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A swimming pool is 20 ft wide and 40 ft long and its bottom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9 ft. If the pool is full of water, find the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool.

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(a) 5625 $\mathrm{lb}$(b) 50625 $\mathrm{lb} \quad$ (or $5.06 \times 10^{4} \mathrm{lb} )$(c) 48750 $\mathrm{lb}$ \quad$($ or $4.88 \times 10^{4} \mathrm{lb}\right)$(d) Net force $=1.347 \times 10^{6}$ Newtons $=3.03 \times 10^{5} \mathrm{lb}$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 3

Applications to Physics and Engineering

Applications of Integration

Missouri State University

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

17:27

A swimming pool is 20 ft w…

01:16

A swimming pool is $20 \ma…

Yeah, nothing. And this question we're asked to we're given a pool and we want to find hydrostatic force in a bunch of specific places at the shallow end of the pool, at the deep end of the pool on one side of it. And at the bottom in order to do that, let's start with finding the hydrostatic force at the shallow end in general. There's this formal here, which is the pressure times area gets the force. So what we should do is, let's reorient our drawing a little bit. Yeah. Let's see your entire drawing, assuming that the water is going to be on the left side. Let's draw that in blue. Yeah. Our pool is going to look something like this. Mhm. I like that. So it'll look something like this. And we're going to be assuming that we're looking at where the 20 ft are. So this height is going to be 40. This is three length. This is a nine length. And let's say we are looking in the direction of 20 ft. Yeah. Mhm. Because all of these dimensions are in feet, right? Secondly, Since we're given, since we're given that we're using imperial units, Okay, We know that we can say that the density is that our density, our unit, our weight density is two rho times G Which we're going to call to be 62.5 mm hmm. This transforms our force equation to delta times D times a this Okay, now we know adult is now we just need to find the distance and the we need to find the distance and the area notice here. This is the distance that we're searching for and in this case it's just having a length of X. So we know that as well. We know that the area in general of some representative strip, which I'll join green. It's going to be something like this. We know that it's, we know that such a rectangle will be three by 20. Okay, Okay. So that would be the area of such a strip of any strip. So the area of any strip here would be up to this rectangle here. We'll call that length delta X. Yeah. So the area at any representative sub point at any representative depth is going to be 62.5 times X. I right Times 20 times delta x. That would be the force at one particular at one particular sub interval. And of course if we take the limit and some all such forces, hey, we arrive at a definite integral. Mhm And it will be going from 0-3, we would have 62.5 Times 20 times x. Dx Doing the simplification, they will lead us with 1250 times X squared divided by two, evaluated from X equals 02, X equals three. And because X zero and a polynomial it vanishes. So I can just plug in three into the, I can just plug in three into X. And when I multiply this, I'm going to get that this is going to be equal to 5006 £25 right now, I want to find it at the deep end. Okay, the setup is almost the same. Okay, the setup is almost the same. Okay, so looking at our diagram, once again, are a width would be here. This would be our depth or a distance. If you want to call it a distance, I call this the one and I'll call this d to say. And we know that distance as well. It's also going to be X. So yeah, it's gonna be so are integral. Are forced into girls want to do the some limit stuff. It's going to be almost it's going to be exactly identical to our first integral. Except the bounds will be going instead from 0-9. Yeah. And when we do this integral, of course we'll have this is equal to 1250 times X squared over two from X equals 02 X equals nine. And of course the same story, the zero will vanish. So I can just plug in nine into this X. And when I evaluate it we will get this is equal to the force at the deep end is 50,006 £25. Now we also want to find the Yeah, yeah. We also want to find the hydrostatic force on one of the sides. Mhm. In order to do that, let's let's draw our coordinates again. Just I'm going to do this. So I don't have to scroll up all the time. Yeah. So to look something like this, Once again we have our usual distances 49 3. And again, we're going to be looking at it from the 20 edge perspective. And let's say we want to look at it from this side here. Yeah, of course, as usual, the depth at any point is just going to be X. However, Okay. However, suppose I want to find the equation of this line here in green. The dotted line. Yeah, we know that. The equation of the starting line Is y equals 20/3 x minus 20. Right. Yes. But what's the significance of that? What we want is we want to find our representative delta X if that makes sense. So essentially what we're doing is we're finding the hydrostatic force up to the point in red and then the hydrostatic force from the red point onward. So, what this means is that we're going to have to be subtracting, we're gonna be subtracting 40 from this wide distance here When we do so we'll get why is you could do 60 this. Okay, now what would be our rectangles for one point? You need to find? Two integral. Mhm. Mhm. Yeah. So starting from the first integral, we can say that we're going to make our rectangles look something like this. We know it has a width of length of 40 