A rod of weight $w$ is supported by two parallel knife edges A and B and is in equilibrium in a horizontal pos

Question

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:

Accepted Answer

Let the normal reactions at knife edges A and B be $R_A$ and $R_B$ respectively.
For rotational equilibrium, the net torque about any point must be zero. Let's take the torque about point B.
The force $R_A$ acts at a distance $d$ from B, producing a clockwise torque: $	au_A = R_A 	imes d$.
The weight $w$ acts downwards at the centre of mass, which is at a distance $(d-x)$ from B. This produces an anticlockwise torque: $	au_w = w 	imes (d-x)$.
Equating the magnitudes of the torques for equilibrium:
$R_A 	imes d = w(d-x)$
$R_A = \frac{w(d-x)}{d}$

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