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

    <h3> Statement </h3>

    <p>
      $4.2.12$
      Determine the force of the tension of the thread connecting two balls of volume 10 cm$^3$  each, if the upper ball floats half-immersed in water. The lower ball is three times heavier than the upper one.      
    </p>
    <center>
      <figure>
        <img src="statement.png"
          loading="lazy" alt="4.2.12" width="200" />
        <figcaption>
          For problem 4.2.12
        </figcaption>
      </figure>
    </center>

    <h3>Solution</h3>
    <p>
      Applying Newton's Second Law...
    </p>
    <center>
        <figure>
          <img src="draw.png"
            loading="lazy" alt="4.2.12" width="200" />
          <figcaption>
            Force analysis
          </figcaption>
        </figure>
      </center>
    <p>
      For upper ball
      $$m_{\uparrow}\~g+T = F_{\uparrow}$$
      $$m_{\uparrow}\~g+T = \rho_0\~g\~\frac{V}{2} \\;(1)$$
      For lower ball
      $$m_{\downarrow}\~g = T+F_{\downarrow}$$
      $$m_{\downarrow}\~g = T+\rho_0\~g\~V \\;(2)$$
      Separating the gravity's force term from $(1)$
      $$m_{\uparrow}\~g = \rho_0\~g\~\frac{V}{2}-T \\;(3)$$
      Dividing $(2)$ by $(3)$, and taking account that $m_{\downarrow}=3\~m_{\uparrow}$
      $$\frac{\rho_0\~g\~V+T}{\rho_0\~g\~\frac{V}{2}-T}=3$$
      and separating $T$
      $$T = \frac{\rho_0\~g\~V}{8}$$
      Taking $\rho_0 = 1\~\rm{g/cm^3} = 10^{-3}\~\rm{kg/cm^3}$ (water's density) and $g=9.8\~\rm{m/s^2}$ (acceleration due to gravity at Earth's surface), 
      $$\boxed{T \simeq 1.2 \cdot 10^{-2}\~\rm{N}}$$
    </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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