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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:

1. Wd/x
2. W(d − x)/x
3. W(d − x)/d
4. Wx/d

Correct answer: W(d − x)/d

Solution

To maintain equilibrium, the sum of vertical forces and the torque about any point must be zero. Taking torque about point B, the normal reaction at A multiplied by d equals the weight W multiplied by its perpendicular distance (d − x). Solving for the normal reaction at A gives W(d − x)/d.

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