Why is it nearly universally understood that force in a liquid against a border is dependent on internal pressure, depth only, rather than weight of the liquid? Thus, for example, that the force from a static liquid against a vertical wall dam is a constant rather than dependent upon the volume of water (except beyond a distance = to the height of the liquid).


Your question seems predicated on a misconception.

The force that a liquid exerts on the wall of a vessel is equal to the product of pressure and area. For the bottom of a vessel with straight walls that means that the force is indeed equal to the weight of the liquid.

This makes sense because in such a vessel the only wall that can provide a force to support the liquid (counteract gravity) is the floor.

If the walls are not straight, then a component of the pressure can result in a vertical force. Again if you sum all these vertical forces they add up to the weight.

Looking now at forces on vertical walls, they do indeed only depend on the pressure - because the only force they are countering is the force due to the same pressure on the opposite wall of the vessel, regardless of the distance between these walls.

Imagine for a moment this would not be the case - in other words, imagine that the pressure did depend on the distance to the opposite wall. What would happen if you inserted a wall halfway across the vessel? You know the pressure on the side walls won't change. But if you divided the bottom of the vessel in two, each half would experience half the force (same pressure, half the area).

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  • $\begingroup$ I do not see a misconception. The computation of force is stated in all books to be a function of only depth and hydrostatic pressure, the remaining geometry of the mass of water being irrelevant. $\endgroup$ – john Apr 30 '17 at 13:50
  • $\begingroup$ You seem to be conflating pressure and force. If a cylindrical vessel has twice the diameter but is filled to the same depth the force on the bottom will be four times greater while the pressure is the same. But the question said "force". $\endgroup$ – Floris Apr 30 '17 at 13:55
  • $\begingroup$ Floris your comment is unclear to me. $\endgroup$ – john Apr 30 '17 at 14:22

Are you referring to formulas like $$p=w h$$, where $p$ is the pressure, $h$ the height and $w$ the weight-density?

In this case the weight-density is given in eg. $\mathrm{kg}\,\mathrm{m}^{-3}$ and the pressure comes out as $\mathrm{kg}\,\mathrm{m}^{-2}$, so you are of course considering the entire weight (heigth times density).

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  • $\begingroup$ A comment would be nice, in case I'm missing something fundamental here ;) $\endgroup$ – Graumagier Feb 22 '15 at 17:25
  • $\begingroup$ I am refeering to force (f) =unit weight (w) x height (h) x width (b) for force on bottom & f=.5whb for force on a vertical flat border. I don't see why it generally understood that volume of water is irrelevant. $\endgroup$ – john Feb 23 '15 at 20:03
  • $\begingroup$ Typo: Formula should be h squared, not h. Also hydrostatic pressure is irrelevant. Mass is relevant. Force is a function of internal pressure in gases not liquids as the vehicle whereby the pressure is expressed as a force is expansion. $\endgroup$ – john Apr 30 '17 at 13:39

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