**Concept:** Overview of Friction

Hey guys. So, in this video I want to introduce friction as well as how to calculate friction, the two types of frictions, we have and in what situations each type will apply. So, let's get started. So, friction happens when two surfaces are in contact, the two surfaces are in contact there has to be a normal force between the two, the basic equation the basic equation for friction is that friction equals mu normal so right away you see how you have to have normal for you to have friction. mu which is a lowercase, lowercase mu, it's a Greek letter, is the quite efficient of friction, coefficient of friction and it has to do with how rough two surfaces are on each other. So, for example, two blocks on ice if you rub them there's not a lot of friction so the coefficient of friction is low mu is a very small number, if you're, if you're rubbing, let's say somehow concrete on concrete, that's weird, but that would be, there would be a lot of friction there. So, mu would be, would have a higher value, okay? There are two types of frictions and they happen in different situations kinetic and Static and it's really important for you to know when do I have kinetic tension and when do I have static friction, at the basic level kinetic friction happens when your velocity is not 0, kinetic means your moving, kinetic means motion, okay? And at the basic level static friction happens when your velocity is 0, static means at rest but it is a little bit more complicated that's why I have these little asterisks here and I'll talk about that in a second. So, the big thing here is this, kinetic friction happens when any object slides skids or slips, any of those words, they are all the same thing. So, if you're sliding on a surface then you will have kinetic friction, like a box, if a box is moving the surface it can only slide on that surface. So, you will have kinetic friction, static friction happens in two different situations, one, the object doesn't move at all. So, it's at rest or two, it moves but it moves without slipping, so the magic word and all of this here is slipping, if it's slipping its kinetic friction, if it's not slipping it is static fiction but the more complicated part here is that it's actually not that the object is moving per se is that the point of contact, the part of the object that touches the surface moves relative to the surface. So, when I said the object is moving I mean the point of contact moves relative to the surface and here the point of contact is at rest relative to the surface. So the best example is the wheel on a car, right? So, if a car moves without, if a wheel rolls without slipping it's because the bottom of the wheel is that rest relative to the ground, right? They move together. So, if a car moves without slipping it's static friction. So, you could move and have static friction and if a car skids through its kinetic friction another way to think about this is, if you have two surfaces that are going like this on each other. So, basically rubbing on each other that's kinetic friction but if you get two surfaces that move together. So, like an object and you put another object on top and then this object in the bottom moves and it causes the object of the top to move with it, they're not rubbing at each other, right? So, that would be static friction. So, you can think of it as slipping or rubbing to determine if it's kinetic or static, alright? So, let's make a kinetic friction, if you're moving with a V to the right your kinetic friction will be to the left.

Now, friction, we usually draw sort of along the floor over here to indicate that that's between the two surfaces, okay? And, this is always, the direction is always going to be opposite to motion, alright? Now, let me go over here, the first situation for static friction is, when the force is not strong enough to get the object moving. So, let's say, I have this object here and I pull with a force F. So, if I'm trying to move this to the right, I'm sure you've experienced this, you pushed something and you just don't have enough strength or you're not pushing hard enough, the reason why the object of you move is because static friction is preventing you from moving, okay? So, the direction here will be opposite to the force trying to move the object, alright? And the idea is that, if you pull hard enough it moves, if you don't pull hard enough it doesn't move, another key idea is that, when you calculate friction, and by the way, I forgot to write these equations here, friction kinetic is mu kinetic normal. Notice how this equation is very similar to the base equation, the only difference is that this is friction kinetic. So, I use the mu kinetic, there are two coefficients of friction one for each type of the two frictions you have, here this equation is a little bit different, it's friction static maximum is mu static normal, okay? And notice I have static, mu static, another thing to notice here is the word maximum, so the idea is that, when you calculate static friction using this equation right here, what you're actually calculating is the maximum amount of static friction, so the idea that static friction is a threshold. So, threshold it's not the actual amount of friction of the object it's the amount of force you have to overcome, amount of friction you have to overcome for this object to start moving. So, if your force F, this guy over here, is less than this threshold the object doesn't move, the object doesn't accelerate, so the acceleration is 0 and if the acceleration is 0 then the actual amount of friction is not your friction static max but