Pressure Due to a Liquid Column

To find the pressure due to a liquid at some depth below its free surface, imagine a vertical cylinder of the liquid, of cross-sectional area A, extending down to a depth h inside a vessel filled with the liquid of density ρ. The weight of the liquid making up this imaginary column is $F = mg$, and this weight acts straight down on the cylinder's base. By the definition of pressure, the pressure exerted by this liquid column on the base of the cylinder is:

Now the mass of the liquid filling the imaginary cylinder is simply its volume times the liquid's density, $m = (\text{volume of cylinder}) \times \rho = (Ah)\rho$. Substituting this in,

So the pressure due to a liquid of density ρ, at rest, at a depth h below its free surface, is simply $h\rho g$. It is worth stressing explicitly that this result does not depend at all on the cross-sectional area A of the imaginary cylinder used to derive it — A cancels out of the formula completely, so the same pressure $h\rho g$ holds at depth h no matter how wide or narrow an imaginary column one imagines there. …