Start learning today, and be successful in your academic & professional career. Start Today!

• ## Related Books

### Start Learning Now

Our free lessons will get you started (Adobe Flash® required).

### Membership Overview

• *Ask questions and get answers from the community and our teachers!
• Practice questions with step-by-step solutions.
• Track your course viewing progress.
• Learn at your own pace... anytime, anywhere!

### Friction

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
• Question 1 0:13
• Question 2 0:47
• Question 3 1:25
• Question 4 2:26
• Question 5 3:43
• Question 6 4:41
• Question 7 5:13
• Question 8 5:50

### Transcription: Friction

Hi everyone and welcome back to Educator.com.0000

This mini-lesson is going to be problem-solving with friction where we are going to go through the APlusPhysics' first page on friction and the link to that is listed down below.0002

Number 1 -- Which vector diagram best represents a cart slowing down as it travels to the right on a horizontal surface?0012

If its velocity is to the right and it is slowing down, it must have an acceleration vector to the left, which means the net force must be to the left.0021

I am going to look for these, where I have a net force to the left.0030

Number 1 has a net force to the right; Number 2 has a friction that is a little bit bigger than that forward force and that has a net force to the left, so that could be slowing down as the object moves to the right.0034

Number 2 -- A student and the waxed skis she is wearing have a combined weight of 850 N.0046

The skier travels down a snow-covered hill and then glides to the east across a snow-covered horizontal surface.0052

Determine the magnitude of the normal force exerted by the snow on the skis as the skier glides across the horizontal surface.0059

In that case, we are going to have weight down, normal force up and those must be balanced because the skier is not accelerating vertically, so the normal force is equal to the weight, which is 850 N.0067

Following on more on this problem -- Calculate the magnitude of the force of friction acting on the skis as the skier glides across the snow-covered horizontal surface.0086

The key here is to find the coefficient of friction and it tells us its waxed skis on snow, so we will need to look that up and because they are gliding across the surface we are going to use the kinetic coefficient of friction.0094

When I look that up, I find out that the kinetic coefficient of friction for waxed skis on snow is somewhere around 0.05, so the frictional force then, is going to be the coefficient of friction times the normal force...0110

...which is 0.05 × 850 N, which gives me a value of about 850 × 0.5 or 42.5 N.0126

A little bit further -- The coefficient of kinetic friction between a 780 N crate in a level warehouse floor is 0.2, so μ = 0.2.0144

Calculate the magnitude of the horizontal force required to move the crate across the floor at constant speed.0156

Let us draw a free-body diagram (FBD). We have the normal force up, the weight down -- and those must be balanced because it is not accelerating up or down, and if it is moving at constant speed horizontally, we must have some applied force and some frictional force and those must be exactly balanced as well.0163

If we want the magnitude of the horizontal force, that is going to be the same as the frictional force.0183

All we have to do is calculate the frictional force, so our applied force is the frictional force, which is μ times the normal force or 0.2 times -- well, our normal force has to be equal to the weight because it is an equilibrium in the y-direction as well.0188

That is 780 N or 0.2 × 780 N, which gives me 156 N as the applied force required to keep that crate moving at a constant speed.0207

Next up we have an ice skater applying a horizontal force to a 20 kg block on a frictionless, level ice, causing the block to accelerate uniformly at 1.4 m/s2 to the right.0224

After the skater stops pushing the block, it slides on to a region of ice covered with a thin layer of sand and there the coefficient of kinetic friction is 0.28.0237

Let us find the magnitude of the force applied to the block by the skier.0247

If it is a 20 kg block -- mass equals 20 kg, then its acceleration is 1.4 m/s2, the force then must just be the mass times the acceleration or 20 kg × 1.4 m/s2, which I believe is right around 28 N .0252

Moving on -- Same problem, but now on the diagram below we are asked to draw a vector to represent the force applied to the block by the skater beginning at (A) and using the scale of 1 cm = 10 N.0280

If 1 cm = 10 N and we need to go 28 N, then that is going to be 2.8 cm, so we need to start at (A) and draw our vector with a length of 2.8 cm to the right, so there is our force.0293

Number 7 -- Determine the magnitude of the normal force acting on the block.0314

In the vertical direction we have the normal force up in the gravitational force, the weight down and those must be equal, otherwise it would accelerate up or down.0318

The normal force is equal to the weight, which is 20 kg × 9.8 m/s2 or 196 N.0328

One last question here -- Calculate the magnitude of the force of friction acting on the block as it slides over the sand-covered ice.0347

Our force of friction is our coefficient of friction times the normal force, which is going to be 0.28 and we just found the normal force was 196 N, so that is going to be 0.28 × 196, which should be right around 55 N.0356

That concludes page 1 of our problem sheet on friction.0384

If everything went well -- Great! -- Keep going on to the AP problems and if it did not, now is a great chance to go back and review that video.0388

Thanks so much for your time and make it a great day.0395