 The dam shown below is a quarter circle 50 meters wide into the paper. Determine the horizontal and vertical components of the hydrostatic force against the dam and the point Cp where the resultant force strikes the dam. So good news and bad news. The bad news is we can't use the same analysis we have been using. We don't have a good way of describing the angle at which the surface lies relative to the surface of the fluid, plus we don't have any good way to try to figure out what the relevant moments of inertia would be. So instead we break this into two chunks. Instead of figuring out the force acting on the dam all at once, we figure out its vertical component and its horizontal component. The vertical component is easy. The vertical component of the weight acting on the dam is just the weight of the water. So I will call that the volume multiplied by the density multiplied by the gravity. The volume of this water multiplied by the density of that water times gravity yields the weight of the water that is in this region, which is going to represent the vertical component of the force acting on the dam. For the horizontal component, we just project the dam over to this line. We consider what the water sees. So we treat this as just a rectangle. And that rectangle is the worst drawing ever. Try that again. And that rectangle has a height of 20 meters and a width of 50 meters. And we can use that same approach that we have been using where we figure out ycp, we figure out the center of gravity, we subtract the two to represent how far it is from the surface, etc. But there's more good news. Because that pressure is always going to act normal to the surface, all we really need is the magnitude. Because our magnitude is going to include a horizontal and vertical component that would fix cp. Makes sense? I think it'll make more sense once we get into it. So the vertical component of the force is just going to be the volume multiplied by the density of water multiplied by gravity. We are assuming standard gravity, the density of water is going to come from table A1 in our textbook. We are assuming standard temperature and pressure because we don't have enough information to deduce otherwise. The density of water at one atmosphere and 20 degrees celsius, which is what we're using as standard temperature and pressure, is 998 kilograms per cubic meter. Everything's in metric, so let's find the used metric, 9.81 meters per second squared. And the volume here is going to be the volume of a quarter of a cylinder that has a radius of 20 meters and a length of 50 meters. So I'm going to say volume of this is one quarter times the area of a circle that has a radius of, again, 20 meters. So pi r squared multiplied by its length, which is 50 meters. That will give us our vertical component of the hydrostatic force. One quarter times pi times 20 squared meters squared times 50 meters multiplied by 998 kilograms per cubic meter multiplied by 9.81 meters per second squared. And since we are talking about a bunch of water, we probably don't want to use newtons as a relevant force unit here. We probably want at least kilonewtons. So a kilonewton is 1,000 newtons. And a newton is a kilogram meter per second squared. Kilograms cancel kilograms, meters, meters, square meters, cancels cubic meters times meters, newtons cancels newtons, second squared cancels second squared, leaving me with kilonewtons. So sum in the mighty calculator to help us out here, we have C. Yes, thank you calculator. We have pi, or which there should be a faster way of typing, times 20 carat two times 50 times 998 times 9.81 divided by four times 1,000. And we get 153,787 kilonewtons, 454 ish mega newtons, which is much more fun sounding. So this isn't our answer, so I won't circle it, but I will underline it so that we know to come back to it. And then for our horizontal component of force, we're just imagining yet another gate problem, is 50 meters long, 20 meters tall. And the force here is just going to be gamma times hcg times area. Gamma is density times gravity, which we already know, hcg is going to be the height from the surface of the fluid, which is the top of the gate in this horizontal analysis. Remember, again, we are considering the horizontal perspective of the fluid. So hcg is just going to be half of the radius, which is 10 meters times the area of effects, which is 20 meters by 50 meters. Therefore, our horizontal component of force is going to be 998 kilograms per cubic meter times 9.81 meters per second squared times 10 meters times 20 meters times 50 meters. And we want kilonewtons again, so a kilonewton is 1000 newtons and a newton is a kilogram meter per second squared. So newtons cancel newtons, meters, meters, meters, meters, cancels, meters, cubic meters, kilograms, cancels, kilograms, second squared, cancels, second squared. And I'm left with kilonewtons. So