If instead of a liquid in those shapes the shape was a rigid body made up of iron or any other thing, Can I still use the same ration $p=ρgh$ to calculate pressure at some height $h$?
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2$\begingroup$ The phase of matter doesn't affect the result, but if $g$ varies throughout the height, as with atmospheric pressure, you need $dp=\rho gdh$. $\endgroup$– J.G.Commented Feb 5, 2023 at 15:42
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2 Answers
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Pressure is only useful in a static fluid. By definition a fluid is a substance that does not support shear stress.
Inside a rigid body, such as a block of iron, pressure is replaced by the more complicated stress tensor
answered Feb 5, 2023 at 16:26
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$\begingroup$ And the pressure at the bottom of those shapes above is different… $\endgroup$ Commented Feb 5, 2023 at 17:05
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$\begingroup$ So, what you're saying is that I can't use p=ρgh to calculate the pressure with solids, Is that what you're saying? $\endgroup$– JackCommented Feb 7, 2023 at 6:33
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1$\begingroup$ Yes. Exactly. The forces in a solid are complicated, anisotropic, and do not obey Pascal's law. $\endgroup$ Commented Feb 7, 2023 at 14:20
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any material, with one rule.
When in any formula you use $V$ as $A . h$ you must keep in mind that the object is a Prism(geometrical shape that has equal top and bottom base), which in this case, fluids have this property.
Another thing, $P = \frac{F}{A}$ so, $P = \frac{F . d}{A . d}$ thus, $P = \frac{W}{V} = \frac{Energy}{Volume}$ and therefore we assume pressure as energy a fluid has(Any kind of energy), and since we want the pressure of a static(stationary)(so it only has potential energy) liquid then we have $U = m.g.h$ and $P = \frac{U}{V} = \frac{m.g.h}{V} = \rho.g.h$
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$\begingroup$ you can use this formula for prism-like objects if you have $\rho$, $g$, and $h$ $\endgroup$– MpH81679Commented Feb 5, 2023 at 17:37