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

Given situation is shown in figure. N1= Normal reaction on AN2= Normal reaction on BW= Weight of the rod  In vertical equilibrium,  N1+N2=W                       . . . .(i)Torque balance about centre of mass of the rod, N1x=N2(d-x)Putting value of  N2 from equation (i)N1x=W-N1(d-x)⇒N1x=Wd-Wx-N1d+N1x⇒N1d=W(d-x)∴  N1=W(d-x)d