A rod of weight W rests horizontally on knife-edge supports A and B separated by d. Its centre of mass is distance x from A. The normal reaction at A is:
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A rod of weight W rests horizontally on knife-edge supports A and B separated by d. Its centre of mass is distance x from A. The normal reaction at A is:

G1 Physics · Accepted Answer

Step 1: For equilibrium, Rₐ+Rᵦ=W. Step 2: Taking torque about A: Rᵦd=Wx, giving Rᵦ=Wx/d. Step 3: Therefore Rₐ=W−Wx/d=W(d−x)/d. Final answer: D — W(d−x)/d.