Solve the following :

Solution:

By Pythagoras theorem,

$(6.5)^{2}=(6)^{2}+B^{2}$

$B=2.5 \mathrm{~m}$

(a) $\tau=F \times d_{\Perp}$

$\mathrm{T}=6 \lg _{6.5}\left(\frac{6.5}{2} \sin \theta\right.$

$=60 \mathrm{~g}\left(\frac{2.5}{6.5}\right)$

$=60 \mathrm{~g}\left(\frac{6.5}{2} \times \frac{2.5}{6.5}\right)$

$\tau=740 \mathrm{~N}-\mathrm{m}$

(b) $N_{1}=60 g$

(Vertical Translatory Equation)

Rotational Equilibrium at $O$

$N_{2}(6.5 \cos \theta)=60 g\left(\frac{6.5}{2} \sin \theta\right)$

$N_{2}=30 g \tan \theta$

$N_{2}=30 g\left(\frac{2.5}{6}\right)$

$N_{2}=120 \mathrm{~N}$