Resultant force on curved surface ABCDE (S = 1.25)

Resultant force on a curved surface = $\sqrt{F_H^2 + F_V^2}$, where $F_H$ is the force on the vertical projection and $F_V$ is the weight of the 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 of 4 m; the surface then dips as a semicircular trough of radius 2 m from B to E through the lowest point D, so D is 6 m below the free surface. $S = 1.25$, $L = 10$ m, $\gamma = 1.25\times 9.81 = 12.26$ kN/m³.

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

Horizontal component

On the trough, the forces on the left and right quarters are equal and opposite horizontally, so they cancel. Only the wall AB (depth 4 m) gives a net horizontal force:

$$F_H = \gamma\frac{h^2}{2}L = 12.26\times\frac{4^2}{2}\times 10 = 981.0\ \text{kN}$$

Vertical component

Volume above the trough = rectangle of 4 m depth over the 4 m width + half-disc of radius 2 m:

$$\begin{aligned} A &= 4\times 4 + \frac{\pi\times 2^2}{2} = 16 + 6.283 = 22.28\ \text{m}^2\\ F_V &= \gamma A L = 12.26\times 22.28\times 10 = 2732.5\ \text{kN (downward)} \end{aligned}$$

Resultant

$$\begin{aligned} F &= \sqrt{981.0^2 + 2732.5^2} = 2903.2\ \text{kN}\\ \theta &= \tan^{-1}\frac{2732.5}{981.0} = 70.3^\circ \text{ below the horizontal} \end{aligned}$$

Answer: $F \approx 2903$ kN, inclined at $70.3^\circ$ to the horizontal; $F_H = 981$ kN, $F_V = 2732.5$ kN.