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For more information, please see full course syllabus of AP Physics C: Mechanics
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Lecture Comments (6)

1 answer

Last reply by: Professor Dan Fullerton
Sat Jan 7, 2017 5:54 PM

Post by Jim Tang on January 7, 2017

Great video! I still can't get sliding vs. skidding stuff with friction around my head. Since sliding is motion, I keep thinking it should be kinetic friction.

1 answer

Last reply by: Professor Dan Fullerton
Tue Nov 15, 2016 6:04 AM

Post by Tuong Tran on November 14, 2016

Hi Professor, for question e of the Free Response, why is the net force in the x direction not F1cos(theta)-friction, but only F1cos(theta)?

1 answer

Last reply by: Cathy Zhao
Sat Aug 13, 2016 9:08 PM

Post by Cathy Zhao on August 13, 2016

Hi Professor Fullerton, I have trouble viewing this video. The video automatically stops at 12:23, so I can't watch any content after 12:23.


  • Kinetic friction acts on objects sliding against each other. Static friction acts on objects that are not sliding.
  • The magnitude of the frictional force is determined by the nature of the surfaces in contact and the normal force acting on the object.
  • The force of kinetic friction is equal to the product of the coefficient of kinetic friction and the normal force.
  • The force of static friction is less than or equal to the product of the coefficient of static friction and the normal force.


Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:06
  • Coefficient of Friction 0:21
    • Coefficient of Friction
    • Approximate Coefficients of Friction
  • Kinetic or Static? 1:21
    • Sled Sliding Down a Snowy Hill
    • Refrigerator at Rest that You Want to Move
    • Car with Tires Rolling Freely
    • Car Skidding Across Pavement
  • Example I: Car Sliding 2:21
  • Example II: Block on Incline 3:04
  • Calculating the Force of Friction 3:33
    • Calculating the Force of Friction
  • Example III: Finding the Frictional Force 4:02
  • Example IV: Box on Wood Surface 5:34
  • Example V: Static vs. Kinetic Friction 7:35
  • Example VI: Drag Force on Airplane 7:58
  • Example VII: Pulling a Sled 8:41
  • Example VIII: AP-C 2007 FR1 13:23
    • Example VIII: Part A
    • Example VIII: Part B
    • Example VIII: Part C
    • Example VIII: Part D
    • Example VIII: Part E

Transcription: Friction

Hello everyone and welcome back to www.educator.com.0000

In this lesson we are going to talk about friction.0003

Our objectives include defining and identifying frictional forces.0006

Explaining the factors that determine the amount of friction between two surfaces and0011

determining the frictional force and the coefficient of friction between two surfaces.0015

Let us start off by talking about the coefficient of friction.0020

The ratio of the frictional force to the normal force provides the coefficient of friction.0024

μ is the coefficient of friction.0029

It is unit less and it is a force of friction divided by the normal force.0031

Friction caused by the interaction of two objects and this coefficient of friction depends on the nature of the surfaces.0035

Some approximate coefficients of friction for different types of friction.0045

Kinetic friction is objects that are sliding against each other.0049

Static friction is objects that are not sliding against each other.0054

Rub around dry concrete for example has a kinetic coefficient of 0.68 and a static coefficient of 0.9.0059

That means once it is moving or once it is sliding, it has less friction than it does and start trying to get it to start sliding.0066

It is pretty common to see a static coefficient of friction larger than the kinetic coefficient of friction.0073

Let us take a look at some situations.0081

Are these situations static or kinetic?0083

A sled sliding down a snowy hill that would be kinetic because we have objects that are sliding.0086

How about a refrigerator at rest that you want to move?0093

That is static because it is not sliding.0096

That is why it takes more force to get something to start sliding to overcome static friction.0099

Once you get it sliding you are dealing with kinetic friction which is less.0104

A car with tires rolling freely, that is static because it is not sliding.0110

If the tires are going down off the pavement in any given point, they are static with respect to the pavement.0116

If you happen to skid across the pavement now you have sliding or kinetic friction.0122

When you are using that coefficient of friction you will use either the kinetic (μ K) or the static (μ S) coefficient.0130

Let us take the example of a car sliding.0141

A car's performance is tested on various horizontal road surfaces.0144

The brakes are applied causing the rubber tires of the car to slide along the road without rolling.0147

Without rolling they are sliding and right away we are thinking kinetic.0153

The tires encounter the greatest force of friction to stop the car on and you have a couple choices here0159

which are dry concrete, dry asphalt, wet concrete, or wet asphalt.0164

I am going to look for the largest coefficient of friction that would be right here on dry concrete and that must be our answer.0172

Largest coefficient of friction, largest amount of frictional force.0179

Here we have a block on an incline.0186

It is sliding down a plane in angle θ.0188

If the angle θ is increased, the coefficient of kinetic friction between the bottom surface of the block and the surface of the incline will?0191

