Q.Why is the moment of a couple independent of the axis of rotation even if the axis is fixed?

[!TLDR] Because the net torque of a couple, computed about ANY axis location, always algebraically reduces to the same value rF -- shown explicitly for an axis between the two forces and an axis outside them, both giving rF. [!ANSWER] Because the torque w …

Section 4.11.1 proves this directly by calculation, not merely by assertion. Placing the fixed axis of rotation BETWEEN the two lines of action of the couple's forces (at perpendicular distances x and y from the two forces, with $x+y=r$ the fixed separation) gives both individual torques acting in the SAME rotational sense, so they ADD: total torque $=xF+yF=(x+y)F=rF$. Placing the axis OUTSIDE both lines of action instead (at perpendicular distances q and p, with $q-p=r$) gives the two individual torques acting in OPPOSITE rotational senses, so they SUBTRACT: total torque $=qF-pF=(q-p)F=rF$. Both calculations, despite using completely different axis positions, arrive at the SAME final value, $rF$, depending only on the force magnitude F and the fixed perpendicular separation r between the two lines of action -- never on where exactly the axis itself was placed. Physically, this happens because a couple has zero net FORCE (the two forces are equal and opposite), so shifting the reference axis by some vector $\vec{d}$ shifts each individual torque contribution by an amount proportional to $\vec{d}\times\vec{F}$ and $\vec{d}\times(-\vec{F})$ res …

Reproduce the two-case algebraic proof from 4.11.1 (axis between the forces; axis outside the forces) and note both give rF, then give t …

Simply asserting 'it's a known property' without giving the actual reasoning -- the axis-independence needs to b …