A metal rod of weight ' $W$ ' is supported by two parallel knife-edges $A$ and $B$. The rod is. in…

A metal rod of weight ' $W$ ' is supported by two parallel knife-edges $A$ and $B$. The rod is. in equilibrium in horizontal position. The distance between two knife-edges is ' $r$ '. The centre of mass of the rod is at a distance ' $x$ ' from A. The normal reaction on A is
  1. $\frac{\mathrm{W} \cdot \mathrm{r}}{\mathrm{x}}$
  2. $\frac{\mathrm{W} \cdot \mathrm{x}}{\mathrm{r}}$
  3. $\frac{W \cdot(r-x)}{x}$
  4. $\frac{\mathrm{W} \cdot(\mathrm{r}-\mathrm{x})}{\mathrm{r}}$

Solution


For equilibrium, $\begin{aligned} & \mathrm{N}_1 \mathrm{r}=\mathrm{W}(\mathrm{r}-\mathrm{x}) \\ \therefore \quad & \mathrm{N}_1=\frac{\mathrm{W}(\mathrm{r}-\mathrm{x})}{\mathrm{r}} \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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