Question

what volume is displaced by a piece of steel weighing 200 kg floating in a lake? ( The density of steel is 7860kgm^-3 )

Answer 1

$$F _ {B} = m g - \rho_ {\text {water}} \cdot V _ {\text {object}} \cdot g$$ $$F _ {B} = \rho_ {\text {object}} \cdot V _ {\text {object}} \cdot g - \rho_ {\text {water}} \cdot V _ {\text {object}} \cdot g$$ $$F _ {B} = \left(\rho_ {\text {object}} - \rho_ {\text {water}}\right) \cdot V _ {\text {object}} \cdot g$$

As $\rho_{\text{object}} > \rho_{\text{water}}$, the piece of steel will not float (its volume will be under the water surface). So:

$$m = \rho_ {\text {object}} \cdot V _ {\text {object}} \Leftrightarrow V _ {\text {object}} = \frac {m}{\rho_ {\text {object}}} = \frac {200}{7860} = 0.02544529 \quad 3 \mathrm {m} ^ {3}$$