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: $\tau_A = R_A \times d$. The weight $w$ acts downwards at the centre of mass, which is at a distance $(d-x)$ from B. This produces an anticlockwise torque: $\tau_w = w \times (d-x)$. Equating the magnitudes of the torques for equilibrium: $R_A \times d = w(d-x)$ $R_A = \frac{w(d-x)}{d}$