Lessons · fluids

Force on a dam: pressure grows with depth, force is the area

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 3 modeling assumptions (author-asserted)

A vertical wall $w = 5$ m wide holds back fresh water ( $\rho = 1000$ kg/m³) to a depth of $H = 4$ m. The water pressure grows with depth as $P = \rho g h$ , so the wall is pushed hardest at the bottom and not at all at the surface. What is the total force on the wall? It is the area under the pressure curve — drag the cursor to raise the water level and watch the force grow as the square of the depth.

Force on the wall

F = \tfrac12\rho g w H^2

(the shaded area)

400000 N

Pressure at the bottom

P = \rho g H

40000 Pa

Average (centroid) pressure

\bar P = \tfrac12\rho g H

20000 Pa

If the pressure were the bottom value everywhere:

F = \rho g H\cdot A

(the rectangle — twice the force)

800000 N

The force is the area under the pressure–depth curve

F = \int_0^H P(h)\,w\,dh

Each horizontal strip of the wall at depth $h$ (width $w$ , height $dh$ ) feels a force $dF = P(h)\,w\,dh$ . The total force is the integral over the depth — the area under the strip-force curve. For a uniform pressure this area is a rectangle ( $P_0 A$ ); when the pressure grows with depth it is the area under a rising line.

check $\tfrac{dF}{dh} = \rho g w h$ ; check $F = \int_0^h \rho g w\,h'\,dh'$ ; recover $F = \bar P A$ ; collapse the uniform-pressure case to $P_0 A$

✓ The force's slope is the strip force: $F'(h) = \rho g w h$ — the area's growth rate is the curve's height. [structural]
✓ The accumulated force is the area: $F(h) = \int_0^h \rho g w\,h'\,dh'$ . [structural]
✓ The memorized $F = \bar P A$ (average pressure $\bar P = \rho g h/2$ at the centroid, area $A = wh$ ) is exactly the area. [structural]
✓ If the pressure did not grow with depth ( $P_0$ uniform), the integral collapses to $F = P_0 A$ — a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “The force on a dam depends on how much water is behind it — a wall holding back a huge lake feels far more force than one holding back a narrow tank of the same depth.”

It does not. The pressure at a given depth is $\rho g h$ regardless of how much water sits at that level — a teaspoon or an ocean. The force on the wall is the area under that pressure profile, $\tfrac12\rho g w H^2$ , set entirely by the depth and the wall's width. This is the hydrostatic paradox: raise the water a little (drag the cursor) and the force jumps because it grows as $H^2$ , but widen the lake behind the wall and nothing changes.

Modeling assumptions — author-asserted, disclosed not discharged

Incompressible fluid of constant density $\rho$ ; gauge pressure (atmospheric pushes equally on both faces of the wall and cancels).
$g = 10$ m/s² as a magnitude, with depth $h$ measured positive downward (the house $g = -10$ convention is for up-positive kinematics; here pressure simply grows with depth).
A flat vertical wall, so each horizontal strip has the same width $w$ ; the force is horizontal and normal to the wall.

The dF/dh–h graph, fully annotated

A static rendering (Matplotlib): the shaded area under dF/dh is the accumulated integral F, and the slope of F is dF/dh. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):