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

    <h3> Statement </h3>

    <p>
      $4.2.14$
      Determine the pressure force of the logs of mass $m$ on the walls of the channel.  The upper log is half submerged in water, and the lower log touches the upper part of the water surface. The logs are equal.
    </p>
    <center>
        <figure>
          <img src="statement.png"
            loading="lazy" alt="4.2.14" width="300" />
          <figcaption>
            For problem 4.2.14 
          </figcaption>
        </figure>
      </center>

    <h3>Solution</h3>
    <p>
      Let's consider the following figure
    </p>
    <center>
        <figure>
          <img src="draw.png"
            loading="lazy" alt="4.2.14" width="550" />
          <figcaption>
            Force analysis
          </figcaption>
        </figure>
      </center>
    <p>
      From right figure
      $$\cos{\theta} = \frac{R}{2R} = 0.5$$
      so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis,
      $$N = N_t\~\sin{\theta} \\;(1)$$
      on $y$-axis
      $$F_{A1} = mg + N_t\~\cos{\theta}$$
      $$\rho_w\~g\~V = mg + N_t\~\cos{\theta} \\;(2)$$
      For half-submerged trunk, on $x$-axis,
      $$N = N_t\~\sin{\theta}$$
      on $y$-axis
      $$F_{A2} + N_t\~\cos{\theta} = mg$$
      $$\rho_w\~g\~\frac{V}{2} + N_t\~\cos{\theta} = mg \\;(3)$$
      Substituting $(2)$ into $(3)$, it is obtained
      $$N_t = \frac{mg}{3\~\cos{\theta}} \\;(4)$$
      Putting $(4)$ into $(1)$
      $$N = \frac{mg}{3}\~\tan{\theta}$$
      $$\boxed{N = \frac{mg}{\sqrt{3}}}$$
      Note: You can prove that both pressure forces generated by trunks are equals.
    </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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