I was doing some problems involving calculating the thrust due to pressure on curved surfaces. One way, of course, was to integrate over the surface to find the thrust.

I noticed that the answer always came out to be the volume above the surface times the density and free-fall acceleration.

I thought of the following argument. The pressure exerts the same force as if there was fluid above it. So the vertical force on it must be the weight of an equivalent fluid column above it.


  1. Can this argument be formalized?
  2. All the problems I tried this on had surfaces that were symmetric about a vertical axis, so the horizontal force was zero by symmetry. Is there a similar/different argument to find the horizontal force without integration?

The example that I have in mind for the second question is an oblique square plane of side length $L$ that makes an angle $\theta$ with the horizontal.

Thanks for any help.


1 Answer 1


What you have found is that to find the force it is easier to use the projected area.

Assume that you have an element of area $dA$ which can be treated as being plane.
The projected area onto a plane which is at an angle $\theta$ to the element is $dA \, \cos \theta$.

Due to the pressure, the force on the element of area $dA$ is $P\,dA$ and that force acts at right angles to the element.
That force due to the pressure can be split into two components one perpendicular to the projected area $P\,dA \sin \theta$ and one at right angles to the projected area $P\,dA \cos\theta$ and you will note that this second component is just the pressure times the projected area.


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