A thin uniform rod AB of mass `m=1.0kg` moves translationally with acceleration `w=2.0m//s^2` due to two antiparallel forces `F_1` and `F_2` (figure). The distance between the points at which these forces are applied is equal to `a=20cm`. Besides, it is known that `F_2=5.0N`. Find the length of the rod.

Explanation

Since, motion of the rod is purely translational, net torque about the C.M. of the rod should be equal to zero.
Thus `F_1l/2=F_2(l/1-a)` or, `F_1/F_2=1-(a)/(l//2)` (1)
For the translational motion of rod.
`F_2-F_1=mv_e` or `1-F_1/F_2=(mw_e)/(F_2)`(2)
From (1) and (2)
`(a)/(l//2)=(mw_c)/(F_2)` or, `l=(2aF_2)/(mw_c)=1m`