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

Equilibrium of Forces – Hydrostatic Pressure

What We’ll Learn

We’ll start from the basic definitions of pressure and density and show how they lead to the familiar hydrostatic formula:

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

Think of a tall glass of water – the deeper you go, the more “push” you feel from the water above.

Key Definitions

QuantitySymbolDefinition
Pressure\$p\$Force acting per unit area: \$p = \dfrac{F}{A}\$
Density\$\rho\$Mass per unit volume: \$\rho = \dfrac{m}{V}\$
Gravitational acceleration\$g\$≈ \$9.81\,\text{m/s}^2\$ on Earth

Step‑by‑Step Derivation

  1. Choose a small slice of fluid.

    Imagine a thin horizontal layer of fluid with thickness \$dh\$, cross‑sectional area \$A\$, and volume \$V = A\,dh\$.

  2. Find the weight of that slice.

    Weight = mass × \$g\$.

    Mass = density × volume: \$m = \rho V = \rho A\,dh\$.

    Therefore, weight \$W = m g = \rho A\,dh\,g\$.

  3. Relate weight to pressure difference.

    The pressure at the bottom of the slice pushes upward on the top surface, while the pressure at the top pushes downward on the bottom surface.

    The net upward force is \$dF = W = \rho A\,dh\,g\$.

    Pressure difference \$dp\$ is this force divided by area:

    \$dp = \frac{dF}{A} = \rho g\,dh\$

  4. Integrate from the surface to depth \$h\$.

    Assume pressure at the surface is \$p_0\$.

    \$\int{p0}^{p} dp = \int_{0}^{h} \rho g\,dh\$

    \$p - p_0 = \rho g h\$

    Thus, the change in pressure from the surface to depth \$h\$ is

    \$\boxed{\Delta p = \rho g \Delta h}\$

💡 Analogy: Think of a stack of books. The bottom book feels a heavier “push” from all the books above it. The deeper you go, the more books (or fluid) are pressing down.

Practical Example

Water in a 2 m deep tank. Density of water: \$\rho = 1000\,\text{kg/m}^3\$.

  • Pressure at the surface: \$p_0 = 0\$ (atmospheric pressure is usually taken as reference).

  • Pressure at the bottom:

    \$\Delta p = \rho g h = 1000\,\text{kg/m}^3 \times 9.81\,\text{m/s}^2 \times 2\,\text{m} = 19\,620\,\text{Pa}\$

  • That’s about 0.2 atm – roughly the pressure you feel when diving a few meters underwater.

Exam Tip Box

Remember:

  • Always start with the definitions: \$p = F/A\$ and \$\rho = m/V\$.

  • Use a thin slice to avoid complications from varying pressure across the slice.

  • When integrating, keep the limits clear: from the surface to the depth of interest.

  • Check units: \$\rho\$ (kg/m³) × \$g\$ (m/s²) × \$h\$ (m) = Pa (N/m²).

  • In multiple‑choice questions, look for the form \$\rho g h\$ – any extra terms usually indicate a trick.

Quick Quiz (Optional)

  1. What would happen to \$\Delta p\$ if the fluid density doubled but the height stayed the same?

  2. How does the hydrostatic pressure change if you go from sea level to the top of a 3 km mountain?

Try answering before looking at the solutions. Good luck! 🚀