Pressure on inclined rectangular plate

For an inclined plane surface: $F = \rho g A\bar h$ and $h^* = \bar h + \dfrac{I_G \sin^2\theta}{A\bar h}$, where $\theta$ is the angle with the horizontal (free surface).

Given: width $b = 0.6$ m, length along the slope $d = 1.2$ m, $\theta = 40^\circ$, top edge 0.70 m below the surface; water.

 free surface ===========================
          : 0.70 m
          +.                 
            \  .   plate 1.2 m long,
             \    .  0.6 m wide
              \ G    .
               \    P   .  40 deg to
                +----------- horizontal

Centroid depth

$$\bar h = 0.70 + \frac{1.2}{2}\sin 40^\circ = 0.70 + 0.6 \times 0.6428 = 1.0857\ \text{m}$$

Area and second moment

$$\begin{aligned} A &= 0.6 \times 1.2 = 0.72\ \text{m}^2 \\ I_G &= \frac{b d^3}{12} = \frac{0.6 \times 1.2^3}{12} = 0.0864\ \text{m}^4 \end{aligned}$$

Total pressure force

$$F = 1000 \times 9.81 \times 0.72 \times 1.0857 = 7668\ \text{N} = 7.67\ \text{kN}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A\bar h} \\ &= 1.0857 + \frac{0.0864 \times 0.4132}{0.72 \times 1.0857} \\ &= 1.0857 + 0.0457 = 1.131\ \text{m} \end{aligned}$$

Measured along the plate from the top edge: $\dfrac{1.131 - 0.70}{\sin 40^\circ} = 0.671$ m (the centroid is at 0.600 m).

Answer: total pressure force $\approx 7.67$ kN; centre of pressure at a vertical depth of 1.131 m below the free surface, i.e. 0.671 m down the plate from its top edge, on the centre line.