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<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>
− $2.8.4$
− Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.
− </p>
− <center>
− <figure>
− <img src="statement.png"
− loading="lazy" alt="2.8.4" width="300" />
− <figcaption>
− For problem 2.8.4
− </figcaption>
− </figure>
− </center>
− <h4>Solution</h4>
− <p class="TxtSolutions">
− Tension in each thread is the same ($T=mg$). Moreover, note that the absolute value of the sum of two any tensions must be equal to the third one which is also $T$ and be directed in opposed sense for maintain the equilibrium. So, consider for example the angle $\gamma$
− </p>
− <center>
− <figure>
− <img src="draw.png"
− loading="lazy" alt="2.8.4" width="300" />
− <figcaption>
− Top view
− </figcaption>
− </figure>
− </center>
− <center>
− $T_R = T = 2T\cos{\frac{\gamma}{2}}$
− </center>
− <p class="TxtSolutions">
− Hence
− </p>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+
+ <p>
+ $2.8.4$
+ Through three holes made in a desktop pass threads united by a common knot. At the free ending of each thread is hanged equal loads. Determine the angles between threads. There is no friction.
+ </p>
<center>
− $\cos{\frac{\gamma}{2}} = 0.5$
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="1.5.3" width="200" />
+ <figcaption>
+ For problem 1.5.3
+ </figcaption>
+ </figure>
</center>
+
+ <h3>Solution</h3>
+ <p>
+ Tension in each thread is the same ($T=mg$). Moreover, note that the absolute value of the sum of two any tensions must be equal to the third one which is also $T$ and be directed in opposed sense for maintain the equilibrium. So, consider for example the angle $\gamma$
+ </p>
+
<center>
− $\gamma = \frac{2\pi}{3} = 120^\circ$
+ <figure>
+ <img src="draw.png"
+ loading="lazy" alt="2.8.4" width="300" />
+ <figcaption>
+ Top view
+ </figcaption>
+ </figure>
</center>
− <p class="TxtSolutions">
+
+ <p>
+ $$T_R = T = 2T\cos{\frac{\gamma}{2}}$$
+ Hence
+ $$\cos{\frac{\gamma}{2}} = 0.5$$
+ $$\gamma = \frac{2\pi}{3} = 120^\circ$$
But, since the above condition of the sum of two any tensions always is true, the three angles must be equal.
+ $$\boxed{\alpha=\beta=\gamma=120^\circ}$$
</p>
− <center>
− $\boxed{\alpha=\beta=\gamma=120^\circ}$
− </center>
− <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">
+
+ <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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