Added Luis's English solution of 2.8.1

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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>
+ $2.8.1$
+ The figure shows structures that hold a load weighing 10 kg. The cables are represented by thin lines, the rod is represented by a double line. Determine the tension force of the cables for case a and the force acting on the rod from the side of the cable thrown over it for case b.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="2.8.1" width="400" />
+ <figcaption>
+ For problem 2.8.1
+ </figcaption>
+ </figure>
+ </center>
+ <h3>Solution</h3>
+ <p>
+ Let's consider the following figure
+ </p>
+ <center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" alt="2.8.1" width="500" />
+ <figcaption>
+ Force analysis
+ </figcaption>
+ </figure>
+ </center>
+ <p>
+ $a$. Applying Newton Second Law for $x$-direction
+ $$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$
+ $$T_1=T_2 \;(1)$$
+ for $y$-direction
+ $$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$
+ Let be $T_1=T_2=T$, so, bak to $(2)$
+ $$2T~\sin{30^\circ}=mg$$
+ $$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$
+ $b$. According figure,
+ $$F^2 = T^2+T^2 = 2T^2$$
+ We consider the same tension $T$ because we suppose the threads are unextensible.
+ $$F = \sqrt{2}~T \;(3)$$
+ Applying Newton Second Law
+ $$T = mg \;(4)$$
+ Putting $(4)$ into $(3)$
+ $$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$
+ Note: Calculations are made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41.
+ </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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