The tank shown in FIGURE CP14.73 is completely filled with a liqu...

Question

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. (a) What minimum spring tension is needed?

Accepted Answer

Hi, everyone in this practice problem, we will have an aquarium containing liquid with a density of row where to maintain the position of the front glass panel and avoid leakage. A spring is fastened to the panel. The springs tension will keep the panel securely close and counteract torque forces that could potentially dislodge the panel. We're being asked to find the equation of the minimum spring tension necessary to ensure that the panel remains in place without causing any sort of leakage. The options given are A F equals to W open rents P knot H minus row G H squared, close parenthesis, B F equals W multiplied by open parenthesis. P no H plus half row G H squared, close parenthesis, C F equals W open parentheses. P dot H minus half row G H squared, close parenthesis. And lastly D F equals W multiplied by open parentheses. P dot H plus row G H squared, close parentheses. So we are also given the um figure or drawing of our aquarium which will help us to determine and help us answer this problem since the pressure of the fluid changes with depth. So we integrate to calculate the total force, we do have to employ integration here because the pressure of the fluid will change with the depth of the liquid in the aquarium. Because of this, the expression for the force is going to be F equals the integral of P multiplied by D A where P is the average pressure of the fluid. And D A is the area of the strip D A or the area of the strip based on this uh figure given is going to equals to W multiple multiplied by D Y. And the fluid pressure or B based on the fluid uh pressures formula will equals to P knot plus row G open parenthesis, H minus Y P knot is the atmospheric or original pressure. And row G H minus Y is going to be the pressure um as a result of, of the depth of the liquid or the point. So if the aquarium is open to the top, then P knot is going to be atmospheric pressure. And thus, we want to substitute this two equation right here into our integration to get started with solving our equation. So then first F will be integrated from zero to point H. So we are looking at integrating from zero to H in the Y direction and P is then going to be substituted with P knot plus row G H minus Y. So B is P dot plus row G open parentheses, H minus Y close parentheses. And D A is going to be W multiplied by D Y just like. So, so next, what we wanna do is we want to pull the W out of our integral based on the integration principle of the integral of C multiplied by D X well equals to C multiplied by the integral of D X where C is just a constant. So W here is also a constant so we can pull it out of the integral to make it easier for us. So F will then equals to W multiplied by the integral from zero to H of open parenthesis P knot plus row G H minus Y close parenthesis multiplied by D Y. So next, we want to employ a different integration principle where we will have the integral of any term on this site, which is A plus B in parenthesis multiplied by D X will equals to the integral of a multiplied by D X plus integral of B multiplied by D X. This is the principle of distribution which we are usually aware of. So we wanna employ that into our uh expression that we have here so that we get F to equals to W multiplied by um open parentheses. First integra from zero to hr P knot multiplied by D Y plus the integral from zero to H of row G H minus Y multiplied by D Y close parentheses. If you guys notice the right term here, row G open multiply open parenthesis, H minus Y, close parentheses can neutralize the same integration principle again to then further simplify this term. So let's employ that one more time so that we will have F to be equals to W multiplied by open parenthesis to integral from zero to H of P no multiplied by D Y plus the integral from zero to H of row G H multiplied by D Y minus the integral from zero to H of row G Y multiplied by D Y close parenthesis. So in this case, P NOTT row G and row G H and row G are all constants. So we can follow the first integration principle that we have previously to then further simplify this. So then we will get F to be equals to W multiplied by open parenthesis. P knot multiplied by the integral from zero to H of D Y plus uh row G H multiplied by the integral from zero to H F D Y minus row G multiplied by the integral from zero to H F Y. Do I close parenthesis? Awesome. Now, we want to employ different integration principle where we have the integral of the X to be just X and the integral of X D X to then be X squared divided by two. We want uh we want to employ this two integration principle into our expression here to then solve for our answer. Don't forget we are evaluating from zero to H. So we want to take a note of that as well. OK. So then we have F to then equals to W multiplied by open parenthesis. P no multiplied by Y evaluated from zero to H plus row G H multiplied by Y evaluated from zero to H minus row G multiplied by A Y squared divided by two mo evaluated from zero to H as well. Next, we want to evaluate each of this integral and doing that, we will then get F to then be equals to W multiplied by P multiplied by H minus zero plus row G H multiplied by H minus zero minus row G H squared, divided by two plus zero, close parenthesis. And that will be the answer. So I'm gonna simplify things a little by just cross out, crossing out all the zeros to make it easier for us to see. And if you notice, we will then have uh the term of row G H squared minus half row G H squared. So then our final answer will then give us F to be equals to W multiplied by open parentheses. P no H plus half row G H squared just like. So, so that will be the final answer for our problem where we have F to then be equals to W multiplied by open parentheses. P dot H plus half row G H squared, close parentheses. And that answer will actually correspond to answer B in our answer choices. So answer B will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you. Mhm.

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