That is going to remain the same.0199

Remember the coefficient of friction just depends on the nature of the surfaces and that does not change as you adjust the angle.0201

How do we calculate the force of friction?0213

The force of friction depends only upon the nature of the surfaces in contact, the μ and the magnitude of the normal force (fn).0216

Or oftentimes I like to write normal force as just (n) so you can say the force of friction is μ × the normal force.0223

Of course friction is fun.0231

You can combine this with Newton’s 2nd law in the free body diagrams like we did in our last lesson to solve more involved problems.0235

Here we go.0243

The diagram below shows a 4kg object accelerating at 10m/s² on a rough, horizontal surface.0244

What is the magnitude of the frictional force (ff) acting on the object?0251

First thing that I am going to do is draw my free body diagram.0257

Let us put our axis over here.0262

As I label my forces we have the normal force.0269

We have the weight of our object (mg).0274

Those must be balanced because it is not accelerating up off the ground or into the ground because that would be goofy.0277

We have a 15 N applied force and we have some frictional force which they are labeling (ff).0283

Let us take a look.0294

Let us apply Newton’s second law in the x direction, the net force in the x direction we will call it to the right positive.0294

We have 15 N - the force of friction = mass × acceleration in the x direction which is 4 kg × 10 m/s².0302

15 N - the force of friction = 40 N.0319

Our force of friction must be 10 N.0324

That is easy and straightforward.0330

Let us take a look at a box on a wooden surface.0334

A horizontal force of 8 N is used to pull a 20 N wooden box.0336

A 20 N wooden box that is not its mass but 20 N is its weight.0341

It is moving toward the right along a horizontal wood surface and it tells us the coefficient of kinetic friction is 0.3.0345

Find the frictional force acting on the box.0352

Remember friction is fun which is going to be 0.3 or coefficient of kinetic friction since it is sliding × the normal force or if we have 20 N as its weight,0356

the normal force n must exactly balance that and be 20 N or otherwise it will accelerate off of the floor.0370

Multiplied by 20 N is going to be 6 N.0377

How about the net force acting on the box?0382

The net force we got 8 N to the right and we got our frictional force to the left0387

and that is going to be 8 N – our frictional force we just found is 6 N.0397

Our net must be 2 N to the right.0403

The mass of the box, all of its weight (mg) is 20 N.0408

The net means its mass must be 20 N over the acceleration due to gravity (G) or 20 N / 10 m/s² is just going to be 2kg.0415

Finally is the acceleration of the box.0428

Net force in the x direction we said was 2 N and that is equal to (ma).0432

A is equal to 2 N over our mass 2kg.0441

It is going to accelerate it to 1 m/s² to the right.0446

Static vs. Kinetic friction.0456

Compared to the force needed to start sliding a crate across a rough level floor, the force needed to keep it sliding once it is moving.0459

Well that is going to be less.0466

You probably know that from practical experience but also our coefficient of kinetic friction is less than our coefficient of static friction.0467

Drag forces.0478

An airplane is moving with the constant velocity in a level flight.0480

Compare the magnitude of the forward force provided by the engines to the magnitude of the backward frictional drag force.0483

There is a key term in this question or one highlight, constant velocity.0490

Whenever you say constant velocity, right away you should be thinking acceleration is 0 and the net force = 0.0497

The forward force of the engines must exactly balance the frictional drag force.0505

Frictional drag force they have to be equal otherwise it would be accelerating so they must be the same.0510

Let us do a sled problem.0521

Susie pulls the handle of a 20kg sled across the yard with the force of 100 N as shown.0524

The yard exerts a frictional force of 25 N on the sled.0530

We are asked to find the coefficient of friction and determine the distance the sled travels.0536

If it starts from rest and Susie maintains her 100 N force for 5 seconds.0542

Let us start by drawing our free body diagram.0548

I am just going to sketch them up here.0551

We have an applied force which is 100 N that is an angle of 30°.0555

We have the normal force from the ground.0566

We have the weight down.0569

We have our frictional force (f) is 25 N.0574

If we wanted to we could draw our pseudo free body diagram.0580

We are going to do that down here in the bottom left.0585

Let us see.0589

Let us put it maybe right there.0591

We still have our normal force up.0597

We have our weight down.0601

We have our frictional force to the left.0605

We got to breakup our 100 N force and angle of 30° in the x and y components.0608

The x component of that is going to be 100 cos 30.0613

I will draw that in here 100 cos 30.0617

The y component is 100 sin 30°.0621

We can start our solutions.0628

Let us start by looking at the net force in the y direction to find the coefficient of friction.0631

Net force in the y direction is going to be (MA y) which we know is going to be 0.0637

The net force in the y we got 100 sin 30 + our normal force - mg = 0.0645

The normal force must equal mg -100 sin 30.0660

Our normal force (mg) that is just going to be our 20kg 10 m/s² - 100 sin 30 that is just going to be 50.0669

