$2.2.24$ 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
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
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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