Rotation Ques 104
- A rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A$. The normal reaction on $A$ is ……… and on $B$ is …
(1997, 2M)
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Answer:
Correct Answer: 104.$\left(\frac{d-x}{d}\right) W, \frac{x W}{d}$
Solution:
Formula:
- Net torque of all the forces about $B$ should be zero.
$$ \begin{aligned} \therefore & W(d-x) & =N _A \cdot d \\ \text { or } & N _A & =\frac{d-x}{d} W \end{aligned} $$
For vertical equilibrium of rod
$$ \begin{aligned} N _A+N _B & =W \\ N _B & =\frac{x}{d} W=W-N _A \\ & =W-\frac{d-x}{d} W=\frac{x}{d} W \end{aligned} $$
BETA
