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Hydrostatic Force on Submerged Surfaces

Intuition

A dam wall doesn't feel one number — it feels a distribution: gentle pressure near the waterline, crushing pressure at the base. Two questions matter for engineering: what is the total force, and where does it effectively act? The answers turn out to involve two old friends from mechanics: the centroid and the moment of inertia of the wetted area.

Mathematical formulation

A plane surface of area \(A\) is submerged, inclined at angle \(\theta\) to the horizontal. Measuring \(y\) along the plate from the fluid surface, the local pressure is

\[P = P_0 + \rho g y \sin\theta.\]

Total (resultant) force — integrate over the plate:

\[F_R = \int_A P\, dA = P_0 A + \rho g \sin\theta \int_A y\, dA\]

Defining the centroid \(y_C = \frac{1}{A}\int_A y\, dA\):

\[\boxed{\,F_R = (P_0 + \rho g\, y_C \sin\theta)\,A = P_C\, A\,}\]

The resultant force equals the pressure at the centroid times the area.

Center of pressure — the point \(y_P\) where \(F_R\) effectively acts, found by moment balance \(y_P F_R = \int_A y P\, dA\):

\[y_P = y_C + \frac{I_{xx,C}}{\left[y_C + P_0/(\rho g \sin\theta)\right] A} \quad\xrightarrow{\,P_0 = 0\,}\quad \boxed{\,y_P = y_C + \frac{I_{xx,C}}{y_C\, A}\,}\]

where \(I_{xx,C}\) is the second moment of the area about a horizontal axis through its centroid (via the parallel-axis theorem \(I_{xx,O} = I_{xx,C} + y_C^2 A\)).

Reference table (area and centroidal moment of inertia):

Shape \(A\) \(I_{xx,C}\)
Rectangle \(a \times b\) \(ab\) \(ab^3/12\)
Circle radius \(R\) \(\pi R^2\) \(\pi R^4/4\)
Ellipse \(a, b\) \(\pi ab\) \(\pi a b^3/4\)
Triangle base \(a\), height \(b\) \(ab/2\) \(ab^3/36\)
Semicircle radius \(R\) \(\pi R^2/2\) \(0.1098\,R^4\)

Worked example: vertical rectangular gate

A vertical gate of width \(w = 2\) m and height \(b = 3\) m has its top edge at the water surface (\(P_0\) ignored — gauge pressure). Find \(F_R\) and \(y_P\).

Centroid depth: \(y_C = b/2 = 1.5\) m, with \(\theta = 90°\).

\[F_R = \rho g y_C A = 1000 \times 9.8 \times 1.5 \times 6 \approx 88\ \text{kN}\]
\[y_P = y_C + \frac{I_{xx,C}}{y_C A} = 1.5 + \frac{2\times 3^3/12}{1.5 \times 6} = 1.5 + 0.5 = 2.0\ \text{m}\]

The force acts at two-thirds depth — the classic result for a triangular pressure distribution.

Physical interpretation

  • \(F_R = P_C A\) works because pressure varies linearly: the mean of a linear distribution over the area is its value at the centroid.
  • \(y_P\) always lies below \(y_C\) (\(I_{xx,C} > 0\)): deeper parts of the plate carry more load, dragging the effective point of action downward. As the plate goes very deep, \(y_P \to y_C\) (pressure becomes nearly uniform).

Common mistakes

  • Placing the resultant at the centroid. The magnitude uses centroid pressure; the location is deeper.
  • Forgetting \(\sin\theta\) for inclined surfaces — depth is \(y\sin\theta\), not \(y\).
  • Mixing up \(I_{xx,C}\) and \(I_{xx,O}\) — the compact formula uses the centroidal moment.

Knowledge graph position

Prerequisites: Hydrostatic equilibrium. Leads to: engineering statics of gates, dams, tanks.

Quiz

Q1 (conceptual). For a fully submerged plate, does doubling the depth of submergence change (i) the resultant force, (ii) the gap \(y_P - y_C\)?

Answer

(i) Yes — \(F_R = \rho g y_C \sin\theta\, A\) grows linearly with centroid depth. (ii) It shrinks: \(y_P - y_C = I_{xx,C}/(y_C A) \propto 1/y_C\). Deep plates feel nearly uniform pressure.

Q2 (computational). A circular porthole of radius 0.5 m has its center 10 m below the surface. Total force?

Answer

\(F_R = \rho g y_C A = 1000\times 9.8 \times 10 \times \pi(0.5)^2 \approx 77\ \text{kN}\) — pressure at the center times area.

Q3 (multiple choice). The center of pressure coincides with the centroid when:

  • (a) the plate is vertical (b) the plate is horizontal (c) never (d) the fluid is a gas
Answer

(b). A horizontal plate sits at one depth — uniform pressure, so the resultant acts at the centroid. ((d) is nearly true in practice but not exactly.)