A uniform rod of length 1 m mass 4 kg is supports on tow knife-edges placed10 cm from each end. A 60 N weight is suspended at 30 cm from one end. The reactions at the knife edges is.

A uniform rod of length 1 m mass 4 kg is supports on tow knife-edges p

1.	A uniform rod of length 1 m mass 4 kg is supports on tow knife-edges placed10 cm from each end. A 60 N weight is suspended at 30 cm from one end. The reactions at the knife edges is.
Answer» a. 60 N, 40 N b. 75 N, 25 N c. 65 N, 35 N d. 55 N, 45 N Solution : AB is the rod. `K_(1)` and `K_(2)` are the two knife EDGES. Since the rod is uniform, therefore its weight acts at its center of GRAVITY G. Let `R_(1)` and `R_(2)` be reactions at the knife edges. For the translational equilibrium of the rod, `R_(1)+R_(2)-60N-40N=0` `R_(1)+R_(2)=60N+40N=100N` For the rotational equilibrium, talking moments about G, we GET `-R_(1)(40)+60(2)+R_(2)(40)=0` `R_(1)-R_(2)=1200/40=30 N`..............(ii) Adding (i) and (ii), we get `2R_(1)=130N` or `R_(1)=65N` Substituting this VALUE in Eq. (i), we get `R_(2)=35N`