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  4. 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

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Q. 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

System of Particles and Rotational Motion

Solution:

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$AB$ 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,
$R_1 + R_2 - 60 N - 40 N = 0$
$R_1 + R_2 = 60 N + 40 N = 100 N \quad ... (i)$
For the rotational equilibrium, taking moments about $G$, we get
$-R_1(40) + 60(20) + R_2(40) = 0$
$R_1 - R_2 = \frac{1200}{40} = 30\,N \quad ...(ii)$
Adding $(i)$ and $(ii)$, we get $2R_1 = 130\, N$ or $R_1 = 65 \,N$ Substituting this value in Eq. $(i)$, we get $R_2 = 35\, N$