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(A) Let the masses be $m_1$ and $m_2$. The pulley is accelerating upwards with $a_0 = g$. In the frame of the pulley, each block experiences a pseudo-

Vedclass · Accepted Answer

$\frac{4 m_1 m_2}{m_1+m_2} g$

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(A) Let the masses be $m_1$ and $m_2$. The pulley is accelerating upwards with $a_0 = g$. In the frame of the pulley, each block experiences a pseudo-force $m_i g$ downwards.For mass $m_1$ (assuming it moves downwards relative to the pulley with acceleration $a$):$m_1 g + m_1 g - T = m_1 a \implies 2m_1 g - T = m_1 a$ $(i)$For mass $m_2$ (moving upwards relative to the pulley with acceleration $a$):$T - m_2 g - m_2 g = m_2 a \implies T - 2m_2 g = m_2 a$ (ii)Adding $(i)$ and (ii):$2m_1 g - 2m_2 g = (m_1 + m_2) a \implies a = \frac{2g(m_1 - m_2)}{m_1 + m_2}$Substitute $a$ into (ii):$T = m_2(a + 2g) = m_2 \left( \frac{2g(m_1 - m_2)}{m_1 + m_2} + 2g \right)$$T = 2m_2 g \left( \frac{m_1 - m_2 + m_1 + m_2}{m_1 + m_2} \right) = 2m_2 g \left( \frac{2m_1}{m_1 + m_2} \right) = \frac{4 m_1 m_2 g}{m_1 + m_2}$

Two blocks of masses $m_1$ and $m_2$ are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with an acceleration $a_0 = g$, then the tension in the string will be

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