From: dr.j.thornburg@gmail-pink.com   
      
   [I'll use '$' to mark inline math symbols and equations, as in a mass   
   $m$.  In the past I've used other symbols like '/' or '|', but those   
   cause ambiguities with division or absolute values.]   
      
   Simplifying Luigi's system a bit, we're taking the laboratory to be an   
   inertial reference frame, and we're considering the horizontal motion of   
   a solid block of ice (assumed uniform in composition) of mass $M_ice$   
   and length $L$.  The ice is being accelerated horizontally to the right   
   by a (horizontal) external force $F_ext$ pushing rightward on the ice's   
   left side.  By Newton's 2nd law, the ice's acceleration is   
   $a = F_ext/M_ice$, i.e., $F_ext = M_ice a$.   
      
   Since we're concerned with the internal compressive forces acting within   
   the ice block, let's consider an imaginary vertical line $vx$ at a distance   
   $x$ from the left side of the ice, dividing the ice into a part $IL$   
   (of length $x$ and mass $m_IL = (x/L) M_ice$) to the left of the line,   
   and a part $IR$ (of length $1-x$ and mass $m_IR = (1-x/L) M_ice$) to   
   the right of the line.   
      
   What (horizontal) forces act across the vertical line $vx$?   
      
   Applying Newton's 2nd law to $IR$, we see that since $IR$ is accelerating   
   to the right with acceleration $a$, there must be a rightward net force   
   of $m_IR a$ acting on $IR$.  The only thing that's pushing (horizontally)   
   on $IR$ is $IL$, i.e., $IL$ must be pushing to the right on $IR$ with a   
   force   
     F5 = m_IR a   
        = (1-x/L) m_ice (F_ext/M_ice)   
        = (1-x/L) F_ext   
      
   That means that $IR$ should also be pushing left on $IL$ with some   
   reaction force $F6$.  (We don't yet know the magnitude of $F6$; we'll   
   work this out below.)   
      
   Applying Newton's 2nd law to $IL$, we see that since $IL$ is accelerating   
   to the right with acceleration $a$, there must be a rightward net force   
   of $m_IL a$ acting on $IL$.  There are two (horizontal) forces acting on   
   $IL$:   
   * an external force $F_ext$   
   * the reaction force $F6$   
   So, we have that   
     F_ext + F6 = m_IL a   
   so   
     F6 = m_IL a - F_ext   
        = (x/L) M_ice (F_ext/M_ice) - F_ext   
        = (x/L) F_ext - F_ext   
        = (x/L-1) F_ext   
        = -F5   
      
   In other words, $F6$ is precisely equal in magnitude and opposite in   
   direction to $F5$, i.e., Newton's 3rd law holds for the opposing   
   horizontal forces acting across line $vx$.   
      
   Notice that the directions of $F5$ and $F6$ are indeed such as to   
   compress the ice.  That is, $IL$ is pushing right on $IR$ with a   
   (compression) force $F5$, and $IR$ is pushing left on $IL$ with a   
   (compression) force $F6 = -F5$.   
      
   Notice that both $F5$ and $F6$ are proportional to $F_ext$.  This means   
   that if $F_ext$ is zero, then $F5 = F6 = 0$.  That is, if there is no   
   external force pushing on the ice (and hence the ice is not accelerating   
   with respect to an inertial reference frame), there are no internal   
   compressive forces $F5$ or $F6$.   
      
   If $F_ext$ is nonzero, then, looking at the equations for $F5$ and $F6$,   
   we see that these compression forces are *non-uniform*, i.e., $F5$ and   
   $F6$ vary with the position $x$: at $x=0$ (the left side of the ice)   
   $F5$ and $F6$ have their maximum magnitudes; at larger values of $x$   
   both forces decrease (linearly with $x$), and at $x=L$ (the right side   
   of the ice) $F5 = F6 = 0$.   
      
