 So this is part two of our segment looking at calculating the forces on submerged planar surfaces. And what we did in the last part, we looked at determining what the magnitude of the force was. And we found that essentially the way to calculate the force was the pressure at the center of area. Now what we're going to do, we're going to try to figure out where does that force act on the plate. Now your gut feeling might be that it acts at the center of area, but that's not true. It actually acts at a location that we referred to as being the center of pressure. And so what we're going to do in this lecture, we're going to dust off your tools of doing statics and static analysis. We're going to figure out where that force is. So in order to do that, what I'm going to do, I'm going to some moments about the center of area of the plate. And remember, the forces are acting on one side here because we're assuming that there is no fluid on the lower side of the plate. But we're going to some moments about the center of area, which we've denoted as being capital C, capital A. So let's sketch out our plate. Here was the plate. And let's say just for sake of argument, the center of area was right there. What we have is we have this force, the resultant force due to the fluid. And it's acting at this location, which we've referred to as being the center of pressure right now. That's unknown. And when we were deriving our equations, what we did is we established the center of area as being the origin of our plate. So that was the origin where the X and Y axis were located. What we're going to do now, let's assume that we have some arbitrary force here. That was DF. And it's acting normal to the plate. It is some distance Y away from the center of area. And recalling from statics, when we sum moments, we have to have a sign for it. So I'll say in the clockwise sense as being positive. And when we sum moments, we're going to say that it is equal to zero. So let's proceed through this. And we'll use this process in order to determine where the center of pressure actually is. So as summing moments, what we have on the left hand side of the equation is zero. On the right hand side, we have this. We have the resultant force multiplied by Ycp. Right now we don't know what that is. That's our unknown that we're trying to solve for. And it's a negative because if you look at the sense of that, it's going to be in this direction, which is counterclockwise. And we add in the integral of Ydf. And that would be this force here. And it is acting in a clockwise sense. And consequently, it is positive. And so I can re-express the differential element force as being the pressure at that particular differential element multiplied by the area. And we'll expand that out. And here again, Pa. If we have a free surface above here, P atmosphere equals Pa. And we have liquid up here. And that is above our plate. So we have that for summing the moments. We'll continue on in the next slide here. Looking back, we have this. So I'm going to expand that term out. And if you're wondering where I got this from, go back and review the last segment because we did a coordinate transformation in order to figure out how to solve for that. So that's where that came from. And now what we can do, let's rewrite this out. So we get this expression here. Now, what we're able to do, if you recall, let's look at our schematic here. We said that the center of area where our coordinate system was based, and consequently, this and this by the definition of the center of area from y, d, a is equal to zero because our origin is about the center of area. And consequently, those two terms disappear. And what we're left with is just this last term here. And it turns out that the integral is a special integral that we have in that equation. And that is the area moment of inertia. And it is with respect to the center of area of the plate. So with that, what we're able to do is we're able to write out our equation again in this form. Rearranging, remember what we're after. We're after the location of the center of pressure. So let's rewrite that. So ixx is being divided by the resultant force. And if you look back in the last lecture segment, what we said was that that was equal to the pressure at the center of area multiplied by the area of the plate. And consequently, this then becomes the y-coordinate of the center of pressure with respect to the center of area of the plate. So that's the y-coordinate. Now let's continue on to try to find the x-coordinate. And for that, I'm just going to give you the result. So we have this expression here. And ixy, again it's the area moment of inertia. And this is with respect to center of area. So be careful. Some books will not necessarily reference it with respect to the center of area in this derivation I have. Sometimes they reference it to the free surface. So if this is your free surface and your plate is down here, sometimes they put their origin up here, which would be quite a bit different from what we're looking at here. But anyways, that's just something to be aware of. In this case, all of the derivation was with respect to the center of area. And we've just determined where the resultant force acts. And that is the center of pressure. So where do you get ixx and ixy? Well, those are listed in any book on fluid mechanics that you look at. And if you're doing pressure and forces on submerged planar surfaces, they will give you values for different types of shapes, circles, semicircles, triangles, rectangles. So you can find those in pretty much any book that you look at on introductory fluid mechanics. And this is for common shapes. So what that concludes is the process of determining the, first of all, the force that acts on a planar submerged surface. And then we followed it up by determining, we got fr to begin with. And then we got xcp and ycp. And what we find is that the center of pressure is different from the center of area. And it turns out the way these equations are formulated, you'll notice we have pressure in the denominator here. The further you go down below the surface, what will happen is the center of pressure and the center of area start to move together. Well, what happens? The center of pressure moves towards the center of area and they become almost the same. But anyways, that gives you some equations that you can do calculations with. In the next segment, we'll take a look at an example problem of applying these equations.