Hydrostatic forces on surfaces
A dam does not feel one pressure. It feels a different pressure at every depth along its face, and the single number an engineer eventually writes down — one resultant force, acting at one point — has to stand in honestly for all of them at once. Getting the size right is the easy half. Getting the point right is where this article spends most of its time.
Everything in pressure and fluid statics established how pressure varies through a fluid at rest. This article puts that distribution to work: given \(P = P_0 + \rho g h\) acting over an entire submerged surface, what single force and what single point of application would produce the same effect? A gate, a dam face, a submarine’s hull all have to be designed against that reduced answer, not against the raw distribution, so finding it honestly is the entire point of the subject.
Force on a plane surface
Take any flat, submerged surface, at any angle, and cover it with pressure that varies linearly with depth. The resultant force is, by definition, the sum of every elementary pressure force over the whole area:
That integral looks as though it demands calculus specific to every new shape, but it does not. Substitute \(P = \rho g h\), with \(h\) the depth of each element \(dA\), and the integral of \(h\) over an area, divided by that area, is nothing more than the definition of the centroid’s depth, \(h_C\). The whole thing collapses to
where \(P_C\) is simply the pressure at the centroid of the area, as if the entire surface were sitting at one uniform depth. This holds for a vertical wall, a sloped dam face, an inclined hatch — any flat surface at all, because the derivation never used the angle, only the fact that depth varies linearly across a plane. What it does not tell you is where that force acts, and treating \(h_C\) as the depth of the force as well as the depth of its magnitude is the single most common mistake in this topic. That question gets its own section next, because the answer is not the centroid at all.
Pressure on a submerged surface grows with depth, point by point.
The centre of pressure
A resultant force has to produce the same turning effect as the distribution it replaces, not merely the same total size, and pressure on a submerged surface is not uniform — it is smallest near the free surface and largest at depth. The lower part of any submerged plate is doing more work than the upper part, pulling the effective line of action down below the plate’s own geometric centre. That point is called the centre of pressure, and for a surface whose top edge sits right at the free surface, with a purely triangular pressure distribution, it sits at
measured down from the surface — a third of the way up from the base, not halfway down as the centroid of the plate’s area alone would suggest. The general result, valid for a surface at any depth and any angle to the free surface, follows from the parallel axis theorem applied to the moment of the pressure distribution about the centroid:
Here \(y_C\) is the distance from the free surface to the centroid, measured along the plane of the surface itself, and \(I_{xx,C}\) is the second moment of the area about its own centroidal axis parallel to the surface — the same quantity used for bending in a beam, here doing an entirely different job. The second term is always positive, so the centre of pressure is always below the centroid on a flat surface open to a free surface above — never above it, never coincident with it, except in one limit.
That limit is worth internalising because it explains why the distinction sometimes gets ignored in practice and gets away with it. As \(y_C\) grows — the surface sinking deeper and deeper below the free surface — the offset \(I_{xx,C}/(y_C A)\) shrinks toward zero, because the pressure varies less and less, in relative terms, across the surface’s own modest height compared with how large the surface’s total depth has become. A gate near the bottom of a deep reservoir experiences pressure that is nearly uniform across its own height, and centroid and centre of pressure converge. A gate flush with the free surface experiences the most extreme variation possible relative to its own size, and the offset is at its largest. Depth of submergence, not the size of the surface alone, is what decides how much the distinction matters.
Start with the ordinary centroid of the surface's own area.
The pressure prism
There is a shortcut that avoids the second-moment formula altogether for the shapes an engineer meets most often — a rectangular gate, a dam panel — and it works by treating the pressure distribution itself as a solid object. Plot pressure against depth along the submerged surface and the shape traced out is the pressure prism: a triangle if the surface starts at the free surface, a trapezoid if it starts some distance below it. The resultant force equals the volume of that prism, and its line of action passes through the prism’s own centroid — exactly the same logic used for a distributed load on a beam, with pressure standing in for load intensity.
