Continuing our discussion of fluid statics. The topic I want to address now is forces on submerged plane surfaces. And, the question is this. Suppose we have some surface which is submerged, say below water surface. For example, the water in this tank here and for example the bottom surface here. If this is horizontal, the pressure there is horizontal, the pressure there is constant rather. So the pressure, the force there is just the pressure multiplied by the area. And this force I can replace with some resultant force which I'll call FR. But, what about the vertical faces here? In this case, the pressure is increasing linearly downwards according to our hydrostatic pressure distribution. And, again, I will have some resultant force here, which is pushing on this plane, and in this case, it's going to be pushing to the left. So the most general question is, if I have some arbitrary plane surface of arbitrary shape, arbitrary inclination, what is the magnitude of the resultant hydrostatic force acting? Where does it act, and what is the direction of the resultant force? And those are the issues we're going to address in this segment. So to do this, I'll consider a completely arbitrary plane surface as shown here. And this arbitrary plane surface, if I look at this in edge view, then I see the view here. And the inclination of this plane surface is inclined at theta to the horizontal with the angle here. And the resultant force I will assume is FR at some location below the free surface. Now, because the pressure forces varying over this object I have to find the force on an element, and then sum the forces over the whole area to get the results in force. So, I'll take an element of the area here, which I'll call dA. Which is submerged below the water surface by a height, h, a depth y Where y is the depth measured down the sloping surface. And the vertical depth of this element below the water surface is H. So at any depth H the pressure is the hydrostatic pressure which is specific weight times H, in other words gamma times H. But pressure is force per unit area, so the force on that element, dF, is gamma h times the area, which is dA. And the total force on that plane's surface is the summation of the forces on these elemental areas, or in the limit, the integral over the area of gamma h dA. Now, we can evaluate that because from simple geometry, the vertical depth, h, is just y times sin theta, so substituting that in, we find that the result in force is the integral over the area of gamma y sin theta dA, and taking out the constants, Gamma sine Theta, we get this expression. But, this integral here, integral over the area of ydA, is by definition, yC times A where yC is one over a, integral ydA, which is the distance of the centroid of the area, below the surface, measured down the sloping incline. Therefore, the resultant force is gamma ayc sine theta, but yc sine theta, this term here It's just the vertical depth of the centroid below the water surface hc. So the resulting force is gamma hc x A, but this term gamma times hc is just the pressure at the centroid. So therefore we arrive at a very simple equation that the result in force is equal to the pressure at the centroid of the plane multiplied by the area of that plane surface. That is the magnitude of the result in force. What is the direction? Well I think intuitively you can see that the direction of this resultant force must be perpendicular to the surface. Why? Because if it wasn't, let's suppose that the resultant force was at some arbitrary angle, then there would be a component of force which is tangential to that surface. Therefore we would have a tangential or a sheer stress and, because of our definition of a fluid, the fluid would stop moving. Therefore, the only way for this to be in static equilibrium is for the resultant force to be perpendicular to the plane. Now the resultant force is the pressure at the centroid multiplied by the area, however, the resultant force does not act at the centroid. It acts at a location which I'll call the center of pressure. So here's the resultant force, FR, and it's acting at the center or pressure here CP. And the center pressure's defined as the point of application of the result and force in other words, where a single equal and opposite force must be added for that plane surface to be in equilibrium. And we find this location by taking moments about this point O, on the free surface. So, the moment of the result in force is Fr times yr, where yr is the location of the center of pressure, measured down the slope from the free surface, and that is equal to the summation of the moments on this elemental area. The force on the element is ydF, the moment arm times force, and the sum of the moments on all of those elements is equal to the integral over the area of ydF. But dF, we've already shown, is gamma y sin theta times dA. So this in turn becomes taking the constants out of the equation, gamma sine theta, integral over the area of Y squared dA. But this quantity, integral over the area of Y squared dA, is equal to the moment of inertia of the service about O which we'll denote by IXO. So therefore, combining those equations we find that the location of the center of pressure of the resultant force is IXO/ycA. But that formula is not very convenient because that gives the moment of inertia about this point. It's much more convenient if we can express the equation in terms of the moment of inertia about the centroid. And to find that, we use the parallel axis theorem, which states that the moment of inertia, about any arbitrary location Ix0 is equal to the moment of inertia about the centroid, plus the area multiplied by the distance squared between the centroid and the point about which the moment of inertia is taken. So combining that, we get this final expression. That yR the location of the central pressure is equal to yC the distance of the centroid plus IXC divided by yC times A where IXC is now the moment of inertia about the centroid. So here's that formula again, yR is yC plus ixc over yc times A and what you notice about this formula is that ixc is always a positive quantity, yC is a positive. A is a positive a quantity. Therefore, this term here is always a positive quantity. In other words, the distance of the center pressure below the centroid is always positive. In other words, the center pressure is always below the centroid, it's deeper. However, if I take this plane and push it downwards. So I take this plane and move it downwards, submerge it more deeply, IxC remains constant, A remains constant, but yC increases. In other words, as I go deeper down, this quantity decreases. It becomes smaller, in other words, the center pressure approaches the centroid as the plate moves down. And the distance here, this distance is the distance of the resulting force below the centroid. Now, this raises another question of course. What is the moment of inertia of a plane's surface? And this typically will look up in tables. This is an extract of the table from FE Reference Handbook and it gives moments of inertia and other properties for various shapes, such as rectangles, triangles, etc. And, this is a detail or a blow up of that table, for triangles and also rectangles. And in the next section we'll do some examples and applications of these surfaces to different types of problems.
