From: dr.j.thornburg@gmail-pink.com   
      
   In article ,   
   Luigi Fortunati  asks   
   > Up to this point, I've been talking about the forces F1 and F2 between   
   > the car and the pin, and between the towbar and the pin.   
   >   
   > And not about the action and reaction between car 1 and the towbar .   
   >   
   > So I ask: is the force F1 of car 1 on the pin equal to the action of   
   > car 1 on the towbar ?   
      
   There is no direct "action of car 1 on the towbar", because car 1 does   
   not apply any forces to the towbar.  Rather, car 1 applies a force (F1)   
   to the pin, and the pin applies a force (let's call it F3) to the towbar.   
      
   If you're asking about the *indirect* action (force) applied to the towbar   
   by virtue of car 1 applying a force to the pin and the pin applying a force   
   to the towbar, i.e., you're asking whether F1 and F3 are equal, see below.   
      
      
   > And is the force F2 of the towbar on the pin equal to the reaction of   
   > the towbar on car 1?   
      
   There is no direct "reaction of the towbar on car 1", because the towbar   
   does not apply any forces to car 1.  Rather, the towbar applies a force   
   (F2) to the pin, and the pin applies a force (let's call it F4) to car 1.   
      
   If you're asking about the *indirect* reaction (force) applied to car 1   
   by virtue of the towbar applying a force to the pin and the pin applying   
   a force to the towbar, i.e., you're asking whether F2 and F4 are equal,   
   see below.   
      
      
   To summarize what we know so far, we have the following forces acting:   
      
   towbar:	force F3 to the right, applied by the pin   
      
   pin:	force F2 to the left, applied by the towbar   
   	force F1 to the right, applied by car 1   
      
   car 1:	force F4 to the left, applied by the pin   
      
   and we know by Newton's 2nd law (applied to the pin) that   
      
     m_pin a = F1 + F2 ,                                            (1a)   
   i.e.,   
     F2 = -F1 + m_pin a and F1 = -F2 + m_pin a ,                    (1b)   
      
   where /a/ is the common acccleration of all three (rigid) bodies.   
   (Hence if /a = 0/ then F2 = -F1, but if /a/ is nonzero then F2 is not   
   equal to -F1.)   
      
      
   You've said in the past that you don't think Newton's 3rd law is always   
   valid, but if we accept it for the moment, it tells us the answer to your   
   questions.   
      
   In particular, if we apply Newton's 3rd law to the towbar/pin interface,   
   it says that F3 = -F2, which by (1b) means F3 = F1 - m_pin a.  So the   
   answer to your first question is that if /a = 0/ then F3 = F1, but if   
   /a/ is nonzero then F3 is not equal to F1.   
      
   And if we apply Newton's 3rd law to the pin/car 1 interface, it says   
   that F4 = -F1, which by (1b) means F4 = F2 - m_pin a.  So the answer   
   to your second question is that if /a = 0/ then F4 = F2, but if /a/   
   is nonzero then F4 is not equal to F2.   
      
      
   If we *don't* want to use Newton's 3rd law, I suspect we can still work   
   out the answers to your questions by using Newton's *2nd* law repeatedly   
   on different combinations of the 3 bodies, in the manner of my post to   
   this newsgroup a few months ago,   
      
     Newsgroups: sci.physics.research   
     Subject: derivation of Newton's 3rd law from 2nd law (was: Re: The   
   experiment)   
     Date: 17 Jun 2025 21:03:56 +0100 (BST)   
     Message-ID:    
      
   I might do this in a future posting.   
      
   ciao,   
      
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