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Question: A tank in the shape of a right circular cone is 4 feet tall and 2 feet in diameter at the top, and is half full of water. How much work is done in pumping the water to the top of the tank?

I believe I solved this problem right but my teacher's answer is different from mine. This was my integral:

$$ work = 62.4\cdot32.2\int_2^4{x\pi\cdot}{(4-x)^2\over16}\cdot d{\bf r}=2630.141 ftlb$$

Where did I go wrong? I used similar triangles to figure out the v(x) and I believe d(x) is simply x since you are pumping the water up. Any help appreciated. Also, what if the conical tank was completely full of oil weighing 50 lbs/ft^3 and the oil has to be pumped to a spot 2 feet above the top of the cone? How much work is done and how would you do that?

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  • $\begingroup$ I assume that $ftlb$ mean foot $\times$ pound (not evident for 90% of people on Earth using metric system) ? $\endgroup$
    – Jean Marie
    Nov 13, 2019 at 18:48

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The issue is that you assumed an INVERTED cone, which I can tell because you used $(4-x)^2$ instead of $\displaystyle \frac{x^2}{16}$.

The base of the cone is at top!

I'm not sure what you mean by pumping to a spot two feet above the top of the cone, but it would be multiplying the integral (perhaps $\displaystyle \int_4^6$cone) by the density of oil $(50)$.

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  • $\begingroup$ Well isn't it an inverted cone for sure if it says that it is 2 feet in diameter at the top? Do you see any issues with it? $\endgroup$ Nov 13, 2019 at 19:11
  • $\begingroup$ @MatinNawabi your equation suggests that the cone's point is at top, but the cone's point is at the bottom $\endgroup$ Nov 13, 2019 at 19:25
  • $\begingroup$ how do I go about getting x^2/16? I simply do not understand how to find that. Does it have to do with similar triangles? $\endgroup$ Nov 13, 2019 at 21:18
  • $\begingroup$ instead of [(4-x)/4]^2, it's just (x/4)^2 $\endgroup$ Nov 14, 2019 at 15:07

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