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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
$60\, N, 40\, N$
$75\, N, 25\, N$
$65\, N, 35\, N$
$55\, N, 45\, N$
Solution
$\mathrm{AB}$ is the rod. $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ are the two knife edges. since the rod is uniform, therefore its weight acts at its centre of gravity $G.$
Let $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ be reactions at the knife edges.
For the translational equilibrium of the rod.
$\mathrm{R}_{1}+\mathrm{R}_{2}-60 \mathrm{N}-40 \mathrm{N}=0$
$\mathrm{R}_{1}+\mathrm{R}_{2}=60 \mathrm{N}+40 \mathrm{N}=100 \mathrm{N}$
For the rotational equilibrium, taking moments about $G.$we get
$-\mathrm{R}_{1}(40)+60(20)+\mathrm{R}_{2}(40)=0$
$\mathrm{R}_{1}-\mathrm{R}_{2}=\frac{1200}{40}=30 \mathrm{N}$
Adding $(i)$ and $(ii),$ we get
$2 \mathrm{R}_{1}=130 \mathrm{N}$ or $\mathrm{R}_{1}=65 \mathrm{N}$
Substituting this value in $\mathrm{Eq}$. $(i).$
we get $\mathrm{R}_{2}=35 \mathrm{N}$
