A rod of weight is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance from each other. The centre of mass of the rod is at distance from A. The normal reaction on A is [201

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Solution

Q.
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 [2015]

1 $\frac{W (d - x)}{x}$

2 $\frac{W (d - x)}{d}$

3 $\frac{W x}{d}$

4 $\frac{W d}{x}$

Ans.
(2)

Given situation is shown in figure.

$N_{1}$ = Normal reaction on A

$N_{2}$ = Normal reaction on B

W = Weight of the rod

In vertical equilibrium,

$N_{1} + N_{2} = W$ ...(i)

Torque balance about centre of mass of the rod,

$N_{1} x = N_{2} (d - x)$

Putting value of $N_{2}$ from equation (i)

$N_{1} x = (W - N_{1}) (d - x) \Rightarrow N_{1} x = W d - W x - N_{1} d + N_{1} x$

$\Rightarrow N_{1} d = W (d - x);$ $∴$ $N_{1} = \frac{W (d - x)}{d}$

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