From Newsgroup: sci.physics.research

In the animation https://www.geogebra.org/classic/uyhpg8ug, there are
two equal and opposite external forces (F1=+100 and F2=-100) acting on particles A and B of body A, which is not accelerating.

By clicking on the appropriate buttons, you can display (or hide) forces
F3 and F4.

My question: does only force F1 act on particle A (and nothing else) or
does force F3 also act?

And does only force F2 act on particle B (and nothing else) or does
force F4 also act?

My question concerns only the forces acting on particles A and B of body
A, not those acting on other bodies not shown in the animation.

Luigi Fortunati
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From Newsgroup: sci.physics.research

Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

In the animation https://www.geogebra.org/classic/uyhpg8ug, there are
two equal and opposite external forces (F1=+100 and F2=-100) acting on particles A and B of body A, which is not accelerating.

By clicking on the appropriate buttons, you can display (or hide) forces
F3 and F4.

My question: does only force F1 act on particle A (and nothing else) or
does force F3 also act?

And does only force F2 act on particle B (and nothing else) or does
force F4 also act?

My question concerns only the forces acting on particles A and B of body
A, not those acting on other bodies not shown in the animation.

Let's assume that A is initially at rest (with respect to the lab inertial reference frame).

If we start with body A with no (horizontal) forces applied to it, and
then apply the forces F1 and F2 at (let's say) t=0, what happens?

Before F1 and F2 were applied (i.e., at times t < 0), mass points A and B
were at rest, and hence unaccelerated (with respect to an inertial reference frame). Hence, by Newton's 2nd law, we infer that the net force acting on
them must be zero. That is, for t < 0, we infer that F3 = F4 = 0.

Immediately after t=0, body A hasn't yet had time to deform (compress)
under the influence of F1 and F2, so the inter-atomic distances within
body A are unchanged from what they were before F1 and F2 were applied.
Since (to an excellent approximation) the inter-atomic forces in a solid
body depend only on inter-atomic distances, these forces are thus also
(at t=0) unchanged from what they were before F1 and F2 were applied.
Hence, immediately after t=0, F3 = F4 = 0.

Therefore, immediately after t=0, there is a net force F1 acting on A
to accelerate A to the right, and a net force F2 acting on B to accelerate
B to the left. In other words, the body will compress. More precisely,
the forces will start compression waves propagating inwards from the left
and right sides of the body.

After a while (this will likely be a time measured in milliseconds for laboratory-sized bodies) we'll reach an equilibrium where all the
compression waves have damped out, every part of the body is stationary
again, and the body is compressed.

Now we can argue that (once the body is in equilibrium compression)
since every part of the body is stationary and hence unaccelerated
(again, with respect to an inertial reference frame), the net force on
every part of the body -- and in particular on point A and on point B --
must be zero. That is,
F1+F3 = 0 and F2+F4 = 0
and hence (once the body is in equilibrium compression) F3 = -F1 and
F4 = -F2.
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"Totalitarianism will not be satisfied to assert, in the face of
contrary facts, that unemployment does not exist; it will abolish
unemployment benefits as part of its propaganda,"
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From Newsgroup: sci.physics.research

Il 24/11/2025 08:38, Jonathan Thornburg [remove -color to reply] <dr.j.thornburg@gmail-pink.com> ha scritto:

Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

In the animation https://www.geogebra.org/classic/uyhpg8ug, there are
two equal and opposite external forces (F1=+100 and F2=-100) acting on
particles A and B of body A, which is not accelerating.

By clicking on the appropriate buttons, you can display (or hide) forces
F3 and F4.

My question: does only force F1 act on particle A (and nothing else) or
does force F3 also act?

And does only force F2 act on particle B (and nothing else) or does
force F4 also act?

My question concerns only the forces acting on particles A and B of body
A, not those acting on other bodies not shown in the animation.

Let's assume that A is initially at rest (with respect to the lab inertial reference frame).