it is your pull F and that's because of action-reaction and the fact that this thing is equilibrium what do I mean? what do I mean by that? So, let's say, if this is a force here is 30 but you pull with 20, if you pull 20 to the right friction opposes you with the same 20, it's not enough to get you moving, friction opposes you with the same 20, so it would look like this F, 20, friction, this is my friction max, would be 30, this is, if my friction max was 30 but I would only be opposed by a friction of 20 because friction cannot oppose you with more force than you're pulling, otherwise if this was friction of 30 then it's object would been moving to the left even though you're pulling to the right and that would make no sense, okay? So, if you're not pulling hard enough friction opposes you with the same force that you're pulling, if you are pulling hard enough, if you are pulling harder than static friction then the acceleration will not be 0, and guess what, the object starts moving, the object starts moving. So, in this case of a box it's going to start sliding, at that point you switch from static friction to kinetic friction. So, as soon as the object starts moving switches to kinetic friction and the amount of friction going up against you is kinetic friction, so notice how friction is actually neither one of these cases, the actual amount of friction that's acting in the box is your Fs max, your friction static max, that's just a threshold to decide do you move or do you not move or do you accelerate or do not accelerate, okay? So, it's an important distinction to make. So, friction static Max is a threshold. Now, friction kinetic, friction kinetic, which you would get with this equation right here, is the actual amount of force opposing motion, is the amount of force opposing motion, in every case where you have kinetic friction your actual friction will be your kinetic friction, it is only static friction that the value you can calculate is a threshold not the actual value. So, I hope I've covered that clearly enough, let me know if you have any questions on that kind, it's kind of a tricky point to understand, I mentioned to you how there are two coefficients of friction, mu k, and mu s, mute kinetic and mu static, one for each type of kinetic, for each type of friction and you should note that there's a, there's a few relationships you should know, first of all mu's are always going to be between 0 and 1, coefficient of friction is always between 0 and 1 and the static friction is always either greater or the same of the kinetic friction, so it kind of looks like this. Alright, so kinetic is never bigger than static and both of them always have to be between 0 and 1, 0 means no friction 1 means maximum friction, okay? The reason why static is bigger than kinetic it's because it's harder to get an object moving then to keep an object moving, you might have experienced this as well, when you're pushing something as soon as it starts moving it gets a little easier to push, there's less resistance. So, there's more resistance to getting it to move, it's harder to overcome static friction, that's why the coefficient friction is bigger, than it is to overcome kinetic friction, okay? The second situation here, which is a little bit more complicated, is for static friction, is when an object moves without skidding or slipping, so the first case here, the object didn't move at all, you didn't pull hard enough, let's say, and the second situation here, the object moves but it moves without skidding or slipping, a good example here, like I said earlier, is a car wheel, if it's accelerating this way without, without slipping on the road, right? It just rolls without slipping then you would have static friction or if you have a box stacked on top of each other and they move together you have static friction as well, the direction isn't opposite to motion or it's not opposite to the force, necessarily opposite to the force that's trying to move the object, it depends on what everything else you have, right? And another idea here is that if you try to accelerate too fast you will skid. So, again the idea of static friction as the threshold, it's a threshold because if you don't

pull hard enough it doesn't actually start moving but if you pull too hard, right? So, if a car will spin to fast it's actually going to skid, anyway, let me give you, I want to go over six different examples just draw some free-body diagrams of different sort of special situations for friction and I want us to try to figure out which type of friction acts on the object, okay? So, it's important that you know this stuff here. So, let's say, here I have a block is pushed to the right and it moves that way. So, I have a block here and it's pushed to the right with a force f, let me draw some other forces here, obviously I also have m, g and normal and they're the same here, it says, it's pushed to the right and it moves that way. So, if the object, let's assume it wasn't moving, so it is now not only moving that way but also accelerating that way, okay? And that means that since it's moving to the right my kinetic friction will be to the left this guy's to the left because V is to the right, kinetic friction opposes V not a, right? So, if I wanted to write, If I wanted to write, calculate the acceleration here, I would write sum of all forces equals m, a, this is my acceleration in the x-axis, I'm only going to look at the x-axis and the forces here, are F positive, that's the pulling force, and friction kinetic negative, and I would be able to find a, x. So, that's the free-body diagram and the equals m, a for