Physics - Rotation Question with Solution | TestHub

Horizontal forces: $N_1={\mu }_2{N}_2$

Vertical forces: $N_2+{\mu }_1{N}_1=mg$

$N_2+{\mu }_1{\mu }_2{N}_2=mg$

$N_2=\frac{mg}{1+{\mu }_1{\mu }_2}$

Torque about the foot of the rod:

$N_1\ell \sin \theta +{\mu }_1{N}_1\ell \cos \theta =mg\frac{\ell }{2}\cos \theta$

$N_1\tan \theta +{\mu }_1{N}_1=\frac{mg}{2}$

Put $N_1={\mu }_2{N}_2=\frac{{\mu }_{2}mg}{1+{\mu }_1{\mu }_2}$

$\tan \theta =\frac{1+{\mu }_1{\mu }_2}{2{\mu }_2}-{\mu }_1$

$\tan \theta =\frac{1-{\mu }_1{\mu }_2}{2{\mu }_2}$

Answer: B