If we start with body A with no (horizontal) forces applied to it, and
then apply the forces F1 and F2 at (let's say) t=0, what happens?

Before F1 and F2 were applied (i.e., at times t < 0), mass points A and B were at rest, and hence unaccelerated (with respect to an inertial reference frame). Hence, by Newton's 2nd law, we infer that the net force acting on them must be zero. That is, for t < 0, we infer that F3 = F4 = 0.

Immediately after t=0, body A hasn't yet had time to deform (compress)
under the influence of F1 and F2, so the inter-atomic distances within
body A are unchanged from what they were before F1 and F2 were applied.
Since (to an excellent approximation) the inter-atomic forces in a solid
body depend only on inter-atomic distances, these forces are thus also
(at t=0) unchanged from what they were before F1 and F2 were applied.
Hence, immediately after t=0, F3 = F4 = 0.

Therefore, immediately after t=0, there is a net force F1 acting on A
to accelerate A to the right, and a net force F2 acting on B to accelerate
B to the left. In other words, the body will compress. More precisely,
the forces will start compression waves propagating inwards from the left
and right sides of the body.

After a while (this will likely be a time measured in milliseconds for laboratory-sized bodies) we'll reach an equilibrium where all the
compression waves have damped out, every part of the body is stationary again, and the body is compressed.

Now we can argue that (once the body is in equilibrium compression)
since every part of the body is stationary and hence unaccelerated
(again, with respect to an inertial reference frame), the net force on
every part of the body -- and in particular on point A and on point B --
must be zero. That is,
F1+F3 = 0 and F2+F4 = 0
and hence (once the body is in equilibrium compression) F3 = -F1 and
F4 = -F2.

Thanks, everything is clear and correct as usual.

So, if I understand correctly, the forces F3 and F4 exist and are as
real as the forces F1 and F2.

I added to my animation the option to choose the external forces F1=+100
and F2=-100 or F1=+100 and F2=-50.

In the first case, what you described so well happens, and in the second
case, after some time, a constant acceleration to the right is reached,
during which what value will the forces F3 and F4 have?

Will the force F3 still be equal to -F1 or will it be different?

And will the force F4 still be equal to -F2 or will it be different?

Luigi Fortunati
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

In article <10g2ihg$2qmg7$1@dont-email.me>, Luigi Fortunati asked

In the animation https://www.geogebra.org/classic/uyhpg8ug,

[[...]]

I added to my animation the option to choose the external forces F1=+100
and F2=-100 or F1=+100 and F2=-50.

So, what happens in the second (asymmetric-forces) case?

To figure this out, we need a more detailed model of the system.
The usual way to model the dynamics of distributed masses like this one
is as a set of point masses interconnected by springs-with-friction.
For this system, I think a nice starting point is to model the body by
3 point masses interconnected by springs-with-friction:

m1 /\/\/\/\/\ m2 /\/\/\/\/\ m3
F1 ----> <---- F2
<---- F3 ----><---- F4 ---->

The external force F1 is applied to m1, and the external force F2 is
applied to m3. This compresses the springs, with compression forces
F3 (between m1 and m2) and F4 (between m2 and m3).

Notice that F3 and F4 are INTERNAL forces -- F3 is NOT the reaction
force of m1 on whatever external object is supplying F1, and F4 is NOT
the reaction force of m3 on whatever external object is supplying F2.

It's important to keep the signs straight:
F1 > 0 means pushing right on m1
F2 < 0 means pushing left on m3
F3 < 0 means pushing left on m1; this force comes from compressing
the m1-m2 spring, which also pushes right on m2
F4 > 0 means pushing right on m3; this force comes from compressing
the m2-m3 spring which also pushes left on m2

To recap, we're considering the case where the friction in the springs-with-friction has damped out all internal motion, i.e., the
distances between the three masses aren't changing any more. This
means that the acceleration (let's call it $a$) of each mass with
respect to an inertial reference frame is the same as that of each
other mass.

