AP Physics 1 & 2 Friction

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This lesson is going to be about friction.0004

Now our objectives are going to be to define and identify frictional forces. Yeah friction!0006

We will explain the factors that determine the amount of friction between two surfaces and determine the frictional force and coefficient of friction between two surfaces.0013

So let us dive in. Friction is a force that opposes motion.0021

Kinetic friction is a type of friction that opposes motion for an object that is sliding along another surface.0027

Kinetic friction is sliding friction.0034

Static friction acts on an object that is not sliding.0036

Now the magnitude of the frictional force is determined by two things: the nature of the surfaces in contact -- and we characterize that with μ -- a variable that refers to the coefficient of friction.0041

Bigger coefficients of friction, bigger μ -- so you are going to have more friction between the two surfaces.0059

Imagine something like -- let us say really flat dress shoes on ice -- very slippery -- compared to two pieces of sandpaper.0065

The sandpaper is going to have a much higher coefficient of friction.0075

The normal force acting on the object is the other item that determines the magnitude of the frictional force.0079

Now as we talk about these types of friction and the magnitudes of these frictional forces, it is important to note that typically, kinetic friction is less than static friction.0086

And you have probably observed that before.0105

Have you ever tried to push something heavy along the floor -- maybe pushing a sofa or a refrigerator or something heavy -- it takes a lot of work to get it started because you have to overcome static friction.0108

Once you have it moving however, now you are into the regime of kinetic friction that usually takes a little bit less force.0120

Kinetic friction is usually smaller than static friction.0125

Now as we talk about these coefficients of friction, we are going to have a different coefficient depending on whether it's sliding or static then.0130

So the coefficient of friction μ -- we are going to talk about the coefficient of kinetic friction μ_k or coefficient of static friction μ_s.0137

So this coefficient of friction is really the ratio of the frictional force and the normal force.0147

Coefficient of friction given by the force of friction divided by the normal force.0152

It depends only on the nature of the surfaces that are in contact.0158

You can look up in many different places approximate coefficients of friction and you can see as we have on the slide here that there are different values for kinetic or static.0163

Rubber on dry concrete has a kinetic coefficient of 0.68.0174

But on static, when it is not sliding, it is 0.9.0179

That means that if you lock your wheels as you are driving down the road on dry concrete -- if there is sliding -- if there is skidding then you have less friction then if they are not sliding.0182

This is the reason for antilock brakes in car.0192

If you are sliding then you are not getting as much stopping force as you would if you were not sliding, so they try and keep cars from sliding with these antilock brakes -- not allowing them to slide.0196

You could look up the coefficient of friction for many different materials.0207

So let us take a look at some examples and try and determine which regime of friction they are in -- kinetic or static.0215

If we have a sled sliding down a snowy hill -- sliding -- there is our key word -- that must be kinetic friction -- we would use the kinetic coefficient.0218

A refrigerator at rest that you want to move -- at rest implies not sliding -- that one is static.0233

A car with the tires rolling freely -- well, we just talked about that -- not skidding, therefore static.0241

If you are skidding across pavement though, you are going to use kinetic coefficient of friction.0249

Let us take an example here. A car's performance is tested on various horizontal road surfaces.0259

The brakes are applied causing the rubber tires of the car to slide along the road without rolling.0264

They are sliding.0271

They encountered the greatest force of friction to stop the car on which of these surfaces -- Dry concrete? Dry asphalt? Wet concrete or wet asphalt?0273

Well, first thing we need to realize is, is if we are sliding, we are looking for the kinetic coefficient.0282

Which one of these is the biggest -- Rubber on concrete, dry and wet? Rubber on asphalt dry and wet? -- 0.68 is our biggest coefficient, so that would have the greatest force of friction -- dry concrete.0288

Another example is we have a block on an incline.0304

The diagram shows the block sliding down a plane, inclined at angle θ -- there is θ.0307

