question_answer

A rod of weight W is supported by two parallel edges A and B and is in equilibrium in 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. Find the normal reactions at the knife edges A and B.

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 \[{{R}_{A}}\times d-W\times (d-x)+{{R}_{B}}\times 0=0\] \[\therefore \]  \[{{R}_{A}}=\frac{d-x}{d}.W.\]