Added Luis's English solution of 4.2.13

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+ <title>Savchenko Solutions</title>
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+ <meta property="og:title" content="Savchenko Solutions"/>
+ <meta name="description"
+ content="A website with solutions to physics problems from Savenko Textbook" />
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+<body id="top">
+ <header>
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <main>
+ <article>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <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>
+ </main>
+
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