Playground
A column of fluid with depth slider; pressure gradient shown as color intensity and a pressure gauge reading.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Pressure at Depthoutput Total pressure at depth h | Pa | M·L⁻¹·T⁻² | 0 – 1000000 | |
| Surface Pressure Pressure at the surface | Pa | M·L⁻¹·T⁻² | 0 – 200000 | |
| Density Fluid density | kg/m³ | M·L⁻³ | 1 – 13600 | |
| Gravity Gravitational acceleration | m/s² | L·T⁻² | 9.8 – 9.8 | |
| Depth Depth below surface | m | L | 0 – 1000 |
Deep dive
Derivation
Consider a thin horizontal slab of fluid at depth h, area A, thickness dh. Force balance: pressure on bottom × A = pressure on top × A + weight of slab. (P + dP)A = PA + ρgA·dh, giving dP/dh = ρg. Integrating from surface (P = P₀) to depth h yields P = P₀ + ρgh.
Experimental verification
Verified by Stevin's hydrostatic paradox demonstration with different-shaped vessels showing equal pressures at equal depths. Modern depth gauges and pressure transducers used in oceanography confirm to ppm precision.
Common misconceptions
- The shape or width of the container does NOT affect the pressure — only depth matters
- Pressure acts equally in all directions at a point, not just downward
- Gauge pressure (ρgh) and absolute pressure (P₀ + ρgh) are different — be careful which one you need
Real-world applications
- Submarine hull design and crush depth ratings
- Dam wall thickness (thicker at the base where pressure is highest)
- Scuba diving depth limits and decompression
- IV drip bag height calculations in hospitals
- Manometers for pressure measurement
Worked examples
Eardrum pressure at 3 m depth
Given:
- ρ:
- 1000
- g:
- 9.8
- h:
- 3
Find: ΔP
Solution
ΔP = ρgh = 1000 × 9.8 × 3 = 29400 Pa ≈ 29.4 kPa
Pressure at the bottom of the Mariana Trench
Given:
- ρ:
- 1025
- g:
- 9.8
- h:
- 11000
Find: ΔP
Solution
ΔP = 1025 × 9.8 × 11000 ≈ 1.10 × 10⁸ Pa ≈ 1090 atm
Scenarios
What if…
- scenario:
- What if you use mercury instead of water?
- answer:
- Mercury is 13.6× denser, so the same depth gives 13.6× more pressure. This is why mercury barometers are short — only ~76 cm to balance one atmosphere.
- scenario:
- What if the container is shaped like an hourglass or wide cone?
- answer:
- The pressure at any depth depends only on h, not shape. This is Stevin's hydrostatic paradox — a tall narrow tube exerts the same pressure on the base as a wide vat at the same depth.
- scenario:
- What if gravity changed (e.g., on the Moon)?
- answer:
- Pressure scales with g. On the Moon (g ≈ 1.6 m/s²), the same depth of water would exert about 1/6 the pressure.
Limiting cases
- condition:
- h = 0
- result:
- P = P_0
- explanation:
- At the surface, pressure equals atmospheric (or whatever sits on top).
- condition:
- ρ → 0
- result:
- P → P_0
- explanation:
- A massless fluid carries no weight — no depth dependence.
- condition:
- h = 10 m of water
- result:
- ΔP ≈ 1 atm
- explanation:
- Every 10 m of water adds roughly one atmosphere — handy diving rule of thumb.
Context
Simon Stevin · 1586
Stevin showed pressure depends only on depth, not container shape — the famous hydrostatic paradox illustrated with vessels of different shapes filled to the same height.
Hook
Why do your ears hurt at the bottom of a swimming pool?
Find the gauge pressure on your eardrum at 3 m below the surface of fresh water.
Dimensions: [ρgh] = (M·L⁻³)(L·T⁻²)(L) = M·L⁻¹·T⁻² = [P] ✓
Validity: Static, incompressible fluid in a uniform gravitational field. Fails for compressible atmospheres over large altitude ranges — use the barometric formula instead.