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