find-the-work-done-in-pumping-the-water-out-of-the-top-of-a-cylindrical-tank-300-ft-in-radius-and-10

Question

Find the work done in pumping the water out of the top of a cylindrical tank 3.00 ft in radius and 10.0 ft high, given that the tank is initially full and water weighs $62.4 \mathrm{lb} / \mathrm{ft}^{3}$. [Hint: If horizontal slices $d x$ ft thick are used, each element weighs $62.4(\pi)\left(3.00^{2}\right) d x$ lb, and each element must be raised $10-x \mathrm{ft}$, if $x$ is the distance from the base to the element (see Fig. 26.66 ). In this way, the force, which is the weight of the slice, and the distance through which the force acts are determined. Thus, the products of force and distance are summed by integration.]

Averell Hause · Accepted Answer

so the card a initially as floats the cylinders and static equilibrium. So we know that the buoyancy force would be equaling the gravitational force. And this would equaling the density of the liquid multiplied by the cross sectional area multiplied by the height multiplied by the acceleration due to gravity. A an external force f pushes, then down on the cylinder, a distance of axe. And so now we have a submerged the length and so the buoyancy force would be equaling the gravitational force, plus this external force that it&#x27;s acting on the top of the cylinder Now. This would be equaling that the density of the liquid multiplied by the cross sectional area of the cylinder, multiplied by H plus ax multiplied by G. And so we know that the density of liquid multiplied by a H G is equal in the gravitational force. So now we&#x27;re left with the external force Equalling the density of the liquid multiplied by the cross sectional area multiplied by G. The acceleration due to gravity times X, the additional distance with additional depth that the cylinder is pushed down into a liquid for part B. Since the amount of work done Ah Would be done by the external force to push the cylinder from X two X plus a, uh, infinite testimony. Small additional distance. Essentially the differential work, the infinite testimony. Small work would then be equaling the force multiplied by the infinite testimony. Small distance, uh, over which the force is applied. This would be equaling the density of the liquid multiplied by a G axe again, DX and so So push the cylinder from zero meters to 10 centimeters. It requires work integrating from the initial X to the final x of the force DX. This is then Equalling to the density of the liquid multiplied by okay, multiplied by G. These were all these air, all constants integrating from zero 2.10 meters x dx. And so we have then the work equaling one house times the density of the liquid multiplied by a G on X initial squared minus X final squared and this is evaluated at zero and at 00.10 meters. And so we have. Then the work Equalling 1/2 multiplied by 10 to the third kilograms per cubic meter multiplied by point 0 to 0 meters quantity squared multiplied by pi multiplied by 9.8 meters per second, squared multiplied by 0.10 meters, quantity squared and the total work done pushing the cylinder would be 0.62 jewels. That is the end of the solution. Thank you for watching.

Mauricio Araiza Canizales · Answer

In the next problem, we want to pump water out of a cylinder which has a circular base of radius Fife it in the height of 200 feet. We might as well also use the fact that the density of water is 62 pounds 62 pounds, first cubic feet. So in this case, let us first determined the volume of oneness lice of water. So the volume of one slice of water Let us just this one over here. So the volume of one slice of water with fitness, the X it&#x27;s gonna be the area of the base times is thickness. And we have already the chairman that the thickness is gonna be the X or a really small number. So the area of the base since we know it&#x27;s a circle, it&#x27;s just gonna be pie. Are square temps T X. Now let us rewrite this Justus 25 pie. Since that radius is just five DX Now, the force involved during this process it&#x27;s gonna be Dellin er than the density of the water times the volume of the water of this slice of water. We&#x27;re here. And so in this case, it&#x27;s just gonna be 62 terms 25 times pi the ex. So we have determined the force that is required to move one slice of water. However, we need to the chairman, the hot the distance that this life is gonna trouble defined their work. So the distance over here, we&#x27;re going to consider their ex to be in the distance to travel. So we&#x27;re gonna sit there. Is zero over here in 200 over here. So as lies that it&#x27;s at the bottom off the cylinder is gonna troll a distance of 200 and its allies that he said to hundreds just gonna be zero. So we can just represent this as the work. Just gonna be the integral from Syria to 200 off. If we simplify, this is gonna be 4550 pie times of the stones, which is just gonna be X the ex. Now we&#x27;re just gonna take Oh, sorry. It&#x27;s 1000. I made a mistake. We&#x27;re here. It was when we were here. It&#x27;s 1000 5 115 So now the work, it&#x27;s just gonna be 1550. Pie him is an integral from 0 to 200 affects the X. In this case, we know that this integral it&#x27;s just X square divided by two violated from 200 using power rule. This simplifies to 725 pie times 200 square. I know zero. And in this case, the work is gonna be equal to 9.37 39 times 10 to the seventh pound foot. And this is a solution to

