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

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, \frac{lb}{ft}$

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:

$V_{oil}=\frac{16\pi h}{2}ft^2=8\pi h \,\,ft^2$

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:

$\rho=\frac{m}{V}$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0\frac{lb}{ft^3})(8\pi h\,\,ft^2)$

$m=480\pi h \frac{lb}{ft}$

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.

$F=480\pi h \,g\,\, \frac{lb}{ft}$