If angle θ is increased -- as that gets steeper -- What happens to the coefficient of kinetic friction between the bottom surface of the block and the surface of the incline?0312

Well, here you have to remember that the coefficient of friction depends on the nature of the surfaces.0321

In this case the surfaces have not changed.0332

Yes, you are going to have some other different effects, but as far as the coefficient of friction goes, the nature of the surfaces has not changed, therefore the coefficient of friction will remain the same.0335

To calculate the force of friction -- again it depends only upon the nature of the surfaces in contact, that coefficient of friction and the magnitude of the normal force.0348

We have a nice direct relationship.0355

Force of friction equals the coefficient of friction μ times the normal force.0358

We can combine this with Newton's Second Law and free-body diagram to solve even more involved problems that we did in our Newton's Second Law discussion.0364

While we are here and talking about friction, let us come back to terminal velocity.0374

Objects following through Earth's atmosphere experience a force of friction that we call air resistance.0379

That is a drag force and as the object goes faster, there is even more of that.0382

Eventually, an object gets going fast enough that the force of friction balances the force of gravity on the object.0388

When that happens you reach what is known as terminal velocity.0394

The net force is zero -- you do not gain any more speed -- the longer you fall.0397

So a graph of velocity versus time for an object that we are now taking into account air resistance -- it is going to start -- say we throw somebody out of an airplane -- their vertical velocity starts at zero and it increases, increases, increases that force of friction.0401

That force of air resistance -- the faster they go gets greater and greater until eventually they hit this asymptote which we know as the terminal velocity.0423

When they do that, at that point where they hit terminal velocity, -- FBD -- the weight of the object and the force of air resistance exactly balance.0435

No net force. No acceleration. Constant velocity.0450

Let us take a look at another example -- finding the frictional force.0456

In the diagram, we have a 4 kg object accelerating at 10 m/s² on a rough horizontal surface.0459

Find the magnitude of the frictional force, (Ff), acting on the object?0467

Let us start with our FBD.0472

We have the normal force, the object's weight, we have this applied force to the right of 50N, and we have a frictional force to the left.0475

All of my forces line up with the axis, so I do not need to draw a P-FBD.0494

Since we are looking for the magnitude of the frictional force, I am going to start by writing Newton's Second Law for the x-direction.0501

I am going to replace now, net force in the x-direction with all the forces acting in the x-direction.0510

I look at my FBD, I have 50N to the right, the applied force, minus force of friction and that must equal my mass- 4 kg times my acceleration 10 m/s².0515

Or 50N minus force of friction equals 40 kg m/s² which is a Newton.0533

Therefore, force of friction must be equal to 10N.0543

Let us take a look at another example. Here we have a box on a wood surface.0551

A horizontal force of 8N is used to pull a 20N wooden box moving toward the right along a horizontal wood surface, where we know that the coefficient of kinetic friction there is 0.3.0559

We are asked to find the frictional force acting on the box, the net force acting on the box, the mass of the box and the acceleration of the box.0569

Well, we will start with our FBD. We have normal force. We have its weight, mg, which it tells us it is a 20N wooden box, so that must be 20.0580

We have a force to the right, an applied force of 8N and we must have some frictional force to the left.0592

If we want to find the frictional force acting on the box, what I am going to write is the force of friction equals μ times the normal force and by the way, look -- friction is F-U-N -- friction's fun.0600

μ is 0.3 and our normal force in this case -- if you look in the y-direction, that must be equal to mg.0616

There is no net force in the y-direction, otherwise that box would spontaneously take up off the table or go through it, and we know that does not happen, they have to be balanced.0625

The normal force here must be 20N, so 0.3 x 20 or 6N.0633

It also asks for the net force acting on the box.0640

Net force in the x-direction is just going to be 8N to the right minus 6N to the left, our frictional force or 2N.0645

How about the mass of the box?0657

Well, we know its weight, mg, is 20N, so if we just divide both sides by g, we should get the mass, which is going to be 20N divided by g(10), -- it is going to be 2 kg.0660

