In the arrangement shown in the figure, pulleys are light, small and smooth. Mass of blocks `A,B` and `C` is `m_(1) =14 kg, m_ (2) =11kg` and `M =52kg` respectively. The block A can slide freely along a vertical rail fixed to left vertical face of block `C` Assuming all the surface to be smooth magntitude of acceleration of block A is `sqrt((10)/(A)` Calculate x
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Answer

5

Explanation

Equations are : `2T -140 = 14a_(A)rarr (1)`
`N_(A) = 14 A rarr(2) 110 -T =11 a_(B) rarr(3)`
`N_(B) =11A rarr(4) T -N_(A) -N_(B) =52 A rarr(5)`
constraint relations is :`a_(A) + a_(A) - a_(B) =0`
`rArr a_(B) =2a_(A)rarr(6)` after solving equations 1 to 6 we get resultant acceleration of `A =sqrt(1+1) = sqrt2 m//s^(2)` resultant acceleration of `B` =
`sqrt(9 +1) =sqrt10 m//s^(2)` resultant acceleration of `C=1m//sec^(2) :.` slipping between `A` & `B` both move with common acceleration `a = (20)/(5) 4m//s^(2)` .