Calculate 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.

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=0^h=3

F = 62.5*(2/3)*[2*3² + 3³/3] = 1125 lb

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

Calculate the fluid force on the side of the right triangle

Solution

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