 Let's take a look at Vérignon's theorem in action. In this case, we have a block that's sitting on some sort of surface. And that block has a weight that's holding it to that surface. And then we're going to apply a force over here on the left. Force that's pushing up a little bit and pushing to the right a little bit. And the question we want to ask is, is the block going to overturn? We're going to assume there's some sort of friction on the surface. And at that moment that the block has begun to lift up, there is a reaction force on that lower right-hand corner, a reaction force in the lower right-hand corner there. But we want to know whether or not the block will overturn. And balancing moments or seeing what kind of moment is around that point, point C, will determine whether or not the block will turn. Actually, we could take the moment about any point, but we're going to keep this simple. We're going to consider the moment around point C, because then we do not have to know what the reaction force at C is, because those will go through that point and will cannot contribute to the moment. So let's begin this by setting a basis. And in this case, the easiest basis is to go ahead and establish a coordinate system where we have vertical and horizontal components. And we're going to go ahead and consider a positive moment to be in the clockwise direction. Now that we've set a basis, let's go ahead and consider the various forces we have in terms of their components in that basis. So if you look over here, let's go ahead and call this force on the right, the force on the right side here, FA, force A. And we'll call this weight here in the middle. We'll just go ahead and call it W. And so our components, FA is going to have two components, FAx and FAy. And we can sort of see that FA is located here in an direction that is described by a 3, 4, 5 triangle. So we see it's a 3, 4 triangle. I'm going to go ahead and I know that if the sides are 3 and 4, then the hypotenuse must have a relative length of 5. And so we can break down FAx to find out that that's the force A times the cosine of an angle. If we define this angle here, we'll call this cosine theta, cosine theta, or 3 fifths of FA. Similarly for FAy, we can recognize that as being the force times the sine of the angle. And that's going to give us 4 fifths of our overall force magnitude FA. And if we wanted to now, we can go ahead and substitute the two Newtons in for FA to figure out the values for each of these. So our component in the x direction is going to be 6 fifths of a Newton. And our component in the y direction is going to be 8 fifths of a Newton. So there are components for our force in that direction. In the case of our weight, we really only have weight in the y direction. And that's going to be the entire five Newtons. And it's pointing down. The arrow is already pointing down. So we're going to define that as being in the direction of the arrow we define. And we notice that the weight in the x direction is going to be 0. It's not going to be existing. So now we want to know what the moments are around point C. So what is the moment around C? Let's do the moments of the weight around point C. That's what I'll use the W and the C for, the moment of the weight around point C. Well, in that particular case, that's going to be equal to the value of the force, which is the weight times the moment arm. Well, what is the moment arm here? In this case, the moment arm is the perpendicular distance. Let's do that with a little dotted line here, shall we? Perpendicular distance to the line of action of that force from point C. And because weights tend to act in the center of gravity or the center of the mass, we'll assume that's halfway across the block. So that's going to be a distance of five centimeters. So our moment around WC is going to be the weight times that change in the X position, or it's going to be five newtons times five centimeters. Now the important thing is we also need to figure out the sign in this particular case to know which direction we're talking about. And if we look, we can see that if we're going around point C here, that the rotation we're going to cause by this weight around point C is going to be in our counterclockwise direction. So we're going to say that that's a negative value. So negative five newtons times five centimeters or negative 25 newton centimeters. That's going to be the moment caused by the weight. Now let's consider the moment caused by the force. We'll say that's moment caused by the force at point A around point C. And that's going to be two pieces. This is where we're going to go ahead and use Varen-Johne's theorem. In this case, we have the Y component. Let me draw it over here on the right. We have the force in the Y direction acting vertically. And we have the force in the X direction acting horizontally. And we're going to consider each of them separately relative to point C. So we need to figure out how much the Y force is displaced in the X direction. So we have the force AY displaced in the X direction. And we have the force FAX displaced in the Y direction. Now the important point here is to figure out which direction each of these is behaving. And you'll notice in the case of FAY around point C, it's pushing it in a clockwise direction. Similarly, FAX acting around point C is also pushing in a clockwise direction. So for both of these, we can define these as being positive moment values. Let's go ahead and substitute in the values that we have. We see that we have 6 fifths of a Newton for FAY times, well, what's delta X? Delta X in this case is the width of the entire block, which is given to us as 10 centimeters. And delta Y is the height of the block, which is 20 centimeters. So did you catch the error here? It's an easy mistake because you're reversing the components. But in this particular case, we've made a mistake of writing down the value for FAY and the place of FAX and the value for FAX and the place of FAY. So in this case, the 6 fifth Newton should be 8 fifths. And the 8 fifth Newton should be 6 fifths. And then when you finish solving the problem, your final answer is actually 40 Newton centimeters and not 44 Newton centimeters. If we do these calculations, we get cancelation here. We get 12 Newton centimeters plus 32 Newton centimeters for a total of 44 Newton centimeters in the positive direction. Well, the sum of the moments around point C is just those two things added together. 44 Newton centimeters and the negative 25 Newton centimeters. Subtracting those two, because the second one is negative, gives us a value of 19 Newton centimeters. And notice we've defined positive as being in a clockwise direction. So this means with this set of forces, the block will begin to rotate clockwise. If the block had a greater weight, let's say, for example, the block instead of having a weight of 5 Newtons had a weight of 10 Newtons, then this value here would have been twice as much or negative 15 Newton centimeters. And notice 44 minus 50 would be negative 6. That means that the block would want to rotate in a counterclockwise direction or it would push itself back down on the table, and then there would be additional normal forces holding it up. That's our example of Vérignon's theorem in action.