Question

A tank containing oil is in the shape of a downward-pointing cone with its vertical axis perpendicular to ground level. (See a graph of the tank here.) In this exercise we will assume that the height of the tank is 10 ft , the circular top of the tank has radius 5 ft, and that the oil inside the tank weighs 56 lb per cubic ft.
How much work does it take to pump oil from the tank to an outlet that is 3 ft above the top of the tank if, prior to pumping, there is only a half-tank of oil? Note: "half-tank" means half the volume in the tank.

Answer 1

(See a graph of the tank here.) <- wheres the graph xD?

Answer 2

First let's find the volume of the the oil. The volume of a cone is\[V=\frac{ \pi \times r^{2} \times h}{3}\]and we have given that h=10 and r=5. Also, we know that it's half full, so the volume of the oil will be\[V_{o}=\frac{\pi \times r^{2} \times h}{6}=\frac{\pi \times 5^{2} \times 10}{6}=\frac{125 \times \pi}{3}\]Now we can use that and the density we were given to find the weight of the oil.\[Weight=Density \times Volume=56 \times \frac{125}{3} \times \pi=\frac{7000}{3} \times \pi\]
Now, this is where the tricky part comes in. From physics, the equation for work is\[W=F \times d\]Gravity is constantly going to be exerting a force on the oil to pull it downwards, and so in order to lift it we need to apply a slightly greater force than gravity does. Applying the same force will result in the oil not gaining any velocity at all. So the best I can do is give you the lower bound for the work.\[W > Weight \times d\]\[W > \frac{7000}{3} \times \pi \times 3\]\[W > 7000\pi\]