Three masses of 1, kg,6, kg,and 3, kg are con | vedclass

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(C) Given: $m_{1} = 1\, kg$ (hanging mass on left),$m_{2} = 6\, kg$ (mass on table),and $m_{3} = 3\, kg$ (hanging mass on right).Let $a$ be the accele

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$2$

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(C) Given: $m_{1} = 1\, kg$ (hanging mass on left),$m_{2} = 6\, kg$ (mass on table),and $m_{3} = 3\, kg$ (hanging mass on right).Let $a$ be the acceleration of the system. The equations of motion for the three masses are:For $m_{3}$ (moving downwards): $m_{3}g - T_{2} = m_{3}a$For $m_{2}$ (moving horizontally): $T_{2} - T_{1} = m_{2}a$For $m_{1}$ (moving upwards): $T_{1} - m_{1}g = m_{1}a$Adding these three equations:$(m_{3}g - T_{2}) + (T_{2} - T_{1}) + (T_{1} - m_{1}g) = (m_{1} + m_{2} + m_{3})a$$(m_{3} - m_{1})g = (m_{1} + m_{2} + m_{3})a$Therefore,the acceleration $a$ is:$a = \frac{(m_{3} - m_{1})g}{m_{1} + m_{2} + m_{3}}$$a = \frac{(3 - 1) 	imes 10}{1 + 6 + 3} = \frac{2 	imes 10}{10} = 2\, ms^{-2}$.

Three masses of $1\, kg$,$6\, kg$,and $3\, kg$ are connected to each other with threads and are placed on a table as shown in the figure. The acceleration with which the system is moving is $....... \, ms^{-2}$. (Take $g = 10\, ms^{-2}$)

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