The work required for the given task of pumping all of the oil out over the top of the tank is 7,500 ft·lb
The known parameters;
The length of the rectangular tank, <em>l</em> = 4 feet
The width of the tank, <em>w</em> = 2 feet
The depth of the tank, <em>h</em> = 5 feet
The weight density of the oil with which the tank is filled, ρ × g = 50 lb/ft³
The unknown parameter
The work required to pump all of the oil out over the top of the tank
Method;
Calculate the force required to lift each slice (layer) of the oil to the top multiplied by the distance, <em>y</em>, the slice moves and summing the result as an integration as follows;
The volume of each slice,
= l × w × dy
The force required to move each slice,
= ρ × g × l × w × dy
The work done,
, in moving the slice a distance, <em>y</em>, is given as follows;
= ρ × g × l × w × y × dy
Therefore, the total work done, W, in pumping all the water located from y = 0, to y = 5, to the top of the tank, is given as follows;
![\mathbf{W = \int\limits^5_0 {(\rho \times g \times l \times w \times y) } \, dy}](https://tex.z-dn.net/?f=%5Cmathbf%7BW%20%3D%20%5Cint%5Climits%5E5_0%20%7B%28%5Crho%20%5Ctimes%20g%20%5Ctimes%20l%20%5Ctimes%20w%20%5Ctimes%20y%29%20%7D%20%5C%2C%20dy%7D)
Therefore;
W = (ρ × g × l × w × y²)/2
Plugging in the values, gives;
W = (50 lb/ft³ × 4 ft. × 3 ft. × (5 ft.)²)/2 = 7,500 ft·lb
The work required to pump all of the oil out over the top of the tank, W = 7,500 ft·lb.
Learn more about the use of integration to calculate the amount of work required for a given task here;
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