For the symmetrical case where F1+F2 = 0, (i.e., F2 = -F1), we saw
that $a = 0$ (i.e., the whole system is unaccelerated with respect
to an inertial reference frame), and Newton's 2nd law applied to m1
gave F3 = -F1, and Newton's 2nd law applied to m3 gave F4 = -F2.

Now let's investigate what happens if we *don't* assume symmetry:

Applying Newton's 2nd law to the entire system, we have
(net force acting on the entire system) = (total mass) a (1)
F1+F2 = (m1+m2+m3) a (2)
a = (F1+F2) / (m1+m2+m3) (3)

Applying Newton's 2nd law to m1, we have
F_net_on_m1 = m1 a (4)
F1+F3 = m1 a (5)
F3 = m1 a - F1 (6)
= m1 (F1+F2) / (m1+m2+m3) - F1 (7)
= (m1/(m1+m2+m3)) (F1+F2) - F1 (8)
= (m1/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2 - F1 (9)
= (m1/(m1+m2+m3) - 1) F1 + (m1/(m1+m2+m3)) F2 (10)
= - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2 (11)

Applying Newton's 2nd law to m3, we have
F_net_on_m3 = m3 a (12)
F2+F4 = m3 a (13)
F4 = m3 a - F2 (14)
= m3 (F1+F2) / (m1+m2+m3) - F2 (15)
= (m3/(m1+m2+m3)) (F1+F2) - F2 (16)
= (m3/(m1+m2+m3)) F1 + (m3/(m1+m2+m3)) F2 - F2 (17)
= (m3/(m1+m2+m3)) F1 + (m3/(m1+m2+m3) - 1) F2 (18)
= (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2 (19)

This gives us the internal forces F3 and F4 in terms of the known applied forces F1 and F2.



As a check, let's look at m2: Given the F3 and F4 calculated above,
we can figure out the net force acting on m2:
F_net_on_m2 = - F3 - F4 (recall F3 < 0 pushes right on m2,
i.e., F3 > 0 would push left on m2,
and F4 > 0 pushes left on m2) (20)
= - (- ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2)
- ((m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2) (21)
= (((m2+m3)/(m1+m2+m3)) F1 - (m1/(m1+m2+m3)) F2)
- ((m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2) (22)
= ((m2+m3)/(m1+m2+m3) - m3/(m1+m2+m3)) F1
+ (-(m1/(m1+m2+m3)) + (m1+m2)/(m1+m2+m3)) F2 (23)
= (m2/(m1+m2+m3)) F1 + (m2/(m1+m2+m3)) F2 (24)
= (m2/(m1+m2+m3)) (F1 + F2) (25)
But Newton's 2nd law applied to m2 gives
(net force acting on m2) = m2 a (26)
= m2 (F1 + F2) / (m1+m2+m3) (27)
= (m2/(m1+m2+m3)) (F1 + F2) (28)
We see that (24) and (27) match, i.e., everything is consistent.



As another check, let's see if we can recover our earlier symmetrical
result (that F3 = -F1 and F4 = -F2) if we set F2 = -F1. Taking
F2 = -F1 in (11), we have that
F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2 (29)
= - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) (- F1) (30)
= - ((m2+m3)/(m1+m2+m3)) F1 - (m1/(m1+m2+m3)) F1 (31)
= - ((m1+m2+m3)/(m1+m2+m3)) F1 (32)
= -F1 (33)
Taking F2 = -F1 in (19), we have that
F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2 (34)
= (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) (- F1) (35)
= (m3/(m1+m2+m3)) F1 + (m1+m2)/(m1+m2+m3) F1 (36)
= ((m1+m2+m3/(m1+m2+m3)) F1 (37)
= F1 = - F2 (38)
so that we have indeed recovered our earlier symmetric analysis.



