Question

A remote ranch has a cylindrical water storage tank. It has a vertical central axis, a diameter of 24 ft, the sides are 5 ft high. The depth of this water is 4 ft. How much work (in ft-lb) would be required to pump all of the water over the upper rim?

Answer 1

Answer:

Explanation:

From the given information:

The diameter of the pool = 24 ft

The radius will be = 24 ft/2 = 12 ft

The volume of water V = πr²(Δx)

V = π× 12²×(Δx)

V = 144π(Δx)

Le's assume water weighs 62.5 lb/ft³;

Then:

the Force (F) will be:

= 144π(Δx) * 62.5

= 9000πΔx  lb

Also, the side of cylindrical water = 5 ft while its depth = 4ft

As such, each slide of water d = 5 - x, and the region is between 0 and 4.

∴

The required work is:

$W = \int^4_0 (5-x) 9000 \ \pi dx \\ \\ W = 9000 \int^4_0 (5-x) \ dx \\ \\ W = 9000 \pi \Big [5x - \dfrac{x^2}{2} \Big]^4_0 \\ \\ W = 9000 \pi \Big [5*4- \dfrac{4^2}{2} \Big] \\ \\ W = 9000 \pi \Big [20-8 \Big] \\ \\ W = 9000 \pi \Big [12 \Big] \\ \\ W = 9000 \pi (12) \\ \\ \mathbf{W = 108000 \pi \ ft.lb}$