1. **Problem Statement:** Calculate the buoyant force and the center of pressure on a vertical rectangular surface submerged in water. 2. **Given:** - Width, $b = 3$ m - Height, $h = 7$ m - Depth of top edge below water surface, $y = 2.5$ m - Density of water, $\rho = 1000$ kg/m$^3$ - Gravitational acceleration, $g = 9.81$ m/s$^2$ 3. **Buoyant Force Calculation:** The buoyant force on a submerged surface is the hydrostatic force due to water pressure. Pressure at depth $y$ is $p = \rho g y$. Since pressure varies linearly with depth, the total force $F$ on the surface is: $$ F = \rho g b \int_y^{y+h} z \, dz $$ where $z$ is depth below water surface. Calculate the integral: $$ \int_y^{y+h} z \, dz = \left[ \frac{z^2}{2} \right]_y^{y+h} = \frac{(y+h)^2 - y^2}{2} $$ Substitute values: $$ F = 1000 \times 9.81 \times 3 \times \frac{(2.5 + 7)^2 - 2.5^2}{2} $$ Calculate inside the fraction: $$ (9.5)^2 - 2.5^2 = 90.25 - 6.25 = 84 $$ So, $$ F = 1000 \times 9.81 \times 3 \times \frac{84}{2} = 1000 \times 9.81 \times 3 \times 42 $$ Calculate: $$ F = 1000 \times 9.81 \times 126 = 1,236,060 \text{ N} $$ 4. **Center of Pressure Calculation:** The center of pressure $y_{cp}$ is the depth where the resultant force acts, given by: $$ y_{cp} = \frac{\int_y^{y+h} z p(z) b \, dz}{F} = \frac{\rho g b \int_y^{y+h} z^2 \, dz}{F} $$ Calculate the integral: $$ \int_y^{y+h} z^2 \, dz = \left[ \frac{z^3}{3} \right]_y^{y+h} = \frac{(y+h)^3 - y^3}{3} $$ Substitute values: $$ \int_y^{y+h} z^2 \, dz = \frac{9.5^3 - 2.5^3}{3} = \frac{857.375 - 15.625}{3} = \frac{841.75}{3} = 280.5833 $$ Calculate numerator: $$ N = 1000 \times 9.81 \times 3 \times 280.5833 = 8,252,000 $$ Finally, $$ y_{cp} = \frac{8,252,000}{1,236,060} \approx 6.68 \text{ m} $$ 5. **Interpretation:** - The buoyant force acting on the surface is approximately $1,236,060$ N. - The center of pressure is located about $6.68$ m below the water surface, which is deeper than the centroid of the surface ($y + h/2 = 2.5 + 3.5 = 6$ m) due to pressure distribution.