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  $M\Rightarrow \text{ Weight }\left(W\right)=Mg$

Initially for the equilibrium $F+F=Mg\Rightarrow F=Mg/2$

When one man withdraws, the torque on the rod $\tau =I\alpha =Mg\frac{l}{2}$

$\Rightarrow \frac{M{l}^2}{3}\alpha =Mg\frac{l}{2}\left[\text{ As }I=M{l}^2/3\right]$

$\Rightarrow \text{ Angular acceleration, }$$\alpha =\frac{3}{2}\frac{g}{l}$

and linear acceleration,

$a=\frac{l}{2}\alpha =\frac{3g}{4}$

Now if the new normal force at A is $F^{\mathrm{\prime }}\text{ then }Mg-F^{\mathrm{\prime }}=Ma$

$\Rightarrow F^{\mathrm{\prime }}=Mg-Ma=Mg-\frac{3Mg}{4}=\frac{Mg}{4}=\frac{W}{4}$