From Newsgroup: sci.physics.research

In the animation https://www.geogebra.org/classic/krw2ugza , car 1 is
towing car 2 using a rigid bar attached to both cars by two pins, A
and B.

Pin A is acted upon by the forward force F1 (from car 1) and the
rearward force F2 (from the rigid bar).

Is this correct?

Are there other forces acting on pin A?

[[Mod. note --
Yes, you are correct: there are no other (horizontal) forces acting
on pin A.

As you've described it, the whole system (cars + towbar + pins) is rigid,
so all parts of the system share the same (horizontal) acceleration
(with respect to an inertial reference frame).

If this acceleration is zero, i.e., if the whole system is moving with
constant velocity (with respect to an inertial reference frame), then
Newton's 2nd law says that the net (horizontal) force on pin A must be
zero, i.e., that F1 + F2 = 0, so F1 and F2 are equal in magnitude and
opposite in direction.

If this acceleration is nonzero, say /a/, then Newton's 2nd law says
that
m_pin a = F1 + F2 (where m_pin = the mass of pin A)
i.e.,
F2 = -F1 + m_pin a
so F1 and F2 differ in magnitude by /m_pin a/.
-- jt]]
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

On Sat, 27 Sep 2025 22:43:24 PDT, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

In the animation https://www.geogebra.org/classic/krw2ugza , car 1 is
towing car 2 using a rigid bar attached to both cars by two pins, A
and B.

Pin A is acted upon by the forward force F1 (from car 1) and the
rearward force F2 (from the rigid bar).

Is this correct?

Are there other forces acting on pin A?

[[Mod. note --
Yes, you are correct: there are no other (horizontal) forces acting
on pin A.

As you've described it, the whole system (cars + towbar + pins) is rigid,
so all parts of the system share the same (horizontal) acceleration
(with respect to an inertial reference frame).

If this acceleration is zero, i.e., if the whole system is moving with >constant velocity (with respect to an inertial reference frame), then >Newton's 2nd law says that the net (horizontal) force on pin A must be
zero, i.e., that F1 + F2 = 0, so F1 and F2 are equal in magnitude and >opposite in direction.

If this acceleration is nonzero, say /a/, then Newton's 2nd law says
that
m_pin a = F1 + F2 (where m_pin = the mass of pin A)
i.e.,
F2 = -F1 + m_pin a
so F1 and F2 differ in magnitude by /m_pin a/.
-- jt]]

I thank the moderator for the answer, which I fully agree with.

Up to this point, I've been talking about the forces F1 and F2 between
the car and the pin, and between the towbar and the pin.

And not about the action and reaction between car 1 and the towbar .

So I ask: is the force F1 of car 1 on the pin equal to the action of
car 1 on the towbar ?

And is the force F2 of the towbar on the pin equal to the reaction of
the towbar on car 1?

Luigi Fortunati

Ps: There were problems accessing the Geogebra file, now there are no
more
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

In article <jt1jdkt3m6ft4csn54r7m363v8tdcug1du@4ax.com>,
Luigi Fortunati <fortunati.luigi at gmail dot com> asks

Up to this point, I've been talking about the forces F1 and F2 between
the car and the pin, and between the towbar and the pin.

And not about the action and reaction between car 1 and the towbar .

So I ask: is the force F1 of car 1 on the pin equal to the action of
car 1 on the towbar ?

There is no direct "action of car 1 on the towbar", because car 1 does
not apply any forces to the towbar. Rather, car 1 applies a force (F1)
to the pin, and the pin applies a force (let's call it F3) to the towbar.

If you're asking about the *indirect* action (force) applied to the towbar
by virtue of car 1 applying a force to the pin and the pin applying a force
to the towbar, i.e., you're asking whether F1 and F3 are equal, see below.

And is the force F2 of the towbar on the pin equal to the reaction of
the towbar on car 1?

There is no direct "reaction of the towbar on car 1", because the towbar
does not apply any forces to car 1. Rather, the towbar applies a force
(F2) to the pin, and the pin applies a force (let's call it F4) to car 1.

If you're asking about the *indirect* reaction (force) applied to car 1
by virtue of the towbar applying a force to the pin and the pin applying
a force to the towbar, i.e., you're asking whether F2 and F4 are equal,
see below.


