Static Friction - Online Tutor, Practice Problems & Exam Prep

Static Friction & Equilibrium

Guys, in the last couple of videos, we talked about kinetic friction. In 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 when we talked about kinetic friction, we said that this happens when the velocity is not equal to 0. You push a book and it's moving, and static, sorry. Kinetic friction tries to stop that object and bring it to a stop. Right? Static friction happens when the velocity is equal to 0. Imagine this book is 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 was 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's 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 oppose you by going to the left. So, this is \( f_s \). Lastly, let's 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. This coefficient of static friction is really just another number, just like \( \mu_k \) is. One thing you should know about this coefficient though is that it's always going to be greater than \( \mu_k \). Normally, they're going to be given to you. So, we've got this 5.1-kilogram block that's at rest on the floor. Now we're given the coefficients of static and kinetic friction. Like we just said, this is \( \mu_s \), and we'll see that it's actually greater than \( \mu_k \). 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. I'm just going to draw a quick sketch of the free body diagram. So, we have our \( mg \) that's downwards. We've already got our applied force, and there's our normal. Whether this object is moving or trying to move, we know that friction is going to oppose the by going to the left. We just don't know what type of friction it is. So which equation are we going to use? Are we going to use \( \mu_f k \), or are we 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 0. And we said that when the velocity is equal to 0, we're going to use the static friction formula. So our \( f_s \) is equal to \( \mu \) static times the normal. So that means our friction force here is going to be 0.6 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 \( n \) is equal to \( mg \). So that just means that we're going to use 5.1 times 9.8, and you'll get a friction force that's equal to 30 Newtons. So let's talk about this. You're pushing with 20 to the right, but the force that we calculated was 30. So, even though you're pushing to the right, the friction force would win, and the book would actually start accelerating to the left, in the direction of the friction. That's crazy. Doesn't make any sense. So, what's happening here? When we use this formula, this \( \mu_s \) times the normal, this is actually called a threshold. This is basically just the amount of force that you have to overcome to get an object to start moving. This times normal is the maximum value of static friction. So, what we do is we actually call this \( f_s \) max, and this is equal to times the normal. So, when we go back here, what we have to do is this static friction formula that we use is actually 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 friction threshold. Basically, we have to figure out whether our force, \( F \), is strong enough to get an object moving. There are really just two options. You either don't or you do. So let's talk about those. If your \( F \) is not strong enough to get an object moving, 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. It's not enough to get it moving. If the object stays at rest, then the friction is just going to be static friction. So basically, what happens is if you haven't yet crossed this threshold, which is kind of just like a number line here, where you have increasing force, then your static friction basically always has to balance out your force. What I mean by this is that if you're pushing with 10, your friction can't oppose you with stronger than you're 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 pull with 20. If you're pulling with 30, static friction just opposes your pull 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, and the acceleration is 0. Now what happens if you actually do overcome that threshold? Basically, if you have a strong enough force to get the object moving, then your force is greater than \( f_s \) max, and what happens here is that the object starts moving. And if it starts moving, then your friction switches from static and it becomes kinetic friction. So, what happens here is that this kinetic friction we already know is just equal to \( \mu_k \) 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 \( F \) to our \( f_s \) 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 is going to be static friction, and it's just going to basically balance out our pull. So, our static friction is going to be 20. So the 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 already know that \( f_s \) 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 \( \mu_k \) 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 here is we've actually crossed that maximum static friction threshold. And so, therefore, the friction that's opposing this book is going to 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's opposing you is just \( f_k \), and so this is just going to be 15. Even if you were to pull a little bit harder with 50 Newtons, it doesn't matter because this \( \mu_k \) times the normal is just a fixed value. So even though you're pulling with 50, kinetic friction would still oppose you with 15. Alright? So that's it for this one, guys. Hopefully, I made sense. Let me know if you have any questions.

