1

concept

## Static Friction & Equilibrium

8m

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Hey, guys. In the last couple videos, we talked about kinetic friction and this video. We're going to talk about the other type of friction, which is called static. They have some similarities, but static friction is a little bit more complicated, so let's check it out. So remember what we talked about kinetic friction. We said that this happens when the velocity is not equal to zero. You push a book, and it's moving and status are kinetic. Friction tries to stop that object and bring it to a stop right. Static friction happens when the velocity is equal to zero. Imagine this book is like, really, really heavy, and it's at rest on the table. What static friction tries to do is it tries to prevent an object from starting to move. So imagine this book is really, really, really heavy. You try to push it, and no matter how much you push, the book doesn't move. That's because of static friction. So the direction of kinetic friction right was always opposite to the direction of motion. Right? So you push this book across the table, it's going to the right kinetic friction opposes with to the left static friction is kind of similar, except the direction is going to be opposite to where the object wants to move or would move without friction. So this heavy book, right? You're pushing it without friction. It would move to the right, so static friction is going to pose you by going to the left. So this is FS. Lastly, let's just talk about the formulas. So the equation for F. K. The kinetic friction is the coefficient of kinetic friction times the normal for static friction. It's very similar, So we're just going to use the coefficient of static friction times the normal. We'll go back to this in just a second here. This coefficient of static friction is really just another number just between zero and one just like him, UK is. One thing you should know about this coefficient, though, is that it's always going to be greater than UK. Normally they're going to be given to you. Uh, you know, an example of this is that we have 0.6 and 0.3, So, actually, let's just get to the example right now. We're gonna go ahead and come back to this in just a second. So we've got this 55.1 kg block that's at rest on the floor like this. Now we're giving the coefficient of static and kinetic friction. Like we just said, This is my us, and we'll see that it's actually greater than UK. And so what we're trying to do in this problem is we're trying to figure out the magnitude of the friction force on the block when we push it with these forces. So this F is 20 and this f is 40. So let's go ahead and get started. I'm just gonna draw really quick sketch of the free body diagram. So we have our MG that's downwards. We've already got our apply force, and then here's our normal. And then whether this object is moving or trying to move, we know that friction is going to oppose that by going to the left. We just don't know what this kind of friction is. So which equation we're gonna use our We're going to use Mu uh f k. Or we're going to use F s. Well, if you think about this, this block is at rest on the floor, which means that the velocity is equal to zero. So we said that the velocity is equal to zero. We're going to use the static friction, um, formula here. So our f s is equal to use static times the normal. So that means our friction force here is going to be 0.6. Remember, that was our arm us times the normal force. Well, if this block is only sliding horizontally and we have two forces in the vertical, that means that they have to cancel. So that means that our and is equal to mg. So that just means that we're gonna use 5.1 times 9.8 and you'll get a friction force that's equal to 30 Newton's. So let's talk about this. You're pushing with 22 the rights. But the friction force that we calculated was 30. So even though you're pushing to the rights, friction force would win and the book would actually start accelerating to the left in the direction of friction. That's crazy. Doesn't make any sense. So what's happening here? What happens is that when we use this formula, this mu s times the normal, this is actually called a threshold this is basically this is basically just the amount of force that you have to overcome to get an object to start moving This mu s times normal is the maximum value of static friction. So we do is we actually call this F. S Max, and this is equal to mu s times the normal. So when we go back here, we have to do Is this static friction formula we use is actually the maximum static friction. This is basically just the threshold that we have to overcome in order to get an object to start moving. So what happens is this threshold is not always the actual friction that's acting on an object to determine whether we're dealing with static friction versus kinetic friction. What we always have to do in problems is we have to compare the forces to that static, that static friction threshold. Basically, we have to figure out whether our force F is strong enough to get an object moving. There's really just two options. You either don't or you do So let's talk about those. If you're f is not strong enough to get an object moving right, that means your force is less than or equal to that static, that maximum static friction. At that point, the object just stays at rest, right? It's not enough to get it moving. And if the object stays at rest and the friction is just gonna be static friction. So basically, what happens is if you haven't yet crossed this threshold, this is kind of just like a number line here where you know you have increasing force, then your static friction basically always has to balance out your force. What I mean by this is that you're if you're pushing with 10, your friction can't oppose you is stronger than your pulling. So that means that the static friction in this case is just 10. If you're pulling harder with 20 static friction opposes your poll with 20. And if you're pulling with 30 static friction just opposes your poll with 30. It always knows how much you're pushing, and it always basically balances out your force so that the object stays at rest in the acceleration zero. Now, what happens if you actually do overcome that threshold? Basically, if you have strong enough force to get the object moving in your force is greater than Fs Max. And what happens here is that the object starts moving and if it starts moving, then your friction switches from kinetic and actually sorry from static, and it becomes kinetic friction. So what happens here is that this kinetic friction we already know is just equal to U K times the normal. So let's go back to our problems here and figure out what's going on. So what we're doing here is we're basically comparing our F two r f s Max. That's how we figure out which kind of friction we're dealing with. So our f is 20 and this is actually less than F. S max, which is equal to 30. So what that means is that our friction force actually is going to be static friction. And it's just basically going to balance out our poll. So are static. Friction is going to be 20. So this static friction here is going to be 20 Newtons. Even though your maximum is 30 now in part B, we don't need to recalculate the maximum. We are. You know that Fs Max is 30 but now we're actually pulling with 40. So basically what happens here is that our f is equal to 40. It's greater than your f s Max, which is equal to 30 which means that the friction becomes kinetic friction. And so we can calculate this by using U K times the normal. So basically, our kinetic friction force is going to be 0.3. That's the coefficient that we were given times 5.1 times 9.8. And if you work this out, you're going to get 15 Newtons. So what happens is here we've actually crossed that maximum static friction threshold. And so therefore, the friction that's opposing this book is actually gonna be kinetic, and it's going to be 15 Newtons. So those are the answers, right? We have 20 Newtons when you're not pulling hard enough and then 15 once you've actually overcome. So basically, what this means is that once you pass this threshold, it actually doesn't matter how hard you pull, because the friction force that supposing you is just gonna be f K. And so this is just gonna be 15. If you if you were to pull a little bit harder with 50 Newtons, it doesn't matter because this mu k times. The normal is it's just a fixed value. So even though you're pulling with 50 kinetic friction would still oppose you with 15. All right, so that's it for this one, Guys, hopefully I made sense. Let me know if you have any question.