Finally, now we have all the right formulas worked out to answer
Luigi's questions:

Consider first the simplest case where the three masses are equal,
m1 = m2 = m3
Then (taking F1 = +100 and F2 = -50 as specified) (11) gives
F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2
= - 2/3 F1 + 1/3 F2
= -83.333
(19) gives
F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2
= (1/3) F1 - (2/3) F2
= +66.667

Notice that there are net forces acting on each body:
F_net_on_m1 = F1+F3
= 100-83.333
= +16.667,
F_net_on_m2 = -F3 - F4
= -(-83.333) - (+66.667) = 83.333 - 66.667
= +16.667,
F_net_on_m3 = F2+F4
= -50+66.667
= +16.667,
These forces are all is nonzero, which is as it should (must!) be
because the whole system is accelerating. In fact, because we're taking
m1 = m2 = m3 and and all three masses have the same common acceleration,
it must be the case that F_net_on_m1 = F_net_on_m2 = F_net_on_m3.


On the other hand, suppose that m1=10, m2=80, and m3=10. Then (11) gives
F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2
= -0.9 F1 + 0.1 F2
= -95
and (19) gives
F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2
= 0.1 F1 - 0.9 F2
= +55
This means the net forces on the three bodies are now
F_net_on_m1 = F1 + F3 = 100-95 = +5
F_net_on_m2 = -F3 - F4 = -(-95) - (55) = 95-55 = +40
F_net_on_m3 = F2 + F4 = -50 +55 = +5
The forces are now different from the equal-mass case (which is not
surprising, since we've changed the mass of each individual body).
Notice that the forces are in the ratio
F_net_on_m1 = F_net_on_m3 = (1/8) F_net_on_m2
which just match the ratios of the masses,
m1 = m3 = (1/8) m2
This is just what we would expect from Newton's 2nd law, since all
all three masses have the same common acceleration.



So, the simple answer to Luigi's questions is "it depends on the body's internal mass distribution".
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"Totalitarianism in power invariably replaces all first-rate talents,
regardless of their sympathies, with those crackpots and fools whose
lack of intelligence and creativity is still the best guarantee of
their loyalty." -- Hannah Arendt, "The Origins of Totalitarianism"
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From Newsgroup: sci.physics.research

Il 28/11/2025 01:33, Jonathan Thornburg [remove -color to reply] <dr.j.thornburg@gmail-pink.com> ha scritto:

In article <10g2ihg$2qmg7$1@dont-email.me>, Luigi Fortunati asked

In the animation https://www.geogebra.org/classic/uyhpg8ug,

[[...]]

I added to my animation the option to choose the external forces F1=+100
and F2=-100 or F1=+100 and F2=-50.

So, what happens in the second (asymmetric-forces) case?

To figure this out, we need a more detailed model of the system.
The usual way to model the dynamics of distributed masses like this one
is as a set of point masses interconnected by springs-with-friction.
For this system, I think a nice starting point is to model the body by
3 point masses interconnected by springs-with-friction:

m1 /\/\/\/\/\ m2 /\/\/\/\/\ m3
F1 ----> <---- F2
<---- F3 ----><---- F4 ---->

The external force F1 is applied to m1, and the external force F2 is
applied to m3. This compresses the springs, with compression forces
F3 (between m1 and m2) and F4 (between m2 and m3).

Notice that F3 and F4 are INTERNAL forces -- F3 is NOT the reaction
force of m1 on whatever external object is supplying F1, and F4 is NOT
the reaction force of m3 on whatever external object is supplying F2.

It's important to keep the signs straight:
F1 > 0 means pushing right on m1
F2 < 0 means pushing left on m3
F3 < 0 means pushing left on m1; this force comes from compressing
the m1-m2 spring, which also pushes right on m2
F4 > 0 means pushing right on m3; this force comes from compressing
the m2-m3 spring which also pushes left on m2

Do these compressions push to the right with the same force as they push
to the left, even when the forces coming from the left are greater (and
not equal) to those coming from the right?

