By Satwik Pasani


2014-02-10 17:40:11 8 Comments

The hydrostatic paradox, states that the

height of the water in any container is independent of the shape of the container.

This does make sense since it is only the height of the fluid relative to ground which determines the pressure, and it is the pressure equilibrium that decides the shape of the fluid. But then consider the (B) container in the figure below.
enter image description here

The pressure at a particular depth say, $h$ is same as in the case of the cylindrical container. But in the cylindrical case (A), any element of fluid is in equilibrium because the difference in pressure of the fluid element above it and that below it exactly balances the weight of the element. Extending this argument to (B), for any element at the edge, the pressure of the fluid above is the same as in the case of the cylindrical container, but below it is the surface of the container. This container, is in equilibrium with the atmospheric pressure outside, and therefore, would be exerting pressure equal to the atmospheric pressure onto the fluid within, which would not be enough to maintain the equilibrium of the fluid element. How then can the equilibrium of the fluid element at the edge be explained?

1 comments

@Emilio Pisanty 2014-02-10 18:05:24

The reason is, quite simply, that the walls of the container exert a force on the fluid given by the pressure at each point, and orthogonal to the container's surface. In your second container, the wall is slanted and therefore provides an upward force which helps stabilize the fluid elements next to it.

If you draw a free-body diagram of the bit of fluid at the edge,

enter image description here

you see that the wall of the container exerts a diagonal force, whose horizontal component is balanced by the pressure exerted by the fluid further in, and whose vertical component is balanced by the atmospheric pressure.

@Alan Rominger 2014-02-10 18:13:17

You have 3 vectors, but you should probably have 4. One to account for the weight for the fluid inside the element itself.

@Emilio Pisanty 2014-02-10 18:28:30

That is correct. You are welcome to edit this into the post.

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