Guys, in previous videos, we've seen how to solve inclined plane problems and we've seen friction problems. So we're really just going to combine these two concepts together in this video because sometimes you're going to run into problems where you have objects on rough inclined planes, meaning inclined planes with friction. So let's go ahead and just jump right into the problem here. There's really nothing new. So we've got this 10 kilogram block that's on a ramp at 37 degrees. We have our coefficients of static and kinetic friction. And in part a, what we want to do is calculate the friction force that's acting on the block when it's released.

So in part a, what we want to do is figure out what is this friction force f. It could be static or kinetic. So to do that, to start things off, we're going to need a free body diagram. So let's get started here. We've got this block, there's going to be an mg that's downwards, right? And then we have no applied forces or tensions, but we are going to have a normal force that points perpendicular to the ramp's surface. And then finally, we have some friction. Now whether the block actually starts to move down the ramp or is trying to move down the ramp, that friction force wants to oppose that motion and so it's going to act up the ramp regardless of whether it's static or kinetic. So that's what we're trying to find here.

Now remember that when we draw free body diagrams for inclined planes, we're going to tilt our coordinate system so that the x axis points down the ramp like this. Basically, this is our plus x and plus y. When we do that, remember our mg basically just gets split up into its components. So we have mgy and then we have mgx. Alright. So those are our mg components here. So now what we have to do is we have to determine what kind of friction we're using, or we have in this problem. Whether it's static or kinetic friction. So that brings us to step number 2. We're going to determine whether this is static or kinetic friction, and there are 2 ways to do this. So in some problems, you'll actually be able to tell from the text. It'll say that the box isn't yet sliding or it is sliding. You'll know whether it's static or kinetic. But in this problem, it's kind of vague. We're supposed to calculate this friction force, but we don't know whether it's still stationary or starts moving. And so what happens is, we're going to have to compare all the non-friction forces. Basically, in this problem here, that's just going to be our mgx. So we're going to have to compare those non-friction forces along the axis of motion to the maximum static friction. We're basically trying to figure out whether our non-friction forces like mgx are strong enough to get this object moving. Right? So we're basically trying to figure out whether we're dealing with fs or fk. And to do that, we're just basically going to calculate mgx and fs max and see which one is bigger.

Alright, so when we're doing mgx, remember it's just mg times the sine of theta, and we can calculate that. We have all those numbers. This is a mass of 10, 9.8, and then we're using the sine of 37. If you go ahead and work this out, you're going to get 59 newtons for mgx. So we know this is equal to 59. Now what we do is we just compare this to the maximum static friction. Remember that's that threshold. Remember that the equation for that is μs times the normal. We know what μs is. It's just 0.6 that was given to us in the problem here. What about this normal force? Well remember on inclined planes, if you have friction and mgx in the x-axis and then only n and mgy in the y-axis, those forces have to cancel. So n is equal to mgy which is equal to mg times the cosine of theta. So when we plug in the normal force we are really just going to do 10 times 9.8 times not sine of 37 but cosine of 37. So if you go ahead and work this number out, what you're going to get is 77 newtons. So what happens here is that our fs max is 47. That's the maximum static friction, but your mgx is stronger than that. So what that means is that if your mgx overcomes the maximum static friction then therefore the actual friction basically just becomes kinetic friction. So this f actually is fk and to calculate this we're just going to use μk times the normal and we can figure that out as well. This Fk here is just going to be μk which is 0.4 and then times the normal force which again is 10 times 9.8 times the cosine of 37. So if you go ahead and work this out, you're going to get 30.8 newtons.

Alright, so basically, we know this is going to be 30.8. MGX is strong enough to get the object moving, that friction becomes kinetic, and we have that kinetic friction force. Alright? So that's the answer to part a. Now what we're doing in part b is trying to figure the block's acceleration. So now in part b, we want to figure out ax. So now if we're trying to figure out ax, now we're going to get into our F equals ma. So we write all the forces in the x-axis equals max, and once we have our direction of positive, we really just have our mgx that's positive, minus our friction kinetic is equal to max. And this is actually really straightforward because we've already calculated all of these values. So what we can do here is we can say that mgx is equal to 59. That's what we calculated over here. Then you're going to subtract the friction force which we just calculated as 30.8. This is going to be 10 times ax. So if you go ahead and solve this, you're going to get 2.82 meters per second squared. You get a positive number for our acceleration, and that makes sense. It just means that it accelerates down the ramp along our direction of positive. Alright. So that's it for this one, guys. Let me know if you have any questions and I'd like to get some practice.