Add forces $A'$, $A''$ and $B'$, $B''$ acting at the centre of mass $G$ of the rod.
Forces $A$ and $A'$ act as a couple of magnitude $Aa$ in a counter-clockwise direction.
Forces $B$ and $B'$ act as a couple of magnitude $Bb$ in a counter-clockwise direction.
So the net couple on the rod is $Aa+Bb$ counter-clockwise and the rod will rotate about the centre of mass $G$ under the influence of the net couple.
You are left with a net force of $A''- B''$ downwards which will accelerate the centre of mass.
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Using the symbols above one can write two equations, one for linear acceleration of the centre of mass $G$, $\dot v_{\rm G}$, and one for the angular acceleration about the centre of mass, $\dot \omega_{\rm G}$.
$B-A= m\,\dot v_{\rm G} $ and $B\, b+A\, a= I_{\rm G}\, \dot \omega_{\rm G}$.
Now the constraint is that $A$ is a pivot and so does not move, thus the upward linear acceleration is equal to the downward linear acceleration, $\dot v_{\rm G}=a\, \dot \omega_{\rm G}$.
This gives the relationship between the magnitude of the applied force, $B$, and the magnitude of the force exerted by the pivot, $A$.
$B(I_{\rm G} -m\,a\,b) = A(I_{\rm G} +m\,a^2) $.
Thus $B>A$ as suggested by @tryingtobeastoic.
Best Answer
You can still have a net torque with the net force being $0$. You do this wherever you turn a door knob, for example. The knob turns about its center, but the center of mass of the knob doesn't accelerate (assuming you don't move the door itself). Such an instance of zero net force and non-zero net torque is called a couple.
There are analogs of Newton's laws for torques and rotations though. For example, for an object rotating about a principle axis we have the analog of $F=ma$ as $\tau=I\alpha$, where $\tau$ is the torque $I$ is the moment of Inertia about the principle axis and $\alpha$ is the angular acceleration (all about the specified principle axis).
A more general form of this comes from the linear momentum $\mathbf p=m\mathbf v$ in the form of $\mathbf L=\hat I\boldsymbol\omega$ where $\mathbf L$ is the angular momentum vector, $\boldsymbol\omega$ is the angular velocity vector, and $\hat I$ is the moment of inertia tensor. Then, just like how $\dot{\mathbf p}=m\dot{\mathbf v}\to\mathbf F=m\mathbf a$, we have $\dot{\mathbf L}=\hat I\dot{\boldsymbol\omega}\to\boldsymbol\tau=\hat I\boldsymbol\alpha$