10 / CHAPTER 02
Hydrostatics and buoyancy
Solve fluids at rest and connect pressure variation to weight.
ANCHOR EQUATION
Hydrostatic pressure
In symbols: p = p₀ + ρgh
Pressure increases with depth in a fluid at rest.
SCRATCH-PAPER LAB
Change one parameter; watch the prediction move.
Increase depth to see why the pressure difference is proportional to ρgh. The line is gauge pressure; atmospheric pressure is not included.
Hydrostatic pressure is the weight above a point
In a fluid at rest, pressure must rise with depth because a lower layer supports the weight of fluid above it. The pressure gradient is a force density, not a statement that pressure is ‘used up.’
Balance a thin fluid slab
Take a horizontal slab of area A and thickness dz, with z upward. Upward pressure force at bottom is pA; downward force at top is (p + dp)A; weight is ρgA dz downward. Equilibrium gives pA − (p+dp)A − ρgA dz = 0, so dp/dz = −ρg. For constant ρ, integrate from surface z=0 with p=p₀: p = p₀ + ρg h where h is depth.
At 3 m depth in water, gauge pressure is ρgh = 998(9.81)(3) = 29.4 kPa. Absolute pressure is about 130.7 kPa if atmospheric pressure is 101.3 kPa.
Density variation matters for gases
The constant-density formula works for liquids over modest heights. In an atmosphere, ρ changes with p, so integrating dp/dz = −ρg produces an exponential-like pressure profile.
Work it: A tank has water depths 0.4 m and 1.2 m at two outlets. Find the gauge-pressure difference and the force difference on equal 2 cm² outlet areas.