this quantity goes 998 times 9.81 times 10 times 20 times 50 all divided by 1000. And we get 97903.8 kilonewtons. So our actual hydrostatic force, that is the answer that we actually care about, is a force, okay, straight line, yeah, is a force which has a vertical component and a horizontal component of 15378953789 kilonewtons and 97903.8 kilonewtons respectively. Therefore, the actual answer, which I'm just going to call f, is going to be the square root of fh squared plus fv squared. So the square root of 153787, the power of 2. Why are you rounding calculator? I don't get you. You shouldn't be rounding. No one is telling you to round. I know that you have a mind of your own and you're very fickle, but what gives? Exploited digits. Float 6. Float 6, calculator. Whatcha doing? Okay, I'm going to change this to exact. Apparently I'm not. You know what, it's fine. Plus, this quantity raised to the power of 2. And we get 182,306 kilonewtons. And of course, there's probably decimals, but my calculator refuses to tell us what they are. 182,306 kilonewtons, or 182.31 meganewtons. Plus, the answer to part A. The answer to part B is a little bit more complicated. We could figure out where the vertical component is acting and where the horizontal component is acting, but it's easier to consider that this vector has an angle. And this angle, which I'll call beta, I don't want to call it theta because I don't want to confuse it with the theta that appears in the YCV and XCV calculations. So I'll just call it beta for now. That beta angle can be calculated because we know the tangent of beta is going to be opposite over adjacent. So 153,789 over 97903.8. Therefore, beta is going to be the arc tangent of that proportion. And I'm going to calculate that just for fun here. 153,789,787. Whoa, whoa, whoa, whoa, whoa. That changes nothing, but I still, yeah, what happened, John? So beta is 153,787 divided by 97903.8. We get one radian, which we could calculate in degrees if we wanted. Yep, thanks calculator. Those are radians, two degrees function, but for any infrabound 57.5 degrees, 57.518 degrees. So the logic here is we know the hydrostatic force is acting normal to the surface regardless of the angle of the surface. So we could figure out the position of point Cp by figuring out the point on the dam where the orthogonal angle relative to horizontal is beta. So we can approach this by considering the surface of the dam as a function on an x and y axis here. And we can describe that line, this function as the equation of a circle that is scooched up, I believe is the technical definition, translated in the y axis by a distance of r, the radius. So the equation for this circle is going to be x minus the horizontal offset, which is zero squared plus y minus, excuse me, y plus the vertical offset, which is r minus vertical offset r squared is equal to r squared. And then this would be, let's see, if I wanted to solve for y, I can write this as y minus r quantity squared is equal to r squared minus x squared. And then I'll take everything to the one half power. So I'll write y minus r is equal to square root of r squared minus x squared. Therefore, y would equal plus r r squared minus x squared plus r. Why are you doing that iPad? And then what we're describing is the point on that function, which has a certain slope. So we can take the derivative of that. For that, I will move on to a new sheet of paper. So we are looking for dy dx, because we're saying we want a certain slope of this line. And dy dx would be the derivative of, okay, so we are calling this derivative of r squared minus x squared dx plus the derivative of a constant x. Okay, and then memory serves. We're going to have to use the chain rule for this. Oh, boy. So fx is going to be the parent function, which is where root of x inside is going to be r squared minus x squared. Yeah, okay. And then the derivative of fx with respect to x is going to be the derivative of x to the one half power. So we're bringing down one half, we're calling this one half times x to the negative one half because we subtracted one, which is one over two times the square root of x. And then dx dx is going to be the derivative of r squared with respect to x minus the derivative of x squared with respect to x derivative of a constant with respect to x is just going to be zero. This is zero minus and then the derivative of x squared with respect to x is going to be two x. Yeah, just two x. And then this is going to be, let's see, f prime of g of x times g prime x. So we're taking f prime, which is the derivative of f with respect to x one over two times the square root of no square root of r squared minus x squared times zero minus two x. So we have zero minus two x, which is just negative two x divided by two times the square root of r squared minus x squared. Then I can cancel the twos for good measure and I have dy dx