Why does net internal torque equal zero if position vectors differ?

[tukms]

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

 

[tukms]

I only try to post Latex but no success :(

again
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

 

[tukms]

though I can't post Latex truly , i guess you would know my confusion , please answer me as soon as possible

thank you :)

 

[Telemachus]

You must use [tex*] and [*/tex] (with out the *) on your latex code.

 

[tiny-tim]

welcome to pf!

hi tukms! welcome to pf!

(try using the X₂ icon just above the Reply box )

the important point in your attachment is "we assume that these forces lie along the line of separation of each pair of particles" …

so if the internal forces are F₁₂ and F₂₁, at positions r₁ and r₂,

then the total moment (about any origin) is r₁ x F₁₂ + r₂ x F₂₁ = (r₁ - r₂) x F₁₂

since r₁ - r₂ is parallel to F₁₂, that comes to zero