 Let's talk a little bit about the principle of moments, also known as Varenjan's theorem. So if we consider a force vector, F, we know that this force vector has a line of action, some extension of that vector that we can consider the forces acting upon. And if we connect that force to a body, we can exert a moment on all of the points that are also on the body, not on that line of action. For example, let's say I have some sort of, I don't know, some sort of widget, some random shape. But any point inside that shape, we can actually exert a moment on, a tendency to make this widget, this piece, spin about that point. And we know that the magnitude of the moment is the magnitude of this force, F, times the moment arm, where the moment arm is the distance of the point we're looking at. Let's label that point P from the line of action. That distance, again, also called the moment arm, is both measured perpendicular to the line of action. And it will be the minimum distance from the line of action to the point. So what Vérignon basically says is the following, that the moment of a sum is the sum of the moments. Now, this is basically paraphrasing. But let's see if we can discuss what that means. Let's consider a force, like the one we drew above, and think about it as being the sum of other forces applied to the same point. Let me go ahead and draw the line of action for our original force. But we'll think about it as potentially being two other forces. Here's our original force. Well, any vector we can consider in 2D is being the sum of two other vectors. Let's think about it as having some random vector here. We'll call this F1. And sum it with another vector that's here, F2. And let's use our idea of a parallelogram to establish what those forces really look like. So if we consider this force as a sum of two other forces applied to the same point, we can write F is equal to F1 plus F2. Use a little squiggle notation to remind us that these are all vectors. Well, according to Vérignon, the moment can be found by summing the moments of each of those other forces using their own moment arms. So let's assume we have some point P out here. And we would notice that our moment around point P, let me erase what Vérignon said here and make it very clear, that our moment around point P would be equal to F times D, where D is the moment arm around the point P compared to the line of action of F. Well, I can similarly take the line of action of F1, in which case it would have its own distance, its own moment arm, let's call that D1. And F2 has its own line of action, and we can measure the distance to F2. We'll call that D2. So each of those components has its own moment arm. Well, all Vérignon is basically saying is that if I wanted to find the moment that corresponds with F times D, another way I could do it is find the sum of moments 1 and 2, where each of those would be F1 times D1 and F2 times D2. In other words, the moment of this sum is equal to the sum of the two separate moments. There are two ways we can take advantage of this. Our first way is we can find the components of a force. In some basis, for example, we might say F of x and F of y in a Cartesian basis. And we can break down and find the moment by using each of those components. For example, here is a force. Here is our component in the x direction. Here is our component in the y direction. Now, if we want to determine what the moment is about some point out here, we'll call this about some point out here, point P, again. What we can do is we can look and say, OK, well, there is a distance. We'll call this DX, a distance x from the point from that y component. And there is a dy, a distance from the x axis there. And we'll notice that if we think about the force y, it's being applied in a clockwise direction, which we'll call positive. So our clockwise direction is Fy times DX. And we have an action in the counterclockwise direction of Fx times dy. So if we want to calculate the moment for a given force, one of the ways to do that is to calculate the moment created by each component. In this case, Fydx and then minus, because it's moving the other direction, Fx dy. Notice it'll be important to pay attention to the orientation of the point relative to where the force vector is to keep your sign straight. So that's one useful result. The second useful result, if we consider the force, apply to a place anywhere along the line of action. And we select a particular place that eliminates a component by making it collinear with the point. Then we can find the moment very simply. That was quite a lot of words to explain it. Let's see if we can show it graphically. If I consider some force, here's my force. And I have a line of action for that force. We can actually consider the force to be applied anywhere along that line of action, and it doesn't change how it creates a moment. Any force applied anywhere along that line of action is going to create the same moment. Well, if I consider my point P here, what I can do is I can take the force and line it up and apply it at a point that is, in this case, perfectly horizontal with P. So I'll reapply my force there. And if I do so, I'll apply my force at that point and then break it down into components f of x and f of y. You'll notice that because f of x runs directly through point P, we can effectively ignore it because it will contribute nothing to the moment. It has no moment arm. And the only thing we care about is this distance. We'll call this dx from point P. So the moment in that particular case around point P is going to be equal to fy times dx, if you figure out the dx at being at this location here. We could do the same thing by applying the point directly above this point P. If we stay on that line of action and consider the forces being applied directly above here, well, now the y component is now meaningless. It gets eliminated because it has no moment arm. And the x component is the only one that counts. So here is f of x. And if we call this distance some dy, we can show that the moment in that case is equal to f of x times that distance y because the other components have been eliminated.