A uniform rod of length 1 m and mass 4 kg is supported on two knife – edges placed 10 cm from each end. A 60 N weight is suspended at 30 cm from  one end. The reactions at the knife edges is

AB is the rod. K1  and K2  are the two knife edges. Since the rod is uniform,  therefore its weight acts at its centre of gravity G.Let R1  and R2  be reactions at the knife edges. For the translational equilibrium of the rod,R1+R2−60N−40  N=0R1+R2=60N+40N=100N     ....... (i)For the rotational equilibrium, taking moments about G, we get−R1(40)+60(20)+R2(40)=0R1−R2=120040=30N       ........ (ii)Adding (i) and (ii), we get 2R1=130N  or R1=65NSubstituting this value in Eq. (i), we get R2=35N