Asked by Armando
An oil tank in the shape of a right circular cylinder, with height 30 meters and a radius of 5 meters is two-thirds full of oil. How much work is required to pump all of the oil over the top of the tank? (The density of oil is 820 kg/m^3 ).
Answers
Answered by
Steve
The weight of oil in the tank is
w = (20*25π m^3) * (820 kg/m^3) * 9.8 = 4018000πN
The center of gravity of the oil is 20m from the top of the tank, so the work required to empty the oil is
4018000π * 20 = 80360000π J
Or, using calculus, the weight of oil in a thin layer of thickness dy is
25π dx * 820 * 9.8 = 200900π N
To lift all the oil, then the work is
∫[10,30] 200900π y dy = 80360000π J
w = (20*25π m^3) * (820 kg/m^3) * 9.8 = 4018000πN
The center of gravity of the oil is 20m from the top of the tank, so the work required to empty the oil is
4018000π * 20 = 80360000π J
Or, using calculus, the weight of oil in a thin layer of thickness dy is
25π dx * 820 * 9.8 = 200900π N
To lift all the oil, then the work is
∫[10,30] 200900π y dy = 80360000π J
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