To recap, we're considering the case where the friction in the springs-with-friction has damped out all internal motion, i.e., the
distances between the three masses aren't changing any more. This
means that the acceleration (let's call it $a$) of each mass with
respect to an inertial reference frame is the same as that of each
other mass.

For the symmetrical case where F1+F2 = 0, (i.e., F2 = -F1), we saw
that $a = 0$ (i.e., the whole system is unaccelerated with respect
to an inertial reference frame), and Newton's 2nd law applied to m1
gave F3 = -F1, and Newton's 2nd law applied to m3 gave F4 = -F2.

Now let's investigate what happens if we *don't* assume symmetry:

Applying Newton's 2nd law to the entire system, we have
(net force acting on the entire system) = (total mass) a (1)
F1+F2 = (m1+m2+m3) a (2)
a = (F1+F2) / (m1+m2+m3) (3)

Applying Newton's 2nd law to m1, we have
F_net_on_m1 = m1 a (4)
F1+F3 = m1 a (5)
F3 = m1 a - F1 (6)
= m1 (F1+F2) / (m1+m2+m3) - F1 (7)
= (m1/(m1+m2+m3)) (F1+F2) - F1 (8)
= (m1/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2 - F1 (9)
= (m1/(m1+m2+m3) - 1) F1 + (m1/(m1+m2+m3)) F2 (10)
= - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2 (11)

Applying Newton's 2nd law to m3, we have
F_net_on_m3 = m3 a (12)
F2+F4 = m3 a (13)
F4 = m3 a - F2 (14)
= m3 (F1+F2) / (m1+m2+m3) - F2 (15)
= (m3/(m1+m2+m3)) (F1+F2) - F2 (16)
= (m3/(m1+m2+m3)) F1 + (m3/(m1+m2+m3)) F2 - F2 (17)
= (m3/(m1+m2+m3)) F1 + (m3/(m1+m2+m3) - 1) F2 (18)
= (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2 (19)

This gives us the internal forces F3 and F4 in terms of the known applied forces F1 and F2.

As a check, let's look at m2: Given the F3 and F4 calculated above,
we can figure out the net force acting on m2:
F_net_on_m2 = - F3 - F4 (recall F3 < 0 pushes right on m2,
i.e., F3 > 0 would push left on m2,
and F4 > 0 pushes left on m2) (20)
= - (- ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2)
- ((m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2) (21)
= (((m2+m3)/(m1+m2+m3)) F1 - (m1/(m1+m2+m3)) F2)
- ((m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2) (22)
= ((m2+m3)/(m1+m2+m3) - m3/(m1+m2+m3)) F1
+ (-(m1/(m1+m2+m3)) + (m1+m2)/(m1+m2+m3)) F2 (23)
= (m2/(m1+m2+m3)) F1 + (m2/(m1+m2+m3)) F2 (24)
= (m2/(m1+m2+m3)) (F1 + F2) (25)
But Newton's 2nd law applied to m2 gives
(net force acting on m2) = m2 a (26)
= m2 (F1 + F2) / (m1+m2+m3) (27)
= (m2/(m1+m2+m3)) (F1 + F2) (28)
We see that (24) and (27) match, i.e., everything is consistent.

As another check, let's see if we can recover our earlier symmetrical
result (that F3 = -F1 and F4 = -F2) if we set F2 = -F1. Taking
F2 = -F1 in (11), we have that
F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2 (29)
= - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) (- F1) (30)
= - ((m2+m3)/(m1+m2+m3)) F1 - (m1/(m1+m2+m3)) F1 (31)
= - ((m1+m2+m3)/(m1+m2+m3)) F1 (32)
= -F1 (33) Taking F2 = -F1 in (19), we have that
F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2 (34)
= (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) (- F1) (35)
= (m3/(m1+m2+m3)) F1 + (m1+m2)/(m1+m2+m3) F1 (36)
= ((m1+m2+m3/(m1+m2+m3)) F1 (37)
= F1 = - F2 (38)
so that we have indeed recovered our earlier symmetric analysis.

