How much torque must the pin exert to keep the rod in the figure from rotating?
Calculate this torque about an axis that passes through the point where the pin enters the rod and is perpendicular to the plane of the figure.
L: 80 cm or .8m
M(of the rod)= 2.0 kg
m(of the weight)= 500g or .5 kg
The weight of the rod with length L acts as the center of the rod.
L/2 from the pin
The torque about the pin is τ = (mg)L +(Mg)L/2
=(0.500kg x 9.8m/s2)0.8m+(2.0kg x 9.8m/s2 x 0.8m) / 2
= 3.92 + 7.84