Added Luis's English solutions of 4.2.6, 4.2.7 and 4.2.14

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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../">$\leftarrow$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>
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