2

example

## Static & Kinetic Friction

4m

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Hey, guys. So let's check out this problem here. I have a 5.1 kg block, and what I want to do is I want to calculate the force that I need to get the block moving. And then I wanted to calculate the force. You need to keep it moving at constant speed. So there's really two different situations here that I'm gonna draw throughout the diagrams for you got 5.1 kg block on the ground. I'm going to be pushing it with some mysterious force here to get it moving. And then in the second case, I've got the block like this. Except now it's moving with some constant speed. So V equals constant here, and I want to figure out how far or how, how hard he to push it to keep it moving at constant speed. So there's two different forces here. All right, so we want to draw a free body diagrams. Let's just draw all the other forces that are acting on these blocks. I've got my weight forest, that's mg. And then I've got the normal force. All right, so in this particular case, the first one where you're trying to push it. You're trying to push it with enough force to get the block moving. So therefore, it's not actually moving just yet, Which means the kind of friction that you're going up against is going to be static friction. And the moment where you actually get the block moving, as we've seen in the previous videos, is that's equal to the F S. Max threshold. Then when you finally actually get it moving when it's moving with some constant speed now you're going up against kinetic friction because the velocity is not zero. So that's really the difference between these two diagrams here in one, you're going up against FS Max and then the other one, you're going up against kinetic friction. All right, so how do we then calculate these forces? Basically just use our F equals M A. So we have our f equals m a here. Now we just pick a direction of positive. It'll be to the rights for both of these diagrams here. So you've got f minus F s Max and then what's the acceleration? Well, the moment right, when I get the block to move, the acceleration is still equal to zero. So we're going to use a equals zero for this. Even though we're actually getting the block to move, we want to figure out the force that you need right before that happens when f is equal to F S, Max. So you have f is equal to F s. Max. Remember that has an equation that's mu static times the normal force. And so the normal force, if you're just looking at a block sliding horizontally, just going to be equal to mg, so basically F is equal to mu static times mg, and so this is going to be 0.7 times 5. times 98 and you're gonna plug this in, you're gonna get 35 Newtons. So basically, assuming that this is the coefficient of static friction, you have to push this block with at least 35 Newtons to get it moving. So then we can figure out the other force here by using basically the exact same method. So now we're gonna do the sum of all forces equals mass times acceleration. Here we know the acceleration has to be zero because the velocity is going to be constant. So your forces are F except now. It's not F s, Max, you're just using f k. So those have to cancel. So your f is equal to F k, which is mu k times normal. So your force is equal to mu k mg, right? Just like we had before. And so this is going to be 0.5 times 5. times 9.8. So now if you plug this in, you're going to get 25 Newtons. So let's talk about that. So we got these two different numbers here, which means that our answer is actually answer choice. See, it takes 35 Newtons to get this block to start moving. Once it actually is moving with some velocity. Then it only takes 25 Newtons. You don't have to push as hard to keep it moving at constant speed. So because the mu static is always greater than or equal to mu kinetic, what that means is that always is harder to get something moving than it is to keep it moving. This is something you've definitely experienced before in everyday life. You push something really, really heavy and you have to push really hard at first. But once you actually get it to move, basically the kinetic friction coefficient decreases. And so it's easier to keep it moving. You don't have to push as hard. All right, so that's it for this one. Guys, let me know if you have any questions.

