From Newsgroup: sci.physics.research

Newton's 2nd law F=ma states that the mass m is acted upon by the
force F and accelerates passively without reacting, also because there
is only one body of mass m and there is no other body on which to
exert the reaction.

However, the 3rd law prohibits an action without a corresponding
reaction.

For this reason, I prepared the animation https://www.geogebra.org/classic/qn5ucjxp where two bodies of mass
m=1kg are pushed to the right by the force F=+1N with an acceleration
of +1m/s^2, exactly as prescribed by Newton's 2nd law.

And I ask: do the bodies acted upon by the blue force F at point X
react with the opposing red force also directed against point X, or
does the red reaction force not exist?

Luigi Fortunati

[[Mod. note --
It's important to realise that it's each body's *center of mass* that's accelerating at +1m/s^2.

Body B is rigid, so it's center of mass is in a fixed location with
respect to the rest of body B, so we can unambibuosly speak of "body B's acceleration".

But body A includes a bunch of water which can move around ("slosh") in
A's water-tank. That means that body A's center of mass can and will move around with respect to the rigid "water tank" parts of body A. Or, to put
it another way, body A's rigid parts move around with respect to body A's center of mass. That means that while the water is sloshing, the
acceleration of body A's rigid "water tank" will be *different* than
+1m/s^2. So, the animation is incorrect in showing A and B always moving side-by-side as they accelerate.

[It's instructive to think about a slightly different
system in which this difference is taken to an extreme:
suppose that we replace body A with body A', which consists
of an 0.5kg box with a small 0.5 kg perfectly-elastic spring,
free to slide horizontally inside the box; the spring is
initially uncompressed at the right end of the box.

Then when we push on the box, the spring initially stays
stationary in the lab (inertial) reference frame, while the
box (mass 0.5kg) accelerates at +2 m/s^2 (with respect to
the lab reference frame). When the left end of the box
strikes the spring, the spring recoils to the right, causing
the box to *slow down* for a short time.]

Fortunately, none of the above affects the answer to the author's question, which is basically asking if Newton's 3rd law still applies to each body.
The answer to that is yes, it does. That is,
* Body A exerts a force of 1N pushing to the left on whatever body
(not showin in the animation) is applying the rightward force on A.
* Body B exerts a force of 1N pushing to the left on whatever body
(not showin in the animation) is applying the rightward force on B
-- jt]]
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From Newsgroup: sci.physics.research

On Sun, 02 Nov 2025 12:52:55 PST, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

Newton's 2nd law F=ma states that the mass m is acted upon by the
force F and accelerates passively without reacting, also because there
is only one body of mass m and there is no other body on which to
exert the reaction.

However, the 3rd law prohibits an action without a corresponding
reaction.

For this reason, I prepared the animation >https://www.geogebra.org/classic/qn5ucjxp where two bodies of mass
m=1kg are pushed to the right by the force F=+1N with an acceleration
of +1m/s^2, exactly as prescribed by Newton's 2nd law.

And I ask: do the bodies acted upon by the blue force F at point X
react with the opposing red force also directed against point X, or
does the red reaction force not exist?

Luigi Fortunati

[[Mod. note --
...
Fortunately, none of the above affects the answer to the author's question, >which is basically asking if Newton's 3rd law still applies to each body.
The answer to that is yes, it does. That is,
* Body A exerts a force of 1N pushing to the left on whatever body
(not showin in the animation) is applying the rightward force on A.
* Body B exerts a force of 1N pushing to the left on whatever body
(not showin in the animation) is applying the rightward force on B
-- jt]]

I asked something else.

I asked about the internal forces F2 and F3 in the animation https://www.geogebra.org/classic/qn5ucjxp, which act on the wall X of
the same body and not on other bodies not shown in the animation.

Do the water in body A and the *rigid* ice in body B exert their
*internal* forces (F2 and F3) against wall X (unbalanced by the
opposing force against wall Y), or do the unbalanced *internal* forces
F2 and F3 not exist and did I invent them?

Luigi Fortunati.

[[Mod. note --
Ok, now I think I understand what you're asking.

