[continued from previous message]   
      
   > so that we have indeed recovered our earlier symmetric analysis.   
   >   
   >   
   >   
   > Finally, now we have all the right formulas worked out to answer   
   > Luigi's questions:   
   >   
   > Consider first the simplest case where the three masses are equal,   
   >    m1 = m2 = m3   
   > Then (taking F1 = +100 and F2 = -50 as specified) (11) gives   
   >    F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2   
   >       = - 2/3 F1 + 1/3 F2   
   >       = -83.333   
   > (19) gives   
   >    F4 = (m3/(m1+m2+m3)) F1 -   (m1+m2)/(m1+m2+m3) F2   
   >       = (1/3) F1 - (2/3) F2   
   >       = +66.667   
   >   
   > Notice that there are net forces acting on each body:   
   >     F_net_on_m1 = F1+F3   
   >                 = 100-83.333   
   >                 = +16.667,   
   >     F_net_on_m2 = -F3 - F4   
   >                 = -(-83.333) - (+66.667) = 83.333 - 66.667   
   >                 = +16.667,   
   >     F_net_on_m3 = F2+F4   
   >                 = -50+66.667   
   >                 = +16.667,   
   > These forces are all is nonzero, which is as it should (must!) be   
   > because the whole system is accelerating.  In fact, because we're taking   
   > m1 = m2 = m3 and and all three masses have the same common acceleration,   
   > it must be the case that F_net_on_m1 = F_net_on_m2 = F_net_on_m3.   
      
   Aside from the one doubt expressed above, everything else is perfect.   
   I updated the animation with your values of F3=-83.33 and F4=+66.67,   
   instead of F3=-??? and F4=+???.   
      
   I also added body X, which is the one exerting the continuous thrust   
   F1=+100 on point A of body A.   
      
   Obviously, body X also receives the external reaction from body A.   
      
   Is this latter reaction (which arrives at point X of body X and not at   
   point A of body A) worth -100 or -83.33?   
      
   Luigi Fortunati   
      
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