From: fortunati.luigi@gmail.com
On Mon, 03 Nov 2025 23:49:17 PST, Luigi Fortunati
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]]
--- SoupGate-DOS v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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