MechanicalFluid MechanicseasyFE ExamCoursework

Hydrostatic Force on a Vertical Gate

A vertical rectangular gate is $b = 2.0\ \text{m}$ wide and $h = 3.0\ \text{m}$ tall. Its top edge is level with the water surface, and water stands against one face only.

Take $\rho = 1000\ \text{kg/m}^3$ and $g = 9.81\ \text{m/s}^2$ .

Find the total hydrostatic force on the gate.

Given

b	2 m	Gate width
h	3 m	Gate height
ρ	1000 kg/m^3	Water density

Hint 1

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

Hint 2

Where is the centroid of a rectangle whose top edge is at the water surface?

Hint 3

h_c = 1.5 m. Do not confuse this with the centre of pressure at 2 m, which is where the force acts.

Worked solution — try the problem first

The resultant hydrostatic force on a plane submerged surface is the pressure at the centroid times the area:

$F = \rho g h_c A$

Centroid depth. With the top edge at the surface, the centroid of the rectangle sits at mid-height:

$h_c = \frac{h}{2} = \frac{3.0}{2} = 1.5\ \text{m}$

Area: $A = bh = 2.0 \times 3.0 = 6.0\ \text{m}^2$

Force: $F = 1000 \times 9.81 \times 1.5 \times 6.0 = 88{,}290\ \text{N}$

$\boxed{F = 88.3\ \text{kN}}$

Note on the line of action. The force is not applied at the centroid — pressure grows linearly with depth, so the resultant acts at the centre of pressure, at $\frac{2}{3}h = 2.0\ \text{m}$ below the surface. That distinction does not change the magnitude asked for here, but it governs the moment about the hinge.

Concepts:HydrostaticsCentre of pressureFluid statics

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