is equal to x over square root of r squared minus x squared. And we are looking for the slope of this line that is equal to this is a 90 degree angle as a horizontal component from the left side that is 57.518 degrees. And then this quantity over here, which I will call alpha, is going to be 180 degrees minus 90 minus 57.518 degrees, which is just going to be 90 minus 57.518 degrees, which is 32.482 degrees 32.482 degrees. So the slope of that would be the rise over the run. Yeah, because I'm talking about slope of this line. So, and that would be rise over run is equal to the tangent of alpha. So we're saying the tangent of this quantity here in degrees, which is 0.63663. So that represents the proportion, which makes sense. And we are solving that using this relationship. So we're saying poorly written dy dx. I'll just rewrite that because we're more zoomed in now. It's fine. dy dx is equal to, and that was x over r squared minus x squared inside of a radical squared minus x squared. And this is equal to 0.6366 x. No, 6, 6, 3. We know our radius is 20 meters. What is x? Well, we can say x is equal to 0.63663 times the square root of r squared minus x squared. And I can write that as 0.63663 times r squared minus x squared to the negative first power. I mean, one half power. Actually, let's let's try this a different way. Let's just call this, okay, we've used alpha, we've used beta gamma mean something. Let's call this delta for now. So we're calling this delta out front. And I am going to get rid of that radical by squaring both sides. So I have x squared is equal to the quantity delta times square root of r squared minus x squared. That's not a square quantity squared, which I can remember that a times b squared is equal to a squared times b squared. So I can write this as x squared is equal to delta squared times the quantity r squared minus x squared. So r squared minus x squared would equal x squared divided by delta squared. So x squared divided by delta squared plus x squared is equal to r squared. We're out of algebra. So let's just move this to a whole different page. Page number two. Always a good sign. No, no, no, that's dotted paper that we want not squared paper. That's better. Okay. Who doesn't love these brief forays into the algebra dimension? Am I right? x squared times one over delta squared plus one is equal to r squared. Therefore x is equal to the square root of r squared over one over delta squared plus one, where r is equal to 20 meters and delta is equal to what was that calculator wake up 0.636629. So we get square root of 2020, 2020 squared divided by one over this quantity squared plus one. And if the calculator would add, we get 5.76812. So this center of pressure is going to be applied at a position that is 5.768 meters in. And since the function of that line was y is equal to square root of radius squared minus x squared plus r, we could calculate that y position as well. So that would be square root of 20 squared, 20 squared minus this value squared plus 20. That doesn't seem right. And we forgot to wrap that in a radical non real result. Cool. Ah, yes, I used the negative sign instead of the subtraction sign. And we get 39.15. And that doesn't seem right. That'd be way above the dam. Oh, it's because, of course, when you're talking about the equation of a circle, there's going to be a top half and a bottom half and the top half would have the positive and the bottom half would have the negative. That's just why we're getting such a high value. So in order to make this the bottom half of the circle, this would have to be minus and then we're talking about and that changes nothing about this except that there is except this is positive instead of negative, which is what I did in the first place. That's interesting. Sometimes two errors will they do cancel out. That's fun. Yeah, because I can just bring out the negative. Anyway, it doesn't affect my x position, it just affects my y position. So if I were to calculate that value, the minus out front, I get a syntax error minus out front, I get 0.89. That seems much more reasonable. So to be clear, I'm not expecting you to actually calculate this information if this problem were on an exam. I would be much more concerned with the magnitude of the forces. So if you weren't following what I was doing in the second half of the problem, don't worry too much about it. Like I said, I'm trying to make the focus of this class, the actual fluid mechanics analysis and not put so much emphasis, especially on a time constrained assessment like an exam on the algebra and the calculus. You should still be able to figure it out and you're still going to have to do enough algebra and calculate to solve problems, but I'm going to try to take the emphasis off of that. So again, if this were an exam problem, I wouldn't be expecting you to find the position of that applied pressure.