this, let's look at the second one, the car accelerates to the right without slipping. Now, I'm going to draw the car wheel here and the idea is that the wheel spins this way and in doing that it pushes against the floor so the wheel pushes against the floor so the floor pushes back against the wheel, action reaction, with a, since you are without slipping, I know that this is static friction because kinetic without slipping is static, alright? The car again is going to move this way and it's going to accelerate this way. Notice how friction is not opposed to velocity in this case, right? Static opposed to velocity, if I write the sum of all forces on the car in the x-axis equals m, a, the only force that pushes the car forward or the only force on the x-axis for that matter is the static friction, I'm going to say that this is the positive direction. Now, there are some internal forces in the wheel but we don't have to worry about that, okay? Equals m, ax. Notice how the only force is going to the right. So, this car does accelerate to the right, okay? That's how we determine the direction by looking at all the forces that exist. Now here, block is pushed against the wall the block does not move. So, here's a wall, here's the block, you push the block with a force F, what else is here? I have m, g and it says, the block doesn't move, the block doesn't slide up and down, the block doesn't slide up and down, it has to be because there has there's a force pulling it up otherwise it would fall, and that's what happens, when you push with an F the wall pushes back, action reaction, this force is called normal, I have two surfaces touching with friction with normal between them. So, there's going to be a friction, friction is going to oppose the attempted direction of motion, if there was no friction this would move down. So, friction is going to pull this thing up, this is friction. Now, because the block does not move, this is static friction, okay? If I wanted to write some equations here, I could say that the forces on the axis, this is a complete equilibrium in the x and y axes. So, F and N have to be the same and friction static and m, g would have to be the same as well and this is the forces cancelling the x, this is the forces cancelling the y axis, let's check out another one here, car goes on a flat curve without slipping. So, without slipping tells me that this could not be kinetic friction, and in fact it's to be static friction. So, first I want to give you a top view, here's a car going around, this is the top view, here's a car going around a circle, a flat curve, what causes the car to go towards the center is static friction, okay? Without getting too much into this, what caused the car going to center is static friction, in fact, if you were to turn, if you were to try to turn without friction you couldn't, you would go straight in, right? So, static friction is what causes the car to go towards the center, there's a centripetal acceleration here, that keeps you towards the center while your instantaneous velocity is up, okay? So, this is the top view, if I wanted to do sort of a back view, you're looking at the car from the back, it would look kind of like this, here's the car and the car has an m, g pulling it down, has a normal pulling it up and the car is being pulled towards the center with a static friction and if you wanted to write the f equals m, a here, be the sum of all forces equals m, a, this car is moving towards the center so this is a centripetal force and static friction is what's responsible for that, it is positive because it goes towards the center and that equals m, ac, okay? We won't get into the details of that but that's the setup, two more and these two are sort of the weird ones here, the more complicated ones, I have to stack blocks I have m2 on top of m1 on top of a surface and they're pushed and accelerate together, they move together. So, I push this, make it a pool here, I pull this with an f and guess what, as a result of that this guy's going to accelerate a1, this guy's going to accelerate a2 but these accelerations are the same because they accelerate together, alright? So, let me draw free-body diagram for each of these, let me start at the top because this guy's by itself, it's a little bit simpler m2, m2 is accelerating to the right so there has to be a force pulling into the right, what is it? Well, m2 accelerates to the right because m1 is pushing it to the right, right? To contact the between causes that but they're accelerating together that means that they can't be slipping relative to each other, in fact, what happens is, this guy in going forward brings m2 with it, okay? So, what causes m2 to move forward is the static friction, the static friction between 1 and 2, aright? That's it, for the one, for m1, m1 is pulled this way with a force f and it's got, there's a floor here, it's going to slide across the floor. So, there is a kinetic, slipping, sliding is kinetic, okay? Friction kinetic of 1, okay? 1 with the floor whatever just 1. Now, this might be the only forces you drew but there's another one here that you have to be careful with and people forget this often and if m1 pushes m2 this way, action reaction, m2 has to push m1 back this way. So, there's an fs between 1 and 2 and these two guys are exactly the same force. So, if you want to calculate some stuff here, you'd write the f equals m, a for both blocks and somehow plug in the equations together, depending on what you're being asked for, okay? So, this is the complete free-body diagram from both of these guys, m2 has one force acting in the x axis and m1 as