The formula

Pressure from force and area
p = F / A
Pressure at depth in a fluid of constant density
p = p₀ + ρgh
Weight, when the force is an object resting on a surface
F = mg

What the symbols mean

SymbolMeaningUnit
pPressurePa (1 Pa = 1 N/m²)
FForce perpendicular to the surfaceN
AArea the force is spread over
p₀Pressure at the surface, usually atmospheric at 1.013 × 10⁵ PaPa
ρDensity of the fluidkg/m³
gAcceleration due to gravity9.80 m/s²
hDepth below the surface of the fluidm
mMass of the objectkg

When it applies

  • Only the component of force perpendicular to the surface counts. A force pushing sideways along the surface contributes nothing.
  • p = F / A gives the average pressure over the contact area, so use the area actually touching.
  • The depth formula assumes a fluid at rest with constant density, which is a good model for water in a tank or a pool.

Worked example

Problem. A 45 kg crate rests on a floor on a base measuring 0.60 m by 0.40 m. What pressure does it exert on the floor?

  1. Find the force. The crate presses down with its weight: F = mg = (45 kg)(9.8 m/s²) = 441 N.
  2. Find the contact area: A = 0.60 m × 0.40 m = 0.24 m².
  3. Divide: p = F / A = 441 N / 0.24 m² = 1837.5 Pa.
  4. Round to two significant figures: p = 1.8 × 10³ Pa, which is 1.8 kPa.

Answer. About 1.8 kPa, roughly one fifty-fifth of atmospheric pressure.

Common mistakes

  • Using the object's whole surface area rather than the part actually in contact.
  • Substituting mass in kilograms where the formula needs force in newtons. Convert with F = mg first.
  • Confusing gauge pressure with absolute pressure. A tire gauge reading 220 kPa means 220 kPa above atmospheric.
  • Thinking pressure at depth depends on the width of the container. It depends only on depth, density and the pressure at the surface.

Related formulas

Sources