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PROBLEM 1.19\n1. The weight of the tower itself is ignored.\n2. Local acceleration of gravity is 32.1 ft/s².\n3. ρwater = 62.4 lb/ft³\n\nThe structure must exert a minimum force equivalent to the weight of the water, which can be expressed as the mass (m) of the water times acceleration of gravity, g.\n\nF = Weight = mg\n\nThe mass of the water can be determined from its density times the volume the water occupies\n\nm = ρV = (62.4 lb/ft³)(1,000,000 gal)(0.13368 ft³)\n | 1 gal | 8,341,632 lb\n\nSubstituting for mass and acceleration of gravity and applying the appropriate conversion factor yield\n\nF = mg = (8,341,632 lb)(32.1 ft/s²) = 8,322,446 lbf\n\nPROBLEM 1.20\n(a) Fgrav = mg\n\n u using Eq. 1.8, m = nM = 0.5 kmol (17.02 kg/kmol) = 8.52 kg\n\n Fgrav = (8.52 kg)(9.81 m/s²)(1 N/0.10197 kg·m/s²)\n = 83.58N\n\n(b) V = Vr/n = 6.60 m³/0.5 kmol = 12 m³ (mole)\n V = Vr = (m/ρ)(6.52 kg/m³) = 0.704 m³\n\nPROBLEM 1.21\n(a)\n\n g = 32.1 ft/s²\n ρ = 356 lb/ft³\n \n V = Vr = π(10²)(12)3/G = 522.4 ft³\n\n Using Eq. 1.9, Vr = MV = (2.8 lb)(1 cu ft)/10 lb/ft³ = 433.39 ft³\n\n(b) Fgrav = mg = (35 lb)(31.0 ft/s²) = 33.7 lb f