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 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 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 going to 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 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 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 going to be static friction. Right? As we're tilting the ramp, basically, the 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 mgx 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 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 m g sin θ = mu _s m g cos θ , 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 Tan θ = mu _s . And the last step we do is just get rid of this tangent by taking the inverse tangent. So θ = tan - 1 mu s , 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 s = tan θ . 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 want to 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 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 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 b