 Okay, so here's part 2. It's a similar problem, really, it's the same thing, but let's suppose I want to design another roadway, and I want it such that no matter what the case, if the cars go at 15 meters per second, it doesn't matter what the friction is, it could be complete ice, they won't fall off, okay. So, you see here, if the car is going 15 miles per hour, 15 meters per second, and there's no friction force, and you want an acceleration that way, there's one solution that you could think of, and that would be to tilt the road a little bit, because in that case, this floor, the force of the road pushing perpendicular to the car will also be pointing perpendicular to the, in the direction of the acceleration, and so you can get this to make your car turn and circle, and not the frictional force. Okay, so I'm going to let you pause, and I want you to change the diagram, and you're going to draw a new picture, and draw the free body diagram. Okay, so I assume you did that. If you didn't, I mean, I might as well come over and do your laundry for you too, right? I'm just kidding. Okay, so let me change this picture some, I want to tilt it, I'm going to just erase that. So I'm going to have it like that, at some angle theta, I don't know how much I need to tilt it, but I need to tilt it, and there's my car, headlights, like that. Okay, there's my new picture, and now I'm going to need to draw my new free body diagram. Gravity still is down, and the normal force is perpendicular to the surface, like that. Now when we drew tilted planes before, a lot of times we tilted the axis too, but in this case I'm not going to do that. Let me draw my x and y axis, here's x, and here's y. And why am I going to do that? I'm going to do that because I want the acceleration to be in just the x or the y direction. You don't want to have to find components of acceleration, it makes it a lot more complicated. So in this case, the acceleration is this way, which is just in the x direction. So what's the next step? I'm going to pause you in just a second. The next step is write down Newton's second law for both the x and the y direction. I hope I'm holding that sign up long enough for you, but I want the video to be not too long too. Okay, so, boom. So I want to write down f net x and f net y. They're going to be different than before. So in order to do this, I only have this force right here that needed an angle for, and that's going to be theta, because you can play around with that if you want. But as you make that go back down to zero, this will be pointing straight up. So that's a good way for you to see that that's the case. So let's pursue the x direction. I'm going to erase this. So here, this is my y component, and that's my x component. So the x component of the normal force is it's going to be equal to n times a sine of theta, and that's going to be equal to the mass times the acceleration of the car, which is going to be v squared over r. And in the y direction, I have this component of the normal force in cosine theta minus mg equals zero. And now here's where you have to be careful. The normal force in this case is not equal to the weight. So many times it is people fall into the trap of always saying that, and it's not the case. Okay, so here's a good pause point. So what you want to do is see if you can solve, if you have enough information to solve for theta, and if you do then do that. Okay, what don't I know? I don't know n, I don't know theta, I don't know the mass. I have three things I don't know. I know the velocity of, I said that was going to be 15 meters per second, and I do know the radius of 57 meters. Okay, so what do I do? I have three variables and two unknowns. Well, let's just, you know, one thing to do if you're not sure what's in that water, just jump in it and find out. So let's jump in the water. I'm going to solve this for n in cosine theta, n equals mg over cosine theta, and I'm going to plug it in up there. So now I get mg over cosine theta, sine theta, equals m v squared over r, and you see I was right. It was okay to jump in that water. There were no sharks, or many eating sea monsters because the mass is canceled. And again, that's good. If the mass didn't cancel, I would have to have a different angle bank for different mass cars, and that would be not nearly as cool. Okay, so I'm trying to solve for theta. So I can divide both sides by g, and I have sine, well you could pause right here. You solve for theta. Why am I doing that? Okay, you did it. Let's see if you did it right. So I can divide both sides by g, sine theta over cosine theta is tangent theta. So I have tangent theta equals v squared over gr. This is a great place to check if you're doing the right thing. Tangent theta is a ratio, so it should have no units. So this should have no units. This is meters squared per second squared, that's meters per second squared times meters, so that has no units, too. So theta is going to be the inverse tangent of v squared over gr. So I know v is 15, so theta equals tangent inverse 15 meters per second squared over 9.8 newtons per kilogram or meters per second squared, they're equivalent times 57 meters. And so what angle does that give me? Okay, let's see, 15 squared. I get 21.9 degrees, so approximately 30 degrees. That's kind of steep, but that's if you want no friction at all. Okay, just take this to the extreme. What happens as you get a bigger and bigger circle, well in that case, r is going to get greater, and so your angle theta is going to be smaller. And if the velocity gets larger, then you're going to need to steep that angle up, increase that angle up. Okay, I mean this is really the same thing they do with the NASCAR racetracks. They tilt it so much. Okay.