1. **Problem Statement:** We are given a circular arc surface AB with radius $r=2$ m and width into the paper of 1 m. The distance from the free surface of water to point E (bottom of AB) is $EB=4$ m. Atmospheric pressure acts at the free surface and the bottom side of AB. We need to find the magnitude and line of action of the hydrostatic force on surface AB. 2. **Knowns:** - Radius $r=2$ m - Width into the paper = 1 m (depth of the submerged surface) - Distance $EB=4$ m (vertical depth from free surface to point E) - The fluid is water with specific weight $\gamma = 9800$ N/m$^3$ - Centroid of area ABC is at $\bar{h} = \frac{4r}{3\pi}$ from the bottom of the circular segment 3. **Step 1: Calculate Depth of Centroid from Free Surface** The distance from free surface to point E is 4 m, so the centroid depth from free surface is: $$ h_c = EB + \bar{h} = 4 + \frac{4 \times 2}{3\pi} = 4 + \frac{8}{3\pi} $$ Calculate numerical value: $$ \frac{8}{3\pi} \approx \frac{8}{9.4248} \approx 0.848 $$ So, $$ h_c \approx 4 + 0.848 = 4.848\ \text{m} $$ 4. **Step 2: Calculate Area of Circular Arc Surface** For a circular arc of radius $r$ and width into paper $b=1$ m, the vertical height of chord $h$ is needed. Since the arc is circular segment with radius 2, and the chord length corresponds to the arc width (not given), we assume the surface area is approximately: $$ A = \text{arc length} \times b $$ The segment length or arc length $L$ for the angle $\theta$ (unknown) cannot be found directly without angle. However, assuming small width into paper, or since only depth and radius provided, we calculate hydrostatic force using given centroid depth and this width (approximate). Alternatively, the problem’s critical information is the location of centroid for the curved surface and radius for hydrostatic calculations. 5. **Step 3: Hydrostatic Force Magnitude** The hydrostatic force on surface AB is: $$ F = \gamma h_c A $$ Where $h_c$ is the depth of the centroid and $A$ is the area. Using width $b=1$ m: The curved surface length $s$ for the circular arc of radius $r=2$ m and chord length $1$ m into the paper is approximately $s=2r\theta$, where $\theta$ is half-angle of arc. But since width into paper is given as 1 m, and radius 2 m, approximate the area as: $$ A = s \times b = s \times 1 $$ Without arc angle specified, the problem references the formula in the other picture (likely the hydrostatic force on curved surfaces uses centroid formula and submerged depth). 6. **Step 4: Line of Action of Hydrostatic Force** For a curved surface submerged vertically: - The horizontal component $F_H$ is equal to the hydrostatic force on the vertical projection of the surface - The vertical component $F_V$ is equal to the weight of the fluid above the curved surface Hence, the resultant force $F_R$ magnitude is: $$ F_R = \sqrt{F_H^2 + F_V^2} $$ And the line of action can be found by moments or centroid of pressure using formulas involving $I_G$, the second moment of area for the projection. 7. **Final answer:** Magnitude of force: $$ F = \gamma h_c A $$ With centroid depth $h_c \approx 4.848$ m, specific weight $\gamma=9800$ N/m$^3$, and width 1 m. Line of action is at: $$ \bar{h} = \frac{4r}{3\pi} $$ from bottom as given. Note: Without the exact formula from the other picture, we use the centroid method and fundamental hydrostatic force formula for curved surface.