Question Number 30984 by rahul 19 last updated on 01/Mar/18
Commented by ajfour last updated on 03/Mar/18
$${Balancing}\:{torque}\:{about}\:{B}\: \\ $$$${Rl}−{mg}\left(\frac{\mathrm{2}{l}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{R}}=\frac{\mathrm{2}\boldsymbol{{mg}}}{\mathrm{3}}\:\:\:\:\:\:\Rightarrow\:\:\:\boldsymbol{{n}}=\mathrm{2}\:\:\:. \\ $$
Commented by ajfour last updated on 03/Mar/18
Commented by rahul 19 last updated on 03/Mar/18
$${thank}\:{u}\:{sir}! \\ $$