As stated, the problem description is self-contradictory: you've said
that body A and body B each have a total mass of 1kg, but you've also
said that their water and ice masses are also 1 kg each. This doesn't
leave any mass for the tank itself (including wall X)! For simplicity,
let's say that each tank (empty) weighs 100 grams, so that that A's water
and B's ice each weigh 900 grams.

Now to your questions:

Yes, the water in body A exerts an internal force F2 pushing to the left
on wall X. Because the water sloshes in the tank, F2 will be time-dependent. Analyzing Body A's motion and forces in detail is hard because of the water sloshing in the tank.

For body B, the answer to your question depends on whether the ice is
stuck to the bottom of B (in which case there's only a very small force
F3), or whether the ice-bottom contact is low-friction (let's idealise
this to say *no* friction). Let's assume the latter.

Apply Newton's 2nd law to wall X: the wall has 2 (horizontal) forces
acing on it, F_external = +1N and F3, so we have
F_net = F_external + F3 = +1N + F3
= m_wall a
= 0.1kg * +1m/s^2
= +0.1N
so we must have F3 = -0.9N. This tells us that there is indeed a force
F3, and tells us what that force is (0.9N pushing left).

Finally, apply Newton's 2nd law to the ice in body B: The ice has only
one force acting on it,
[I'm assuming that the ice isn't stuck to wall Y,
i.e., that wall Y isn't exerting any horizontal forcw
on the ice.]
a reaction force (let's call it F4) applied by the wall to the ice.
Newton's 2nd law applied to the ice gives
F4 = m_ice a
= 0.9kg * +1m/s^2
= +0.9N
This tells us that wall X pushes right on the ice with a force of 0.9N.

Notice F3 = -F4, i.e., Newton's 3rd law also holds here.
-- jt]]
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From Newsgroup: sci.physics.research

On 2025-11-02 12:52:55 +0000, Luigi Fortunati said:

Newton's 2nd law F=3Dma states that the mass m is acted upon by the
force F and accelerates passively without reacting, also because there
is only one body of mass m and there is no other body on which to
exert the reaction.
=20
However, the 3rd law prohibits an action without a corresponding
reaction.

That's right. A force does not exisst as a separate being. Force is
simply a number that quantifies an interaction between two bodies.
If there is a force on a body there is also another body somewhere
because one body alone cannot have an interaction. The forces on
each body have equal magnitudes and opposite directions.

--=20
Mikko
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From Newsgroup: sci.physics.research

On Mon, 03 Nov 2025 23:49:17 PST, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

I asked about the internal forces F2 and F3 in the animation >https://www.geogebra.org/classic/qn5ucjxp, which act on the wall X of
the same body and not on other bodies not shown in the animation.

Do the water in body A and the *rigid* ice in body B exert their
*internal* forces (F2 and F3) against wall X (unbalanced by the
opposing force against wall Y), or do the unbalanced *internal* forces
F2 and F3 not exist and did I invent them?

Luigi Fortunati.

[[Mod. note --
Ok, now I think I understand what you're asking.

As stated, the problem description is self-contradictory: you've said
that body A and body B each have a total mass of 1kg, but you've also
said that their water and ice masses are also 1 kg each. This doesn't
leave any mass for the tank itself (including wall X)! For simplicity,
let's say that each tank (empty) weighs 100 grams, so that that A's water
and B's ice each weigh 900 grams.

Now to your questions:

Yes, the water in body A exerts an internal force F2 pushing to the left
on wall X. Because the water sloshes in the tank, F2 will be time-dependent. >Analyzing Body A's motion and forces in detail is hard because of the water >sloshing in the tank.

For body B, the answer to your question depends on whether the ice is
stuck to the bottom of B (in which case there's only a very small force
F3), or whether the ice-bottom contact is low-friction (let's idealise
this to say *no* friction). Let's assume the latter.

Apply Newton's 2nd law to wall X: the wall has 2 (horizontal) forces
acing on it, F_external = +1N and F3, so we have
F_net = F_external + F3 = +1N + F3
= m_wall a
= 0.1kg * +1m/s^2
= +0.1N
so we must have F3 = -0.9N. This tells us that there is indeed a force
F3, and tells us what that force is (0.9N pushing left).

