1. **Problem statement:** Find the magnitude and line of action of the hydrostatic force acting on surface AB, a circular arc with radius 2 m and width 1 m, submerged in water with atmospheric pressure prevailing. 2. **Calculate hydrostatic force components:** - Force on circular arc segment (F1): $$F_1 = \rho g \times \text{Area} \times \text{depth centroid} = 1000 \times 9.81 \times \frac{1}{4}\pi \times 2^2 \times 1 = 30819\,\text{N}$$ - Force on rectangular surface below arc (F2): $$F_2 = 1000 \times 9.81 \times 5 \times (1 \times 2) = 98100\,\text{N}$$ - Force on bottom rectangular surface (F3): $$F_3 = 1000 \times 9.81 \times 4 \times (1 \times 2) = 78480\,\text{N}$$ 3. **Resolve force components by direction:** - Forces F1 and F3 act at angle 270° (vertical down), so components are $(F_1 \cos 270°,F_1 \sin 270°) = (0,-30819)$ and $(F_3 \cos 270°,F_3 \sin 270°) = (0,-78480)$ - Force F2 acts at angle 0° (horizontal right), components $(98100,0)$ 4. **Sum horizontal and vertical forces:** - Horizontal force: $$F_H = 0 + 98100 + 0 = 98100\,\text{N}$$ - Vertical force: $$F_V = -30819 + 0 - 78480 = -109299\,\text{N}$$ 5. **Calculate resultant force magnitude:** $$F_R = \sqrt{F_H^2 + F_V^2} = \sqrt{98100^2 + (-109299)^2} = 146086\,\text{N}$$ 6. **Calculate angle of the resultant force relative to horizontal:** $$\theta = \tan^{-1} \left( \frac{F_V}{F_H} \right) = \tan^{-1} \left( \frac{-109299}{98100} \right) = -48.09^\circ$$ **Final answer:** The hydrostatic force magnitude is approximately $146086$ N acting at an angle of $-48.09^\circ$ relative to the horizontal.