Question

Calculate the buoyant force on 10,000 metric tons of solid steel completely submerged in water and compare this with the steel's weight. State if the steel will remain submerged in water. Take density of steel as 7.8 x 10 kg/m

Answer 1

Answer:

$F_{buoyant}= 1,26*10^7Newtons$

The steel density $7.8 x 10^3 kg/m^3$ is bigger that the water density  $1x 10^3 kg/m^3$, then the steel remain submerged in water (does not float)

Explanation:

First of all, density of steel is: $7.8 x 10^3 kg/m^3$

Mass, m, of the solid steel, with density, rho, and volume, Vol:

$m=\rho_{steel}*Vol$

$Vol=m/\rho_{steel}$

$m=10000ton=10^7Kg$

Archimedes principle, buoyant force

$F_{buoyant}=Vol*\pho_{water}*g=m*\pho_{water}*g/\rho_{steel}=10^{7}*1000*9.81/(7.8*10^3)= 1,26*10^7Newtons$

The steel density $7.8 x 10^3 kg/m^3$ is bigger that the water density  $1*10^3 kg/m^3$, then the steel remain submerged in water (does not float)