Mike Gaerlan · Answer

Okay, so in this problem, we have a cylinder of water. Okay, The height is 200 feet up with a radius of five feet. Okay, so if we have a slice of water here Ah, we need to bring it up by, say, a height of H. Um, okay. So the the the mass of this slice, though, is going to be so if we have TV, BP is going to be pie r squared times the h like so. So that&#x27;s going to be five square pie. Th there&#x27;s going to be 25 pie th okay Not to move that. Huh? Uh, DW It&#x27;s going to be the mass. Um, where The road devi And then times, uh, the distance we bring it up. Okay, so this is going to be 62 times 25 pie times h d h. And we&#x27;re taking the integral from. So we start at our top height zero down to 200. We go from 0 to 200 here, and then when you work on this integral, you get, um so 62 times 25 high h squared over to prom, 0 to 200. Um, and just using a calculator to get this number, we get 9.74 times 10 to the seventh, uh, foot pounds and we are done.

Tony Cheung · Answer

for this question were given the work formula for full tank with a height of 10 and we are asked to find a work formula for only the top half of a tank. So we need to understand how this integral works. We see that there is a lower bound of zero and an upper bound of 10 which means we are emptying this tank from the height of 10 all the way down to zero. So if we want to find the formula for only half top half of the tank, we would still start from 10. And we&#x27;re only going down to five and the rest of the formula will remain the same. So So it would just be the same exactly the same here.

Adnan Gill · Answer

So let&#x27;s start by drawing a picture off the situation. So we have a cylindrical tank and at the bottom of the tank we can say wise, equal to zero at the top off the tank, we can say why is equal to four readers and then above the top off the tank is another point, which is one year long, and the water is gonna come up off there. So this tank, we will take an elemental volume which is off thickness D white and the radius is R. So this volume DV were people to apply times r squared d y or happen people to since our is equal to two meters, Therefore we can say that they want him off. This would be four pi times the like. Now there force the F which is needed to lift this volume up will be equal to the weight off this. So that would be equal to four pi times 9800 which is the weight of one meter cube off water. So therefore we can say that DF is equal to oh 3 39,200 party times D y. So this needs to be lifted through a distance off. Why four minus y meters, But then it also used to be lifted through another one meter. It&#x27;s on the total distance through which intestine lifted is four minus y plus one meter. So we can now say that you worked on To move this elemental mass would be to 71,200 pie times four minus y plus one d y or we can say VW is equal to 42,200 pi times five minus. Why Lee y well, now that we know the work done to needed to move this elemental test then were picking now is apply the integral over the limit from 0 to 4 to get the total worker. So work were people to any girl from 0 to 4 39,200 hi times five minus y d like and that would be cool to 39,200 pie time. Five Y minus Y squared over two over the limit from 0 to 4 when the lower limit would give me zero close bicycle to zero would make the whole technique because you&#x27;re So this will now be 39,200 pie and then times five times four minus four square over to minus zero. So that comes out to be called before 70400 High jewels of work is needed.

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