
Let T be the tension in cable.
Since the system is at equilibrium, the net torque, about an axis perpendicular to the plane and passing through the point of contact of the rod and the wall, is zero. Therefore, torque about point C
Tsin30∘×60=mg×50+8g×100
2T×60=20×50+80×100
⇒3T=100+800
⇒300N.
