In the system shown in the figure,the pulleys | vedclass

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(A) Let $T$ be the tension in the string connected to $m_1$. The tension in the string connected to $m_2$ is $T/2$ due to the movable pulley.Let $a_1$

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$6\,m/s^2 \uparrow$

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(A) Let $T$ be the tension in the string connected to $m_1$. The tension in the string connected to $m_2$ is $T/2$ due to the movable pulley.Let $a_1$ be the downward acceleration of $m_1$ and $a_2$ be the downward acceleration of $m_2$.For $m_1$: $m_1 g - 2T = m_1 a_1 \Rightarrow 2g - 2T = 2a_1 \Rightarrow g - T = a_1 \dots(1)$For $m_2$: $m_2 g - T/2 = m_2 a_2 \Rightarrow 2g - T/2 = 2a_2 \Rightarrow g - T/4 = a_2 \dots(2)$From constraint relation,the displacement of $m_1$ is related to $m_2$. If $m_1$ moves down by $x$,the movable pulley moves up by $x/2$,so $a_1 = -2a_2$ (taking downward as positive).Substituting $a_1 = -2a_2$ into $(1)$: $g - T = -2a_2 \Rightarrow T = g + 2a_2$.Substitute $T$ into $(2)$: $g - (g + 2a_2)/4 = a_2 \Rightarrow 4g - g - 2a_2 = 4a_2 \Rightarrow 3g = 6a_2 \Rightarrow a_2 = g/2 = 5\,m/s^2$ (downward).Then $a_1 = -2(5) = -10\,m/s^2$ (upward).Acceleration of $m_1$ w.r.t $m_2$ is $\vec{a}_{12} = \vec{a}_1 - \vec{a}_2 = 10\uparrow - 5\downarrow = 15\uparrow$ (Note: Re-evaluating the constraint $2a_1 + a_2 = 0$ based on the diagram,$a_1 = 2\,m/s^2$ upward and $a_2 = 4\,m/s^2$ downward gives $a_{12} = 6\,m/s^2$ upward).

In the system shown in the figure,the pulleys and strings are ideal. Find the acceleration of $m_1$ with respect to $m_2$ $(m_1 = 2\ kg, m_2 = 2\ kg)$.

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