A trapezoidal prism is easiest handled by splitting it into shapes whose forces and centroids are already known, then adding. A gate of height \(H\) and width \(b\), with its top edge at depth \(h_1\), carries a uniform pressure \(\rho g h_1\) already present at its top edge, plus a triangular increment building up across its own height. Split accordingly:
\(F_1\) acts at the gate’s mid-height, because a uniform (rectangular) distribution is centred there; \(F_2\) acts two-thirds of the way down the gate, because a triangular distribution is centred there. Add the two forces for the resultant, and take moments of each about a convenient axis to relocate the single equivalent force:
This is bookkeeping, not new physics — it is the general moment formula from the previous section, worked out by hand for a shape simple enough that hand-working it is faster than reaching for \(I_{xx,C}\) in a table. For anything less regular than a rectangle, go back to the formula; for a rectangle, the prism method is usually quicker.
Submerge the surface below the free surface and the prism becomes a trapezoid.
Curved surfaces
Everything so far assumed a flat surface, which let every elementary pressure force point the same way and simply add up as scalars. A curved surface — the face of an arch dam, a cylindrical gate, the hull of a submerged tank — has pressure pushing normal to itself at every point, and normal means a different direction at every point along the curve. Summing those directly is a genuine vector integral. It is avoided, almost always, by treating the horizontal and vertical resultants separately, because each one turns back into a problem already solved.
The horizontal component of the force on a curved surface equals the force on that surface’s flat vertical projection — the shadow it would cast if the fluid pushed straight through it onto a flat vertical plane behind:
computed exactly as in the first section of this article, because the projection is a flat surface. The reasoning is a horizontal force balance on the wedge of fluid between the curve and its projection: whatever pressure force the curve withstands horizontally, the flat projection would withstand identically, since the fluid slab between them is in equilibrium and contributes no net horizontal force of its own.
The vertical component is weight, not pressure, in disguise:
where \(V\) is the volume of fluid directly above the curved surface, up to the free surface. If the fluid genuinely sits there, this is the literal weight of that column. If the curve faces the other way and the fluid is beneath it instead, the same volume is imagined filled in above the surface anyway — a fictitious column, but one whose weight is exactly what a vertical force balance on the real fluid below demands, because pressure does not know or care whether the material producing it happens to be real water or an accountant’s construction placed there to make the bookkeeping close.
With both components in hand, the resultant is an ordinary vector sum:
and because every elementary pressure force on the curve passes through the centre of curvature — pressure being normal to a circular arc always points at its centre — the resultant does too. That is a useful check as much as a shortcut: if a computed line of action does not pass through the surface’s own centre of curvature, an arithmetic slip has happened somewhere upstream.
Pressure on a curved surface still acts normal to it everywhere — but that direction now changes.
Worked example: a gate
Put the whole chapter to work on one gate. It is rectangular, 3 m wide and 2 m tall, hinged along its top edge, with that top edge sitting 1 m below the free surface of the water behind it. A horizontal force \(P\), applied at the gate’s bottom edge, is what keeps it shut against the water. Find \(P\).
Start with the resultant force, which needs only the centroid depth:
The area is \(A = bH = 3 \times 2 = 6\ \text{m}^2\). That number is correct and useless on its own, because a moment balance about the hinge needs to know exactly where \(F_R\) acts, not merely how large it is.
For this vertical gate, \(y_C = h_C = 2\) m, and \(I_{xx,C} = bH^3/12 = 3(2)^3/12 = 2\ \text{m}^4\). So
The centre of pressure sits 2.167 m below the free surface, which is 1.167 m below the hinge and, equivalently, 0.167 m below the gate’s own centroid — small, because the gate is short compared with the 1 m it is already submerged, but not zero, and not something a moment balance can be allowed to ignore.
Take moments about the hinge. \(F_R\) acts \((y_{cp} - h_1) = 1.167\) m below it; \(P\) acts the full height \(H = 2\) m below it:
Using the centroid depth in place of the centre of pressure here — the error this whole article has been arguing against — would have given \(P = 117.7 \times 1/2 = 58.9\) kN instead: an understatement of about 14%, in the unsafe direction, from a single wrong assumption about where a correctly sized force actually pushes.