Question

Bonus-10 Points] An irrigation canal gate is hinged at the top and opened by means of cable. The fluid is water with a density of 62.4 lbm/ft3. The local gravitational acceleration is 31.7 ft/s. The gate is rectangular with a height of 3 feet and a width of 4 feet. Determine the required force in the cable in units of [lbr] to open the gate. a Hinge 3 зн 35°

Answer 1

Now the distance of centre of gravity of the gate from the free surface is, h^bar = 1 + 3/2 h^bar = 1 + 1.5 = 2.5 ft so, pressure force of water on the gate, is F_c = rho g A h^bar Now, rho = 62.4 lb/ft^3, g = 31.7 ft/s^2, A = 3 times 4 = 12 ft^2 so, F_c = 62.4 times 31.7 times 12 times 2.5 lb ft/s^2 = 62.4 times 12 times 2.5 (lbf) F_c = 1872 lbf Now, point where this force acts is the centre of pressure, \r\n And, centre of pressure location from free surface, is, h_c = h^bar + I_G/Ah^bar = 2.5 + 1/12 times 4 times 3^3/12 times 2.5 h_c = 2.8 ft from the free surface so, now, drawing the FBD of the gate \r\n and now, (sigma)_hinge = 0 implies F cos theta times 3 = F_c times \r\n