three forces acting the x axis, don't forget this guy here. Now, the last one is stacked blocks, one is push the other one is tied. So, it's kind of like this, there's a wall here, okay? And, I'm going to push here with a force m, one is going to accelerate, so I have an a1 this way but 2 won't, the velocity 0, the acceleration 0, okay? So, it stays in place. So, let's look at what happens with block2, just like in the previous example it's being pushed to the right by the friction between the two guys here. So, friction is pushing it to the right, however, I don't know what kind of friction just yet, this one tries to move it, however there is a tension here that holds it in place. So, because of that the objects will slide relative to each other, the surfaces will slide or slip on each other. So, this has to actually be a kinetic friction between 1 and 2 and that's it, those two forces cancel, for object 1 I have f pulling it this way, object1 was slide on the surface. So, there is a friction kinetic of object1 and this friction, since 1 pushes 2 to the right with kinetic friction then 2 has to apply a kinetic friction to the left on one right here, but this is a friction kinetic between 1 and 2 these two are the same. So, what's different about these guys, I have a tension here, which I didn't have there, and then these frictions here are kinetic instead of static because they're slipping with each other, anyway, that's the basic difference between static friction kinetic friction, we're now going to look into how to calculate these into some problems but I just wanted to give you a really good foundation of trying to figure out what kind of friction I have and what the big differences between those two frictions are, and if you understand that I think you'll find that solving these problems isn't that hard, right? Let me know if you guys have any questions, hope this make sense.

**Concept:** Static & Kinetic Friction

Hey guys. In this video I want to work out a few examples of problems involving both static and kinetic friction to help us better understand how they work, what the difference between them is and how to apply, let's check it out. So, remember, if you pull with the force F on an object. So, for example here, that force F has to overcome, has to be greater than the static friction, the maximum static friction, the threshold of friction for the object to accelerate, okay? You have to overcome your static maximum static friction to accelerate and in case the initial velocity was 0 that means that now that you have an acceleration you would begin moving, okay? So, let's check this out, if your force however is not greater than the threshold Fs max your acceleration will be 0, okay? So, let's say, if your force is less or equal to Fs max then what's going to happen is your acceleration is 0, acceleration 0 means equilibrium, means forces are cancelling, right? Means forces are cancelling therefore your actual friction, which will be static, but the actual amount of friction will be simply your F because these two guys have to cancel, okay? For you to have an acceleration of 0, if your Fs max, let's say, was 40 and I pull with a 30, that's not enough but friction cannot oppose me with 40 because then I would have 40 to the left, 30 to the right and this block would be moving towards friction, which is crazy, right? what happens is that friction opposes me with just the amount of force that I am pulling with, okay? That's because these two numbers have to cancel, these two yellow numbers over here have to cancel so that my acceleration is 0, okay? So, if you're not pulling hard enough acceleration 0 and the actual friction on you is the same as the pulling force, which here I call big F, if you are pulling hard enough the acceleration will not be 0 and this means you're accelerating so you're certainly moving, and then that means you're actually going up against kinetic friction. So, if you pull with a force that is greater than Fs max it means you're moving, you're accelerating to the right and you're accelerating to the right, which means you have some sort of velocity, in this case, let's say you started from rest. So, you're also moving this way. So, friction would be this way but it would be a kinetic friction, okay? This would be a kinetic friction because you are moving, okay? So, it's important to point out that Fs max serves only for you to determine whether or not you're moving but the actual friction will be either your pull, if you don't move, if you don't accelerate or your kinetic friction if you do move, okay? So, let's work out some problems now, I'm going to this first one, I want you guys to try this practice one over here. So, a 10 kilogram block is at rest, the coefficients of friction are these two over here and we're going to use for all these problems g equals tangent. So, it's a little faster, okay? So, m, k is 0.4, Ms is 0.6, I want to know the maximum static friction that can act on the block, maximum static friction that can act on the block is just Fs max, which is mu static normal, right? I had this up here for you guys as well and mu static is 0.6 and then I just have to find normal. So, if I have a block at rest, looks like this, m, g, normal will equal, m, g in this case the mass is 10 times the gravity, the gravity is 10. So, 10, 10, is 100, so this is 100 as well and that's because this object has no acceleration the y-axis just sits there so the forces have to cancel, okay? So, 100 is what goes in here and the maximum static friction this thing is capable of or that could act on the block is just 60, okay? Just plug in, plug and chug.

