Question

If kerosene has a specific gravity of 0.820, what force will be exerted on the circular bottom of a cylindrical kerosene tank that has a diameter of 12-ft and a height of 30 ft?

Answer 1

Answer:

$F = 774146.534\,N$

Explanation:

The pressure at the bottom of the tank is:

$P_{bottom} = (0.820)\cdot (1000\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot (30\,ft)\cdot (\frac{0.305\,m}{1\,ft} )$

$P_{bottom} = 73581.921\,Pa$

The force exerted on the circular bottom is:

$F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}$

$F = 774146.534\,N$