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Work required to extract fluid

by Metaleer
Tags: extract, fluid, required, work
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Apr30-07, 01:29 PM
P: 123
1. The problem statement, all variables and given/known data

A right circular cone with water has a height of 6 ft and the upper circular base has a radius of 8 ft. How much work is needed to empty said container if the extraction is done from above?

2. Relevant equations

3. The attempt at a solution

First of all, I tried dividing the cone shaped container into thin slices and calculate the amount of work needed to lift each and every one of these thin slices out of the container via antiderivatives...

But I'm not sure how to continue. Any help would be appreciated. :)
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Apr30-07, 02:36 PM
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The spirit of your 'attempt' is correct. So what did you get for the integral and what's the problem from there?
Apr30-07, 05:30 PM
P: 688
is this question for a math class of physics class?

if it is for physics, then simply treat it as a particle at its center of mass, and calculate the work... that reduces the problem to finding the center of mass of a cone.

Apr30-07, 05:32 PM
P: 688
Work required to extract fluid

or you can set up an integral......look at each differential disk, what is the mass of that disk and how much does it have to be lift?

and then,
[tex]W=\Delta E=mg\Delta h[/tex]

just a side note... does the question say whether not the density of that cone is uniform? if not then you'll get some additional complications.
what do you get for the integrand?
May1-07, 09:04 AM
P: 123
Dick, the integrand is somewhat complicated to put here, Mathematica won't let me copy it for some reason, but the problem was, ok I have a differential disk... its mass would be calculated with density = mass/volume?

tim_lou, the first option seems easier... I determined that the center of mass is a quarter of the total height from the top, so it's at a height of 4.5 ft.

Note that I converted to Internation System units. The negative work means that an external force has to act on the water to get it out, as the water won't obviously won't drain out on its own. What do you think?

Thanks again, Dick and tim_lou. I believe this method is easier than using differential disks.

EDIT: I think it would be better to consider where the center of mass originally was as it having 0 potential energy... this way W = - m*g* 1.3716

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