WEBVTT physics/ap-physics-1-2/fullerton 00:00:00.000 --> 00:00:04.000 Hi everyone and welcome back to Educator.com 00:00:04.000 --> 00:00:06.000 This lesson is going to be about friction. 00:00:06.000 --> 00:00:13.000 Now our objectives are going to be to define and identify frictional forces. Yeah friction! 00:00:13.000 --> 00:00:21.000 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. 00:00:21.000 --> 00:00:27.000 So let us dive in. Friction is a force that opposes motion. 00:00:27.000 --> 00:00:34.000 Kinetic friction is a type of friction that opposes motion for an object that is sliding along another surface. 00:00:34.000 --> 00:00:36.000 Kinetic friction is sliding friction. 00:00:36.000 --> 00:00:41.000 Static friction acts on an object that is not sliding. 00:00:41.000 --> 00:00:59.000 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. 00:00:59.000 --> 00:01:05.000 Bigger coefficients of friction, bigger μ -- so you are going to have more friction between the two surfaces. 00:01:05.000 --> 00:01:15.000 Imagine something like -- let us say really flat dress shoes on ice -- very slippery -- compared to two pieces of sandpaper. 00:01:15.000 --> 00:01:19.000 The sandpaper is going to have a much higher coefficient of friction. 00:01:19.000 --> 00:01:26.000 The normal force acting on the object is the other item that determines the magnitude of the frictional force. 00:01:26.000 --> 00:01:45.000 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. 00:01:45.000 --> 00:01:48.000 And you have probably observed that before. 00:01:48.000 --> 00:02:00.000 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. 00:02:00.000 --> 00:02:05.000 Once you have it moving however, now you are into the regime of kinetic friction that usually takes a little bit less force. 00:02:05.000 --> 00:02:10.000 Kinetic friction is usually smaller than static friction. 00:02:10.000 --> 00:02:17.000 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. 00:02:17.000 --> 00:02:27.000 So the coefficient of friction μ -- we are going to talk about the coefficient of kinetic friction μk or coefficient of static friction μs. 00:02:27.000 --> 00:02:32.000 So this coefficient of friction is really the ratio of the frictional force and the normal force. 00:02:32.000 --> 00:02:38.000 Coefficient of friction given by the force of friction divided by the normal force. 00:02:38.000 --> 00:02:43.000 It depends only on the nature of the surfaces that are in contact. 00:02:43.000 --> 00:02:54.000 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. 00:02:54.000 --> 00:02:59.000 Rubber on dry concrete has a kinetic coefficient of 0.68. 00:02:59.000 --> 00:03:02.000 But on static, when it is not sliding, it is 0.9. 00:03:02.000 --> 00:03:12.000 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. 00:03:12.000 --> 00:03:16.000 This is the reason for antilock brakes in car. 00:03:16.000 --> 00:03:27.000 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. 00:03:27.000 --> 00:03:35.000 You could look up the coefficient of friction for many different materials. 00:03:35.000 --> 00:03:38.000 So let us take a look at some examples and try and determine which regime of friction they are in -- kinetic or static. 00:03:38.000 --> 00:03:53.000 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. 00:03:53.000 --> 00:04:01.000 A refrigerator at rest that you want to move -- at rest implies not sliding -- that one is static. 00:04:01.000 --> 00:04:09.000 A car with the tires rolling freely -- well, we just talked about that -- not skidding, therefore static. 00:04:09.000 --> 00:04:19.000 If you are skidding across pavement though, you are going to use kinetic coefficient of friction. 00:04:19.000 --> 00:04:24.000 Let us take an example here. A car's performance is tested on various horizontal road surfaces. 00:04:24.000 --> 00:04:31.000 The brakes are applied causing the rubber tires of the car to slide along the road without rolling. 00:04:31.000 --> 00:04:33.000 They are sliding. 00:04:33.000 --> 00:04:42.000 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? 00:04:42.000 --> 00:04:48.000 Well, first thing we need to realize is, is if we are sliding, we are looking for the kinetic coefficient. 00:04:48.000 --> 00:05:04.000 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. 