Finally, now we have all the right formulas worked out to answer
Luigi's questions:

Consider first the simplest case where the three masses are equal,
m1 = m2 = m3
Then (taking F1 = +100 and F2 = -50 as specified) (11) gives
F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2
= - 2/3 F1 + 1/3 F2
= -83.333
(19) gives
F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2
= (1/3) F1 - (2/3) F2
= +66.667

Notice that there are net forces acting on each body:
F_net_on_m1 = F1+F3
= 100-83.333
= +16.667,
F_net_on_m2 = -F3 - F4
= -(-83.333) - (+66.667) = 83.333 - 66.667
= +16.667,
F_net_on_m3 = F2+F4
= -50+66.667
= +16.667,
These forces are all is nonzero, which is as it should (must!) be
because the whole system is accelerating. In fact, because we're taking
m1 = m2 = m3 and and all three masses have the same common acceleration,
it must be the case that F_net_on_m1 = F_net_on_m2 = F_net_on_m3.

Aside from the one doubt expressed above, everything else is perfect.
I updated the animation with your values of F3=-83.33 and F4=+66.67,
instead of F3=-??? and F4=+???.

I also added body X, which is the one exerting the continuous thrust
F1=+100 on point A of body A.

Obviously, body X also receives the external reaction from body A.

Is this latter reaction (which arrives at point X of body X and not at
point A of body A) worth -100 or -83.33?

Luigi Fortunati
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From Newsgroup: sci.physics.research

Il 29/11/2025 09:08, Luigi Fortunati ha scritto:

Il 28/11/2025 01:33, Jonathan Thornburg [remove -color to reply] <dr.j.thornburg@gmail-pink.com> ha scritto:

In article <10g2ihg$2qmg7$1@dont-email.me>, Luigi Fortunati asked

In the animation https://www.geogebra.org/classic/uyhpg8ug,

[[...]]

I added to my animation the option to choose the external forces F1=+100 >>> and F2=-100 or F1=+100 and F2=-50.

So, what happens in the second (asymmetric-forces) case?

To figure this out, we need a more detailed model of the system.
The usual way to model the dynamics of distributed masses like this one
is as a set of point masses interconnected by springs-with-friction.
For this system, I think a nice starting point is to model the body by
3 point masses interconnected by springs-with-friction:

m1 /\/\/\/\/\ m2 /\/\/\/\/\ m3
F1 ----> <---- F2
<---- F3 ----><---- F4 ---->

The external force F1 is applied to m1, and the external force F2 is
applied to m3. This compresses the springs, with compression forces
F3 (between m1 and m2) and F4 (between m2 and m3).

Notice that F3 and F4 are INTERNAL forces -- F3 is NOT the reaction
force of m1 on whatever external object is supplying F1, and F4 is NOT
the reaction force of m3 on whatever external object is supplying F2.

It's important to keep the signs straight:
F1 > 0 means pushing right on m1
F2 < 0 means pushing left on m3
F3 < 0 means pushing left on m1; this force comes from compressing
the m1-m2 spring, which also pushes right on m2
F4 > 0 means pushing right on m3; this force comes from compressing
the m2-m3 spring which also pushes left on m2

Do these compressions push to the right with the same force as they push
to the left, even when the forces coming from the left are greater (and
not equal) to those coming from the right?

[[Mod. note -- 83 lines snipped here. -- jt]]

Finally, now we have all the right formulas worked out to answer
Luigi's questions:

Consider first the simplest case where the three masses are equal,
m1 = m2 = m3
Then (taking F1 = +100 and F2 = -50 as specified) (11) gives
F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2
= - 2/3 F1 + 1/3 F2
= -83.333
(19) gives
F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2
= (1/3) F1 - (2/3) F2
= +66.667

If the forces F4 and F3 inside body A are +66.667 and -83.333, then they
are not zero!

Luigi Fortunati
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