Length of rod AB = 1 m N1 = normal reaction of the wall N2 = normal reaction of the floor The frictional force (f) between the rod and the floor acts along AD. The weight mg of the rod acts at its midpoint (centre of mass) C so that AC = AB/2. For translational equilibrium N2 = mg and N1 = f Taking moments of forces about A, we have for rotational equilibrium N2 × 0 - mg × AD + N1 × BE = 0 ⇒ 0 - N2 × AD + f × BE = 0 (∴ mg = N2 , N1 = f) ⇒