   Summing up, the answer to Luigi's question   
      
   > How could the block of ice (or any other material) compress if, as you   
   > say, there is no opposing force?   
      
   is that because the ice is accelerating, at any position in the block   
   the inertia of the part of the block to the right of that position ($IR$)   
   provides the compressive force pushing left on, i.e., acting to compress,   
   the part of the block to the left of that position ($IL$).   
      
      
      
   It's instructive to compare the "horizontal ice" system to a different   
   system, which I'll call the "vertical ice" system:  Suppose now our   
   block of ice (still with mass $m_ice$) is oriented vertically, resting   
   (stationary) on a table (so that $L$ is the ice's height, and $x$   
   measures height from the bottom of the block) in a uniform (vertical)   
   gravitational field with gravitational acceleration $g$.   
      
   In the vertical-ice system, consider an imaginary horizontal line $hx$   
   at the position (height) $x$, dividing the ice into a lower part $Ilow$   
   (of height $x$ and mass $m_Ilow (x/L) m_ice$ and an upper part $Ihigh$   
   (of height $L-x$ and mass $m_Ihigh = (1-x/L) m_ice$.   
      
   What vertical forces act across the horizontal line $hx$?   
      
   Observe that $Ihigh$ is stationary, and hence unaccelerated vertically.   
   Applying Newton's 2nd law to $Ihigh$, that means that the net vertical   
   force on $Ihigh$ must be zero, so $ILo$ must be pushing up on $Ihigh$ with   
   a force $vF5$ which just balances $Ihigh$'s weight, i.e.,   
     vF5 = m_Ihigh g   
         = (1-x/L) m_ice g.   
      
   That means that $Ihigh$ should also be pushing down on $Ilow$ with some   
   reaction force $vF6$.  (We don't yet know the magnitude of $vF6$; we'll   
   work this out below.)   
      
   There are two ways to work out $vF6$:   
      
   (1)  We can observe that $vF6$ is just $Ihigh$'s weight, i.e.,   
          vF6 = -m_Ihigh g                  (the - sign is to denote that   
                                           $vF6$ is pushing *down*)   
              = -(1-x/L) m_ice g   
      
   or   
      
   (2)  We can apply Newton's 2nd law twice:   
        Consider first the entire ice block.  There are two vertical forces   
        acting on it:   
        * the block's weight, -m_ice g (pushing down)   
        * some reaction force $F_table$ applied by the table   
        Since the block is stationary and hence unaccelerated, we know   
        by Newton's 2nd law applied to the entire ice block that the sum   
        of these two forces must be zero, i.e., we must have   
        $F_table = m_ice g$ (pointing up).   
        Now apply Newton's 2nd law to $Ilow$.  There are three vertical   
        forces acting on it:   
        * F6 pushing down   
        * $Ilow$'s own weight, -m_Ilow g = (x/L) m_ice g$ (pushing down)   
        * $F_table$ pushing up   
        Since $Ilow$ is stationary and hence unaccelerated, we know   
        by Newton's 2nd law applied to $Ilow$ that the sum   
        of these two forces must be zero, i.e., we must have   
           F6 - m_Ilow g + F_table = 0   
        so that   
           F6 = m_Ilow g - F_table   
              = (x/L) m_ice g - m_ice g   
              = (x/L-1) m_ice g   
              = -(1-x/L) m_ice g   
      
   We get the same answer for $vF6$ either way, $F6 = -(1-x/L) m_ice g$.   
   Notice that this is precisely equal in magnitude and opposite in direction   
   to $vF5$, i.e., we've shown that Newton's 3rd law holds for the opposing   
   vertical forces across the horizontal line $hx$.   
      
   Notice that both $vF5$ and $vF6$ are proportional to the gravitational   
   acceleration $g$.  This means that if $g$ is zero (i.e., if we're actually   
   in free-fall), then $F5 = F6 = 0$.  That is, if there is no external   
   gravitational force pushing on the ice, there are no internal compressive   
   forces $F5$ or $F6$.   
      
   Notice also that the directions of $vF5$ and $vF6$ are such as to compress   
      
   [continued in next message]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)