Answer:
As shown in Fig. let \[{{R}_{A}}\] and \[{{R}_{B}}\] be the normal reactions at the edges. Since the rod is in equilibrium, the sum of moments of the forces about either knife edge must be zero. Taking moments of forces about point A, we get \[{{R}_{A}}\times 0+W\times x-{{R}_{B}}\times d=0\] \[\therefore \] \[{{R}_{B}}=\frac{x}{d}.W\] Taking moments of forces about point B, we get \[{{R}_{A}}\times d-W\times (d-x)+{{R}_{B}}\times 0=0\] \[\therefore \] \[{{R}_{A}}=\frac{d-x}{d}.W.\]
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