The tank shown in FIGURE CP14.73 is completely filled with a liquid of density p. The right face is not permanently attached to the tank but, instead, is held against a rubber seal by the tension in a spring. To prevent leakage, the spring must both pull with sufficient strength and prevent a torque from pushing the bottom of the right face out. (b) If the spring has the minimum tension, at what height d from the bottom must it be attached?

Video transcript

Hello, everyone. Let's go through this practice problem in an underwater themed restaurant. A large decorative aquarium displays a captivating liquid with a density row. This tank has the shape of a rectangular prism with dimensions L for length, capital H for height and W for width. As shown in the figure A spring is used to prevent the front glass panel of the aquarium from being pushed out at the bottom by counteracting the torque on the panel, determine the minimum height above the bottom at which the spring should be fixed given that it has the minimum tension note, the tank has an opening to the atmosphere. We have four, multiple choice options. Option A one half of H option B, 1/6 of H squared, option C one third of a of H or H divided by three and option DH cubed divided by two, right? So I'm gonna be honest, this is a pretty tricky problem that involves some concepts that don't come up a whole lot. But the premise is fairly simple. We simply need to find the height, small H at which the spring needs to be positioned in order to keep the, the the panel of the tank in place. Since the tank is filled with water, there's going to be a force due to the pressure of the water pushing the tank outwards. So, so the spring needs to counteract that force from the pressure. So in order to determine the best possible height for the spring to be at, we essentially need to find the height at which where the center of the force of distribution is. And that's, and that can be a little tricky to figure out. But we have used similar concepts when we've talked about center of mass. Remember that center of mass basically tells us the center of a distribution of mass. And in the case of this problem, we essentially want to find the center of force, the center of the force distribution. So to find that center of force, we can use a method in an equation very similar to what we do when we analyze the center of mass of a of a more complicated system. So first, let's recall our center of mass formula. Now, in its most general form, the center of mass, the position of a center of mass XC uh sub CM is equal to the integral of whichever dimension we're looking at with respect to the mass all divided by that system's mass. That is the most general formula for finding the center of mass of a system. So we're gonna want to come up with a formula that's analogous to this but applies to a force distribution. But for our purposes, since we're looking for a height, that means that the dimension that we're looking at is the Z dimension. So the formula we're going to use, we're gonna say that the height we're looking for H is equal to the integral of the Z dimension with respect to the, for, with, to the force that is being acted on the panel from the pressure all divided by the total force on the panel due to the water. So at its core, this is the equation that our solution is going to be based around. But you can probably start to see now why this is going to get tricky because we're going to have to do an integral. And since pressure is variable dependent on depth into the water, just finding this force term is going to involve another integral of its own. And that's why this problem can be time consuming and lengthy. So let's get started. First, let's find some formulas that we're going to use for the force and the pressure now recall that the force due to pressure is equal to the magnitude of that pressure multiplied by the surface area over which that pressure is acting. Also recall that that pressure is a function of height. It's a function of the depth into a fluid. Specifically that pressure is equal to the density of the fluid row multiplied by the acceleration due to gravity, G multiplied by the depth into the fluid that we're looking at. So the depth from the top of the fluid. So for example, if we're looking at the pressure at some point halfway through the fluid, the depth is going to be the distance between the water's surface and at that point. So if we're going to look at some depth Z into the page along the axis where Z is measured from the floor of the tank, then that means that our height, our depth can be measured as capital H minus Z because the depth at any point is just going to be the same as the total depth of the tank minus the distance from the bottom that we're currently looking at. This means that in order to find the total force on the panel will have to do an integral of the pressure with respect to the depth in the tank. So this is the step where we want to make sure that we know how to properly set up an integral. So we want to choose an area element to integrate along the depth of the tank. And I'm going to choose that as I'm gonna choose that to be a little strip of area going across the depth of the tank. And I'll call this area D A, the differential element that we can integrate across this strip is basically a rectangle which has a width of W since that's the depth of the tank and it has a height of DZ which is in which is an infinitesimally small element on its own. So if we're going to integrate this element along the height of the tank, and that means that our force equation is actually an integral, it's going to be an integral of P with respect to A. Since the D A element is what we're moving up and down along the height of a tank, we also need to have the bounds of the integral. Since the integral is going to be done across the height of a tank, that means our lower bound is Z or zero and our upper bound is capital H the full height of the tank. Now, as we mentioned earlier, the D A element is essentially a rectangle, which means it has an area of length times width of the rectangle for our purposes. That is the width of the rectangle, which is just the depth into the page W multiplied by the height of it, which we discussed is the infinite decimal DZ. We can expand our force equation by plugging this W multiplied by DZ in for the D A. And then we can plug in our pressure formula into P so that we have something to integrate. So our expanded our new force equation is the integral from zero to H of P which I'm now going to write as row G multiplied by H minus Z multiplied by D A or W multiplied by DZ. So now we