Question

A force F is applied to a uniform, thin rod of mass 4 kg and length l = 50 cm. The rod has pure translational motion in the vertical plane along a smooth, horizontal surface as shown. If F = 60 N, the angle $\alpha$ for translation of rod in the given orientation comes to be $\alpha =ta{n}^{-1}\left(\frac{2}{x}\right)$, where x is

Answer 1

Since the rod is in translation, so angular acceleration is zero.
$\therefore$ torque about O,

$F\frac{l}{2}sin\alpha -N\frac{l}{2}cos\alpha =0\Rightarrow Fsin\alpha =Ncos\alpha \Rightarrow tan\alpha =\frac{N}{F}.......\left(1\right)\text{Also},N=mg.......\left(2\right)So,\Rightarrow tan\alpha =\frac{mg}{F}=\frac{40}{60}=\frac{2}{3}\therefore \alpha =ta{n}^{-1}\left(\frac{2}{3}\right)x=3$