Решение на момент правки #4675 от , автор astrosander. Это не текущая версия.
For problem $2.8.1$
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.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.8">$\leftarrow$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, back 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 were 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>
    </p>



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