Pressure on rectangular plate 3 m × 5 m

The plate is vertical with the 3 m side horizontal, so width $b = 3$ m and depth $d = 5$ m. Use $F = \rho g A\bar h$ and $h^* = \bar h + \dfrac{I_G}{A\bar h}$.

$$A = 3 \times 5 = 15\ \text{m}^2, \qquad I_G = \frac{b d^3}{12} = \frac{3 \times 5^3}{12} = 31.25\ \text{m}^4$$

(i) Top edge flush with the water surface

$$\begin{aligned} \bar h &= \frac{5}{2} = 2.5\ \text{m} \\ F &= 1000 \times 9.81 \times 15 \times 2.5 = 367\,875\ \text{N} = 367.88\ \text{kN} \\ h^* &= 2.5 + \frac{31.25}{15 \times 2.5} = 2.5 + 0.8333 = 3.333\ \text{m} \end{aligned}$$

(Check: $h^* = \frac{2}{3}d = \frac{2}{3}\times 5 = 3.333$ m.)

(ii) Top edge 2 m below the water surface

$$\begin{aligned} \bar h &= 2 + 2.5 = 4.5\ \text{m} \\ F &= 1000 \times 9.81 \times 15 \times 4.5 = 662\,175\ \text{N} = 662.18\ \text{kN} \\ h^* &= 4.5 + \frac{31.25}{15 \times 4.5} = 4.5 + 0.4630 = 4.963\ \text{m} \end{aligned}$$

Answer: (i) $F = 367.88$ kN at 3.333 m depth; (ii) $F = 662.18$ kN at 4.963 m depth. The deeper the plate, the closer the centre of pressure comes to the centroid.