Scene #

Description

Narration

1

A definition for the force of friction is showing on the screen. The narrator reads the definition.

The narrator writes the equation for the force of friction on the screen and explains it.

You're probably already aware of the force of friction in your daily life. But now we want to look at a way to quantify that in a mathematical way. The force of friction, which is often abbreviated F sub F, is a force that opposes motion, converting kinetic energy, energy of motion, into heat in the process.

The force of friction is defined by the equation F sub F equals mu F sub n. And that means the force of friction is equal to the coefficient of friction times the normal force. Now a couple of things to quickly go over about the force of friction.

2

Below the equation the narrator draws a green arrow pointing east to represent the velocity. A blue arrow is drawn going in the opposite direction to show the force of friction. Another blue arrow is drawn pointing north and labels it Fn.

First, it always opposes motion. So if your velocity vector looks like that, that means the force of friction is going to be the opposite direction, like this. Along with being in the opposite direction is the velocity vector. It's also perpendicular to the normal force.

Now we talk broadly about the force of friction. But there are actually two types of frictional force. The first type of frictional force is called static friction.

3

In red the narrator writes about static friction and kinetic friction. He explains it as he writes about it.

Static friction occurs when there is no motion. So think about trying to push a heavy box. When you're pushing against it, and it's not moving at all, the type of friction you're fighting against at that point is static friction. And the symbol for the coefficient of static friction is mu sub s. The s, of course, standing for static.

So again, imagine you're pushing on that heavy box, trying to overcome the static friction, at some point you do overcome that. And the box becomes much easier to push. And at that point, you're now dealing with kinetic friction.

Kinetic friction applies when there is motion. And the symbol for the coefficient of kinetic friction is mu sub k. Thinking about the example of pushing a box and trying to get it going, it's always going to be harder to overcome static friction than it is to overcome kinetic friction. And that's because for all objects, the coefficient of static friction will always be larger than the coefficient of kinetic friction.

Now let's look at a couple of examples the deal with static and kinetic friction.

4

An example of kinetic friction problem is showing on the screen. The narrator reads it out loud.

The narrator describes how to draw a free body diagram, he draws it as he describes it.

Let's look at an example of kinetic friction. Linda is pushing a 65 kilogram cart filled with laboratory equipment with a force of 81 newtons. The coefficient of friction between the cart and the floor is 0.10. How fast does the cart accelerate?

Let's begin by drawing a complete free body diagram of the cart that Linda is pushing. We'll draw the cart as a dot. We know that the force of gravity is acting downward. And that's equal to the cart's mass times gravity.

We know that the normal force is acting upward, F sub n. We know that there's a forward push by Linda. We'll call that F sub p. And we know that there's a force of friction opposing that motion. We'll call that F sub f.

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To the right of the free body diagram the narrator discusses the variables being used and records them in red on the screen.

All right, let's write down the variables that we know. We know mu sub k, the kinetic coefficient of friction, is 0.10. We know that the mass m is 65 kilograms. And we know that the force of the push, Fp, is equal to 81 newtons.

We know that the forces are balanced in the y direction because the cart isn't accelerating up above the ground or down into the ground. So the normal force is also going to be equal to mg. The force of the push is equal to 81 newtons. All we need to solve for is the force of friction.

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The narrator writes in blue the equation for the Force of Friction. As he describes the equation and the steps to solve the problem he writes them down.

To do that, we can use the equation for the force of friction, which says force of friction is equal to mu times the normal force. Or another way of writing that would be mu times mass times gravity in this case. All right, let's put in the values that we know.

Mu is 0.10. And there's no units on that. Mass is 65 kilograms. And gravity is 9.81 meters per second squared.

Now normally we would put negative 9.81. But since with the force of friction, we already know the direction it's pointing from our free body diagram. We know it's going to be pointing in the direction opposite of motion. And so we're just going to put the magnitude of the acceleration of gravity, 9.81 meters per second squared.

Multiply this out, and you'll find that the force of friction is 64 newtons.

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In green the narrator writes the sum of all forces and writes the steps down to solve this portion of the problem as he says them.

Now to find the acceleration of the cart, we're going to apply Newton's second law in the x direction. Newton's second law tells us that the sum of all forces is equal to mass times acceleration.

So let's write that down. Sum of all forces in the x direction is equal to our object's mass, which is 65 kilograms times our object's acceleration in the x direction. The sum of all forces in the x direction-- well, we have a positive 81 newton force. And then in the negative direction, we have the force of friction, so minus 64 newtons.

Simplifying this down a little bit, we get 65 kilograms times the acceleration in the x direction is equal to 17 newtons. If we divide both sides of this equation by 65 kilograms, we get that the acceleration in the x direction is equal to 0.26 meters per second squared. So that was an example of how we can take kinetic friction into account in a kinematics problem.

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A problem on static friction is showing. The narrator reads it out loud.

Now let's look at a problem that uses static friction. This problem asks, Glenn is pushing his 141 kilogram refrigerator to a new location. Only when he pushes with a force of 429 newtons does the refrigerator begin to move. What is the coefficient of static friction between the refrigerator and the floor?

So in this situation, before Glenn could even get the refrigerator moving, he had to overcome that static frictional force. And after that, the pushing will become easier. But first, you have to overcome that static friction.

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The narrator describes how to draw a free body diagram for the problem as he describes it he draws it on the screen.

As normal, let's begin by drawing the free body diagram for this situation. We'll represent the refrigerator as a dot. There's the force of gravity pulling downwards, F sub g. And that's equal to mass times gravity.

There's the upward normal force, F sub n. And again, because we know that these forces are balanced and there's no acceleration upward or downward, we know that that's equal to mg as well.

There's the force of the push, we'll call F sub p. And that's equal to 429 newtons. And then there's the backward force of friction. And we know because there was no motion at this point, that that is also equal to 429 newtons.

Those two forces canceled out right up until that moment.

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The narrator writes the equation to solve for the force of friction in blue. As he describes the steps to solve the problem he writes the steps down.

Now let's use our definition of friction to solve for the coefficient of friction. The force of friction is equal to mu times the normal force. In this case, it's mu sub s, the coefficient of static friction.

The force friction we know is 429 newtons. That's equal to the coefficient of static friction, which we're solving for times the normal force, which in this case was equal to mass 141 kilograms times gravity, 9.81 meters per second squared. And again, because we're only dealing with magnitudes here, we already know the direction. We don't need to worry about the sign on gravity.

Multiply this out and you get 429 newtons equals the coefficient of static friction times 1,383 newtons. Divide both sides of the equation by 1,383 newtons and you get the coefficient of static friction is equal to 0.31. And again, there were no units on coefficients of friction.