and a width of Delta X. So such and so such a representative. Some a representative rectangle will have to be summed up and arranged like we had in the formulas above. So it would look something like this would be the integral from 0-3. You would have 62.5 Times X Times 40 dx. Right? Which when we multiply this out, This will be equal to 62.5 times 40 times X squared divided by two from X equals 02, X equals three. Obviously because of polynomial can't take zero, can take zero values, it becomes zero here and we can say that this force here, It's going to be equal to if I plug in three, this will be equal to 11,000 to 50 lb. No. Now as for the second force we're going to need to make our approach a little bit different. If I was to take a representative point here for instance, let me scroll up a little bit more notice the height is going to be just the height of this rectangle. That is We would have 60 -20/3 X would be this is the height of such a rectangle at any point in this interval We would know that the forest then would be going an integral from 3-9 Except we have to have a 62.5 times X times 60 -20/3 x dx not a delta X. That's a dx Yeah. After you do all this summing things. Yeah, We can factor things out of this integral that if I factor out a 20 for instance And all of the constants would have 62.5 times 20 Times the integral from 3 to 9 of 20. Or we would have three x. Yes, we would have three x. Okay. Three x minus. And we'll also distribute that exit to it minus one third. X squared. Yeah, pretty easy to evaluate this integral because all it does is require the power. So when we do so we'll have three X squared divided by two minus 1/9 X cubed. This will be evaluated from X equals 32 X equals nine. Okay, when we do so we'll do our substitution is now so we'll have three times nine squared over two minus nine cubed over nine minus 62.5 times 20 times three times three squared over two minus one. Our three cubed over nine. So that would be our substitution. Yeah. Now, when we do all of this evaluation, We're going to get this is equal to £37,500. So the total force would be equal to the sum of both of these two total forces. We found that is this number and this number here. Okay, So you can add them together and we're going to get this will be equal to £48,750. Now the last part is we want it at the bottom. Yeah. Yeah we want the hydrostatic force at the bottom of the pool. Once again I'm going to redraw the diagram diagram will look something like this and once again yeah. Once again let's focus on something else. Well maybe not like that. So it'll look more like it'll look more like this. Okay. Yeah. Yes. Now let's let's divide it here at the point where the where the pool would begin to slope and let's consider an angle data and as well as some small distance dx D. L. Yeah we would want the high park. Basically what we want to do is we know we have a rectangular strip. Its length is D. L. Yeah and its height or width or whatever it's with is 20. So the area is just 20 DL but what is dl well let's consider a triangle. Let's consider the triangle that we had drawn earlier where we had the angle theta. Dx and D. L. Wow. Clearly the coastline represents this because I can form a nice little trig and metric relationship here. So we have a coastline of feta would be equal to dx D. L. Yeah. Which means that if I solve for D. L. Would be equal to the second of data. Dx. Yeah. Yeah. Now obviously obviously here if we have the large if we use some similar triangles, what we have is this same angle theta. We have the base of it being 40 and the height being six. So what we can obviously find from this from this angle here we have the tangent of theta would be equal to would be equal to Uh huh. six divided by 40. Which means that data would be equal to would be equal to the arc tangent of 6/40. And let me just call that angle alpha so that we can work with that later. So we'll keep that in mind. So substituting this back into here, we have that D. L. Would be equal to the second of alpha dx. The next thing we need to do is we need to find the depth. We need to find some depths of water. Okay, so let's draw this. Let's consider another triangle or not? Let's not another triangle. Let's label some points with some arbitrary ones. Let's call it A B A C B. And then arbitrarily let's just call that point D. Let's call that point E. Well obviously if you notice, you know that A C A B A C, it's the same thing. Has to be proportional to B. D. D. E. Okay. And if we label these arbitrary side lengths X. And why, respectively. Okay, what this means? Is that What this means is that we would have 40/6 is proportional two x over y. And solving that we get an equation, we get a linear equation for a line. Yes. And since this is always going to be less than three, more than three ft deep at any point in time, We can say that the depth at any point is going to be three plus. And if I simplify this, this would be 3/20. Yes. Now with that being said, We're now allowed to set up our force integral. It would be the integral from 0 to 40 of Delta, which is 62.5 times D three plus 3/20 X Times 20 times the second of alpha dx. Yeah. Yeah. And evaluating this is very straightforward Because all we need to do is bring out constants. So it would have 62.5 times 20 times the second of alpha would be equal to would be multiplied by three X plus three X squared divided by 40 Evaluated from x equals 0, 2 x equals 40. And obviously if I plug in a polynomial, A zero into a polynomial, it will just cancel. So if I plug in 40 into everything and substitute back our value of alpha, you'll get we will get that our force is equal to £303,003 55. That's how you do this question

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