Thank you.

So, to my question, "Do the forces F2 and F3 really exist, or did I
invent them?" the answer is: the force F3 (exerted by the ice against
wall X) actually exists and pushes to the left with a force of -0.9N.

How does the mass of the ice (which is a scalar quantity) react with
the force F3 (which is a vector)?

[[Mod. note -- Newton's 2nd law applied to the ice says
F2(vector) = m_ice(scalar) * a (vector).
Then Newton's 3rd law says
F3(vector) = - F2(vector)
= - m_ice(scalar) * a(vector)
-- jt]]


Does it use its own inertia or not?

In other words: is the force F3 = -0.9N (or is it not) the inertia of
the block of ice itself, which is transformed into an effective force
to the left against wall X, which is accelerating it?

[[Mod. note --
Force and inertia are two quite different things, so a force can't
possibly "be" the inertia of something, any more than (say) a lake
can be a thunderstorm.

So far as we know, inertia is an inherent property of mass (mass-energy
in relativity). So if a mass (the ice) is accelerating (with respect to
an inertial reference frame), there must be a net force applied to the
mass. Looking at this system, the only thing that could be applying that
force is wall X. As noted above, Newton's 2nd law gives the magnitude and direction of that force F2 (vector), and then Newton's 3rd law gives the magnitude and direction of the reaction force F3 applied by the ice to wall
X.
-- jt]]

Luigi Fortunati.
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From Newsgroup: sci.physics.research

On Mon, 03 Nov 2025 23:49:17 PST, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

[[Mod. note --
Ok, now I think I understand what you're asking.

As stated, the problem description is self-contradictory: you've said
that body A and body B each have a total mass of 1kg, but you've also
said that their water and ice masses are also 1 kg each. This doesn't
leave any mass for the tank itself (including wall X)! For simplicity,
let's say that each tank (empty) weighs 100 grams, so that that A's water
and B's ice each weigh 900 grams.

Now to your questions:

Yes, the water in body A exerts an internal force F2 pushing to the left
on wall X. Because the water sloshes in the tank, F2 will be time-dependent. >Analyzing Body A's motion and forces in detail is hard because of the water >sloshing in the tank.

For body B, the answer to your question depends on whether the ice is
stuck to the bottom of B (in which case there's only a very small force
F3), or whether the ice-bottom contact is low-friction (let's idealise
this to say *no* friction). Let's assume the latter.

Apply Newton's 2nd law to wall X: the wall has 2 (horizontal) forces
acing on it, F_external = +1N and F3, so we have
F_net = F_external + F3 = +1N + F3
= m_wall a
= 0.1kg * +1m/s^2
= +0.1N
so we must have F3 = -0.9N. This tells us that there is indeed a force
F3, and tells us what that force is (0.9N pushing left).

Finally, apply Newton's 2nd law to the ice in body B: The ice has only
one force acting on it,
[I'm assuming that the ice isn't stuck to wall Y,
i.e., that wall Y isn't exerting any horizontal forcw
on the ice.]
a reaction force (let's call it F4) applied by the wall to the ice.
Newton's 2nd law applied to the ice gives
F4 = m_ice a
= 0.9kg * +1m/s^2
= +0.9N
This tells us that wall X pushes right on the ice with a force of 0.9N.

Notice F3 = -F4, i.e., Newton's 3rd law also holds here.
-- jt]]

Based on your own numbers, wall X experiences two forces: the external
force F = +1N directed to the right and the force F3 = -0.9N directed
to the left.

The difference between the two opposing forces (F and F3) is the net
force +0.1N that accelerates wall X without compressing it, while the
remaining opposing forces (+0.9 and -0.9) compress it without
accelerating it.

One force accelerates it and the other two compress it.

How do we detect the net force +0.1N? We measure the acceleration of
wall X.

How do we detect opposing forces on the same body? We measure its
compression.

And now let's move on to the block of ice.

Based on your analysis, the block of ice experiences only one force,
the external force F4 = +0.9N, and nothing else.

If this were indeed the case, the block of ice should accelerate
without experiencing any compression.

