Final answer:
The hydrostatic force against a submerged hemispherical plate is found by integrating water pressure over the area of the plate submerged. The integral takes into account the varying depth and the curvature of the plate. The result gives the total force exerted by the water on one side of the plate.
Step-by-step explanation:
The hydrostatic force against one side of a submerged hemispherical plate can be calculated by integrating the pressure over the area of the plate that is in contact with the water. The pressure at a depth is given by P = ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth. Since the hemispherical plate has a diameter of 6 ft, its radius r is 3 ft. The depth below the surface varies from 2 ft at the top to 5 ft at the bottom of the hemisphere (2 ft below the surface plus the 3 ft radius).
Establish a coordinate system where y varies from 0 at the top of the hemisphere to -3 ft at the bottom. The pressure will then be P(y) = 62.4 lb/ft3 × (2 - y) since the weight density of water is 62.4 lb/ft3. The differential area dA on a horizontal strip at depth y is the circumference of the horizontal circle times the thickness dy, thus dA = 2π ∙ √(r2 - y2) ∙ dy. The differential force dF is the pressure at y times dA, dF = P(y) × dA. The total force F is the integral of dF from y = 0 to y = -3 ft.
Therefore, the hydrostatic force F is represented by the integral:
F = ∫0-3 62.4 (2 - y) 2π √(9 - y2) dy. Evaluating this integral gives the total hydrostatic force on one side of the plate in pounds, which can be rounded to the nearest whole number.