7. Friction, Inclines, Systems
Inclined Planes with Friction
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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 or beginning to slide or the block slides at constant speed. And really these two phrases that are pretty common are related to 2 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 these points in just a second here. I wanna 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 gonna 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 gonna calculate that angle. In the second problem, it's gonna be very similar, except now, we're gonna calculate the angle where it already is moving, but now it slides down at constant velocity like that's and we wanna 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 gonna just draw off 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 mgy components and an mgx 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 gonna 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 MGX 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. Friction. So really, this m g 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 wanna calculate this theta critical s. And so how do we do that? We're just gonna use our f equals m a. 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 we think the object is gonna slide, so down the ramp like this. That's our direction of positive, which means that when we expand out our forces, we have mgx 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 0. Remember, we're trying to calculate this 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 mgx is equal to f s max. So later on, we're also gonna see that when we're sliding in constant speed, the acceleration also is going to be 0. 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 mgx is mgsine of theta, and then my f s max has an equation. Remember, this is just mu static times the normal. So this is gonna be mu static times the normal force. And remember that in these problems, your normal force is really just equal to your mgy, which is equal to your mg times the cosine of theta. So what we can do here is we have mg times the sine of theta is equal to mu static times mg cosine of theta, which means that the mgs on both sides of the equation actually will just cancel out. Right? If m g cancels out m g. So I'm basically left with 2 variables, theta and also my mu static. So I want I want to basically figure out an expression for this theta here. So I'm gonna move this to the other side because I have my mu static here on the right. So I'm gonna divide out the cosine, and I'm gonna move it to the other side so we can basically just collapse it to a single trig function. When you take sine over cosine, remember that becomes a tangent. So tangent theta 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 theta critical mu static, and that's the equation. Basically, these are your 2, critical angle equations for theta critical s. So you have theta critical s equals tangent inverse of mu static. If you have mu static, you can figure out theta critical, but you also have the other way around. Mu mu static 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. Alright. So basically, if we wanna figure out this theta critical for our problem, this is we're just gonna take the inverse tangents of the mu static that we were given, which is 0.75. So if you do this, what you're gonna get is 37 degrees. So this is the angle that you have to tilt the ramp at so that the block starts sliding. Alright. So now in the second problem, we wanna 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 it slides down a constant velocity. If we kept this thing at 37 degrees, it would to accelerate down the ramp. So we're gonna have to adjust a little bit basically, like, you know, tilt it, tilt it down a little bit so that the block starts sliding at constant velocity. Alright? So what happens is now we're we're calculating theta critical k, and the entire solution is gonna 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 gonna 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 MGX is equal to your maximum static friction, and we know that again because of equilibrium. Here where the block slides at constant speed, we have MGX is equal to kinetic friction instead. Alright. 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. Alright. So this is k. Just use mu k, and then finally, our two equations. So basically, what happens is when you rewrite all of these equations here, your theta critical k is gonna be the inverse tangents, not of mus, but of mu k. And your mu k is gonna be the tangents, not of theta critical s, but of theta critical k. So that's the angle where it slides down at constant speed. So now if we just calculate this, our theta critical k is gonna be the inverse tangents of and now we're just gonna use our mu k 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 got 37 and 17. It's kind of related to the idea that it's harder to get something moving than 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 2 different angles. Alright. So the last point I want just wanna mention is that notice how both of these equations here, theta critical s and k, depend only on the coefficients 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, that 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.
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