SATHEE: Rotation 5 Question 20

Rotation 5 Question 20

32. 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: 32. $(\frac{d - x}{d}) W, \frac{x W}{d}$

Solution:

Net torque of all the forces about $B$ should be zero.

$\begin{aligned} ∴ & W (d - x) & = N_{A} \cdot d \\ or & N_{A} & = \frac{d - x}{d} W \end{aligned}$

For vertical equilibrium of rod

$\begin{aligned} N_{A} + N_{B} & = W \\ N_{B} & = \frac{x}{d} W = W - N_{A} \\ = W - \frac{d - x}{d} W = \frac{x}{d} W \end{aligned}$

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Rotation 5 Question 20

32. 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 …

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32. 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 …