A uniform rod of length 1, m and mass 4, kg i | vedclass

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Question

$\mathrm{AB}$ 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 ce

Vedclass · Accepted Answer

$65\, N, 35\, N$

Vedclass · Answer

$\mathrm{AB}$ 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 $\mathrm{R}_{2}=35 \mathrm{N}$

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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