Is Net Internal Torque Really Zero in All Cases?

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The discussion centers on the concept of net internal torque in rigid bodies, questioning the textbook conclusion that it equals zero. It highlights that while internal forces between two particles are equal and opposite (F12 = -F21), the torque generated by these forces depends on the position vectors (r1 and r2). The net torque is derived from the equation Γnet = r1 × F12 + r2 × F21, which simplifies to Γnet = (r1 - r2) × F12. Since the vector (r1 - r2) is aligned with the force's line of action, the angle is zero, resulting in no net torque. This analysis clarifies why net internal torque is considered zero in this context.
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I have a confusion of a conclusion in the textbook ( I posted it in the attach file )

Net internal torque equals zero ( similarly to conclusion in the Newton’s third law ) but I myself reckon that torque is defined as
${\rm{r \times F}}$
And maybe there occurs the case below :
${\rm{F}}_{{\rm{21}}} {\rm{ = - F}}_{{\rm{12}}} $
But
${\rm{r}}_{21} {\rm{F}}_{{\rm{21}}} \ne {\rm{r}}_{12} {\rm{ - F}}_{{\rm{12}}} $

Could someone help me analyze this situation , I think that the conclusion in textbook is true but it is still fuzzy for me
Thank you in advance
 
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I wish your code made more sense.
Torque of internal forces is not accounted because the net resultant of the radius vectors is along the line of action of the force.
Considering two particles 1 and 2 on the rigid body. Let their position vectors be r1 an r2.Let no external torque act for simplification.
Then Γnet = r1 × F12 + r2 × F21
You know F12=-F21 by Newton's Third Law.
Hence Γnet =(r1-r2) × F12
r1-r2 will be along the line joining the two particles and this is also the line of action of the force b/w the two particles[F12]. Since angle is 0 there is no torque.
I hope this is useful.
 
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