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

`F = 774146.534\,N`

Step-by-step explanation:

The pressure at the bottom of the tank is:

`P_(bottom) = (0.820)\cdot (1000\,(kg)/(m^(3)))\cdot (9.807\,(m)/(s^(2)))\cdot (30\,ft)\cdot ((0.305\,m)/(1\,ft) )`

`P_(bottom) = 73581.921\,Pa`

The force exerted on the circular bottom is:

`F=(73581.921\,Pa)\cdot ((\pi)/(4) )\cdot [(12\,ft)\cdot ((0.305\,m)/(1\,ft) )]^(2)`

`F = 774146.534\,N`