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Water pressure is simply $pgh/2$ for a vertical wall parallel to the depth of the water at exactly the half the water's depth. Suppose the wall is tapered (angled) slightly at around 2 degrees from the y-axis (parallel to the wall). How would you measure the pressure at this point? Would it be the contributing pressure from both the vertical and horizontal pressure? As such: $pgh \sin (\theta) + 1/2pgh\cos(\theta)$.

I'm assuming pressures can be summed as forces. This may be false.

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Firstly, the pressure you are talking about is the average pressure exerted on the wall. Secondly, does this water have air above it? If so the average pressure will be $P_0+\frac{1}{2}\rho gh$, which I agree is the pressure at a point halfway down the wall ($P_0$ is the atmospheric pressure). Finally, the average pressure is independent of the angle of the wall, what is changing is the direction of the normal force (as it is always perpendicular to the wall). Below I show a detailed calculation.


The pressure at any point along the wall is given by $P = P_0 + \rho g x$, where $x$ is the vertical distance from the surface of the water. Assuming the wall has a constant width of, say $a$, the total force on the wall is given by (given $A = \frac{ah}{\cos\theta}$ we have $dA = \frac{a}{\cos\theta}dx$) \begin{eqnarray} F &=& \int P~ dA = \frac{a}{\cos\theta}\int^h_0 dx \left(P_0 + \rho g x\right)\\ &=& \frac{ah}{\cos\theta} P_0 + \frac{1}{2}\rho g h\frac{ah}{\cos\theta} \end{eqnarray} (Note that this force is pointing normal to the surface of the wall. We also see that its magnitude is dependent on $\theta$, however as we will see, the average pressure is independent of $\theta$). The area of the surface is $\frac{ah}{\cos\theta}$ so the average pressure is just $$P_{avg} = \frac{F}{A} = P_0 + \frac{1}{2}\rho g h$$ So we see that the average pressure is independent of the angle of our wall.

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Nice! I don't have enough reputation to up-vote it but I accepted the answer. – ono Jun 11 '13 at 16:51

Suppose you have a vertical wall and your slanted wall. Now take a horizontal pipe and put one end at a depth of h/2 by the straight wall and the other end at a depth of h/2 by the slanted wall, so the pipe is horizontal.


It should be obvious that the pressure by the straight wall, P$_1$, is the same as the pressure by the slanted wall, P$_2$, otherwise water would flow along the pipe and you'd have a perpetual motion device.

The pressure is only dependant on the water depth and not on the angle of the wall.

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Great explanation! – ono Jun 11 '13 at 16:49
I'll up-vote it when I have enough reputation – ono Jun 11 '13 at 17:04

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