 In this video we provide the solution to question number four for practice exam number three for math 1220 In which case we have a hydrostatic force integral that we have to set up We do not have to evaluate it, but we do have to set it up So what's the information we're given we have a trough that is filled with a liquid of density 840 kilograms per meters cubed some things to note here is that this liquid is not water because that is not the density of water But the units are also kilograms per meters cubed That means in order to calculate the weight of this liquid We are going to have to incorporate the acceleration due to gravity which is 9.8 9.8 meters per second squared, which is left. It's mentioned here as a reminder in case you forgot that Um, we aren't we aren't going to be told the length of the trough because when it comes to hydrostatic force problems It really doesn't matter How much water is behind it? Because of laws of fluid dynamics And so that the this trough has a equilateral Triangle on both sides for which each side of the triangle is 8 feet long and the vertex is at the bottom So it's like I have drawn here on the board on the slide. I should say It'll be relevant to us in just a moment that we're going to need to know the height of this thing Let me try that again Again, my diagram is not a perfect equilateral triangle, but we're going to be very interested in the height of this thing Be aware that because this is an equilateral triangle this altitude Which meets the other side at a right angle does cut the other side in half So that this distance right here is 4 Also, since it's an equilateral triangle, each of the angles is 60 degrees for the whole triangle This is a right angle like we said it was an altitude and by symmetry of equilateral triangles It cuts this in half so that 60 degree angle gets bisected into a 32 into two 30 degree angles And so in particular this triangle right here is a 30 60 90 triangle So using some simple trigonometric properties if this side is four and this side is eight Like you can use the Pythagorean theorem for example We're going to see that the other side is four times the square root of three that'll be useful to us in a little bit That's that's about as extensive as the trigonometry is going to be for us Now remember when we do hydrostatic force problems, we're using the fact that force is equal to pressure times area The pressure for these problems is always fairly simple The pressure is going to be these constant numbers. We have this 840 kilograms per meters cubed Times by the 9.8 Meters per second square then we have to multiply that by the depth So depending on how we set this thing up the depth could be a couple different things The way I like to handle hydrostatic force problems is I'm going to put X I'm going to call my variable x. I usually just like to call it x x equals zero And then I'm going to orient my axis so that x equals zero is at the surface of the liquid And then I'm going to point it downward so that depth is actually x in that situation Here's our x-axis so we observe that depth is just equal to x In this situation and therefore our pressure is just going to be this 840 times 9.8 times x now you can actually multiply 840 times 9.8 if you want to but if you leave it factored I'm not going to have any problem with that in this integral whatsoever So that takes care of the pressure part then the area, okay when it comes to area We look at a typical cross section at depth x right here. So the distance from the surface downward is x The thickness of this is going to be a small change of x. That's going to be a dx So then what we're left with is the width of this thing. We're going to call that w for a moment So the area is equal to w times dx Putting this all together. We see that the hydrostatic force is going to be the integral of pressure times depth Excuse me pressure times area that the pressure was this 840 times 9.8 times x The area was then w times dx like so We're integrating the respect to x do make sure that you have this dx as part of the integral that is necessary for full credit As we're integrating the respect to x we then ask ourselves What is the depth of this thing like how did how far do we go? We go from the top to the bottom and so we're going to integrate from 0 to 4 times the square root of 3 So those that bound did come into play there Um, so this is pretty good. The the next thing we need to deal with of course is the w itself I mean you could factor you can pull the 840 times 9.8 out in front All right, but we still have to deal with that w here. So I'm actually gonna I'm going to move this down here. I'm going to pull this out 840 times 9.8 Times the integral from 0 to times 2 4 times root 3 you're going to have that x there So what's left here is then that The w how do you express w? To handle w we're going to play around with some similar triangles If you look at the whole triangle and actually I don't necessarily want to do the whole triangle Well, no, no, we we'll do the whole triangle. You have of course this whole triangle right here You cut in half if you want to look at this triangle, that's fine as well If you look at the whole triangle, you're going to have that on the top four Excuse me eight is the whole length coincides to the height, which is four root three On the other side The whole length right here coincides with w So we get w there and that coincides with its height Its height right here, which notice the whole distance is four root three This distance is x so the difference will actually be four root three minus x So if we solve for w, you'll get w equals Well eight over four root three there you can cancel you get two over the square root of three times four root three minus x like so we're going to plug that in for w And so we end up with two x over the square root of three times four root three minus x like so If you want to distribute that two x over the square root of three through you can do that You can multiply these things out too as well, but those won't none of those things will give you extra points This we could argue is a set up and sufficiently simplified integral that will measure the hydrostatic force against one side of the One of the triangular sides of this trough