Resultant force on curved surface ABCDE (S = 1.1)

The horizontal component of the force equals the force on the vertical projection of the surface; the vertical component equals the weight of liquid vertically above the surface up to the free surface.

Assumed geometry (the figure is not fully clear): A is at the free surface, AB is a vertical wall 4 m high, BC is a quarter circle of radius 2 m (concave upward), and CDE is a horizontal bed 4 m long at 6 m depth. $S = 1.1$, $L = 20$ m, $\gamma = 9.81\times 1.1 = 10.79$ kN/m³.

 A |~~~~~~~~~~~~~~~~~~~~~~  free surface
   |
   | 4 m
 B |
    \   R = 2 m
     '-.
 C ------ D ------ E   (depth 6 m)
   |<---- 4 m ---->|

Horizontal component (vertical projection = 6 m, free surface to the bed):

$$\begin{aligned} F_H &= \gamma\,\frac{h^2}{2}\,L = 10.79\times\frac{6^2}{2}\times 20\\ &= 3884.8\ \text{kN} \end{aligned}$$

Vertical component (area of liquid above the surface, per metre length):

• Above arc BC: the depth of the arc below the free surface is $4 + \sqrt{R^2-(x-2)^2}$, so the area is a $2\times 4$ rectangle plus the quarter disc of area $\pi R^2/4 = \pi$: $8 + \pi = 11.14$ m²
• Above bed CE: $4\times 6 = 24$ m²

$$\begin{aligned} A_{total} &= 8 + \pi + 24 = 35.14\ \text{m}^2\\ F_V &= \gamma A_{total} L = 10.79\times 35.14\times 20 = 7584.3\ \text{kN} \end{aligned}$$

Resultant

$$\begin{aligned} F &= \sqrt{3884.8^2 + 7584.3^2} = 8521.3\ \text{kN}\\ \theta &= \tan^{-1}\frac{7584.3}{3884.8} = 62.9^\circ \text{ below the horizontal} \end{aligned}$$

Answer: $F \approx 8521$ kN (8.52 MN), at $62.9^\circ$ to the horizontal; $F_H = 3884.8$ kN, $F_V = 7584.3$ kN (downward).