Homework SourseCalculate the fluid force on the side of the right triangle
Calculate the fluid force on the side of the right triangle of height 3 feet and base 2 feet submerged in water vertically, with upper vertex located at a depth of 4 feet. The density of water =62.5 ft/s^2.
Solution
Pressure at distance h from top vertex = g(4+h)
Taking a strip of thickness dh at depth h from the top vertex. Its width = h(W/H) where W and H are dimensions of triangle.
Area of strip = dh*h(W/H)
Force = pressure *Area = g(4+h)*h(W/H)*dh
Force in lb = (4+h)*h(W/H)*dh
Integrating it from h = 0 to h = 3, Force = (W/H) [2h2 + h3/3]h=0h=3
F = 62.5*(2/3)*[2*32 + 33/3] = 1125 lb
