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September 16, 2014

September 16, 2014

Posted by **Claire** on Wednesday, November 3, 2010 at 6:02am.

The answers should be 27.9kN, 1.26m from the top.

Can someone please explain how to work it out?

- Fluid Mechanics -
**MathMate**, Wednesday, November 3, 2010 at 11:12amFor a uniform liquid, the pressure at a depth of h equals ρh.

For a tank of radius r (length does not matter) containing a mixture of oil and water, the hydrostatic pressure at any point h below the top is

p1(h)=0.8h for 0<h≤r, and

p2(h)=0.8h+(h-r)(1-0.8) for r<h≤2r

The width of the tank at any height h from the top is

w(h)=sqrt(r²-(r-h)²)

The magnitude of the hydrostatic force, F, is the sum of the integrals

F=∫ 0 r p1(h)dh + ∫r 2r p2(h)dh

=2.646g N

The total moment about the top:

Fh=∫ 0 r x*p1(h)dh + ∫r 2r x*p2(h)dh

=3.353g N-m

The position can be found by dividing Fh/F = 1.267 m from the top.

Note that 2.646g gives 26.0 N as opposed to the 27.9 N given in your answer.

Check my calculations/the answer given in the book.

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