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I am really confused on how to approach the following question and how to set it up. So the question is:

An open tank has the shape of a frustrum of a right circular cone. The tank is $4\:\mathrm{ft}$ across the top and $8\:\mathrm{ft}$ across the bottom and the height is 6ft. How much work is done to fill the tank by pumping the water over the edge? Water= $62.4\frac{lbs}{ft^3}$.

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  • $\begingroup$ Is forces really the right tag for this question? I'd at least use energy as well as homework question. $\endgroup$ Commented Oct 31, 2015 at 23:57
  • $\begingroup$ Also, from what elevation is the water being pumped from? $\endgroup$ Commented Oct 31, 2015 at 23:58
  • $\begingroup$ elevation? I am not sure what you mean by this, this question is from my calc 2 class. $\endgroup$
    – Carlos V
    Commented Nov 1, 2015 at 0:00
  • $\begingroup$ Ah... it's a calc question. Well, they probably intended for the water to originally be at the level of the bottom of the frustum. But to be honest, that's really lazy of them, it's essentially like leaving out the +C in an integral. $\endgroup$ Commented Nov 1, 2015 at 0:02
  • $\begingroup$ Oh ok so without that value it is not possible to do? $\endgroup$
    – Carlos V
    Commented Nov 1, 2015 at 0:05

2 Answers 2

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The question, as written, has no need for calculus.

Find the volume of water needed to fill the frustum. http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html

Find the weight of that water

Find the work done to pump all that weight up 6 feet over the top of the wall of the tank, letting it splash down into the tank.

Done.

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  • $\begingroup$ Ha ha ha. Wow, good point. It really is a poorly written question, all in all. $\endgroup$ Commented Nov 1, 2015 at 6:30
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We know that the gravitational potential energy of an object is $mgh$. Initially, $h=0$ for the water before it is pumped in, so the potential energy is 0 as well. From here, this is a related rates problem, the rates you want to be relating is the rate of increase of the gravitational potential energy of a horizontal layer of the water filling the tank to the rate of increase of the height of the container. This way, you can integrate over the height of the tank to find the total energy/work. I don't want to give the entire problem away if I don't have to, so I'll stop here if you understand the premise and what you need to do now. If you don't, just let me know.

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  • $\begingroup$ yeah im not sure what to do still, and just to let you know this problem is from my calc class. $\endgroup$
    – Carlos V
    Commented Nov 1, 2015 at 0:17
  • $\begingroup$ You're doing related rates right now, right? $\endgroup$ Commented Nov 1, 2015 at 0:18
  • $\begingroup$ Let me put it this way. $E_g = mgh$. You want to relate $dE_g$ to $dh$. $g$ is a constant. $h$ is clearly a function of $h$. Now you just need to express $m$ as a function of $h$. And just to clarify, $h$ isn't the maximum height of the tank, but a variable that goes from $0$ to the maximum height of the tank, $H$. $\endgroup$ Commented Nov 1, 2015 at 0:28
  • $\begingroup$ Actually no this is the physics part of my calculus class. $\endgroup$
    – Carlos V
    Commented Nov 1, 2015 at 5:31

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