Q.
A uniform rod of length 1m and mass 4kg is supported on two knife-edges placed 10cm from each end. A 60N weight is suspended at 30cm from
one end. The reactions at the knife edges is
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System of Particles and Rotational Motion
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Solution:
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−40N=0 R1+R2=60N+40N=100N...(i)
For the rotational equilibrium, taking moments about G, we get −R1(40)+60(20)+R2(40)=0 R1−R2=401200=30N...(ii)
Adding (i) and (ii), we get 2R1=130N or R1=65N
Substituting this value in Eq. (i), we get R2=35N