have an integral that includes a Z term, the height the along the Z axis. And it's all being integrated with respect to Z. And this is a fairly simple integral to evaluate which makes our lives much easier. So the first step in evaluating this integral is just going to be to pull out the constants that we're not going to use in the integral itself. So the row, the G and the W can all come out, this would be multiplied by the integral from zero to H of H minus Z with respect to DZ. And this is a case of a fairly simple integral where we're going to just want to use the integral power rule. So this integral is going to become row GW and then multiplied by, in parentheses, the H becomes H multiplied by Z then minus. And then we apply the integral power rule to integrate the Z. So recall that the integral power rule involves us raising the exponent by one and then dividing the term by the new exponent. So the Z will become Z squared since one plus one is two divided by two. And now we'll have to apply the fundamental theorem of calculus to get the bounds to be accounted for. So recall that to apply the fundamental theorem of calculus, we need to plug the upper bound in for the variable of integration. So in this case, we're gonna plug in capital H in for Z. So big H multiplied by Z or big H, in this case equals H squared minus then capital H and for Z so H squared divided by two and then we subtract the same thing. But with the lower bound of integration plugged in, so H multiplied by zero, it just goes to zero minus zero squared divided by two. That also just goes to zero. So we're just left, we're just left with the simple term of row multiplied by G multiplied by W multiplied by H squared minus H squared divided by two, which can be simplified pretty simply as row GW multiplied by H squared divided by two. So it can basically just be written as one half row GW H squared. That's about as much as we can simplify it. And this gives us a value for F and since F is a piece of the H formula that we discussed earlier, we're gonna want to hold on to this for now we'll move on because another, another piece of the H formula that we need to establish is the DF, the integral of Z with respect to F. So we want to find an expression for that as well before we start plugging these terms into our H formula. Remember as we discussed earlier that DF is equal to P, the pressure multiplied the by the area element D A which can further be expanded by plugging in our terms for P and D A as row multiplied by G multiplied by H minus Z multiplied by W multiplied by DZ. Now, in our H formula that we discussed earlier, we're taking the integral of Z with respect to F, the integral of Z multiplied by DF. And that is the numerator of our H formula. So I'm going to, for the sake of simplicity, I'm going to separately evaluate the contents of the numerator. So let's separately take the integral of Z with respect to DF. So that way we can just plug in whatever we solve for this as the numerator for our H formula when we finally go to solve the problem. So the bounds are going to be the same as usual, we're integrating across the height of the tank. So we're integrating from zero to capital H and this integral is equal to the integral of zero to H of Z row G H minus Z multiplied by WDZ. So we're now basically going to solve pretty much the same integral as before except the only difference being that we have an additional Z term right here. So let's get to work solving this as usual. The first thing we're going to do is pull all the constants out. So the row comes out, the G comes out, the W comes out and we're left with the integral from zero to H of Z multiplied by H minus Z with respect to Z. Now, to get this into a, a form that's more easily integrable, we can distribute the Z to both the H and the other Z. So we end up with a row GW multiplied by the integral of H multiplied by Z minus Z squared. So now we've made not simplified this a bit. And now to actually solve the integral, we just use the integral power rule again, just like you did with the previous one. So row GW this becomes, and then we apply the power rule. So H multiplied by Z squared and then raised divided by two just like we did before and then minus uh the originals A Z squared. So applying the integral power rule, this becomes a Z cube divided by three. And we once again apply the fundamental theorem of calculus just as we did before. So this becomes row multiplied by G multiplied by W multiplied by big brackets. We plug the H in for the Z. So that's H cubed divided by two minus the H cubed again divided by three, then minus for the lower bounds. So zero in for the Z that's H multiplied by zero squared divided by two that just goes to zero minus zero cubed divided by three that also goes to zero. So the whole thing we're left with is row multiplied by G multiplied by W multiplied by H cube divided by two minus H cube divided by three, we can simplify this further by taking the least common denominator. So the least common denominator for two and three is just six. So this can be written as row GW multiplied by then three H cubed divided by six minus. And then converting the next one into six that becomes two H cube divided by six, which can be written even more simply as row GW multiplied by just H cube divided by six. So the whole thing we're left with just reduces to 1/6 multiplied by row gwh cubed. And this is what the numerator of our earlier H equation will be. So now all that stuff for us to do is plug in our H equation into the H equation. We'll plug in the two things we found. So as we discussed earlier, H is equal to the integral of Z with respect to F divided by F. And as we just discussed earlier, the integral of Z with respect to F is 1/6 multiplied by G row W big H cubed. And this is being divided by F which as we discussed earlier is one half multiplied by row GWH squared. So now a bunch of things are gonna cancel out the rows, cancel out. The G is cancel out the WS, cancel out two of the HS will cancel out. And we're just lucky with a single H in the numerator. And then we divide 1/6 by one half and find that 16 divided by one half is just equal to one third. So what we're left with is one third H or H divided by three. And so that is the answer to our problem. And if you look back at our multiple choice options, we can see that this is what option C says. So option C is the answer to this problem and that is it for this video. I hope this video helped you out if it did and you'd like more practice. Please consider checking out some of our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you all have a lovely day. Bye bye.