A thin uniform rod AB of mass `m=1kg` moves translationally with acceleration `a=2m//s^(2)` and to two anitiparallel forces `F_(1)` and `F_(2)`. The distance between the points at which these forces are applied is equal to `l=20cm` besides it is known that `F_(2)=5N` find the length of the rod.

Explanation

Let x be the distnace of centre
point C of rod from D. then
`F_(2)-F_(1)=ma`
or `F_(1)=3N`
Further `tau_(c)=0`
`thereforeF_(2)x=F_(1)(0.2+x)`
`5x=F_(1)(0.2+x)`
`therefore5x=3(0.2+x)`
or `x=0.3m`
`therefore` length of rod `=2(x+0.2)=1.0`m