quot;A tank has the shape of an inverted circular cone (point at the bottom) with height 10 feet and radius 4 feet. The tank is full of water. We pump out water (to a pipe at the top of the tank) until the water level is 5 feet from the bottom. The work W required to do this is given byquot;I know how to find it if it's a right circular cylinder, but how do I adjust for the changing area? Also would a and b = 0 and 5? Thanks ahead of time!
calculate the amount of water U start with by computing volume of cone: V = 1/3πr?h {r = 4', h = 10'} V = 0.333π(4)?(10) = 167.55 ft? calculate the amount of water remaining in cone after pumping is done: V = 0.333π(2)?(5) = 20.944 ft? calculate the volume of water pumped: 167.55 - 20.944 = 146.6 ft? 1 ft? of water weighs 62.4 lb, calculate weight of water removed: 146.6(62.4) = 9148 lbs The Center of Mass of the water in the cone (when h = 10') is at 1/4 of 10 ft = 2.5 ft high The Center of Mass of the water in the cone (when h = 5') is at 1/4 of 5 ft = 1.25 ft height The Center of C.of M. change = 1.25 + (2.5 - 1.25)/2 = 1.25+0.625 = 1.875 ft If we consider that the Center of Mass of the water removed was at 1.875 ft, All the water removed was then lifted an average of 10 - 1.875 = 8.125 ft Work done = W = (9148)(8.125) = 74,300 ft-lbs ANS
