Question

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answer 1

Assuming h as the height of the cylindrical tank

`F=480\pi h \,g\,\, (lb)/(ft)`

Step-by-step explanation:

Assuming that the height is
`h` we can find the volume of the cylindrical tank, then:

`V=\pi*r^2*h`

The diameter is 8.00 ft then
`r=4.00 ft` the total volume of the tank is:

`V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2`

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

`V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2`

We know the density of the oil
`\rho=60.0\,lb/ft^3`, with this we can fing the mass of oil that we have because:

`\rho=(m)/(V)` then
`m=\rho V`

Then the mass of oil that we have is:

`m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)`

`m=480\pi h (lb)/(ft)`

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that
`F=mg` then we just need to multiply the mass by the gravity.

`F=480\pi h \,g\,\, (lb)/(ft)`