A 10ft trough filled with water has a semicircular cross section of diameter 4ft.?

Question

How much work is done in pumping all the water over the edge of the trough? Assume the water weighs 62.5lb/ft^3

Answer 1

The area, A, of the cross section is 1/2(pi x r^2) where r = 2; pi = 3.14159. Hence A = 1/2(4 * 3.14159) = 6.28318 ft^2 . Since the length of the trough is 10 feet, the volume of water will be 62.8318 ft^3. The total weight of the water will be 62.5 x 62.8318 lb or 3926.9875 lb
Work is given by W = f * d where f is a force and d is a distance over which the force acts. In this case f is due to the weight of water, 3926.9875 lb. This force will act over the distance from the top of the trough to the bottom in order to extract all the water. This distance is 2 feet. Therefore the work required will be 7853.975 ft lbs.

Answer 2

V = (1/2)(1/4)πd^2L
F= (62.5)(1/8)π(4^2)(10)
W = Fs
W = 2(62.5)(1/8)π(4^2)(10)
W = (62.5)π(4)(10) ft.lb.
W = 7,854 ft.lb.