Question

A uniform rod of length `1 m` mass `4 kg` is supports on tow 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.
A. 60 N, 40 N
B. 75 N, 25 N
C. 65 N, 35 N
D. 55 N, 45 N

Answer 1

Correct Answer - c

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`