Professor Anderson

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>> OK. Christian just had a great question about the g sine theta and when we might use the cosine component instead of the sine component. And that's going to come into play when we start thinking about forces. So, this is a little preview of what's to come. If I have a box on an incline and I want to figure out what forces are acting on the box, how do I do it? Well, any time we're figuring out forces, we always go to something called a free body diagram. And a free body diagram just means make the object of interest a dot and let's draw some arrows. That's all it means. So, what are the forces that are going to act on the box? Gravity is straight down. Gravity has a force of mg. There is a normal force, n, from the incline pushing on the box and there is a frictional force, f sub k. If the box is sliding, then it is a kinetic frictional force. Now, when you're faced with this sort of problem, you just have to add up vectors. Forces are just vectors like any other vector, acceleration, or velocity. You add them up according to vector rules. Which looks a little tricky when they're not at right angles to each other. The normal force and the frictional force are at a right angle to each other, but mg is somewhere in-between the two and so it would be much nicer to have everything pointing along orthogonal axes. And so let's redraw our free body diagram like this. Here's my object. I'm going to keep my normal force in that direction. I'm going to keep my frictional force in that direction, but I'm going to break up mg into two components. One of those components is that way. One of those components is that way. What is this component of mg? Is it mg cosine theta or is it mg sine theta? This one is mg sine theta, which we just proved with our triangles. Or, looking at the limits as theta goes to zero. As theta goes to zero, this thing is on a flat surface. It doesn't seem like we should have any horizontal force on it if it's on a flat surface and sine is the one that goes to zero as theta goes to zero and that means that this thing right here is mg cosine theta. So, now when we get to forces, now we have orthogonal coordinates here and if I think about coordinates x and y that are tilted, now all we have to worry about is the forces in the x direction. What are they? They're mg sine theta minus f sub k. All of that is going to be equal to the mass times the acceleration in the x direction. The forces in the y direction are going to be n minus mg cosine theta. All of that is equal to the mass times the acceleration in the y direction, but if it's moving along an incline, then we say that's zero. And then there's only one more equation that we need, which is f sub k, the frictional force, is in fact equal to mu sub k times n where mu sub k is the coefficient of kinetic friction. And now with those three equations you can solve for things like the acceleration of this thing down the plane. OK? So, that's when you would need this cosine theta because it's going to be buried in here, causing a frictional force. OK? Good. If that's not clear, come see me in office hours. Cheers.

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>> OK. Christian just had a great question about the g sine theta and when we might use the cosine component instead of the sine component. And that's going to come into play when we start thinking about forces. So, this is a little preview of what's to come. If I have a box on an incline and I want to figure out what forces are acting on the box, how do I do it? Well, any time we're figuring out forces, we always go to something called a free body diagram. And a free body diagram just means make the object of interest a dot and let's draw some arrows. That's all it means. So, what are the forces that are going to act on the box? Gravity is straight down. Gravity has a force of mg. There is a normal force, n, from the incline pushing on the box and there is a frictional force, f sub k. If the box is sliding, then it is a kinetic frictional force. Now, when you're faced with this sort of problem, you just have to add up vectors. Forces are just vectors like any other vector, acceleration, or velocity. You add them up according to vector rules. Which looks a little tricky when they're not at right angles to each other. The normal force and the frictional force are at a right angle to each other, but mg is somewhere in-between the two and so it would be much nicer to have everything pointing along orthogonal axes. And so let's redraw our free body diagram like this. Here's my object. I'm going to keep my normal force in that direction. I'm going to keep my frictional force in that direction, but I'm going to break up mg into two components. One of those components is that way. One of those components is that way. What is this component of mg? Is it mg cosine theta or is it mg sine theta? This one is mg sine theta, which we just proved with our triangles. Or, looking at the limits as theta goes to zero. As theta goes to zero, this thing is on a flat surface. It doesn't seem like we should have any horizontal force on it if it's on a flat surface and sine is the one that goes to zero as theta goes to zero and that means that this thing right here is mg cosine theta. So, now when we get to forces, now we have orthogonal coordinates here and if I think about coordinates x and y that are tilted, now all we have to worry about is the forces in the x direction. What are they? They're mg sine theta minus f sub k. All of that is going to be equal to the mass times the acceleration in the x direction. The forces in the y direction are going to be n minus mg cosine theta. All of that is equal to the mass times the acceleration in the y direction, but if it's moving along an incline, then we say that's zero. And then there's only one more equation that we need, which is f sub k, the frictional force, is in fact equal to mu sub k times n where mu sub k is the coefficient of kinetic friction. And now with those three equations you can solve for things like the acceleration of this thing down the plane. OK? So, that's when you would need this cosine theta because it's going to be buried in here, causing a frictional force. OK? Good. If that's not clear, come see me in office hours. Cheers.