Is Net Internal Torque Really Zero in All Cases?

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The discussion clarifies that net internal torque equals zero in a system of two particles due to the alignment of the radius vectors with the line of action of the internal forces. Specifically, when considering two particles with forces F12 and F21, the net torque is expressed as Γnet = (r1 - r2) × F12. Since F12 = -F21 by Newton's Third Law, and the angle between the radius vector and force vector is zero, the resulting torque is indeed zero. This conclusion aligns with the principles outlined in classical mechanics.

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tukms
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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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