Solution

Since the moment of inertia and the clockwise moment with respect to the center of mass are $ml^2/12$ and $Na/2$, respectivly, the angular acceleration is

$$\frac{Na/2}{ml^2/12}=\frac{6Na}{ml^2}.$$

Similarly, since the moment of inertia and the clockwise moment with respect to the remaining support are $ml^2/12+m(a/2)^2$ and $mga/2$, respectively, the angular acceleration is

$$\frac{mga/2}{ml^2/12+m(a/2)^2}=\frac{6ga}{l^2+3a^)}.$$

Equating the two expressions for the angular acceleration, we have

$$\frac{6Na}{ml^2}=\frac{6ga}{l^2+3a^2}\qquad\Rightarrow\qquad N=\frac{mgl^2}{l^2+3a^2}.$$