When a fluid is at rest (or moving at constant velocity, where acceleration is zero), forces balance and the fluid is in static equilibrium. In this state, pressure varies only with depth, not with horizontal position or the shape of the container. This principle is the foundation of hydrostatics.
Deriving pressure as a function of depth
Consider a small cylinder of fluid immersed in a larger body at depth \( z \) below the surface. Let the cylinder have cross-sectional area \( A \) and height \( dz \). The mass of this element is \( dm = \rho A \, dz \), where \( \rho \) is the fluid density.
Forces on the element:
- Upward: pressure force from below = \( P(z) \cdot A \)
- Downward: pressure force from above = \( P(z+dz) \cdot A \), plus weight \( = dW = dm \cdot g = \rho A g \, dz \)
For equilibrium: \[ P(z) \cdot A = P(z+dz) \cdot A + \rho A g \, dz \]
Rearranging: \[ P(z) - P(z+dz) = \rho g \, dz \]
Or: \[ \frac{dP}{dz} = -\rho g \]
(Negative because pressure decreases as we go up.)
Integrating from the surface (where \( P = P_0 \) at \( z = 0 \)) to depth \( h \) below the surface: \[ P(h) = P_0 + \rho g h \]
This is the hydrostatic pressure formula: pressure increases linearly with depth.
Pressure at different depths
For water (\( \rho \approx 1000 \) kg/m³, \( g = 9.8 \) m/s²):
- Surface (0 m): \( P_0 = 101,325 \) Pa (atmospheric)
- 10 m depth: \( P = 101,325 + 1000 \times 9.8 \times 10 = 199,325 \) Pa \( \approx 2 \) atm
- 100 m depth (deep diving): \( P = 101,325 + 1000 \times 9.8 \times 100 = 1,080,325 \) Pa \( \approx 10.7 \) atm
Pressure doubles roughly every 10 meters of water depth.
Worked example: Water pressure in a dam
A concrete dam holds back a reservoir of fresh water. The water surface is at height \( h = 50 \) m above the base of the dam. What is the pressure on the dam wall at the base?
Given:
- Depth: \( h = 50 \) m
- Water density: \( \rho = 1000 \) kg/m³
- Atmospheric pressure: \( P_0 = 101,325 \) Pa
- \( g = 9.8 \) m/s²
Solution: \[ P_{\text{bottom}} = P_0 + \rho g h \] \[ P_{\text{bottom}} = 101,325 + (1000)(9.8)(50) \] \[ P_{\text{bottom}} = 101,325 + 490,000 = 591,325 \text{ Pa} \approx 5.8 \text{ atm} \]
The gauge pressure (relative to atmosphere) at the base is \( \rho g h = 490 \) kPa. This enormous pressure means the dam must be thick and strong to withstand the force. For a dam section of width 100 m, the total force on a 50 m height is:
\[ F = \int_0^{50} P(h) \times 100 \, dh = \int_0^{50} (101,325 + 1000 \times 9.8 \times h) \times 100 \, dh \]
The force is not simply \( P_{\text{avg}} \times A \) because pressure varies with depth, but rather \( \rho g \times (\text{water volume}) \) distributed across the height.
Pascal's principle
In a static fluid, a pressure change at any point is transmitted uniformly throughout the fluid. If you increase pressure on a piston in a closed hydraulic system, that pressure boost appears instantly everywhere in the connected fluid. This principle enables hydraulic force amplification: a small force on a small piston can lift a large weight on a large piston.
Pressure in gases vs liquids
For gases, \( \rho \) is much smaller (air: \( \rho \approx 1.2 \) kg/m³). Pressure variation with altitude is therefore gentler:
\[ P(h) = P_0 \exp\left(-\frac{\rho g h}{P_0}\right) \] (exponential, not linear)
This is why atmospheric pressure drops more gradually with altitude than water pressure with depth. At 10 km altitude, atmospheric pressure is roughly 0.25 atm; at 10 m water depth, it's 2 atm.
Continuity of pressure across interfaces
At a boundary between two fluids (e.g., oil floating on water), pressure must be continuous. If the oil-water interface is level, both fluids experience the same pressure at that point. However, the pressure variation with depth below the interface differs for each fluid, depending on \( \rho_{\text{oil}} \) vs \( \rho_{\text{water}} \).
What you should be able to do
- Derive the hydrostatic pressure formula \( P = P_0 + \rho g h \)
- Calculate pressure at any depth in a fluid
- Apply Pascal's principle to hydraulic systems
- Understand why pressure variation with depth is linear for liquids but exponential for gases
- Estimate pressure magnitudes at practical depths (ocean, dam, atmosphere)
Preview of §14.3
Archimedes' principle: a body immersed in a fluid experiences an upward buoyant force equal to the weight of displaced fluid.
📚 See also: Halliday Vol 2, Ch 14, §14.2.
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