3

example

## Preventing a Block From Moving

4m

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All right, guys, let's check this one out. We have this 15 kg block that's at rest were given the coefficient of static friction. And we want to figure out how hard you have to push down on this block in order to keep it from moving. So, basically, let's sketch this out. We have a 15 kg block, this on a flat surface, right? And basically what we're gonna be doing is pushing down. I'm gonna call this Force f down. That's what we're trying to find. And we want to push down hard enough so that we can generate enough friction so that a horizontal force which we know is 300 cannot get it moving. All right, So we want to do is we want to, uh, first start off with the free body diagram. So let's go ahead and do that. So we've got our box, we've got RMG that's downwards. And then we also have any applied forces. We know there's two r f down is what we're trying to solve for, and we also have an applied force. It acts to the rights. This is f equals 300. Now there's also going to be some normal force because they're on the floor. And then finally, what happens is without friction. The box would start moving to the right. But if we're going to keep this thing for removing, that's because there has to be some static friction that is opposing this. So there is some friction here that is opposing that, right? So it actually goes. So that actually brings us to the second step, which we've already just talked about. Here. We have to determine whether this friction is static or connected based on the problem text or by looking at all the forces involved. And what we've seen here is that we're pushing down on the block to keep this box from moving basically. So what happens is we know that this friction is going to be static. All right, so let's go ahead and now, right? R f equals m A. So I'm gonna write out my ethical dilemma here. I'm going to just pick a direction of positive up and to the rights now. Usually we would start with the X axis, but because we're looking at a force that's in the Y axis, we're gonna go and start with our Y axis first four, uh, first. So we've got our some of all forces equals m A y. Now this box isn't going up or down in the vertical axis. We know the acceleration is going to be equal to zero. So therefore we expand our forces, remember, our normal is positive minus M g minus f down, and this is equal to zero. Those forces have to cancel. So here's f down. And if I go ahead and just solve for this variable over here by bringing it to the other side and flipping the equation around f down is really just equal to the normal minus mg. Okay, so this f down the force that I need to get this object to prevent this object for moving is going to be equal to the normal minus mg. But the problem is, I actually don't know what this normal force is. So to figure this out whenever I get stuck in one axis, I usually just go to the other access to solve. So I'm gonna go to the X axis forces to now solve for this. So I've got f equals M A and the X axis. So what are our forces? We have our f, which is the 300 minus your F s. And what's the acceleration? Well, here's the kicker, guys, if we're trying to figure out how much we need to push on the block to prevent it from moving, we're basically trying to figure out what is the minimum force that we need so that the static friction exactly balances out the 300. And so what happens is this is gonna be F s, Max. So the minimum force is going to be where the F S Max is going to balance out the 300 if 300 was just a little bit more than it would actually get the object to start moving. So this FS here is actually maximum static friction. And so because of that, because the object doesn't move, then that means that the acceleration has to be equal to zero. These forces have to balance. So that means that your f is equal to your F S max, which is equal to remember mu static times the normal. So remember we came to the X axis because we were stuck and we wanted to figure out what that normal force is. Now we can figure it out. Our normal forces really just going to be equal to your f divided by mu static. So your f is 300 your mu static 0.7, you'll get a normal force of 429. So your normal force is 429. Basically, if the normals 429 then your f S Max is going to be 300 to balance out the 300 that you're pushing it with. So now we have our normal force here, which means that F down is just equal to minus your mg, which is 15 times 9.8. So if you guys go ahead and plug this in your calculators, you're gonna get 282 Newtons. So if you look at our answer choices, that's answer choice, see? All right, guys, let's take this one

4

Problem

A 36N force is needed to start a 7.0 kg box moving across the floor. If the 36.0 N force continues, the box accelerates at 0.70 m/s^{2}. What are the coefficients of static and kinetic friction?

A

μs = 0.52 ; μk = 0.64

B

μs = 0.64 ; μk = 0.64

C

μs = 0.52 ; μk = 0.45

D

μs = 0.45 ; μk = 0.32

Additional resources for Static Friction

PRACTICE PROBLEMS AND ACTIVITIES (4)

- A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the...
- A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the...
- A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the...
- A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if...