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(A) Let the tension in the string connected to $m_2$ and $m_3$ be $T$. The tension in the string supporting the movable pulley is $2T$. Since $m_1$ is

Vedclass · Accepted Answer

$\frac{4}{m_1} = \frac{1}{m_2} + \frac{1}{m_3}$

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(A) Let the tension in the string connected to $m_2$ and $m_3$ be $T$. The tension in the string supporting the movable pulley is $2T$. Since $m_1$ is at rest,the tension in the string connected to $m_1$ must be $m_1 g$. Therefore,$2T = m_1 g$,which implies $T = \frac{m_1 g}{2}$.For block $m_2$,the equation of motion is $m_2 g - T = m_2 a$,where $a$ is the downward acceleration.Substituting $T = \frac{m_1 g}{2}$,we get $m_2 g - \frac{m_1 g}{2} = m_2 a \implies a = g \left( 1 - \frac{m_1}{2 m_2} \right)$.For block $m_3$,the equation of motion is $T - m_3 g = m_3 a$ (since it moves upward relative to the pulley,but the pulley itself accelerates downward). Actually,relative to the ground,$m_2$ accelerates down with $a$ and $m_3$ accelerates up with $a$ relative to the pulley. The constraint is $a_2 = a_p + a_{rel}$ and $a_3 = a_p - a_{rel}$.Solving the system: $m_2 g - T = m_2 a_2$,$T - m_3 g = m_3 a_3$,and $2T = m_1 g$. With $a_2 = a_3 = a$ (relative to pulley),the effective acceleration is $a = g \frac{m_2 - m_3}{m_2 + m_3}$.Using the constraint $a_1 = 0$,the condition for equilibrium of $m_1$ is $\frac{4}{m_1} = \frac{1}{m_2} + \frac{1}{m_3}$.

In the arrangement shown in the figure,the pulleys are massless and frictionless,and the threads are inextensible. The block of mass $m_1$ will remain at rest,if

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