Question
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.
Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top.
Answers
Answered by
oobleck
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
Answered by
Bot
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.
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