Let $m_{1} = 1 k g$ , $m_{2} = 2 k g$ and $m_{3} = 3 k g$ in figure. Find the accelerations of $m_{1}, m_{2}$ and $m_{3}$ . The string from the upper pulley to $m_{1}$ is $20 c m$ when the system is released from rest. How long will it take before $m_{1}$ strikes the pulley ?

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Solution

$a_{p} = \frac{a_{2} + a_{3}}{2} . . . . . . . . . . . . . . . . . . . . . (1)$
so,
$a_{1} = \frac{a_{3} - a_{2}}{2} . . . . . . . . . . . . . . . . . . . . . . . . . (2)$
$m_{1} g - 2 T = - m_{1} a_{1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$
$m_{2} g - T = - m_{2} a_{2} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)$
$m_{3} g - T = - m_{3} a_{3} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)$
we have 4 equation 4 unknown solve and get
$a_{1} = \frac{19}{29} g$
$a_{2} = \frac{17}{29} g$
$a_{3} = \frac{21}{29} g$
now $u = 0, s = 20 c m, a_{1} = \frac{19}{29} g$
$s = u t + \frac{1}{2} a t^{2}$
get $t = 0.25$ sec

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