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September 13th, 2015, 04:03 AM   #1
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Height of feed in a trapezoidal trough.

A trough has a cross-section in the form of a trapezium. Its base has a length of 1m, and the sides slope out at 45 degrees. to the horizontal. The trough is filled with feed to a depth of x metres. Find the value of x given that the centre of mass of the contents of the trough is 0.5m above the base.

Centroid of a trapezium is (h/3)[(b+2a)/(b+a)], where h is the height and b is the longest (base) and a is the shortest base (although in this problem, it is a on the horizontal).

The height of the trough is sin(45), if we take the length of the sloped edge to be 1m, as well.

The answer should be 0.866
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September 15th, 2015, 09:47 AM   #2
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$\displaystyle \bar{y} = \frac{h}{3} \left(\frac{2a+b}{a+b}\right)$

note from the diagram ...

$b=1$, $a-2h=1 \implies a=2h+1$

$\displaystyle \frac{1}{2} = \frac{h}{3} \left[\frac{2(2h+1)+1}{(2h+1)+1}\right]$

$\displaystyle \frac{1}{2} = \frac{h}{3} \left(\frac{4h+3}{2h+2}\right)$

$\displaystyle \frac{3}{2h} = \frac{4h+3}{2h+2}$

$\displaystyle h = \frac{\sqrt{3}}{2} \approx 0.866$
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September 16th, 2015, 02:54 AM   #3
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It's near to accurate.

Last edited by skipjack; September 16th, 2015 at 04:16 AM.
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September 16th, 2015, 04:28 PM   #4
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Thanks for the reply. It is a good answer and the diagram is very clear.
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