as “Fa” in trial three.
Instructor
Before we can make any conclusions from this part of the lab, we need to go back to our free-body
diagram and see what the applied force and weight tell us.
(force diagram on screen)
Since the block slides uniformly, what do you know about the force of friction? Tell your teacher.
Instructor
Did you say that the force of friction equals the applied force? You’re right. And since the block lies
flat on the table, what do you know about the normal force pressing on the block? Tell your teacher.
On a horizontal surface, the normal force equals the weight. Hope you said that.
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Now go back to your data table and fill in the blanks at the top of the third and fourth columns. While
you’re at it, circle “Ff” and “FN” since those will help us make our conclusions.
(table on screen)
VO
Table A should look like this so far. Now let’s make some conclusions. Use your data to answer some
questions. Your teacher will provide you with the questions and will give you time to answer them.
Then we’ll come back and talk about the answers.
(text on screen)
VO
Local Teachers: Turn off tape and give students problem set number one from facilitator's guide.
(Pause Tape Now graphic)
(table on screen)
VO
To answer number (1a), you should have looked at trials one and two. As the normal force increased,
the force of friction also increased. This should make common sense to you since pressing the surfaces
together with more force makes the surfaces interlock more.
To answer (1b), you should have looked at trials one and three. The area of contact between the
surfaces had no effect on the force of friction.
Instructor
Why doesn’t the area of contact affect friction? You already know how the normal force pressing the
objects together affects friction, so let’s look at this.
(diagram on screen)
Since normal force is simply a reaction to the block pressing down on the table, let’s look at the force
with which the blocks in these two pictures press down on the table beneath them. Imagine the block in
the first picture consisting of four small blocks. Each presses down with a small force.
In the second picture, the block consists of only two blocks, but each one weighs more. So the total
force pressing the surfaces together is the same. And that means the force of friction will be the same
on the two blocks.
(lab apparatus on screen)
VO
Now let’s look at question number two. In every trial, the scale reading was greater when the block
first started to move. That’s because static friction must be overcome first, and static friction is greater
than kinetic friction. Once the block is in motion, only kinetic friction has to be overcome.
(table on screen)
VO
Finally, why is it easier to slide a refrigerator across a floor than to lift it and carry it to the other side of
the room? To answer this, just look at any trial. Notice that the force of friction is always less than the
block’s weight. Well, to slide the refrigerator, all you have to do is apply enough force to equal the
force of friction between the refrigerator and floor. And that’s always less than the weight. Got it?
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Instructor
Our lab results showed us that the force of friction between two surfaces is not affected by the area of
contact between them but it does depend on the normal force pressing the surfaces together. Remember
that the normal force acting on an object is a reaction force pushing upward on the object perpendicular
to the surface.
Physicists have discovered that there is a definite ratio between the force of friction and normal force
acting on the surfaces. Let’s go back to our lab data and see if we agree.
(table on screen)
VO
Here is a data table similar to yours, but showing only trials one and two. We’ve also added another
column, which you should add to your table. At the top, write “Ff” divided by “FN” like this. We want
to find the ratio between the force of friction and the normal force. For trial one, we divide 0.60 N by
2.0 N. The answer is 0.30. The units cancel out, leaving no unit for this ratio.
For trial two, we divide 0.90 N by 3.0 N, and get 0.30 again. When normal force increased, so did the
force of friction. The ratio remains constant for these two surfaces.
(green chalkboard on screen)
VO
There are two factors that affect the force of friction between two surfaces. The first is the normal force
pressing the two surfaces together.
The second factor affecting friction is the nature of the materials making up the two surfaces in contact.
The ratio of the force of friction to the normal force for two particular surfaces is a constant. This ratio
is called the coefficient of friction and is represented by the Greek symbol, mew.
Mew depends only on the nature of the surfaces in contact. This includes the composition of the
materials and qualities such as smoothness, temperature, and so forth.
The coefficient of static friction, “µ s” is greater than coefficient of kinetic friction, “µ k”
Instructor
You already know that the coefficient of kinetic friction of wood on the countertop material was 0.30.
In the lab, we’re going to determine mew for wood on these four surfaces: cardboard, cork, sandpaper,
and rubber. Just for fun, predict which will have the largest and the smallest values for µ. Tell your
teacher. And while the tape is paused, make a data table for collecting the data we need to determine
“µ k”for wood on each surface.
