A Thin uniform rod `AB` of mass `m=1.0kg` moves translationally with acceleration `a=2.0 m//s^(2)` due to two antiparallel forces `F_(1)` and `F_(2)`. The distance between the points at which these forces are applied is equal to `d=20 cm`. Besides, it is known that `F_(2)=5.0 N`. Find the length of the rod.

Explanation

Let the length of the rod be `L` and `C` is the center of rod.

The rod is moving with acceleration `a=2 m//s^(2)`
`5-F_(1)=ma=1xx2implies F_(1) = 3N`
The rod is in rotational equilibrium, taking torque about the center of rod
`Sigma tau_(C)=0impliesF_(1)(L)/(2)=5((L)/(2)-d)`
`3(L)/(2)=5((L)/(2)-0.2)`
`L=1 m`