A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to

Let the mass of the rod is  M⇒ Weight (W)=MgInitially for the equilibrium F+F=Mg⇒F=Mg/2When one man withdraws, the torque on the rod τ=Iα=Mgl2⇒ Ml23α=Mgl2  As I=Ml2/3⇒ Angular acceleration, α=32gland linear acceleration,a=l2α=3g4Now if the new normal force at A is F′ then Mg−F′=Ma⇒ F′=Mg−Ma=Mg−3Mg4=Mg4=W4