Force to hold hinged rectangular gate

Find the water force and its line of action, then take moments about the hinge.

Given: gate 2 m wide, 4 m high (vertical), water 6 m deep above the bottom of the gate, hinge 0.1 m below the centroid G.

 free surface ============
          | 2 m
     +----------+  top (2 m depth)
     |          |
     |    G     |  4 m   (G at 4 m depth)
     |  hinge o |  4.1 m depth
     |     P    |  4.333 m depth
     |          |
     +----------+  bottom (6 m depth)
              <-- P_b (applied force)

Water force

$$\begin{aligned} A &= 4 \times 2 = 8\ \text{m}^2, \quad \bar h = 2 + 2 = 4\ \text{m} \\ F &= \rho g A\bar h = 9810 \times 8 \times 4 = 313\,920\ \text{N} = 313.92\ \text{kN} \end{aligned}$$

Centre of pressure

$$\begin{aligned} I_G &= \frac{2 \times 4^3}{12} = 10.667\ \text{m}^4 \\ h^* &= 4 + \frac{10.667}{8 \times 4} = 4 + 0.3333 = 4.333\ \text{m} \end{aligned}$$

Moments about the hinge

• Hinge depth $= 4 + 0.1 = 4.1$ m.
• Water force acts $4.333 - 4.1 = 0.2333$ m below the hinge, so it tends to turn the gate open at the bottom.
• Force $P_b$ at the bottom acts $6 - 4.1 = 1.9$ m below the hinge, opposite to the water force.

$$\begin{aligned} P_b \times 1.9 &= 313.92 \times 0.2333 \\ P_b &= \frac{73.25}{1.9} = 38.55\ \text{kN} \end{aligned}$$

Answer: a horizontal force of about 38.55 kN must be applied at the bottom of the gate, towards the water, to keep it vertical.