Hydrostatic Force on a Submerged Surface

Pressure grows linearly with depth, so the load on a vertical wall is a triangular distribution rather than a uniform push. That splits the problem in two: how large is the resultant, and where does it act. Answering only the first is the standard mistake.

Because pressure varies linearly, its average over the surface is exactly its value at the centroid. So F = ρg·h̄·A with no integration needed, where h̄ is the centroid's depth.

This holds for a plane surface at any angle, submerged at any depth. Only the centroid's *vertical* depth enters — the surface's inclination affects the area and the geometry, not the formula.

A 3 m deep, 10 m wide vertical gate: h̄ = 1.5 m, A = 30 m², so F = 1000 × 9.81 × 1.5 × 30 = 441 kN — about the weight of forty cars.

A triangular load has its resultant at the centroid of the *triangle*, a third of the way up from the base. For a vertical rectangle extending from the free surface, that is two thirds of the depth down — not halfway.

In general, taking moments of the pressure distribution gives h_cp = h̄ + I_G/(h̄·A), where I_G is the surface's second moment of area about its own centroidal axis.

• Every term in the correction is positive, so the centre of pressure is always below the centroid.
• As h̄ grows, the correction shrinks — a deep sluice sees near-uniform pressure and the two points converge.
• For a surface breaking the free surface, the correction is at its largest and cannot be ignored.

Assume the centroid and the moment about a dam's toe comes out about 33% light. That is the difference between a stable design and an overturning one.

On a curved surface the normal direction rotates from point to point, so direct integration is unpleasant. Resolve into components instead, and each becomes a problem already solved.

1. 1Horizontal component: equal to the force on the surface's vertical projection — a plane-surface problem.
2. 2Vertical component: equal to the weight of the fluid directly above the surface, real or imagined.
3. 3Combine the two at the end. They need not pass through the same point, so take moments carefully.

The imagined volume works because pressure at a point depends only on depth, not on what occupies the space above. Fluid beneath a curved gate pushes up by exactly the weight of the column that *would* sit above it. Extend the same argument to a fully enclosed body and it becomes Archimedes' principle.

Real dam failures are overturning or sliding, not the concrete crushing. So the moment arm — the centre of pressure — matters more than the force. Silt raises the effective density of the retained material, and uplift pressure under the base reduces the friction holding it down; both are judgements about the loading rather than outputs of the formula.