Static & Kinetic Friction

So let's check out this problem here. I have a 5.1 kilogram 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 want to calculate the force needed to keep it moving at constant speed. So there are really two different situations here that I'm going to draw out the diagrams for. We've got a 5.1 kilogram block on the ground, and 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 hard you need to push it to keep it moving at constant speed. So there are two different forces here. Alright. So we want to draw the free body diagrams. Let's just draw all the other forces that are acting on these blocks. I've got my weight force. That's \( mg \), and then I've got the normal force. Alright. 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 \( f_s_{\text{max}} \) 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 0. So that's really the difference between these two diagrams here. In one you're going up against \( f_s_{\text{max}} \) and then the other one you're going up against kinetic friction. Alright. So how do we then calculate these forces? Basically, just use our \( f = ma \). So we have our \( f = ma\) here. Now we just pick a direction of positive. It'll be to the right for both of these diagrams here. So you've got \( f - f_s_{\text{max}} \). And then what's the acceleration? Well, the moment right when I get the block to move, the acceleration is still equal to 0. So we're gonna use \( a = 0 \) 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 \( f = f_s_{\text{max}} \). So you have \( f = f_s_{\text{max}} \). 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 gonna be equal to mg So basically \( f = \mu_{\text{static}} \times mg \) and so this is gonna be \( 0.7 \times 5.1 \times 9.8 \), 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 0 because the velocity is gonna be So your forces are \( f \), except now it's not \( f_s_{\text{max}} \), you're just using \( f_k \). So those have to cancel. So your \( f = f_k \) which is mu k times normal. So your force is equal to \( \mu_k \times mg \), right, just like we had before. And so this is gonna be \( 0.5 \times 5.1 \times 9.8 \). So now if you plug this in, you're gonna 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 C. 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 it 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. Alright. So that's it for this one, guys. Let me know if you have any questions.

Preventing a Block From Moving

Guys, let's check this one out. We have a 15-kilogram block that's at rest. We're given the coefficient of static friction, and we want to figure out how hard you have to push down on this block to keep it from moving. So basically, let's sketch this out. We have a 15-kilogram block that's on a flat surface. Right? And basically, what we're going to be doing is pushing down. I'm going to 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. Alright? So, what we want to do is first start off with a free body diagram. So let's go ahead and do that. So, we've got our box, we've got our mg that's downwards, and then we also have any applied forces. We know there are 2. Our F_down is what we're trying to solve for. And we also have an applied force that acts to the right. This is F = 300. Now there is also going to be some normal force because you'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 from moving, that's because there has to be some static friction that is opposing this. So there's some friction here that is opposing that. Right? So that 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. Alright? So let's go ahead and now write our F = ma. So I'm going to write out my F = ma here. I'm going to just pick a direction of positive up and to the right. 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 going to go and start with our y-axis first. So we've got our sum of all forces equals ma_y. Now this box isn't going up or down in the vertical axis, so we know the acceleration is going to be equal to 0. So, therefore, when we expand our forces remember our normal is positive minus mg minus F_down and this is equal to 0. 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 from 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 axis to solve. So I'm going to go to the x-axis forces to now solve for this. So I've got F = ma in the x-axis. So what are our forces? We have our F, which is 300 minus your Fs. 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 going to be Fs_max. So the minimum force is going to be where the Fs_max is going to balance out the 300. If 300 were just a little bit more, then 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 0 and these forces have to balance. So that means that your F is equal to your Fs_max which is equal to, remember, μ_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 force is really just going to be equal to your F divided by μ_static. So your F is 300 Your μ_static is 0.7 You'll get a normal force of 429. So your normal force is 429. Basically, if the normal is 429 then your Fs_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 429 minus your mg, which is 15 times 9.8. So if you guys go ahead and plug this into your calculators, you're going to get 282 newtons. So if we look at our answer choices, that's answer choice C. Alright, guys. Let's hit this one.

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/s². 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

Here’s what students ask on this topic:

What is static friction and how does it differ from kinetic friction?

Static friction occurs when an object is at rest and opposes the force trying to initiate movement. It is defined by the equation $f_s={\mu }_{s}N$, where $N$ is the normal force. Kinetic friction, on the other hand, occurs when an object is already in motion and is defined by $f_k={\mu }_{k}N$. The coefficient of static friction (${\mu }_s$) is typically greater than that of kinetic friction (${\mu }_k$), meaning it takes more force to start moving an object than to keep it moving.

How do you calculate the maximum static friction force?

The maximum static friction force can be calculated using the equation $f_s={\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction and $N$ is the normal force. The normal force is usually the weight of the object, calculated as $mg$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity (approximately 9.8 m/s²).

What happens when the applied force exceeds the maximum static friction?

When the applied force exceeds the maximum static friction, the object begins to move. At this point, the friction force transitions from static friction to kinetic friction. The kinetic friction force is then calculated using the equation $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction and $N$ is the normal force.

Why is the coefficient of static friction usually greater than the coefficient of kinetic friction?

The coefficient of static friction is usually greater than the coefficient of kinetic friction because it generally takes more force to overcome the initial resistance to motion (static friction) than to keep an object moving (kinetic friction). This is due to the microscopic interactions and interlocking between the surfaces in contact, which are more pronounced when the object is at rest.

How do you determine whether to use static or kinetic friction in a problem?

To determine whether to use static or kinetic friction in a problem, you need to compare the applied force to the maximum static friction force. If the applied force is less than or equal to the maximum static friction force, the object remains at rest, and you use static friction. If the applied force exceeds the maximum static friction force, the object starts to move, and you switch to using kinetic friction.