Question

Example ロAn aquariumn contains seawater (γ- 64.0 lb/ft3) to a depth of 1 ft as shown in the figure. To repair some damage to one corner of the tank, a triangular section is replaced with a new section. 0.3 ft 0. 2.5 ft Determine: the magnitude of the force of the seawater on this triangular area the location of this force

Answer 1

As distances are given in figure, Surface of interest lies in vertical plane y_c = h_c = 0.9 Ft Magnitude of Force is F_R = gamma h_c A F_R = 64 times 0.9 times 0.3/2 F_12 = 2.59 lb y-coordinate of center of pressure y_R = I_xc/y_c A + y_c y_R = (0.0081/36)/6.9 times 0.9/2 + 0.9 I_xc = 0.3 times (0.3)^3/36 thus y_R = 0.00556 + 0.9 = 0.906 Ft x_R = I_xyc/y_cA + x_c So I_xyc = (0.3)(0.3)^2/72 = 0.0081/72 x_R = (0.0081/72)/(0.9)(0.09/2) = 0.00278 Ft x_c = 0 thus coordinates (x_R, y_R) = (0.00278, 0.906)