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A tank has a fixed vertical retaining wall. On one side of the wall, the tank is filled with a liquid of density $\rho$, to a height of $1$ m. On the other side of the wall, the tank is filled with a different liquid of density $0.5\rho$, to a height $h$ m. Let $p_A$ be the atmospheric pressure.

(a) For each liquid, determine the pressure at a height $d$ below the surface.

(b) If the liquids exert forces of equal magnitude on the retaining wall, find the value of $h$.

My attempt:

(a) For the first liquid, we have $p_1 = p_A + dg\rho$, where $g$ is the gravitational acceleration. Similarly, for the second $p_2 = p_A + 0.5dg\rho$.

(b) How would I go about answering this one? Would I have to integrate the equations for pressure with respect to the area of the wall in contact with the liquids?

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closed as off-topic by John Rennie, Jon Custer, Kyle Kanos, heather, David Z Jan 31 '17 at 21:18

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Your intuition is good on the first one.

For the second one, force will be directly proportional to pressure for both.

You might have to "integrate" but if you draw the pressure (force) distribution out on each wall you will see they vary only vertically; and the variation is linear. You could even do the integration by inspection if you notice they are triangles.

(Also, you can cancel out the atmospheric pressure, it will be acting on both walls regardless of if there is liquid there, and it makes the integration look more complicated)

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    $\begingroup$ So would it be as simple as drawing a graph of pressure minus atmospheric pressure on the $y$-axis of each graph against vertical distance below the surface on the $x$-axis, then calculating the area of the triangle for each? $\endgroup$ – user143811 Jan 30 '17 at 22:01
  • $\begingroup$ I think you've got it. $\endgroup$ – JMac Jan 30 '17 at 22:02

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