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The question asks, For which design will the water pressure at the base of the dam wall be the greatest? The answer is all are equal.

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This doesn't make sense to me.

Using the formula: P (pressure) = p (atmospheric pressure) + rho (density) * g (gravity) * y (depth below water surface)

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What is confusing me is the y value, the depth below the surface. Yes both red points are the same depth below the water surface. But doesn't the red point in dam III only "see" a smaller column of water above it and so shouldn't it have less pressure on it than the red point in dam II?

Also, I know the formula would give the points at the base of dam I and dam II equal. But why doesn't the extra water in dam II (shaded in black) cause it to have a higher pressure at the base? enter image description here

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  • $\begingroup$ This is the nature of hydrostatic pressure! $\endgroup$ – Dr Xorile Sep 19 '16 at 22:12
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    $\begingroup$ Consider a thin piece of paper. It does not matter whether it is inside or outside, vertical or horizontal etc. It experiences the same air pressure on both sides. $\endgroup$ – Dr Xorile Sep 19 '16 at 22:16
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    $\begingroup$ Pascal's law: en.wikipedia.org/wiki/Pascal%27s_law $\endgroup$ – Gert Sep 19 '16 at 22:43
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    $\begingroup$ In the case of figure III, even though the vertical columns are smaller, the overhanging rock from the dam is exerting a downward force on the water below (that turns out to be equal to the missing column of water above). Also, as Gert points out, Pascal's law tells us that the pressure doesn't vary in the horizontal direction. Pressure gets transmitted horizontally, without any change. So, in all three cases, the pressure in the vicinity of the dam at a given depth is the same as the pressure at the same depth away from the dam. $\endgroup$ – Chet Miller Sep 19 '16 at 23:39
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    $\begingroup$ No. A solid like rock is able to exert shear forces on vertical planes to support the weight of the rock. Unlike the case of a liquid, it doesn't have to depend exclusively on normal stresses like pressure to support the weight. $\endgroup$ – Chet Miller Sep 19 '16 at 23:46
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When you have a body of water at each point is the water the net force is zero - you have static equilibrium.

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Imagine a beaker ABCD with a hole in the bottom BC being held partially immersed in some water as shown in the left hand diagram.
The water pressure at the bottom of the beaker is determined by the vertical height of water.
There is static equilibrium at each point in the water and the beaker.

Close the hole - nothing changes with the pressure due to the water at the bottom of the beaker still the same.

Remove the water surrounding the beaker.
Nothing inside the beaker changes.
How is it that the water stays in the beaker?
The walls of the beaker exert forces on the water in the beaker which before were exerted by the surrounding water.
The pressure of the water on the base of the beaker is still determined by the vertical height from the base of the beaker to the surface of the water.

So now go through the same sequence for the vessel with the shape EFG to show that the water pressure at F is determined by the vertical height between F and the surface of the water.
With the water removed from the outside of the container the walls of the container exert forces on the enclosed water equal to those which had been exerted by the surrounding water.

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Take a tube of toothpaste with the top off, and hold it vertically. Now squeeze it from the sides, i.e. horizontally.

Does any toothpaste come out the top (vertically)? Why?

Because in any fluid, pressure "goes around corners".

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The water pressure below the surface is independent of the way you reach that point under the surface. So you can imagine going straight down to the bottom anywhere in the reservoir and then slide along the bottom to the point of interest. The pressure in the end is only given by the vertical depth. The overhanging dam does not change this. You can visualize the hydrostatic pressure as being due to the weight of a column of water of cross section of unit area pressing on the flat ground. But this is misleading in the case with the overhanging dam. The pressure at a given depth has no direction and is independent of the way you reach a certain point at a given depth.

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