From Newsgroup: sci.physics.research
[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
the ice. That is, $Ilow$ is pushing up on $Ihigh$ with a (compression)
force $vF5$, and $Ihigh$ is pushing down on $ILow$ with a (compression)
force $F6 = -F5$.
If $g$ is nonzero (e.g., if we're on the Earth's surface), 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 height $x$: at $x=0$
(the bottom of the ice) $F5$ and $F6$ have their maximum magnitudes; at
larger heights $x$ both forces decrease (linearly with the height $x$),
and at $x=L$ (the top of the ice) $F5 = F6 = 0$.
As you can see, the analysis of the vertical-ice system was basically
identical to the analysis of the horizontal-ice system, with the
vertical-ice gravitational acceleration $g$ playing the same role as
the horizontal-ice acceleration $a = F_ext/m_ice$. This is "just" an
example of Einstein's equivalence principle at work: an accelerated
reference frame can equivalently be viewed as stationary in a uniform gravitational field.
--
-- "Jonathan Thornburg [remove -color to reply]" <
dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"Liz Holmes didn't go to jail because she deceived people. She went to
jail because she deceived the wrong people."
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