4.2.6. A piece of iron weighs $9.8 \mathrm{~N}$ in water. Determine its volume. The density of iron is $7.8 \cdot 10^3 \mathrm{m^3}$.
Solution
Let's consider the following figure
Force analysis
The normal force $N$ is what the equipment measures as apparent weight. From Newton Second Law $$N = mg-F_A$$$$N = \rho~g~V-\rho_w~g~V$$$$N = (\rho -\rho_w)~g~V$$ and taking $\rho_w = 1000$ kg/m$^3$$$\boxed{V = \frac{N}{(\rho -\rho_w)~g}=147 \rm{cm}^3}$$