use the equation ∆p = ρg∆h

Equilibrium of Forces

Learning Objective

By the end of this lesson you should be able to apply the hydrostatic pressure relationship

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

to analyse situations where fluids are in static equilibrium and to determine the forces acting on submerged surfaces.

Key Concepts

• Static equilibrium: The condition where the vector sum of all forces acting on a body is zero, i.e. \$\sum \vec F = \mathbf{0}\$.
• Pressure in a fluid: The normal force per unit area exerted by the fluid, increasing with depth.
• Hydrostatic pressure equation: \$\Delta p = \rho g \Delta h\$, where

  • \$\Delta p\$ = change in pressure between two points (Pa)
  • \$\rho\$ = density of the fluid (kg m⁻³)
  • \$g\$ = acceleration due to gravity (≈9.81 m s⁻²)
  • \$\Delta h\$ = vertical depth difference (m)

• Force on a submerged surface: \$F = p A\$, where \$p\$ is the pressure at the surface and \$A\$ is the area.

Derivation of the Hydrostatic Pressure Equation

Consider a column of fluid of cross‑sectional area \$A\$ and height \$\Delta h\$. The weight of the fluid column is \$W = \rho A \Delta h \, g\$. This weight is balanced by the pressure difference between the top and bottom of the column:

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

Canceling \$A\$ gives the familiar result:

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

Application to a Submerged Plane Surface

For a vertical rectangular plate of height \$h\$ submerged in a fluid, the pressure varies linearly from \$p1 = \rho g h1\$ at the top to \$p2 = \rho g h2\$ at the bottom. The resultant force acts at the centre of pressure, located at \$\frac{2}{3}\$ of the depth for a vertical plate.

Worked Example

1. A rectangular gate 2.0 m high and 1.5 m wide is hinged at its top edge. The gate is submerged in water so that its top edge is 0.5 m below the free surface. Calculate the total hydrostatic force on the gate and the point of action of this force.
2. Given: \$\rho_{\text{water}} = 1000\ \text{kg m}^{-3}\$, \$g = 9.81\ \text{m s}^{-2}\$.
3. Solution:

   • Depth at top: \$h1 = 0.5\ \text{m}\$, at bottom: \$h2 = 0.5 + 2.0 = 2.5\ \text{m}\$.
   • Pressure at top: \$p1 = \rho g h1 = 1000 \times 9.81 \times 0.5 = 4.905 \times 10^{3}\ \text{Pa}\$.
   • Pressure at bottom: \$p2 = \rho g h2 = 1000 \times 9.81 \times 2.5 = 2.4525 \times 10^{4}\ \text{Pa}\$.
   • Resultant force on a vertical plate:

     \$F = \frac{(p1 + p2)}{2} \times A = \frac{(4.905\times10^{3} + 2.4525\times10^{4})}{2} \times (2.0 \times 1.5)\$

     \$F = \frac{2.9430\times10^{4}}{2} \times 3.0 = 1.4715\times10^{4}\ \text{N}\$

   • Centre of pressure for a vertical plate:

     \$y{cp} = \frac{h1 + \frac{h}{3}}{2} = \frac{0.5 + \frac{2.0}{3}}{2} = 0.5 + 0.667 = 1.167\ \text{m}\$

     measured from the free surface.

Summary Table of \cdot ariables

Practice Questions

1. Derive the expression for the centre of pressure on a plane surface inclined at an angle \$\theta\$ to the vertical.
2. A cylindrical tank of radius 1.2 m is filled with oil (\$\rho = 850\ \text{kg m}^{-3}\$) to a depth of 3.5 m. Calculate the pressure at the bottom and the total force on a circular hatch of diameter 0.5 m located at the bottom.
3. Explain why a fluid at rest cannot support shear stress, and relate this to the condition of equilibrium.

Key Take‑aways

• In static equilibrium the net force on a body is zero.
• The hydrostatic pressure increase with depth is linear and given by \$\Delta p = \rho g \Delta h\$.
• Resultant forces on submerged surfaces can be found by integrating pressure over the area, often simplified using average pressure for planar surfaces.
• Understanding the location of the centre of pressure is essential for designing hinges, supports, and safety checks in engineering applications.