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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>
+ $2.2.24$
+ 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:
+ $$
+ v_{1y} = -v_Ay \quad (v_{by} = 0)$$
+ $$
+ 2v_{1y} + v_{2y} + v_{ky} = 0$$
+ $$
+ 2v_{1y} + v_{2y} + v_{3y} + u_y = 0
+ $$
+ $$
+ 4v_{1y} = -u_y$$
+ $$
+ v_{1y} = -\frac{u_y}{4}
+ $$
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