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

Equilibrium of Forces – Hydrostatic Pressure

Objective (AO1)

• Derive, from the definitions of pressure and density, the hydrostatic‑pressure relation

  \[\Delta p = \rho\,g\,\Delta h\]

  and recognise the limits of its applicability.

• Apply the relation to fluids at rest, to real‑world devices and to the calculation of buoyant force.

Key Definitions (AO1)

Assumptions for the Derivation (AO1)

1. The fluid is incompressible – its density \(\rho\) is constant with depth.
2. The fluid is at rest (static equilibrium).
3. Only the weight of the fluid column between the two points contributes to the pressure difference.

Derivation of \(\displaystyle \Delta p = \rho g \Delta h\)

Consider a rectangular element of fluid of cross‑sectional area \(A\) and height \(\Delta h\).

Pressures on the top and bottom faces are \(p1\) (upper) and \(p2\) (lower).

• Upward force from the top surface: \(F{\text{top}} = p1A\)
• Downward force from the bottom surface: \(F{\text{bottom}} = p2A\)
• Weight of the element: \(W = \rho g A\Delta h\)

Static equilibrium ⇒ net force = 0 (upward taken as positive):

\[

p1A \;-\; p2A \;-\; \rho g A\Delta h = 0

\]

Dividing by \(A\) gives

\[

p1 - p2 = -\rho g\Delta h

\]

Re‑arranging, the pressure at the lower point is

\[

p2 = p1 + \rho g\Delta h

\]

Thus the pressure difference between two depths separated by \(\Delta h\) is

\[

\boxed{\Delta p = p2 - p1 = \rho g \Delta h}

\]

Interpretation (AO2)

• The pressure in an incompressible fluid increases linearly with depth.
• The factor \(\rho g\) is the hydrostatic pressure gradient (units Pa m\(^{-1}\)).
• If the fluid is compressible (e.g. a gas) the density varies with height and the simple linear law no longer holds.

  For an ideal gas of constant temperature,

  \[

  p(h) = p0\,e^{-\,\frac{\rho0 g}{p_0}h}

  \quad\text{or}\quad

  p(h)=p_0\exp\!\left(-\frac{M g h}{RT}\right)

  \]

  where \(M\) is the molar mass and \(RT\) the gas constant product.

Real‑World Applications (AO2)

• Barometer – the height of a mercury column balances atmospheric pressure: \(p{\text{atm}} = \rho{\text{Hg}} g h_{\text{Hg}}\).
• U‑tube manometer – pressure differences are measured by the difference in liquid heights on the two arms.
• Blood‑pressure cuff – inflation creates a gauge pressure that must exceed the arterial pressure to stop blood flow.
• Hydraulic lift – the force transmitted through an incompressible fluid follows \(F = pA\); the lift’s mechanical advantage is the ratio of piston areas.
• Scuba diving & swimming pools – the increase in pressure with depth determines required structural strength and affects the human body.

Link to Upthrust (Buoyant Force) (AO2)

Archimedes’ principle follows directly from the hydrostatic‑pressure relation.

1. Consider a submerged body of volume \(V\). The pressure on its bottom face is higher than on its top face by \(\Delta p = \rho g h\), where \(h\) is the vertical span of the body.
2. The net upward force (upthrust) is the pressure difference multiplied by the horizontal area, which after integration gives

   \[

   F{\text{upthrust}} = \rho{\text{fluid}}\,g\,V

   \]

   i.e. the weight of the displaced fluid.

This relation is required when answering syllabus questions on floating and sinking objects.

Practical Investigation (AO3)

Uncertainty Propagation (AO3)

When \(\Delta p = \rho g \Delta h\) is used to calculate a pressure difference, the relative uncertainty is

\[

\frac{\delta (\Delta p)}{\Delta p}

= \sqrt{\left(\frac{\delta\rho}{\rho}\right)^{2}

+\left(\frac{\delta h}{\Delta h}\right)^{2}

+\left(\frac{\delta g}{g}\right)^{2}}

\]

In most school‑level work \(\delta g\) is negligible; the dominant contributions are from the density of the liquid (temperature dependent) and the measured height.

Worked Examples (AO2)

Example 1 – Pressure increase in fresh water

Find the gauge pressure at a depth of \(5\;\text{m}\) in a lake ( \(\rho = 1000\;\text{kg m}^{-3}\) ).

\[

\Delta p = \rho g \Delta h

= (1000)(9.81)(5)

= 4.905\times10^{4}\;\text{Pa}

\]

≈ 0.49 atm higher than the surface pressure.

Example 2 – Upthrust on a submerged block

A wooden block of volume \(0.020\;\text{m}^{3}\) is fully immersed in seawater (\(\rho_{\text{sea}} = 1025\;\text{kg m}^{-3}\)). Calculate the buoyant force.

\[

F{\text{upthrust}} = \rho{\text{sea}} g V

= (1025)(9.81)(0.020)

= 2.01\times10^{2}\;\text{N}

\]

Summary (AO1 & AO2)

• For an incompressible fluid at rest, pressure varies with depth as \(\Delta p = \rho g \Delta h\).
• The derivation follows directly from force equilibrium on a thin fluid element.
• Gauge pressure = absolute pressure – atmospheric pressure; always state which is being used.
• The same principle underlies barometers, manometers, hydraulic devices, and the calculation of buoyant force (Archimedes’ principle).
• When the fluid is compressible (gases) the exponential barometric formula must be used instead of the linear law.
• Practical measurement of \(\Delta p\) can be performed with a U‑tube manometer; uncertainties in \(\rho\) and \(\Delta h\) dominate the error budget.