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October 22nd, 2014, 01:06 PM   #1
Joined: Oct 2013

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Pulling up water from an Inverted Cone - a few points of confusion

Hi all,

Given an inverted cone tank of water (full), the radius is 1.5m at the top, and the length, from top to bottom is 12m - We need to measure the work from pulling up all the water. Density and gravity are just standard here (1000, 9. eight [posted a smiley face when I put "eight"]).

I have two points of confusion here:

1) When finding out the radius, I set up similar triangles. 1.5/12 = r/12-y. This is incorrect, because you get a y value, and a constant. The exercise uses just plain old point-slope, with the bottom of this tank at the origin. I see what the correct method is, but I don't understand why my use of sim. triangles here is incorrect.

*The correct answer for the radius is y/8.

2) Once I have the area for the circle (pi y^2 / 64), my instinct is to sum up the areas of all the circles, when in fact we have to multiply that by 12-y. I am confused as to why we can't sum up the circles with pi y^2 / 64, as well as, obviously, the constants.

Thanks for looking.

Last edited by leo255; October 22nd, 2014 at 01:09 PM.
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October 22nd, 2014, 01:40 PM   #2
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work =$\displaystyle \int$ WALT

W = weight density
A = cross-sectional area of a horizontal slice of liquid w/r to y
L = lift distance of a slice in terms of y
T - slice thickness (dy)

using a set of coordinate axes, sketch a line from the origin (bottom of cone) to the point (1.5,12) ... equation of this line is

$\displaystyle y = 8x \implies x = \frac{y}{8}$

this line forms the cone when rotated about the y-axis.

$\displaystyle r = x = \frac{y}{8}$

$\displaystyle A = \pi x^2 = \pi \left(\frac{y}{8}\right)^2$

$\displaystyle L = 12-y$

work = $\displaystyle W\pi \int_0^{12} \left(\frac{y}{8}\right)^2(12-y) \, dy$

Last edited by skeeter; October 22nd, 2014 at 02:15 PM.
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