# Work problem

• Jul 7th 2012, 06:51 PM
failblogjunkie677
Work problem
You have a oil tank in the shape of a cone tip side down the diameter is 10m and the height is 6m the oil density is 900kg/m^3 find the work to pump all of the oil out the top. (the tank is full, 9.8m/s^2 as the acceration due to gravity F=ma & W=Fd)
• Jul 8th 2012, 05:06 AM
HappyJoe
Re: Work problem
Usually work is defined in terms of moving only _a point mass_ through some distance with some constantly applied force.

So in this case, I suggest you find the center of mass of your oil tank, and find the distance from this center of mass to the top of the tank, using this as your "d".
• Jul 8th 2012, 08:12 AM
skeeter
Re: Work problem
Quote:

Originally Posted by failblogjunkie677
You have a oil tank in the shape of a cone tip side down the diameter is 10m and the height is 6m the oil density is 900kg/m^3 find the work to pump all of the oil out the top. (the tank is full, 9.8m/s^2 as the acceration due to gravity F=ma & W=Fd)

work = $\displaystyle \int WALT$

where ...

W = weight density in Newtons per cubic meter
A = area of a representative horizontal cross-sectional slice in terms of y
L = vertical lift distance of a representative horizontal cross-section slice in terms of y
T = horizontal cross-section slice thickness in terms of y

let the bottom of the conical tank (the apex) be at the origin. one side of the cone is defined by the line $\displaystyle y = \frac{6x}{5}$

$\displaystyle W = 900 \cdot 9.8$

$\displaystyle A = \pi \left(\frac{5y}{6}\right)^2$

$\displaystyle L = 6 - y$

$\displaystyle T = dy$

work = $\displaystyle \int_0^6 8820 \cdot \pi \left(\frac{5y}{6}\right)^2 \cdot (6-y) \, dy$