1. ## Volume/Work word problem?

When work is defined to be force times distance, and that the weight (force) of a liquid is equal to its volume times its density. A fish tank has a rectangular base of width 2 feet and length 6 feet and sides of height 5 feet. If the tank is filled with water weighing 62.5 lb per cubic foot, the work required to pump all the water just over the top of the tank is what?

When I spent about an hour on this (I know way too long) I got 9,375ft/lb, is this what I was supposed to get? Did I do something wrong?

2. ## Re: Volume/Work word problem?

Let x be the distance from the top of the tank. The amount of work required to pump out a layer of water dx thick is weight times distance. Weight is $\rho=62.5$ times volume, volume is length=6 times width=2 times height=dx. The height goes from 0 (top of the tank) to 5 (bottom of the tank). So it's $\int_0^5 62.5 \cdot 6 \cdot 2 \cdot x \, dx$.

And the units will be ft-lb, not ft per lb.

- Hollywood

3. ## Re: Volume/Work word problem?

Originally Posted by hollywood
Let x be the distance from the top of the tank. The amount of work required to pump out a layer of water dx thick is weight times distance. Weight is $\rho=62.5$ times volume, volume is length=6 times width=2 times height=dx. The height goes from 0 (top of the tank) to 5 (bottom of the tank). So it's $\int_0^5 62.5 \cdot 6 \cdot 2 \cdot x \, dx$.

And the units will be ft-lb, not ft per lb.

- Hollywood
Okay I think that's similar to the way I integrated it, because when I did that one I still got 9,375. So it's 9,375ft-lb! Thanks