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Anderson Video - Work and Dot Product

Professor Anderson
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 >> So dot product is a mathematical tool for dealing with vectors. So if I have a vector A, and I have some other vector B, and I want to calculate the dot product, how do I visualize it? First off, I put the vectors together. And then I figure out the angle between them. The dot product is this -- A dot B is defined as magnitude of A times magnitude of B times cosine of the angle between them. It's the projection of one vector onto the other. This is a scalar. So this is known as the scalar product. So if you hear those words, it's the same thing as dot product -- dot product or scalar product -- because there's no vectors on the right-hand side here, okay. You took two vectors and got rid of the vector signs by writing it as this scalar product. So back to the question that we just had. If we have C dot C, what do we get? We get magnitude of C times magnitude of C times cosine of the angle between them. What's the angle between C and C? Zero. And cosine of 0 is 1, and so we do just get C squared. So dot products we're going to use a lot in terms of work. Later on, you're going to learn about cross products, which becomes AB sine theta. Let's think about Cartesian coordinate system. If we have our good, old Cartesian coordinate system, xyz, and we think about the unit vectors, the one along x we call the i hat. The one along y we call j hat. The one along z we call k hat. Those things have magnitude of 1, but they have different directions. So if I think about the dot product, i hat dot j hat, what do I get? I get the magnitude of i hat, which is 1. I get the magnitude of j hat, which is 1. And then I get cosine of the angle between them. What's the angle between i hat and j hat? Ninety degrees. Cosine of 90 degrees is 0. And so i hat dot j hat is 0. Likewise, j hat dot k hat is 0. k hat dot i hat is 0. It's good to remember these things because when you do dot products between vectors, it's going to simplify quite a bit because a lot of those terms are going to be zero. So let's go back to our definition for work. Work we said was f cosine theta delta x. But that looks a lot like this dot product right here. So can we just write it like this -- f dot delta x -- where I put vector signs on top of those things? And the answer is, of course, yes, we can do that. Displacement's a vector. Force is a vector. There's an angle between them, and so we can calculate theta and write down the dot product for those two. Okay, so this is getting more general to the definition of work, but we need to do one more step, which is if the force is varying over that distance, we have to take that into consideration. And so in general, work is the integral of f dot dx. They might write it with an s in your book, ds. It's whatever path you're taking, okay. So the force applied over the particular path. Okay, let's see how that works for something that we understand, like gravity. So gravity says objects are pulled towards the center of the earth. If I want to lift an object, I'm going to have to apply a force to it. And let's say we lifted a height h by applying a lifting force. And that lifting force doesn't have to be any bigger than mg, right. Here's my object. I'm going to lift it by applying mg to it, and it's going to go up a distance h. Let's calculate the work now by this definition. The work due to lifting it is going to be the integral of F lift dotted with, why don't we call it y since we're going vertically, okay. So we just rewrote this definition right here. All right. But we know what F lift is equal to. F lift is just going to be a magnitude of mg, but let's figure out what the angle is to put right here. What's the angle between my lifting force and the displacement? Well, the displacement is up. The lifting force is also up. And so the angle between those two is zero degrees. Cosine of 0 is 1. And so we're just left with integral F lift dy. And now we know exactly what that is. F lift is mg. We still have our dy, and we're going to lift it from zero to h. mg is a constant, and so that comes out in front of the integral. And if I integrate y, if I integrate dy from zero to h, I just get y evaluated zero to h, and that, of course, becomes mgh, which we already knew, right. We knew that gravitational potential energy was mgh. So what does this mean? What it means is I've done work lifting this box, but that work is energy. It has to go into something. What does it go into? It goes into the gravitational potential energy of the box such that if I let the box go, it will fall back down, okay. All right, good.