The solution before revision #1818 of , by astrosander. This is not the current version.
For problem $2.2.29$
Savchenko Solutions

Solutions of Savchenko Physics Textbook

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>

    <p>
      $2.2.24$
      A monkey of mass $m$ is balanced by a counterweight on block $A$. Block $A$ is balanced by a weight of $2m$ on block $B$. The system is stationary. How will the load move if the monkey starts to evenly select the rope at a speed $u$ relative to itself? Ignore the mass of blocks and friction
    </p>
    <center>
      <figure>
        <img src="statement.png"
          loading="lazy"  width="120" />
        <figcaption>
          For problem 2.2.24
        </figcaption>
      </figure>
    </center>

    <h3>Solution</h3>
    <center>
      <figure>
        <img src="draw.png"
          loading="lazy" width="180" />
        <figcaption>
            Forces acting on the system
        </figcaption>
      </figure>
    </center>
    <p>
      Newton's second law of motion for y-axis:
      $$
      \begin{cases}
          2T - 2mg = 2ma_{1y} \\
          T - mg = ma_{2y} \\
          T - mg = ma_{3y}
      \end{cases}
      $$
      From these three equations, it's evident that:
      $$
      a_1 = a_{2y} = a_{3y}$$
      $$\quad \int_0^{v_{1y}} dv_{1y} = \int_0^{v_{2y}} dv_{2y} = \int_0^{v_{3y}} dv_{3y}$$
      $$v_{1y} = v_{2y} = v_{3y}
      $$
      The lengths of two ropes:
      $$
      \begin{cases}
          L_1 = 2y_b - y_1 - y_a \\
          L_2 = 2y_A - y_2 - y_k
      \end{cases}
      $$
      Taking the derivative of the rope lengths with respect to time, we obtain these relationships:
      $$
      \begin{cases}
          0 = 2v_{by} - v_{1y} - v_{Ay} \\
          0 = 2v_{Ay} - v_{2y} - v_{ky}
      \end{cases}
      $$
      From velocity-addition formula:
      $$
      \vec{v} = \vec{v}^\prime + \vec{u}$$
      $
      v_{ky} = v_{3y} + u_y$ from the conditions of the problem
    </p>
    <p>
      Eventually:
      $$
      v_{1y} = -v_Ay \quad (v_{by} = 0)$$
      $$
      2v_{1y} + v_{2y} + v_{ky} = 0$$
      $$
      2v_{1y} + v_{2y} + v_{3y} + u_y = 0
      $$
      $$
      4v_{1y} = -u_y$$
      $$
      v_{1y} = -\frac{u_y}{4}
      $$
  </p>

  <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p>
    


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
    </p>
</footer>