STEP II 1992 question 14 solution

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STEP II 1992 question 14 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 14 solution

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$\displaystyle \text{Newton's second law on }P_2:\; m_2g-T=m_2a \\ \text{Newton's second law on }P_1:\; T-m_1g=m_1a\\ \text{Dividing, }\frac{m_2g-T}{T-m_1g} = \frac{m_2}{m_1}\\ m_1m_2g - m_1T = m_2T - m_1m_2g\\ T(m_1+m_2) = 2m_1m_2g\\ T = \frac{2m_1m_2g}{m_1+m_2}.\\ P_3 \text{ remains at rest if there is no acceleration} .\\ \Leftrightarrow m_3g = 2T.\\ m_3g = \frac{4m_1m_2g}{m_1+m_2} \text{and result follows} .$