And finally, the acceleration of the box.0675

Well, acceleration is net force divided by mass. We just determined the net force here was 2N.0678

We determined the mass was 2 kg, so the acceleration must be 1 m/s².0689

Let us take a look at an example where we explore the difference between static and kinetic friction?0705

Compared the force needed to start sliding a crate across a rough level floor, the force needed to keep it sliding once it is moving is -- well if needed to start you need to overcome static friction -- once its sliding, it is kinetic.0710

Kinetic is less than static, therefore it is going to be less.0727

Let us take a look at a drag force.0734

An airplane is moving with a constant velocity in level flight. We have an airplane moving with constant velocity in level flight.0737

Compare the magnitude of the forward force provided by the engines -- we typically call that thrust -- to the backward frictional drag force.0749

Well, let us draw our FBD.0758

There is our airplane. We have some thrust forward.0761

We have a drag force backwards -- force that is pulling it up, we call lift and we have its weight.0770

Now if it is moving at constant velocity in level flight, everything must balance out -- they must be equal.0781

So the force of the thrust or the force of the engines must balance the force of drag, therefore they must be equal.0789

Another example, have Suzie over here pulling a sled.0800

She pulls the handle of a 20 kg sled across the yard with a force of 100N and that is at an angle of 30 degrees.0805

The yard exerts a force of 25N on the sled due to friction.0812

We are asked to find the coefficient of friction between the sled and the yard and determine the distance the sled travels if it starts from rest and Suzie maintains her 100N force for 5 s.0816

Well, let us start off with our FBD -- y, x.0828

There is our sled.0839

We have weight down -- force of friction to the left 25N.0840

We have the normal force from the ground up and we have the applied force of Suzie, which is 100N at an angle of 30 degrees.0846

So my P-FBD -- I will draw that down here.0858

Right away, let us put in our red vectors -- the ones that are already lined up with the axis.0865

We have mg, force of friction, and normal force.0870

Now we have to break that up into components, so the x-component of Suzie's applied force is going to be 100N cosine 30 and the y-component 100N sine 30 degrees.0878

Now we can go start to solve our questions.0900

Find the coefficient of friction between the sled in the yard.0904

I am going to start by writing Newton's Second Law and I am going to look in the y-direction-- equals ma_y.0906

I am going to replace the net force in the y-direction with all the different things I see here.0915

I have 100 sine 30 and that is going to be 50, plus the normal force, F_n minus mg.0919

And I know -- common sense tells me -- that sled is not going to go spontaneously accelerating up off the ground so that must all equal 0 -- acceleration in the y is 0.0931

So I can solve for the normal force -- the normal force then must be mg - 50, which is going to be mass(20)kg x 10 - 50 or 150N.0941

μ then, the coefficient of friction, is the force of friction divided by the normal force which is 25 over 150 or 0.167 -- there is our coefficient of friction.0961

Now then, it also tells us to determine the distance the sled travels if it starts from rest and Suzie maintains her 100N force for 5 s.0978

Well to do that, it would be nice to know the acceleration of the sled in the x-direction.0987

Let us go to Newton's Second Law in the x-direction.0992

F_netx is going to be 86.6N 100 cosine 30 minus the force of friction, 25N or 61.6N.0996

Therefore, acceleration which is the net force divided by the mass is 61.6N/20 Kg -- it is going to accelerate at about 3.1 m/s².1007

So, if we want to find out how far it goes in that 5 s-interval, we can go back to our kinematics.1024

Δx = V initial T + 1/2AT² and V initial here -- if it starts from rest is 0.1032

So that is just going to be 1/2 x a(3.1) m/s² x the square of our time, 5 s² or 38.8 m.1041

Hopefully that gives you a good start with friction, the coefficient of friction.1062

Thanks for watching Educator.com. Make it a great day.1065