Three masses $m_{1} = 4 ext{ }kg,m_{2} = 4 ext{ }kg$ and $m_{3} = 6 ext{ }kg$ are suspended from

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Three masses $m_{1} = 4	ext{ }kg,m_{2} = 4	ext{ }kg$ and $m_{3} = 6	ext{ }kg$ are suspended from a fixed smooth frictionless pully as shown in the figure below. The value of $T_{1}/T_{2}$ is $\_\_\_\_$ . (take $g = 10	ext{ }m/s^{2}$ )

Accepted Answer

Sol. [Key Concept: Write Newton's second law for each mass over the pulley, solve for the acceleration, then get $T_{1}$ and $T_{2}$.]$${60 - T_{2} = 6a }{T_{2} + 40 - T_{1} = 4a }{60 = 14a }{a = \frac{60}{14} = \frac{30}{7} }{T_{1} = 40 + 4 	imes \frac{30}{7} = \frac{280 + 120}{7} }{T_{1} = \frac{400}{7} }{60 - T_{2} = 66 	imes \frac{30}{7} }{60 - \frac{180}{7} = T_{2} }{\frac{420 - 180}{7} = T_{2} = \frac{240}{7} }{\frac{T_{1}}{{	ext{ }T}_{2}} = \frac{5}{3} }$$

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