From: fortunati.luigi@gmail.com   
      
   On Sat, 11 Oct 2025 17:10:16 PDT, Luigi Fortunati   
    wrote:   
      
   >Jonathan Thornburg [remove -color to reply] wrote on 02/10/2025   
   >10:03:51:   
   >> ...   
   >> Hopefully that clarifies things!   
   >   
   >Yes, you did.   
   >   
   >F_Y_on_X = - F_X_on_Y for any value of F_ext_on_X and F_ext_on_Y.   
   >   
   >If I substitute numbers for the letters...   
   >F_ext_on_X=+6N   
   >F_ext_on_Y=-3N   
   >m_X=6kg   
   >m_Y=3kg   
   >   
   >Is it correct that they are...   
   >F_Y_on_X =-4N   
   >F_X_on_Y=+4N?   
   >   
   >Hi, Luigi   
   >   
   >   
   >[[Mod. note --   
   >Yes.   
   >-- jt]]   
      
   If the only negative force applied to the system is the external force   
   F_ext_on_Y=-3N, how does the force F_Y_on_X =-4N exist?   
      
   Where does he get that extra -1 from? Who is providing it to him?   
      
      
   [[Mod. note --   
   For convenience of discussion, let's take the motion as being horizontal,   
   with the + direction being to the right.   
      
   The only negative *external* force applied to the system is the external   
   force F_ext_on_Y=-3N (pushing left on Y with a force of 3N).  But because   
   there's a larger-in-magnitude force F_ext_on_X (pushing right on X with a   
   force of 6N), the net force on the whole system is to the right, so the   
   whole system is accelerating to the right with an acceleration (with   
   respect to an inertial reference frame) of   
     a = (F_ext_on_X + F_ext_on_Y) / (m_X + m_Y)   
       = +3N / 9kg   
       = +1/3 m/s^2   
      
   The answer to Luigi's question is that X and Y are both *accelerating*   
   (to the right), there are also internal forces acting between X and Y:   
   X is pushing rightward on Y (this is F_X_on_Y), and Y is pushing leftward   
   on X to the left (this is F_Y_on_X).   
      
   So F_ext_on_Y is *not* the only negative force -- there is also F_Y_on_X.   
      
   We can use Newton's 2nd law to work out what these forces are:   
      
   If we apply Newton's 2nd law to X, we see that   
     F_net_on_X = m_X a   
                = 6kg * +1/3 m/s^2   
                = +2N to the right   
   But F_net_on_X is just the sum of F_ext_on_X and F_Y_on_X, so that means   
     F_ext_on_X + F_Y_on_X = +2N   
   Since we know that F_ext_on_X = +6N, this means that F_Y_on_X must be -4N.   
      
   If we apply Newton's 2nd law to Y, we see that   
     F_net_on_Y = m_Y a   
                = 3kg * +1/3 m/s^2   
                = +1N to the right   
   But F_net_on_Y is just the sum  of F_ext_on_Y and F_X_on_Y, so that means   
     F_ext_on_Y + F_X_on_Y = +1N   
   Since we know that F_ext_on_Y = -3N, this means F_X_on_Y must be +4N.   
      
   *IF* X and Y were *NOT* accelerating (with respect to an inertial reference   
   frame), then it would be true that F_Y_on_X = F_ext_on_Y.  But (given the   
   specified external forces) they actually *are* accelerating, so there's no   
   reason to expect F_Y_on_X to be the same as F_ext_on_Y.   
      
   In fact, the difference between F_ext_on_Y and F_Y_on_X is just the net   
   force on Y, so by Newton's 2nd law this difference must be nonzero (i.e.,   
   those two forces must differ) if (as is the case here) X and Y are   
   accelerating.   
   -- jt]]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)