Pressure, Manometers & Hydrostatic Force
The hydrostatic law, absolute vs gauge pressure, piezometric head, and computing resultant force + centre of pressure on submerged plane surfaces.
- Apply the hydrostatic law to find pressure at any depth in a static fluid
- Distinguish absolute, gauge and vacuum pressure; compute with manometers
- Locate the centre of pressure and resultant hydrostatic force on a plane surface
A dam wall two metres thick at the base and a submarine hull a few centimetres thick obey the same pressure-versus-depth law. Get the hydrostatic integral wrong and the structure fails; get it right and you can design tanks, manometers and hydraulic presses with the same three lines of algebra.
Hydrostatic law: dp/dz = −ρg (z measured upward). For constant ρ:
- p = p₀ + ρgh (pressure increases with depth)
- Two points at the same elevation in a connected static fluid of the same density have the same pressure (Pascal / hydrostatic paradox).
Pressure measures:
- Absolute: p_abs = p_atm + p_gauge
- Vacuum: p_vac = p_atm − p_abs when p_abs < p_atm (quoted as positive)
- Piezometric head: p/(ρg) + z = constant along a vertical in a static fluid — the workhorse of groundwater and manometer problems.
Manometers: balance pressure columns. U-tube with a gauge fluid of density ρ_m: walk from one leg to the other adding ρgh when descending, subtracting when ascending, set equal at the end.
Hydrostatic force on a plane submerged surface:
- Resultant magnitude: F = p_c × A, where p_c is pressure at the centroid of the submerged area A (works even if the surface is not horizontal, for constant Ϸ).
- Centre of pressure (where the resultant effectively acts): for a rectangular gate of width b hinged along the top edge submerged to depth h₁ with height h₂ below the hinge: y_cp = y_c + I_xx,c/(y_c × A) where y is measured along the surface from the free surface, y_c is centroid distance, I_xx,c is second moment about the centroidal axis parallel to the free surface.
- For a vertical rectangle from the surface: y_cp = (2/3)×height — always below the centroid, because pressure grows with depth.
Curved surfaces: resolve into horizontal component (force on the vertical projection, acting at its centroid-pressure point) and vertical component (weight of the fluid column above the curved surface, real or imaginary).
| Formula / rule | |
|---|---|
| Hydrostatic law | dp/dz = −ρg ⇒ p = p₀ + ρgh |
| Absolute vs gauge | p_abs = p_atm + p_gauge |
| Resultant on plane | F = p_c × A |
| Centre of pressure | y_cp = y_c + I_xx,c/(y_c A) |
| Vertical rect (from surface) | y_cp = (2/3) × height |
For a full-width rectangle whose top edge is at the free surface, y_c = h/2 and y_cp = 2h/3 — memorise the 2/3 result; the integral derivation is the proof.
F = p_c × A is exact for constant ρ, but the centre of pressure is NOT the centroid unless the surface is horizontal.
- 1
Centroid depth
Gate spans depth 1 m to 4 m; centroid at y_c = 1 + 3/2 = 2.5 m.
- 2
Area
A = 2 × 3 = 6 m².
- 3
Resultant force
F = ρg y_c A = 1000 × 9.81 × 2.5 × 6 ≈ 147.2 kN.
- 4
Second moment
I_xx,c = b h³/12 = 2 × 3³/12 = 4.5 m⁴ (about centroidal axis parallel to surface).
- 5
Centre of pressure
y_cp = y_c + I_xx,c/(y_c A) = 2.5 + 4.5/(2.5 × 6) = 2.5 + 0.3 = 2.8 m below the surface.
- 6
Sanity
y_cp > y_c as expected (always below centroid for pressure growing with depth).
Solution
F ≈ 147 kN; centre of pressure at 2.8 m below the free surface (0.3 m below the centroid).
Same gate but with its BOTTOM edge at the free surface (extending upward is impossible — so flip: bottom edge just below surface, extending downward 3 m). Where is y_cp relative to the centroid now vs the previous case?
Hint: Top edge at 0, bottom at 3 m — compute y_c and compare.
Using p = ρgh with h measured from the bottom of the tank
h is always depth below the FREE SURFACE (or the piezometric reference) — fix the datum first.
Setting y_cp = y_c for a vertical or tilted gate
Only a horizontal surface has y_cp = y_c; everywhere else the deeper, higher-pressure part pulls the resultant down.
Define centre of pressure and state why it does not coincide with the centroid of a vertical submerged plane.
Centre of pressure is the point where the resultant hydrostatic force effectively acts. For a vertical surface, pressure increases linearly with depth, so the lower half contributes more force than the upper half, shifting the resultant below the centroid (y_cp = y_c + I_xx,c/(y_c A) > y_c).
Source: Standard definition + reasoning question.
- dp/dz = −ρg; p = p₀ + ρgh with h from the free surface.
- F on a plane = p_c × A; y_cp always ≥ y_c for non-horizontal surfaces.
- Curved surfaces decompose into horizontal (projected vertical area) + vertical (fluid weight) components.
