Is a Couple always needed to produce torque? In my book, its written that:

When we produce torque on a pivoted body, we are always producing a couple as the reaction force from the pivot provides an equal, opposite and parallel force.

So suppose I have a body like this:
Here, two parallel, opposite but unequal forces $A$ and $B$ are acting on a body where magnitude of $A$>$B$.

*

*Will the body rotate?

*If yes, how will it do so?

*Where is the couple here, and the pivot?

*The net force of $A$ and $B$ is acting downwards outside the body towards $A$, so what kind of motion will be exhibited by the body?

 A: 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.
