An aquarium of length L , width (front to back) W , and depth D is filled to the top with liquid of density p . (b) Find an expression for the force of the liquid on the front window of the aquarium.

Video transcript

Hey, everyone in this problem, we have a rectangular tank. So I'm gonna draw that out here. So that's the main face of it. And then here it's de here's its depth into the page. There we go. OK. And the problem tells us that the tank has a length of X. So the length is X as a width of why bit of the Y and that as a height of Z, we're also told that the tank has, it's filled with salty water with a density of row. The problem asks us to find an expression for the force experienced by a vertical face measuring Y by Z. So we're just looking for the force on the, on the area Y by Z. And we have four multiple choice option option, a one half row G Z Y, option B row G Z squared, Y, option C one half row G Z squared Y and option D two row G Z squared Y. OK. Now, normally when we're looking for the force due to a pressure, in more simplified cases, we can use the formula stating that the force due to a pressure is equal to the magnitude of that pressure multiplied by the surface area, the area of the surface that it's touching in this case. However, we can't just use that formula on its own because the, because the pressure in this case varies as a function of height, so the pressure is going to be equal to row G multiplied by the depth into the fluid into the tank, which in this case for us is Z. So row G Z A is what the force is going to be. So because the force is always changing depending on where we are along Z, that means that in order to solve this problem, we're going to have to do an integral, we're going to have to take some area of this face and then integrate it along the height of the tank. So what I'm going to do is I'm gonna take a little small strip, a little small, infinitesimally thin strip of area. And I'm gonna call it D Z since it's an element of the height of Z and that's gonna be the height of the strip. And because the width into the page of the tank is constant, it's just Y that means that the width into the page of the strip is also going to be Y. So that means that the area of this element D A is just going to be Y multiplied by D Z. So because F is what's changing, we're going to have to integrate ad F which is just going to be an elemental version of the equation that we wrote earlier. So row G Z multiplied by D A, the area element which expanding that out a bit is gonna be row G Z multiplied by Y D Z. So the force, the total force on the phase, we can find it by integrating it gonna be equal to F is it's gonna be equal to the integral of D F. And the bounds of the integral are just going to be along the height of the tank. So it's gonna go from a height of zero to height of Z because that's how high the tank is. So expanding this out a bit further, F is gonna be equal to the integral of row G Z Y D Z. So all we've got to do now is integrate this with respect to Z. So we got, got a Z term here. So the integral power rule states that we're just gonna have to raise it by one power and then divide it by that power. So that means that the Z is going to become a Z squared, then the whole term is getting multiplied by a one half. So one half G Z squared Y and then we have to apply the bounds from zero to A Z using the fundamental theorem of calculus. So if we do with that, then we have one half row G Z squared Y and then minus. And then we're gonna put in the same formula, but instead plugging in zero for a Z. So one half row G zero squared Y and because we're multiplying by zero squared, that entire formula disappears and goes to zero. So that means that our force is just one half row G Z squared Y and that is going to be it, that is the integral that we've been trying to solve for. And if you look at our options, we'll notice that option C is exactly that one half row G Z squared multiplied by Y. This was a problem that involved an integral. So I hope this video helped you out. And if you'd like to learn more, please check out some of our other tutoring videos that will hopefully give you more practice with these types of problems. That's all for now. And I hope you all have a lovely day. Bye bye.