derive, from the definitions of pressure and density, the equation for hydrostatic pressure ∆p = ρg∆h

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Equilibrium of Forces

Equilibrium of Forces – Hydrostatic Pressure

Objective

Derive, from the definitions of pressure and density, the equation for hydrostatic pressure

\$\Delta p = \rho g \Delta h\$

and understand its application in fluids at rest.

Key Definitions

  • Pressure (\$p\$): Force exerted per unit area, \$p = \dfrac{F}{A}\$ (SI unit: pascal, Pa).

  • Density (\$\rho\$): Mass per unit volume, \$\rho = \dfrac{m}{V}\$ (SI unit: kg·m⁻³).

  • Gravitational acceleration (\$g\$): Acceleration due to Earth’s gravity, \$g \approx 9.81\ \text{m·s}^{-2}\$.

Assumptions for the Derivation

  1. The fluid is incompressible (constant \$\rho\$).

  2. The fluid is at rest (static equilibrium).

  3. Only the weight of the fluid column contributes to the pressure difference.

Derivation of Hydrostatic Pressure

Consider a vertical column of fluid of cross‑sectional area \$A\$ and height \$\Delta h\$ (see figure below).

Suggested diagram: A vertical fluid column of height \$\Delta h\$ and area \$A\$, with forces \$p1A\$ at the top, \$p2A\$ at the bottom, and weight \$ \rho g A \Delta h \$ acting downward.

For the column to be in equilibrium, the net force must be zero. The forces acting are:

  • Upward force due to pressure at the top: \$F{\text{top}} = p1 A\$.

  • Downward force due to pressure at the bottom: \$F{\text{bottom}} = p2 A\$.

  • Weight of the fluid column: \$W = \rho \cdot g = \rho (A\Delta h) g\$.

Setting the sum of forces to zero (taking upward as positive):

\$p1 A - p2 A - \rho g A \Delta h = 0\$

Dividing through by \$A\$ gives:

\$p1 - p2 = -\rho g \Delta h\$

Rearranging, the pressure difference between the bottom and top of the column is

\$\Delta p = p2 - p1 = \rho g \Delta h\$

Interpretation of the Result

The equation shows that pressure increases linearly with depth in a fluid of constant density. The term \$\rho g\$ is often called the hydrostatic pressure gradient.

Table of Symbols

SymbolQuantityUnitsDefinition
\$p\$PressurePa (N·m⁻²)Force per unit area, \$p = F/A\$
\$\rho\$Densitykg·m⁻³Mass per unit volume, \$\rho = m/V\$
\$g\$Gravitational accelerationm·s⁻²Acceleration due to gravity, ≈ 9.81 m·s⁻²
\$\Delta h\$Depth differencemVertical distance between two points in the fluid
\$\Delta p\$Pressure differencePaChange in pressure between two depths

Example Application

Calculate the increase in pressure at a depth of 5 m in water (\$\rho_{\text{water}} = 1000\ \text{kg·m}^{-3}\$).

\$\Delta p = \rho g \Delta h = (1000\ \text{kg·m}^{-3})(9.81\ \text{m·s}^{-2})(5\ \text{m}) = 4.905\times10^{4}\ \text{Pa}\$

Thus, the pressure is about \$4.9\times10^{4}\ \text{Pa}\$ (≈ 0.49 atm) greater than at the surface.

Summary

  • Pressure in a static fluid increases with depth according to \$\Delta p = \rho g \Delta h\$.

  • The derivation follows directly from equilibrium of forces on a fluid element.

  • This relationship is fundamental for problems involving fluid statics, such as barometers, manometers, and pressure at depth.