From Newsgroup: sci.physics.research

The dynamometer directly measures the force of its own contraction
and, indirectly, also measures the equality of the forces received.

The first measurement is direct and corresponds to what the
dynamometer itself reports.

The second, however, can only be obtained indirectly by observing from
the outside what happens to its motion.

If we place the dynamometer at point X(x=0) in the animation https://www.geogebra.org/classic/zedqvfcx, it will receive the action
of body A (force F1) from the left and the reaction of body B (force
F2) from the right.

The combined action of these two opposing forces crushes the
dynamometer, which measures this compression.

But the dynamometer, with its motion, also gives us another indication
of the two opposing forces it experiences: if it remains stationary
after the collision (final velocity unchanged
vi_dynamometer=vf_dynamometer=0 m/s), this means that the opposing
forces F1 and F2 are equal. However, if it *doesn't* remain stationary
at x=0 (vi_dynamometer=0 m/s and vf_dynamometer=+0.33 m/s), this means
that the forces F1 and F2 it experiences are different.

Newton's second law F=ma (which causes the dynamometer to accelerate
only if the force F1 is different from F2) contradicts the third law,
which, absurdly, always requires the eternal equality of F1 and F2.

I will soon also provide the general equation that governs the
relationship between action and reaction during the collision.

[[Mod. note --
Three comments:

Comment #1:
I think that instead of a dynamometer (which measures torque and power),
what you actually want at x=0 want is a strain gauge, calibrated to measure
the forces between bodies A1 & B1.

Comment #2:
For the A1+A2 collision with B1, this animation seems to violate conservation of momentum: the initial state has a net momentum to the right (A1 & A2 both moving to the right, B1 moving to the left), but the final state has a net momentum to the left (A1 & A2 both moving to the left, B1 moving to the right).

Comment #3:
Newton's 3rd law says nothing about whether
force exerted on dynamometer by A1
and
force exerted on dynamometer by B1
are or are not equal.
-- jt]]
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From Newsgroup: sci.physics.research

On Thu, 07 Aug 2025 11:46:25 PDT, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

[[Mod. note --
Three comments:

Comment #1:
I think that instead of a dynamometer (which measures torque and power),
what you actually want at x=0 want is a strain gauge, calibrated to measure >the forces between bodies A1 & B1.

Whatever it is, it remains stationary if it experiences two equal
opposing forces and accelerates if it experiences two unequal opposing
forces (Newton's first and second laws).

Comment #2:
For the A1+A2 collision with B1, this animation seems
to violate conservation of momentum:

Momentum conservation is perfectly preserved; just read the values of
the forces to see it: F1+F2+F3+F5=+1.33-1+0.33-0.67=0.

the initial state has a net momentum to the right (A1 & A2 both
moving to the right, B1 moving to the left), but the final state has a net >momentum to the left (A1 & A2 both moving to the left, B1 moving to the right).

No, even in the final state, the net momentum is to the right because
they all move to the right (including the dynamometer): the collision
is inelastic!

That small return to the left is the second compression-decompression
phase that also exists in an inelastic collision, during which the
forces that slow the opposing momentum of the other body act (on both
bodies).

After these two phases, there are no more forces and no more
accelerations: finally, they all move as one body.

To the right.

The final velocities (which I reported in the animation itself) are
all positive and equal to +1/3 (including the dynamometer).

Comment #3:
Newton's 3rd law says nothing about whether
force exerted on dynamometer by A1
and
force exerted on dynamometer by B1
are or are not equal.

The action (the force F1) exerted by body A on the dynamometer (of
negligible mass) is exactly equal to the action the dynamometer then
transmits directly to body B (why should it transmit less?).

The reaction (the force F2) exerted by body B on the dynamometer (of
negligible mass) is exactly equal to the reaction the dynamometer then transmits directly to body B (why should it transmit less?).
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