For Part B I want to know the friction on the block if it was moving, right? So, if the block is moving, this means the velocity is not 0, it has a velocity and I would then have kinetic friction, which is just mu k, normal, and I already have these numbers, mu k is 0.4 and normal is 100. So, friction kinetic is just 40 Newtons. So, this object is moving, it would have be going up against the friction of 40 newtons, alright? So. what I want you to do now, is to pause the video and try to figure this one out here, practice one, I'm going to keep going, hopefully you tried it and I have a 5 kilogram block on the floor. So, there's an m, g of 50, okay? Using gravity as a 10 to make it faster, you figure out that it takes a force of 35 to get moving. Now, force of 35 to get moving a horizontal force like this. So, force to get moving, you need to overcome static friction to get moving, so if it takes you a force of 35 to get moving that 35 is the static friction threshold. So, 35 to get moving is the way of telling that Fs max is your 35, okay? So, and then I also tell you that it takes 25 to keep moving, okay? 25 to keep moving, so the idea of keep moving is that there's some friction here, you're moving, let's say this way. So, there's a kinetic friction here and you want to keep moving, right? If there's kinetic friction and I don't pull on this block it's going to get slower and slower and slower, but if I want to keep moving I need my F to cancel, to exactly cancel my friction, in other words, I want my pulling force F2 equal Fk, this means that they cancel, the acceleration is 0 and in this object keeps moving at the same pace. So, if I tell you that it takes 25 Newtons to keep it moving that means that this is 25 right here, and then that also means that this guy is 25 as well. So, I'm basically saying that Fs max is 35 and Fk is 25. Now, if I expand Fs max, which is mu static, normal, I am able to solve for mu static, same thing here, if I expand this into mu kinetic, normal equals 25 I am able to solve from mu kinetic, all I need to know is normal, okay? if you look at this block over here, m, g is going down with 50, there are no other forces in the y axis. So, normal has to cancel out that 50, normal is 50 over here, because of m, g. So, I just have to plug this in, N is 35 over, I'm sorry, mu normal, mu static is 35 over normal, which is 50 and this gives us a 0.7 and mu kinetic is 25 over normal, which is 50, this gives us a 0.5, okay? And that is the, these two here are the two final answers, hopefully you got it, let me go on to example two. So, here it says, you pull on the block in example one, right here, with various horizontal forces F, for each value of F, fill the cells below. So, for each one of these values here I want to fill in these cells and see what's happening, okay? So, this is a 10 kilogram block that is at rest, right? it says it's at rest and the coefficients of friction are 0.4 and 0.6, okay? So, the initial velocity here is 0 and the coefficient of friction mu k is 0.4 and mu static is 0.6 and I'm going to pull on it this way with various forces. Now, the first force is 0, which means an actually not pulling at all. So, if you don't pull obviously, this one's kind of silly, it doesn't move, okay? What kind of friction is working up against it? Remember, if you're not moving there's no kinetic friction and static friction is a friction that is, in case you don't move, going up against your F, right? So, let's say pull with an F of 10 and that's not enough to move, then the static friction will exactly cancel that 10 right there, okay? So, if I'm not, if I'm pulling with 0 the friction that's opposing me is 0, which means it doesn't exist. So, there's no friction, okay? Non or maybe like non-applicable, there's no friction here, okay? And the actual friction against you is 0 because there's nobody, there's no friction going up against you and the acceleration is 0 as well because you don't move, everything just cancels, alright? Now, I'm being pulled with 30, okay? And by the way, I want to I want to rework this here, remind you that the kinetic friction on this, we calculated to be 40 and the maximum static friction, we