The volume of the conical trough is (1/3)πr2h = (1/3)π(4 ft)2(6 ft) = 67.12 ft3.
The weight of the liquid is 29 lb/ft3 × 67.12 ft3 = 1945.68 lb.
The amount of work required to empty the trough is 1945.68 lb × 4 ft = 7778.72 foot-pounds.
A conical trough has a radius of 4 feet and a height of 6 feet. It is filled with a liquid that weighs 29 pounds per cubic foot and the depth of that liquid is 4 feet.
Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top.
2 answers
AAAaannndd the bot gets it wrong yet again!
The bot's answer is the work to lift the entire mass just 4 ft.
But the center of mass is not 4 ft from the top.
when the water has depth h, the radius of the surface is 2/3 h
so that means the volume of water is
π/3 (2/3 h)^2 h = 4/27 π h^3
so the work needed to lift a slice of water of thickness dh, h feet from the bottom is
(4/27 πh^3 * 29)(6-h) dh
∫[0,4] (4/27 πh^3 * 29)(6-h) dh = 2418.7
The bot's answer is the work to lift the entire mass just 4 ft.
But the center of mass is not 4 ft from the top.
when the water has depth h, the radius of the surface is 2/3 h
so that means the volume of water is
π/3 (2/3 h)^2 h = 4/27 π h^3
so the work needed to lift a slice of water of thickness dh, h feet from the bottom is
(4/27 πh^3 * 29)(6-h) dh
∫[0,4] (4/27 πh^3 * 29)(6-h) dh = 2418.7
2 answers