00:05:04.000 --> 00:05:07.000 Another example is we have a block on an incline. 00:05:07.000 --> 00:05:12.000 The diagram shows the block sliding down a plane, inclined at angle θ -- there is θ. 00:05:12.000 --> 00:05:21.000 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? 00:05:21.000 --> 00:05:32.000 Well, here you have to remember that the coefficient of friction depends on the nature of the surfaces. 00:05:32.000 --> 00:05:35.000 In this case the surfaces have not changed. 00:05:35.000 --> 00:05:48.000 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. 00:05:48.000 --> 00:05:55.000 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. 00:05:55.000 --> 00:05:58.000 We have a nice direct relationship. 00:05:58.000 --> 00:06:04.000 Force of friction equals the coefficient of friction μ times the normal force. 00:06:04.000 --> 00:06:14.000 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. 00:06:14.000 --> 00:06:19.000 While we are here and talking about friction, let us come back to terminal velocity. 00:06:19.000 --> 00:06:22.000 Objects following through Earth's atmosphere experience a force of friction that we call air resistance. 00:06:22.000 --> 00:06:28.000 That is a drag force and as the object goes faster, there is even more of that. 00:06:28.000 --> 00:06:34.000 Eventually, an object gets going fast enough that the force of friction balances the force of gravity on the object. 00:06:34.000 --> 00:06:37.000 When that happens you reach what is known as terminal velocity. 00:06:37.000 --> 00:06:41.000 The net force is zero -- you do not gain any more speed -- the longer you fall. 00:06:41.000 --> 00:07:03.000 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. 00:07:03.000 --> 00:07:15.000 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. 00:07:15.000 --> 00:07:30.000 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. 00:07:30.000 --> 00:07:36.000 No net force. No acceleration. Constant velocity. 00:07:36.000 --> 00:07:39.000 Let us take a look at another example -- finding the frictional force. 00:07:39.000 --> 00:07:47.000 In the diagram, we have a 4 kg object accelerating at 10 m/s² on a rough horizontal surface. 00:07:47.000 --> 00:07:52.000 Find the magnitude of the frictional force, (Ff), acting on the object? 00:07:52.000 --> 00:07:55.000 Let us start with our FBD. 00:07:55.000 --> 00:08:14.000 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. 00:08:14.000 --> 00:08:21.000 All of my forces line up with the axis, so I do not need to draw a P-FBD. 00:08:21.000 --> 00:08:30.000 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. 00:08:30.000 --> 00:08:35.000 I am going to replace now, net force in the x-direction with all the forces acting in the x-direction. 00:08:35.000 --> 00:08:53.000 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. 00:08:53.000 --> 00:09:03.000 Or 50N minus force of friction equals 40 kg m/s² which is a Newton. 00:09:03.000 --> 00:09:11.000 Therefore, force of friction must be equal to 10N. 00:09:11.000 --> 00:09:19.000 Let us take a look at another example. Here we have a box on a wood surface. 00:09:19.000 --> 00:09:29.000 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. 00:09:29.000 --> 00:09:40.000 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. 00:09:40.000 --> 00:09:52.000 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. 00:09:52.000 --> 00:10:00.000 We have a force to the right, an applied force of 8N and we must have some frictional force to the left. 00:10:00.000 --> 00:10:16.000 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. 00:10:16.000 --> 00:10:25.000 μ is 0.3 and our normal force in this case -- if you look in the y-direction, that must be equal to mg. 00:10:25.000 --> 00:10:33.000 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. 00:10:33.000 --> 00:10:40.000 The normal force here must be 20N, so 0.3 x 20 or 6N. 00:10:40.000 --> 00:10:45.000 It also asks for the net force acting on the box. 00:10:45.000 --> 00:10:57.000 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. 00:10:57.000 --> 00:11:00.000 How about the mass of the box? 00:11:00.000 --> 00:11:15.000 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. 