If 200 - 50 = 150 N for our normal force.0685

Our force of friction then (μ) is our coefficient of friction is our frictional force divided by normal force.0690

Our frictional force it gives us 25 N and our normal force we just said was 150 N.0700

Our coefficient of friction must be 0.167.0707

We want to see how far it travels given this information and that looks like a kinematics problem but I think first I need to find the acceleration.0716

Let us take a look at the net force in the x direction that is going to be 100 cos 30 that is 86 .6 N - our frictional force which we just said was 25 N0726

and that is going to be 86.6 -25 that 61.6 N.0739

Our acceleration is our net force divided by the mass or 61.6 N / 20kg is about 3.1 m/s².0746

I know our acceleration and our initial velocity and I know the time.0762

I can use my kinematic equations now to find how far it goes.0767

Δ x = V initial t + ½ (at)².0772

It is a horizontal problem.0778

Our initial velocity is 0 so this is just ½ × our acceleration 3.1 m/s² × our time 5s² or about 38.8 meters.0780

Let us finish off by looking in an old AP free response problem.0800

What we are going to do is have you go to this link to find the 2007 released mechanics exam free response number 1.0804

I recommend you go to the site, look it over, print it out, even try it on your own first while you hit pause on the video and see how you do.0813

Then come back here hit play on the video and check your answers as you go.0820

If you get stuck use my solution to help get you going again through to the problem.0824

I have that here the 2007 mechanics free response 1.0830

We are given the box and we are applying a force at some angle.0837

We are asked on the figure to draw a free body diagram showing all the forces on the block.0840

We will start with the block and we should be pretty good at this by now.0845

Our forces, we have the normal force.0849

We have its weight.0855

We have this force (f1).0858

And we have our frictional force (ff).0863

They are forces.0871

Part B, just derive an expression for the normal force exerted by the surface on the block.0874

If we want the normal force and I am going to look in the y direction with Newton’s 2nd law.0881

Net force in the y direction must be n + we got to use the y component of (f1) that is going to be f1 sin our angle θ.0887

And down we have - mg and all of that has to equal 0.0901

Therefore, if I solve for the normal force that is going to be mg - f1 sin θ.0907

Part C, derive an expression for the coefficient of kinetic friction between the block and the surface.0922

To do that, I need to find first the force of friction.0929

We will go to Newton’s 2nd law in the x direction.0934

My x is going to be f1 cos θ - our force of friction.0939

And because we know force of friction is μ × the normal force, we can write that as f1 cos θ - μ × our normal force = ma1 our acceleration.0947

Therefore μ × the normal force must equal f1 cos θ –ma1 which implies then that μ must equal f1 cos θ –ma1/ the normal force.0970

We already determined that the normal force is mg - f1 sin θ so this implies then that μ must equal (f1 cos θ - ma1) ÷ mg - f1 sin θ.0996

Let see we now have got the part D.1029

Let us go to the next page to give ourselves some room.1031

It looks like we got some graphs to make.1034

Let us see if we can draw these in here first.1037

For part D, we are going to have a couple of graphs where we have a velocity graph and a position graph.1040

For part D, we have velocity vs. time and we have x vs. time displacement.1061

If we want to graph these from what we already know we must have a constantly increasing velocity because our object is accelerating.1073

The T graph of that if we start from rest and tells us we do must look kind of like that.1083

In our position time graph as it says, it starts at position 0 must have that shape.1092

There are our graphs for part D.1102

Part E, if the applied force is large enough the block will lose contact with the surface.1106

Derive an expression for the magnitude of the greatest acceleration the block can have and still maintain contact with the ground.1113

As we do that one, I am going to realize that when that first breaks contact with the ground, right at that point the normal force is 0.1122

The ground is no longer pushing up on it and that is going to be my condition that is going to help me figure out where this occurs.1130

If the normal force is 0 that implies them that f1 sin θ = mg or f1 is going to be equal to mg / sin (θ).1136

We also know from our f net equation that force in the x direction is going to be f1 cos θ = ma.1151

There is no friction because there is no normal force.1164

Acceleration = f1 cos θ / m.1167

We will call that equation 2 then equation 1 and as we combine them 1 and 2,1174

that means that our acceleration (A) which is f1 cos θ / m must be equal to how we replace f1 with mg / sin (θ).1182

We have mg / sin (θ).1197

We still have a cos (θ) here as well as an (m).1202

We can do a couple of simplifications m / m makes a ratio of 1.1207

Sin / cos is tan (θ).1211

I would write this as g / tan (θ).1213

There is our acceleration right when it starts to leave ground.1223

Right at the point where it is just barely touching the ground g / 10 θ.1226

Hopefully that gets you a good start on friction.1233

Thank you for watching educator.com.1236

I look forward to seeing you again soon. Make it a great day everyone.1238