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Statement

$4.2.7$ A body in water weighs three times less than in air. What is the density of the body?

Solution

Let's consider the following figure

These normal forces $N_i$ is what the equipment measures as weight in each case. From Newton Second Law for air $$N_a = mg$$ Since $m=\rho~V$, $$N_a = \rho~V~g \;(1)$$ Applying Newton Second Law for water $$mg = N_w + F_A$$ $$\rho~V~g = N_w + \rho_w~V~g$$ $$N_w = (\rho-\rho_w)~V~g \;(2)$$ Dividing $(1)$ by $(2)$, taking account that $\frac{N_a}{N_w} = 3$ $$\rho = \frac{3}{2}\rho_w$$ as $\rho_w = 1$ g/cm$^3$ $$\boxed{\rho = 1.5~\rm{g/cm^3}}$$

BSc. Luis Daniel Fernández Quintana
Physics Department (FCNE)
Universidad de Oriente, Cuba