Two men of unequal masses hold on to the two | vedclass

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(D) Let the masses of the two men be $m_1$ and $m_2$ where $m_1 > m_2$. Let $T$ be the tension in the rope.$1$. If the lighter man $(m_2)$ is stationa

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All of the above.

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(D) Let the masses of the two men be $m_1$ and $m_2$ where $m_1 > m_2$. Let $T$ be the tension in the rope.$1$. If the lighter man $(m_2)$ is stationary,the tension $T = m_2 g$. Since $m_1 > m_2$,the net force on the heavier man is $m_1 g - T = m_1 g - m_2 g = (m_1 - m_2)g$. Thus,the heavier man accelerates downwards with $a = (m_1 - m_2)g / m_1$. This is possible.$2$. If the heavier man $(m_1)$ is stationary,the tension $T = m_1 g$. The net force on the lighter man is $T - m_2 g = m_1 g - m_2 g = (m_1 - m_2)g$. Thus,the lighter man accelerates upwards with $a = (m_1 - m_2)g / m_2$. This is possible.$3$. If both men climb or slide such that they exert forces to maintain equal and opposite accelerations,the system dynamics allow for both to move with the same magnitude of acceleration in opposite directions. This is also possible.Since all scenarios are physically possible,the correct option is $D$.

Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following are possible?

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