>> Professor Anderson: "Can you please go over how to correctly draw a free-body diagram into sine-cosine components?" Ah-ha. Excellent question. So this is something that a lot of students struggle with, is breaking down the free-body diagram into sine and cosine components. So let's do it together right here. Okay, let's say we have a box on an inclined plane. Okay, probably the most basic problem in physics, right? Box on inclined plane, and let's see if we can understand the forces that are at work here, all right? Now anytime you have an object, the free body always looks like this. It's a dot. Okay, take your box. You turn it into a dot. What are the forces that are acting on this box? Mg is, of course, down. What other forces might be acting on the box? Jorge, what do you think? >> Static friction. >> Professor Anderson: Static friction. Which way it static friction? >> It would be going up and to your left. >> Professor Anderson: Okay, static friction, up and to the left. So we'll say that the box is at rest, right? If it's static friction, then it has to be at rest. Jorge, what other forces might be acting on the box? What you guys hand the mic to Jorge, and let's -- let's make sure we can hear him. Okay, so Jorge, we've got gravity going down. You said that static friction is going up to the left. What other forces are going to act on this box? >> What else will be the weight of the box. >> Professor Anderson: Okay, we've got that. That's mg, right? >> Okay. >> Professor Anderson: What else? >> Gravity's already taken. >> Professor Anderson: Maybe ask one of your neighbors if they have any thoughts. Simon? >> Would it be normal force? >> Professor Anderson: Okay, normal force, right? If there is an inclined plane here, there is some surface that has to push up on the box, and that's what we call the normal force n, right? Anytime something is in contact with the surface, it's pushing on it. Is that it? Yes. That's it. Okay, those are all the forces that are acting on the box. Now this looks nice, except these things are not all at right angles to each other, and that's the problem, right? We would like to redraw this such that they are all at right angles to each other. So normal force, F sub s, those are at right angles to each other. We can leave them alone. But mg is the one that's causing the problem. Mg is not aligned to either of those, but I can always break a vector into two components. That's one and that one, and this triangle is a right triangle, and we need to figure out if this is theta or that is theta, and the way I like to do it is just take a guess and see if you're right. So let's take a guess. Let's say that theta is right there. If that one is theta, then this side of the triangle has to be mg sine theta, and this side of the triangle has to be mg cosine theta. Okay, you can review your trig and convince yourself that that's right, but remember Sohcahtoa, sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent would be opposite over adjacent. So this force down the incline is mg sine theta, and this force into the incline is mg cosine theta, but remember, we just took a guess, right? We weren't really sure if that theta corresponded to that theta, and so now we need to decide are we right? Is this correct? Anybody remember how we do that, how we decide if that guess is correct or not? It has to do with looking at the limits, right? And the limit that I like to look at is the following. Let theta go to 0. If theta goes to 0 degrees, that would be this condition right here. It's on a nearly horizontal surface, and therefore, the force down should be mg. Which would be the force down? This one right here, right. This was our surface. The whole thing's going to rotate. Mg cosine theta should go to mg. Is that right? Ah-ha! Cosine of 0 is, in fact, 1, and so we say we did it right. This is the proper free-body diagram, okay? You can go through a little bit of trig and convince yourself, a little bit of geometry to convince yourself that this theta does, in fact, correspond to that theta. It's a simple, little proof that takes you about a half a page. But I like the idea of just taking a guess and then looking at the limits, and whenever you look at the limits, that will tell you whether your guess was right or wrong. If I had made that one theta, then this limit would immediately tell me that it's not right, because mg sine of 0 would go to in that case, and we know that can't be. Okay? Good question.