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

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#4.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>

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

    <h3>Solution</h3>
    <p>
      Let's consider the following figure
    </p>
    <center>
        <figure>
          <img src="draw.png"
            loading="lazy" alt="4.2.7" width="400" />
          <figcaption>
            Force analysis
          </figcaption>
        </figure>
      </center>
    <p>
      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$^3$
      $$\boxed{\rho = 1.5\~\rm{g/cm^3}}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      BSc. Luis Daniel Fernández Quintana<br>
      Physics Department (FCNE)<br>
      Universidad de Oriente, Cuba<br>
    </p>



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