calculated to be 60. So, this means that for this object to move you have to pull with at least 60, here I'm pulling with 30. So, I got a 30 and 30 is not enough to overcome 60. So, this object will not move, okay? And it doesn't move because it's going up against static friction that prevents it from moving, how much static friction? It's going up against 30 Newtons of static friction, again, if I pull with 30 and it doesn't move this thing has to oppose me with 30 so that they both cancel, this is not of 60 here, if this was a 60, 30 to the right 60 to the left you'll be accelerating to the left and that makes no sense, this 60 is just a threshold and it doesn't accelerate, right? Now, I pull 50, again not enough, I have to pull with at least 60 for this thing to start moving, still not enough, still going up against static friction and the static friction now instead of 30 it's 50 because it matches my pull. Remember, I wrote here, that if you're not pulling hard enough the actual friction is the same as the pulling force, okay? Lighting that up, and it still doesn't accelerate. Now, we're going to move because I'm pulling harder, I'm pulling with a 70, okay? So, I can write here, since F is greater than the Fs max this object will accelerate, acceleration will not be 0, it will be some acceleration to the right, this object will move. Now, that it's moving guess what, because it's moving, okay? Acceleration is not 0 and we're going to be going up against kinetic friction, it's like the friction switches instantaneously from one type to the other once you break that barrier, let's see what happens. So, here you're actually going up, kinetic friction, how much friction not 70, right? It only counters your friction if you're not moving going up against static friction but actually the constant number for kinetic friction it's always the same, 40 Newtons, okay? The acceleration will not be 0 and we can calculate it using F equals m, a, sum of all forces equals m, a. Now, just to draw this real quick, what's happening here is I'm pulling with an F of 70 and I'm going up against now a kinetic friction since I overcame static friction of 40, okay? So, the sum of all forces will be positive 70 plus negative 40 because they're going opposite directions, I'm saying they are going to the right as positive, the mass is 10 and if I solve for acceleration here, I get 30 equals 10a and I get that the acceleration is 3 meters per second squared, okay? And that goes right here, cool? So, hopefully this example helps illustrate basically what it takes to get moving, what happens when you don't move, what happens when you do move, okay? One last point here and I want you guys to try this practice problem, sometimes you're only going to be given one coefficient of friction and what you're supposed to do is basically assume that that number is just both of them, right? The coefficient of friction could be the same, you just can't have the kinetic be greater than the static but they could be the same, so the problem says the coefficient of friction is 0.5 but it doesn't say if it's static or kinetic, you'll assume that it's both. So, in this case they are the same, right? If two coefficients are given, sometimes you get this, they'll give you two coefficients but what they won't tell you, the problem will tell you, which one is which, and then they're given but not identified. So, you just have to know that the greater one will be the static, okay? The static coefficient of friction, for example, here I'm giving you a 0.5 and 0.6. So, I'll just let you know this means that the 0.6 is the static and the 0.5, ups I wrote 0.5 here, 0.6 is the static, and the 0.5 is the kinetic, okay? Because the kinetic can't be greater than the static or either the same or the static is bigger, alright? So, I want you guys to try this practice problem out and hopefully you get it.

**Problem:** You push horizontally on a 10-kg box so that it moves on a flat surface with a constant 2 m/s. The coefficients of friction between the box and the surface are 0.5 and 0.6.

(a) What force is needed to keep the box at 2 m/s?

(b) If you stop pushing, what will be the acceleration of the box?