**Step 1: Calculate the Hydrostatic Force on the Wall** The pressure at depth \( h \) in a fluid is given by \( P = \rho g h \), where: - \( \rho = 1000 \) kg/m³ (density of water) - \( g = 9.81 \) m/s² (acceleration due to gravity) - \( h \) is the depth below the water surface. The total hydrostatic force \( F \) on a vertical wall is: \[ F = \rho g A \overline{h} \] where: - \( A \) is the area of the submerged wall, - \( \overline{h} \) is the depth to the centroid of the submerged area. For a wall 25 m wide and 4 m high, \[ A = 25 \times 4 = 100 \text{ m}^2 \] The centroid is at half the height: \[ \overline{h} = \frac{4}{2} = 2 \text{ m} \] Now, calculate the force: \[ F = 1000 \times 9.81 \times 100 \times 2 = 1,962,000 \text{ N} \] **Explanation:** This step determines the total horizontal force due to water against the wall by integrating the pressure distribution over the submerged surface. --- **Step 2: Calculate the Required Weight \( W \) Using Moment Balance** The hydrostatic force acts at a depth of \( \frac{4}{3} \) from the bottom of the wall (centroid of pressure for a rectangle). Taking moments about the pinned point at the bottom: - Moment arm of hydrostatic force: \( \frac{4}{3} \) m above the base - Moment arm of the weight \( W \): 3 m from the wall (given in the diagram) Set the clockwise moment due to water force equal to the counterclockwise moment due to the weight: \[ F \times \frac{4}{3} = W \times 3 \] \[ 1,962,000 \times \frac{4}{3} = W \times 3 \] \[ 2,616,000 = W \times 3 \] \[ W = \frac{2,616,000}{3} = 872,000 \text{ N} \] **Explanation:** This step ensures rotational equilibrium by balancing moments around the pinned base, using the calculated force and its line of action. --- **Final Answer Summary:** The required weight \( W \) to hold the wall is **872,000 N**. This value maintains equilibrium by counteracting the hydrostatic moment produced by water pressure on the submerged wall surface.