I’ll bet you’re expecting your teacher to give you a table to copy. Wrong. You know what we need to
measure, so design your own table.
(Pause Tape Now graphic)
(students on screen)
VO
We’re using the same block that we used in the Part one of the lab, so the weight will remain the same,
2.0 N. Record this in your data table for every trial.
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We’ll start with wood on cardboard. The applied force needed to slide the wood uniformly across the
cardboard is 0.4 N.
When we pull the wood uniformly across the cork surface, the scale reading is 1.0 N.
For the rubber surface, an applied force of 2.4 N was required.
Finally, we tested the sandpaper and measured an applied force of 1.4 N.
Now use the data to calculate the coefficient of kinetic friction for wood and each surface. Your
teacher will check your results and see who was right in their predictions. We’ll talk about some other
important coefficients of friction when we come back.
(Pause Tape Now graphic)
Instructor
Here are the results. Were you surprised that the coefficient of friction between wood and rubber was
greater than wood and sandpaper? I was. But I guess that’s why we make our tires out of rubber. It
has really good friction and we need that for gripping the road to turn and to stop.
Now here’s a challenge question for you. Why is important to have good treads on your tires? Think
about friction and then tell your teacher. I’ll wait.
As long as the weather is nice, it’s not important to have good treads. But when it rains, the tread is
vital. The coefficient of kinetic friction between rubber and dry concrete is 0.8 and between rubber and
wet concrete is 0.5. For rubber and a sheet of water, mew would be much, much less. The tread gives
the water a place to go so that the rubber can meet the road and create some friction. Without friction,
the tires would slide and the car’s inertia would keep it moving in a straight line when you might like to
turn or slow down.
Now it’s time to solve some force problems involving friction. Let’s start with a couple of examples.
(text on screen)
VO
The problem is “A 260 newton crate initially at rest on a horizontal floor requires a horizontal force of
slightly more than 79 newtons to start it moving. We want to calculate the coefficient of static friction
between the crate and floor. First, we are asked to draw a free-body diagram for the crate. Concentrate
on the forces acting on the crate, not on any other object, like the earth or the floor. There are four
forces acting on the crate.
Let’s assume that the crate moves from left to right. So we would draw the applied force to the right
and label it “Fa” This means that the maximum force of static friction between the crate and floor is 79
newtons to the left.
Finally, there’s one more force acting on the crate. The earth pulls down on the crate and that makes
the crate push down on the floor. As a reaction, the floor pushes up on the crate with a normal force
equal to its weight. We’ll label it “FN”and remind ourselves that it’s equal to the weight.
The coefficient of friction is the force of friction divided by the normal force. So we plug and chug 79
newtons divided by 260 newtons. The coefficient of static friction is 0.30.
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(text on screen)
VO
Now your teacher will put these two example problems on the board or overhead and give you time to
try them.
Local Teachers: Turn off tape and give students problem set number two from facilitator's guide.
(Pause Tape Now graphic)
(problem on screen)
VO
In the first problem, the normal force is equal to the weight of the chair, and the applied force is equal
to the force of friction because the chair is moving uniformly.
The normal force is the mass of the chair, 32 kg, times g which is 9.8 m/s2. This equals 314 N.
The coefficient of kinetic friction equals the force of friction, 66 N, divided by 314 newtons, or 0.21,
rounded to two significant digits.
(refrigerator problem on screen)
VO
In the next problem, we want to find the applied force. And that is equal to the force of friction. Since
we know both the coefficients of friction and the normal force, we can calculate the force of friction,
using the equation, “Ff” equals mew times “FN.”
The force of friction equals 0.35 times 1800 newtons, or 630 newtons. That’s the amount of force we
must apply to slide the refrigerator uniformly across the floor.
Instructor
Well, you’ve learned a lot about friction today, and you have some practicing to do. But now, it’s time
to …SHOW YOU THE TRICK!!
Bet you thought that was going to be a Show What You Know, didn’t you? Well, we’ve run out of
time, so your teacher will do that for us. But I did promise to show you the trick with the cork and
flask. Here it is.
I have a cloth napkin here. I’m going to push it, corner first, into the flask and form a little pocket to
hold the cork. When the cork is covered by the cloth, all I have to do is pull and out they come
together.
That’s because the force of friction between the cloth and cork is large, keeping them together, while
the force of friction between the glass and cloth is small, letting the napkin slide easily. Once again,
physics to the rescue.
See you next time.
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