Hey, guys. So now that we've taken a look at objects on rough inclined planes, occasionally in these problems you'll have to calculate something when the object is doing something very specific, like it's starting to slide, are beginning to slide or the block slides at constant speed. And really, these two phrases that are pretty common related to two special angles in these rough inclined plane problems. These angles are called critical angles, and I'm gonna show you how we calculate them in this video. So we're gonna actually come back to these points in just a second here. I want to just jump into the problem because it's very visual to understand. So here we're gonna place a block on an adjustable ramp. So I literally have a block on an adjustable ramp like this. And the whole idea is that we're going to tilt the ramp slowly until the block suddenly starts sliding. So basically what happens is I'm gonna tilt this block right into the point where the block starts moving right there, and we're going to calculate that angle in the second problem. It's going to be very similar, except now we're going to calculate the angle where it already is moving. But now it slides down at constant velocity like that. We want to figure out that angle as well. And really, those are the critical angles. So let's go ahead and check this out here. We're going to just draw some free body diagrams and then use our F equals M A, just like any other forces problem. So if we take a look at the forces that are acting on this block while it's on the ramp, right, so when it's tilted like this, we're gonna have some MG that wants to pull it down the ramp, which means that we have an M G Y components and an MG X components. Why doesn't it start sliding yet even as we start tilting the ramp? That's because there's some friction that's preventing it. So there's some friction that acts up the ramp, and because it's not yet moving, this is going to be static friction, right? As we're tilting the ramp, basically, static friction prevents the object from sliding. Now we know that as we tilt it higher and higher and higher, your MG X gets stronger and stronger right, The force that's pulling it down, the ramp gets stronger, and basically your static friction in order to prevent it from sliding has to also increase. Well, basically, the point right where the block starts to slide. That's where you've overcome the maximum static friction. So really, this mg X at the angle where it starts sliding is actually balancing out the F S. Max. So that's what we're going up against here. So we want to calculate this fate a critical s. And so how do we do that? We're just going to use our F equals, Emma. So we've got f equals m A here and now we just pick a direction of positive, so usually that's just gonna be the direction. But we think the object is going to slide. So down the ramp like this, that's our direction of positive, which means that when we expanded our forces, we have MG X, and then we have F S Max. So what's our acceleration? Well, in these problems, even though we're calculating the moment where it starts to slide, the acceleration is actually equal to zero. Remember, we're trying to calculate the special angle right at the moment before it starts sliding, right? So even though the block starts to slide, we want the angle right before that actually happens, right where MG X is equal to F S. Max. So later on, we're also going to see that when we're sliding in constant speed, acceleration also is going to be zero. So really, both of these are just equilibrium problems. Alright, So basically, what I'm gonna do is I'm going to move my f S Max over to the other side and I'm gonna expand out both of these terms. I know this MG X is mg sine of data and then my f s Max has an equation. Remember, this is just mu static times the normal. So this is going to be a new static times the normal force. And remember that in these problems, your normal force is really just equal to your G y, which is equal to your mg times. The cosine of theta so we can do here is we have mg times. The sine of theta is equal to, um use static times mg co sign of data, which means that the mgs on both sides of the equation actually we'll just cancel out right? You have MG cancels out mg. So I'm basically left with two variables feta and also my mu static. So I want I want to basically figure out an expression for this data here. So I'm gonna move this to the other side because I have my new static here on the rights. So I'm gonna divide out the co sign. I'm gonna move it to the other side so we can basically just collapsed into a single trig function. When you take sine over cosine, remember that becomes a tangent. So tangent data critical s is really just equal to mu static. And the last step we do is just get rid of this tangent by taking the inverse tangent. So tha tha critical s is equal to the inverse tangent of metastatic. And that's the equation. Basically, these are your two critical angle equations for theta critical s. So you have to to critical s equals tangent, inverse of metastatic. If you have a static, you can figure out that a critical. But you also have the other way around. Uh, metastatic is equal to the tangent of theta critical. Basically, it's just two sides of the same equation. If you have one, you can always figure out the other. All right, So basically, if we want to figure out this theater critical for our problem, this is we're just going to take the inverse tangents of the Meuse static that we were given, which is 0.75. So if you do this, you're gonna get is 37 degrees. So this is the angle that you have to tilt the ramp at so that the blocks starts sliding. All right, so now in the second problem, we want to calculate this angle here again, where the block is sliding down at constant speed. So basically, once the object starts moving, we're gonna have to adjust the ramp so that it slides out in constant velocity. If we kept this thing at 37 degrees, it would continue to accelerate down the ramp. So we're gonna have to adjust a little bit, basically like tilted, tilted down a little bit so that the block starts sliding a constant velocity. Alright, so what happens is now we're calculating tha tha critical K and the entire solution is going to be almost the exact same thing that we did on the left side here. I'm gonna copy most of this stuff over because most of it is going to be the same. So we copy this diagram over. The only thing that's different about this problem is that because you are sliding with some velocity already, you're not going up against maximum static friction anymore. You're actually going up against kinetic friction. That's really the only difference. So we're just basically going to replace every single s in the solution with just K's. So here what happens is we have When the block starts sliding, your MG X is equal to your maximum static friction. And we know that again because of equilibrium here, where the block slides a constant speed, we have MG X is equal to kinetic friction instead. All right, so what I'm gonna do is I'm gonna copy this whole entire thing here, and we're basically just going to replace every single s with a K. So I'm gonna go ahead and do that. All right, So this is Kay just using UK, and then finally are two equations. So basically what happens is when you rewrite all of these equations here. Your thought a critical K is going to be the inverse tangents, not of us. But if the U. K and your UK is going to be the tangents, not of fate a critical s but a fatal critical. Okay, so that's the angle where it slides down at constant speed. So now, if we just calculate this Arthur to critical K is going to be the inverse tangents of, and now we're just gonna use arm UK, which is given to us 0.31. And if you go ahead and work this out, you're gonna get 17 degrees. So let's talk about these two angles. We've got 37 17, and it's kind of related to the idea that it's harder to get something moving that it is to keep it moving. So the idea here is that you have to tilt this thing at 37 degrees to overcome the maximum static friction. Once you do that, then you only have to lower it to 17 degrees so that the block continues sliding here at constant velocity. That's why we got these two different angles. All right, so the last point I just want to mention is that notice how both of these equations here, data critical s NK depend only on the coefficient of static and kinetic friction. They don't depend on any other variables. Like, for instance, the mass, which might be counterintuitive. You might think that if the box was heavier, the angle would be different. But actually, it doesn't matter. It doesn't affect the angle at all, because again, this MG cancels out when you do the F equals m a right. So hopefully that makes sense. Guys, let me know if you have any questions.