Intermediate · 12 min
Fluid pressure and buoyancy
Calculate pressure changes with depth and the upward force from displaced fluid.
Pressure increases with depth
In a fluid at rest with nearly constant density, the pressure increase with depth is Δp = ρ × g × h. Here h is vertical depth below the reference level. This is a pressure difference. For absolute pressure below an open liquid surface, add atmospheric pressure to it. At the same depth in a connected fluid at rest, pressure is the same regardless of the container's shape.
Worked example
At 2 m below a water surface, using ρ = 1,000 kg/m³ and g = 10 m/s²: Pressure increase = 1,000 × 10 × 2 = 20,000 Pa
The buoyant force
Fluid pressure acts on all sides of an immersed object and is greater lower down. The resulting upward buoyant force equals the weight of the displaced fluid: Fᵦ = ρfluid × g × Vdisplaced. Use the immersed volume, which may be less than the object's total volume. For an object floating at rest with no other vertical forces, buoyant force balances weight.
Worked example
An object displaces 0.002 m³ of water. Buoyant force = 1,000 × 10 × 0.002 = 20 N
Model a floating object
- An object floats at rest and weighs 40 N. Find the buoyant force and the volume of water it displaces, using density 1,000 kg/m³ and g = 10 m/s².
- Explain why you need the displaced volume rather than the object's entire volume when it is only partly immersed.