Question

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

• A. 60 N, 40 N
• B. 75 N,25 N
• C. 65 N, 35 N
• D. 55 N, 45 N

Answer 1

Answer: (C)

65 N, 35 N

Answer 2

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 centre of gravity G.

Let $R_{1}$and $R_{2}$be reactions at the knife edges.

For the translational equilibrium of the rod.

$R_{1} + R_{2} - 60N - 40N = 0$

$R_{1} + R_{2} = 60N + 40N = 100N$ …..(i)

For the rotational equilibrium, takings moments about G, we get

$- R_{1}(40) + 60(20) + R_{2}(40) = 0$

$R_{1} - R_{2} = \frac{1200}{40} = 30N$ ….. (ii)

Adding (i) and (ii), we get

$2R_{1} = 130NorR_{1} = 65N$

Substituting this value in Eq. (i),

We get $R_{2} = 35N$