a cattle trough has a trapezoidal cross section with a height of 1 m and horizontal sides of width 1/2 m and 1 m assume the length of the trough is 10 m. A. how much work is required to pump out the water in the trough? B. if the length is doubled does the amount of work needed double? explain.

Answer 1

The trapezoidal cross-section for water at depth 1/2 + h/2 = (1+h)/2

So, the surface area of the water at depth h is 10*(1+h)/2 = 5(1+h)

So, the volume of the sheet of water is

5(1+h) dh

That means that the weight of the water (at 1000kg/m^3) layer is

5(1+h) dh * 1000 * 9.8

Since work = force (weight) * distance, the work required to lift the layer of water to the top of the trough is

5(1+h) dh * 1000 * 9.8 * (1-h)
= 49000 (1-h^2) dh

Now just integrate that from 0 to y where y is the depth of the water, and you have your answer.

Note that since the length of the trough is a constant, doubling the length just doubles the work.