 Okay, what we're going to do in this segment, we're going to solve an example problem involving a planar submerged surface and we're going to be looking at the forces on that surface. Let me begin by drawing out a schematic and then I'll discuss what the problem is and what we're looking for. Okay, so I think this specifies the schematic for our problem. What we have is a plate submerged below a surface. Within that surface we have water, so the water is distributed around here. On the other side we have atmospheric pressure, so there's air on the other side. And what we are told to look for, we are asked to find the force required to hold this gate closed. And so the force is going to be acting right here. And we're told to find, so we're told to find the force F required to hold this gate shut. And consequently you can imagine you're going to have pressure build up here, that pressure is going to be acting on this side of the gate and that's going to be trying to pop it open. We have to hold it closed. And so that's what we're trying to solve here. How are we going to do it? Okay, we're going to use the equations that we just arrived for determining the force that is on a submerged planar surface. We have a planar surface here, so that will work fine and we'll also use the equations telling us where that force acts, which acts at the center of pressure of the plate. We're going to begin by drawing out a free body diagram. Again, fluid statics is a lot like statics that you've looked at in other courses. So we begin with the free body diagram. We have the hinge and if you're called from statics, hinges have two different components of forces on them. And then we have our plate. We have the force trying to hold the plate shut. What other forces do we have on here? Let me draw. This here is the center of area and we also have yet to be determined the center of pressure and that is where the force due to the fluid is acting and we call that force FR. So that is the force. I'm drawing it normal. It is going outwards and so that is the free body diagram XCP. We have not yet determined but that would be XCP, YCP for the center of pressure. Okay, so let's take a look at the gate area. We're told that it is a square gate and consequently we have the center of the area located symmetric at that location. B is two meters wide and it's two meters in the vertical and I'll define those as being B and L. So we know through geometry four meters squared. Now the other thing that we're going to need, we're going to need the moments of inertia about X and Y and so go to your fluids book, look those up for a square plate and what you'll find is the following. And IXY in this case is zero because it is symmetric but what we'll do we'll plug in the values for IXX and when we do that we get four over three and the units of that are going to be meters to the fourth. So we'll carry that through in our calculation when we're looking for XCP and YCP. So let's continue on now and what we're going to begin with is trying to resolve what is the force due to the hydrostatics and if you recall the force itself is equal to the pressure at the center of area however that force acts at the center of pressure. So the pressure at the center of area is equal to the atmospheric pressure plus RhoGHH down to the center of area of the plate. I can pull the Rho and the G out, let's look back at our diagram, I'll go two back. So what we're trying to do, we're trying to find, the center of area of the plate is right there, it's symmetric, we're trying to find the height to that location. So really what we're trying to solve for is this HCA. So we can see we have one meter plus whatever it translates into to give us the center of that location. So let's go back to our calculation here, we have the one meter plus and now we're going to use some trigonometry looking back. We have the angle here, we know that that's 30 degrees so we can apply trigonometry and figure out what this distance is here and that turns out to be 2 sine 30 and I'll divide by two because we're going to the middle of the plate and the plate is two meters long. So with that, what do we get? We get PCA, so the pressure at the center of area is pressure atmosphere plus RhoGHH times 1.5. Now the thing that we said was that P atmosphere is on the outside of the plate so the P atmosphere is just going to drop out and with that what we're ending with for the pressure at the center of area is RhoGHH times 1.5. Now that pressure, we now need to figure out the force or the location of the force and that force is going to be acting not at the center of area but at the center of pressure. So let's work on the center of pressure now. So I've introduced the values for the y-axis center of pressure location and with that what we then obtain is yCp is minus 0.111 meters and remember this is with respect to the center of area. So if we look at our schematic which was back here, the fact that it's coming out negative, the y direction was in that way. The fact that it's negative that means that it's below. This was y positive and this is y negative so it's down below in terms of the location for xCp. This is an easier one but I'll write it out explicitly anyways. We have Ixy Pcaa and we found that Ixy was equal to 0 because the plate is symmetric and consequently xCp is equal to 0 meters. That is with respect to the center of area that just means that it's at the same location as the center of area for the y-coordinate. So we've now determined where the force is acting. We figured out that it is located here, 0.111 meters below the center of area. So with that what we can do, let's go back and determine first of all the hydrostatic force because where was our free body diagram? Okay, our free body diagram is here. What we're trying to figure out is what is this force required? We still haven't determined that. That's the force due to the fluid, FR. So let's calculate the hydrostatic force. So we get this equation. We plug in the density of water, 1,009.81 for the gravitational constant and then add the other terms which was h, that was the depth, and then 4 was the area of the plate. And what we get is 58.86 kilonewtons is the force. That is the force due to the hydrostatic water above the plate. Now what we want to do, if you recall, the thing that we're after is we're after this force here in order to keep the plate shut. So we're trying to find this force. This was the unknown thing. We know everything else in the problem. So what we can do now, we're going to go to statics again, and we're going to sum moments about the hinge. Because if you look back here, why are we doing that? Well think about with statics. Whenever you do statics, we don't know hxhy. I mean, we probably could solve for them, but we're lazy. We don't want to do that. We're going to sum moments about that. We don't have to worry about hxhy. We know FR. We don't know F, but from this, by summing moments about h, we're then able to figure out what that restoring force is or the force in order to hold it shut. So let's do that on a new slide just to make it cleaner. We're going to sum moments about the hinge on the plate. Okay, so that's the geometry for the problem and the forces there. So we're going to sum moments. What I'm going to do, I'm going to do positive for being counterclockwise this time. Remember in statics, just as long as you're consistent, that's the main thing. So if we're looking counterclockwise as being positive, that means that FR is going to generate a positive moment and how far away is it from h, the point where we're summing? It's 1 plus 0.111. And then the restoring force or the force that's holding the plate shut is going to be operating in a clockwise manner, and so it will be negative. And how far away is it? Well, it's the length of the plate, which is 2 meters. We get that equation there. We can then solve for the force required to hold the plate shut, which is this one, the only unknown because we've determined that already. And what we get is 32.7 kilonewtons. And if you really want to be specific, you could say down, but that should be pretty down and to the left in case somebody gets upset that you haven't shown it as being a vector. But it should be pretty intuitive. You have to push it down in order to keep that plate from opening up due to the hydrostatic pressure on the other side. So that is an example problem of using the forces on a planar surface. We've used the forces. We've used determining where the center of pressure is, and we found that it's different than the center of area. And it gives you an example of how to apply it.