Question

"an open tank has the shape of a right circular cone. the tank is 6 feet across the top and 5 feet high. how much work is done in emptying the tank by pumping the water over the top edge? note: the density of water is 62.4 lbs per cubic foot."

Answer 1

The amount of workdone by in emptying the tank by pumping the water over the top edge is
`\mathbf{ 1169.99 \pi \ ft-lbs}`

Step-by-step explanation:

Given that the tank is 6 feet across the top and 5 feet high.

Using the similar triangles.

`(3)/(5) = (r)/(y)`

5r = 3y

`r = (3)/(5) y`

Thus; each disc is a circle with area

A =
`\pi ( (3)/(5) y)^2`

The weight of each disc is ;

`m = \rho_(water)A`

= 62.4 ×
`\pi ( (3)/(5) y)^2`

=
`(561.6)/(25) \pi y^2`

The distance pumped is 5-y

Thus; the workdone in pumping the tank by pumping the water over the top edge is :

`W = (561.6 \pi)/(25) \int\limits^5_0 {(5-y)} \, y^2dy`

`W = (561.6 \pi)/(25) \int\limits^5_0 {(5y^2-y^3)} dy`

`W = (561.6 \pi)/(25)[52.083]`

`\mathbf{W = 1169.99 \pi \ ft-lbs}`

The amount of workdone by in emptying the tank by pumping the water over the top edge is
`\mathbf{ 1169.99 \pi \ ft-lbs}`