And yet, for the entire duration of the acceleration, the block of ice undergoes a compression that can be measured by highly precise
instruments.

How could the block of ice (or any other material) compress if, as you
say, there is no opposing force?

Luigi Fortunati

[[Mod. note --
Consider an imaginary vertical line dividing the block of ice into a
part /IL/ to the left of the line and a part /IR/ to the right of the
line. The position of the vertical line is such that /IL/ contains a
fraction /f/ of the ice, where /f/ is some as-yet-unspecified number
in the range [0,1], i.e., /IL/ has a mass /f m_ice/ and IR has a mass
/(1-f) m_ice/.

/IR/ is accelerating to the right, so by Newton's 2nd law there must be
a net force /F5/ to the right acting on IR. This force can only come
from /IL/ being compressed and pushing right on /IR/. By Newton's 3rd
law, /IR/ also pushes left on /IL/ with a force /F6 = -F5/

Since uniform acceleration is equivalent to a gravitational field, this
system is very analogous to a rope hanging vertically, where the tension
varies along the rope.

In a following posting I'll work out this analysis in a bit more detail.
-- jt]]
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From Newsgroup: sci.physics.research

On Fri, 07 Nov 2025 12:49:08 PST, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

Based on your own numbers, wall X experiences two forces: the external
force F = +1N directed to the right and the force F3 = -0.9N directed
to the left.

The difference between the two opposing forces (F and F3) is the net
force +0.1N that accelerates wall X without compressing it, while the >remaining opposing forces (+0.9 and -0.9) compress it without
accelerating it.

One force accelerates it and the other two compress it.

How do we detect the net force +0.1N? We measure the acceleration of
wall X.

How do we detect opposing forces on the same body? We measure its >compression.

And now let's move on to the block of ice.

Based on your analysis, the block of ice experiences only one force,
the external force F4 = +0.9N, and nothing else.

If this were indeed the case, the block of ice should accelerate
without experiencing any compression.

And yet, for the entire duration of the acceleration, the block of ice >undergoes a compression that can be measured by highly precise
instruments.

How could the block of ice (or any other material) compress if, as you
say, there is no opposing force?

Luigi Fortunati

[[Mod. note --
Consider an imaginary vertical line dividing the block of ice into a
part /IL/ to the left of the line and a part /IR/ to the right of the
line. The position of the vertical line is such that /IL/ contains a >fraction /f/ of the ice, where /f/ is some as-yet-unspecified number
in the range [0,1], i.e., /IL/ has a mass /f m_ice/ and IR has a mass
/(1-f) m_ice/.

/IR/ is accelerating to the right, so by Newton's 2nd law there must be
a net force /F5/ to the right acting on IR. This force can only come
from /IL/ being compressed and pushing right on /IR/. By Newton's 3rd
law, /IR/ also pushes left on /IL/ with a force /F6 = -F5/

Yes, but only if the third law is correct.

And mind you, I'm not saying the two opposing forces don't exist, I'm
just saying they're not equal.

Since uniform acceleration is equivalent to a gravitational field, this >system is very analogous to a rope hanging vertically, where the tension >varies along the rope.

Exactly, acceleration is (almost perfectly) equivalent to the
gravitational field (equivalence principle).

But the variation in tension along the rope indicates a direction and
a sense: it is a tension that is also a vector!

In a following posting I'll work out this analysis in a bit more detail.

I look forward to your next post with great interest.

Luigi Fortunati

P.S. You haven't answered my question: if the block of ice, as you
say, is subject only to the external force F4 = +0.9N, why is it under
tension throughout the acceleration? If there is tension, it means
that there are also opposing forces, not just net forces.
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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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From Newsgroup: sci.physics.research

In article <mt2.1.4-36528-1762831347@gold.bkis-orchard.net>, I wrote

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 $m_Ihigh = (1-x/L) m_ice$.

Oops, sorry, I was missing an equals sign in the 2nd-to-last line;
it should read
(of height $x$ and mass $m_Ilow = (x/L) m_ice$ and an upper part $Ihigh$
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; 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."
-- commenter on /arstechnica.com/, 2025-10-07
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