Problem — Derivation of Stevin's law

Problem

Show, by considering a column of liquid of constant cross-section $A$ and height $h$, that the hydrostatic pressure at the base is $p = \rho g h$, independent of the shape of the container. State which assumptions you are using.

Solution

Step 1 — Assumptions. We assume the liquid is:

• incompressible, i.e. with uniform and constant density $\rho$;
• at rest (hydrostatic equilibrium, no motion);
• immersed in a uniform gravitational field of acceleration $g$.

Step 2 — Isolating the column. We ideally consider a vertical column of liquid of cross-section $A$ and height $h$, extending from the free surface down to the base. Its volume is $V = A\,h$ and its mass:

$m = \rho V = \rho A h$

Step 3 — Weight of the column. The weight of the column, directed downward, is:

$P = mg = \rho A h\,g$

Step 4 — Equilibrium at the base. The column is at rest: the force that the liquid below (or the bottom) exerts upward on its base must balance its weight. This force $F$ acts on the surface $A$, so the pressure at the base is force over area:

$p = \frac{F}{A} = \frac{P}{A} = \frac{\rho A h\,g}{A} = \rho g h$

Result.

$\ev{p = \rho g h}$

Step 5 — Independence from shape. The cross-section $A$ cancels: the pressure does not depend on the area of the column, nor consequently on the shape of the container, but only on the depth $h$ below the free surface (hydrostatic paradox). At equal $h$, a narrow container and a wide one give the same pressure at the bottom.

Dimensional check. $[\rho][g][h] = \dfrac{\text{kg}}{\text{m}^3}\cdot\dfrac{\text{m}}{\text{s}^2}\cdot\text{m} = \dfrac{\text{kg}}{\text{m}\cdot\text{s}^2} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$, consistent with a pressure. ✓

Linked atoms

Argomenti: Fluid statics and dynamics Concetti: Pressure · Stevin’s law Competenze: Dimensional analysis