A 10-ft trough filled with water has semicircular cross section of diameter 4 ft. How much work is done in pumping all the water over the edge of the trough? Assume that water weighs 62.5 lb/ft3.
The more work shown the better.
2006-08-26 10:51:18 · 2 answers · asked by Yogi_Bear_79 3 in Science & Mathematics ➔ Mathematics
x^2+y^2=2 is the eq. for a circle.
x=sqrt(1-y^2)
integrate from the top of the trough to the bottom
int{0-2ft}10*62.5*2x*dy
int{0-2ft}1250*sqrt(1-y^2)dy
It was a long time since I integrated, so please check the next step extra carefully.
{0-2ft}1250*(1-y^2)^(3/2) / 3y=546
I'm not used to other than SI-units, so the unit for energy is ?
edit. I forgot to multiply by g=9.81 inside the integral, so I guess my answer is 5356
2006-08-26 11:35:21 · answer #1 · answered by Anonymous · 1⤊ 0⤋
I don't know what you mean by work done, but I guess that depends on the volume. So you know that the formula for the volume of half a cylinder is:
V = pi * r ^ 2 * h / 2
Just plug that in to the numbers you already have:
V = pi * 2 ^ 2 * 10 / 2
V = 20 * pi
That's the total volume in feet, and the weight of water is 62.5 lb/cubic foot. That means that the total weight of the water in the trough is:
62.5 * 20 * pi =
3960 pounds.
2006-08-26 18:14:35 · answer #2 · answered by Anonymous · 0⤊ 0⤋