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.