00:11:15.000 --> 00:11:18.000 And finally, the acceleration of the box. 00:11:18.000 --> 00:11:29.000 Well, acceleration is net force divided by mass. We just determined the net force here was 2N. 00:11:29.000 --> 00:11:45.000 We determined the mass was 2 kg, so the acceleration must be 1 m/s². 00:11:45.000 --> 00:11:50.000 Let us take a look at an example where we explore the difference between static and kinetic friction? 00:11:50.000 --> 00:12:07.000 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. 00:12:07.000 --> 00:12:14.000 Kinetic is less than static, therefore it is going to be less. 00:12:14.000 --> 00:12:17.000 Let us take a look at a drag force. 00:12:17.000 --> 00:12:29.000 An airplane is moving with a constant velocity in level flight. We have an airplane moving with constant velocity in level flight. 00:12:29.000 --> 00:12:38.000 Compare the magnitude of the forward force provided by the engines -- we typically call that thrust -- to the backward frictional drag force. 00:12:38.000 --> 00:12:41.000 Well, let us draw our FBD. 00:12:41.000 --> 00:12:50.000 There is our airplane. We have some thrust forward. 00:12:50.000 --> 00:13:01.000 We have a drag force backwards -- force that is pulling it up, we call lift and we have its weight. 00:13:01.000 --> 00:13:09.000 Now if it is moving at constant velocity in level flight, everything must balance out -- they must be equal. 00:13:09.000 --> 00:13:20.000 So the force of the thrust or the force of the engines must balance the force of drag, therefore they must be equal. 00:13:20.000 --> 00:13:25.000 Another example, have Suzie over here pulling a sled. 00:13:25.000 --> 00:13:32.000 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. 00:13:32.000 --> 00:13:36.000 The yard exerts a force of 25N on the sled due to friction. 00:13:36.000 --> 00:13:48.000 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. 00:13:48.000 --> 00:13:59.000 Well, let us start off with our FBD -- y, x. 00:13:59.000 --> 00:14:00.000 There is our sled. 00:14:00.000 --> 00:14:06.000 We have weight down -- force of friction to the left 25N. 00:14:06.000 --> 00:14:18.000 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. 00:14:18.000 --> 00:14:25.000 So my P-FBD -- I will draw that down here. 00:14:25.000 --> 00:14:30.000 Right away, let us put in our red vectors -- the ones that are already lined up with the axis. 00:14:30.000 --> 00:14:38.000 We have mg, force of friction, and normal force. 00:14:38.000 --> 00:15:00.000 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. 00:15:00.000 --> 00:15:04.000 Now we can go start to solve our questions. 00:15:04.000 --> 00:15:06.000 Find the coefficient of friction between the sled in the yard. 00:15:06.000 --> 00:15:15.000 I am going to start by writing Newton's Second Law and I am going to look in the y-direction-- equals may. 00:15:15.000 --> 00:15:19.000 I am going to replace the net force in the y-direction with all the different things I see here. 00:15:19.000 --> 00:15:31.000 I have 100 sine 30 and that is going to be 50, plus the normal force, Fn minus mg. 00:15:31.000 --> 00:15:41.000 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. 00:15:41.000 --> 00:16:01.000 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. 00:16:01.000 --> 00:16:18.000 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. 00:16:18.000 --> 00:16:27.000 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. 00:16:27.000 --> 00:16:32.000 Well to do that, it would be nice to know the acceleration of the sled in the x-direction. 00:16:32.000 --> 00:16:36.000 Let us go to Newton's Second Law in the x-direction. 00:16:36.000 --> 00:16:47.000 Fnetx is going to be 86.6N 100 cosine 30 minus the force of friction, 25N or 61.6N. 00:16:47.000 --> 00:17:04.000 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. 00:17:04.000 --> 00:17:12.000 So, if we want to find out how far it goes in that 5 s-interval, we can go back to our kinematics. 00:17:12.000 --> 00:17:21.000 Δx = V initial T + 1/2AT² and V initial here -- if it starts from rest is 0. 00:17:21.000 --> 00:17:42.000 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. 00:17:42.000 --> 00:17:45.000 Hopefully that gives you a good start with friction, the coefficient of friction. 00:17:45.000 --> 00:17:49.000 Thanks for watching Educator.com. Make it a great day.