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  <h4>Statement</h4>
  <p>
    $4.2.13$
    Two equal trunks are placed as the figure shows. The lower trunk is linked to the vertical wall by wires that form an angle of 45$^\circ$ with it. The upper trunk is half submerged in water. Find the density of the trunks.   
  </p>
  <center>
    <figure>
      <img src="statement.png"
        loading="lazy" alt="4.2.13" width="300" />
      <figcaption>
        For problem 4.2.13
      </figcaption>
    </figure>
  </center>
  <h4>Solution</h4>
  <p class="TxtSolutions">
    Let's analyze the following force diagram
  </p>
  <center>
    <figure>
      <img src="diagram.png"
        loading="lazy" alt="4.2.13" width="300" />
      <figcaption>
        Force diagram for each trunk
      </figcaption>
    </figure>
  </center>
  <p class ="TxtSolutions">
    Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction,
  </p>
  <center>
    $-N'\cos{45^\circ}+N = 0 \\;(1)$
  </center>
  and on $y$-axis
  <center>
    $mg = \rho_0 g V_s + N'\sin{45^\circ} = \rho_0 g V_{s1} + N'\frac{\sqrt{2}}{2}$
  </center>
  <p calss="TxtSolutions">
    as trunk 1 is half submerged, $V_{s1} = \frac{\pi R^2 L}{2}$ and its mass is $m = \rho V$, so separating $\rho$
  </p>
  <center>
    $\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \\;(2)$
  </center>
  <p class="TxtSolutions">
    Applying Newton Second Law for cylinder 2, on $x$-axis
  </p>
  <center>
    $N'\cos{45^\circ}-T\cos{45^\circ}=0$
  </center>
  <center>
    $N' = T \\;(3)$
  </center>
  <p calss="TxtSolutions">
    and for $y$-axis
  </p>
  <center>
    $\rho_0 g V_{s2} = T\sin{45^\circ}+N'\sin{45^\circ}+mg$
  </center>
  <p class="TxtSolutions">
    in this case $V_{s2} = \pi R^2 L$ and considering that $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$,
  </p>
  <center>
    $\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \\;(4)$
  </center>
  <p class="TxtSolutions">
    Putting $(3)$ int $(4)$
  </p>
  <center>
    $(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \\;(5)$
  </center>
  <p class="TxtSolutions">
    Substituting $(5)$ into $(2)$ and developing algebraically
  </p>
  <center>
    $\rho = \frac{2}{3}\rho_0$
  </center>
  <p class="TxtSolutions">
    as the water's density is $\rho_0=1$ g/cm$^3$,
  </p>
  <center>
    $\boxed{\rho=\frac{2}{3}\~\rm{g/cm^3}}$
  </center>
    <p class="TxtSolutions" 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>
</article>