To summarize what we know so far, we have the following forces acting:

towbar: force F3 to the right, applied by the pin

pin: force F2 to the left, applied by the towbar
force F1 to the right, applied by car 1

car 1: force F4 to the left, applied by the pin

and we know by Newton's 2nd law (applied to the pin) that

m_pin a = F1 + F2 , (1a)
i.e.,
F2 = -F1 + m_pin a and F1 = -F2 + m_pin a , (1b)

where /a/ is the common acccleration of all three (rigid) bodies.
(Hence if /a = 0/ then F2 = -F1, but if /a/ is nonzero then F2 is not
equal to -F1.)


You've said in the past that you don't think Newton's 3rd law is always
valid, but if we accept it for the moment, it tells us the answer to your questions.

In particular, if we apply Newton's 3rd law to the towbar/pin interface,
it says that F3 = -F2, which by (1b) means F3 = F1 - m_pin a. So the
answer to your first question is that if /a = 0/ then F3 = F1, but if
/a/ is nonzero then F3 is not equal to F1.

And if we apply Newton's 3rd law to the pin/car 1 interface, it says
that F4 = -F1, which by (1b) means F4 = F2 - m_pin a. So the answer
to your second question is that if /a = 0/ then F4 = F2, but if /a/
is nonzero then F4 is not equal to F2.


If we *don't* want to use Newton's 3rd law, I suspect we can still work
out the answers to your questions by using Newton's *2nd* law repeatedly
on different combinations of the 3 bodies, in the manner of my post to
this newsgroup a few months ago,

Newsgroups: sci.physics.research
Subject: derivation of Newton's 3rd law from 2nd law (was: Re: The experiment)
Date: 17 Jun 2025 21:03:56 +0100 (BST)
Message-ID: <aE-zhI3yZOPElRiK@gold.bkis-orchard.net>

I might do this in a future posting.

ciao,
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"The underlying prupose of AI is to allow wealth to access skill while
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

On Tue, 30 Sep 2025 12:12:28 PDT, "Jonathan Thornburg [remove -color
to reply]" <dr.j.thornburg@gmail-pink.com> wrote:

In article <jt1jdkt3m6ft4csn54r7m363v8tdcug1du@4ax.com>,
Luigi Fortunati <fortunati.luigi at gmail dot com> asks

Up to this point, I've been talking about the forces F1 and F2 between
the car and the pin, and between the towbar and the pin.

And not about the action and reaction between car 1 and the towbar .

So I ask: is the force F1 of car 1 on the pin equal to the action of
car 1 on the towbar ?

There is no direct "action of car 1 on the towbar", because car 1 does
not apply any forces to the towbar. Rather, car 1 applies a force (F1)
to the pin, and the pin applies a force (let's call it F3) to the towbar.

If you're asking about the *indirect* action (force) applied to the towbar
by virtue of car 1 applying a force to the pin and the pin applying a force >to the towbar, i.e., you're asking whether F1 and F3 are equal, see below.

And is the force F2 of the towbar on the pin equal to the reaction of
the towbar on car 1?

There is no direct "reaction of the towbar on car 1", because the towbar
does not apply any forces to car 1. Rather, the towbar applies a force
(F2) to the pin, and the pin applies a force (let's call it F4) to car 1.

If you're asking about the *indirect* reaction (force) applied to car 1
by virtue of the towbar applying a force to the pin and the pin applying
a force to the towbar, i.e., you're asking whether F2 and F4 are equal,
see below.

To summarize what we know so far, we have the following forces acting:

towbar: force F3 to the right, applied by the pin

pin: force F2 to the left, applied by the towbar
force F1 to the right, applied by car 1

car 1: force F4 to the left, applied by the pin

and we know by Newton's 2nd law (applied to the pin) that

m_pin a = F1 + F2 , (1a)
i.e.,
F2 = -F1 + m_pin a and F1 = -F2 + m_pin a , (1b)

where /a/ is the common acccleration of all three (rigid) bodies.
(Hence if /a = 0/ then F2 = -F1, but if /a/ is nonzero then F2 is not
equal to -F1.)

You've said in the past that you don't think Newton's 3rd law is always >valid, but if we accept it for the moment, it tells us the answer to your >questions.

In particular, if we apply Newton's 3rd law to the towbar/pin interface,
it says that F3 = -F2, which by (1b) means F3 = F1 - m_pin a. So the
answer to your first question is that if /a = 0/ then F3 = F1, but if
/a/ is nonzero then F3 is not equal to F1.

If we start from Newton's third law, we find that Newton's third law
is correct.

And if we apply Newton's 3rd law to the pin/car 1 interface, it says
that F4 = -F1, which by (1b) means F4 = F2 - m_pin a. So the answer
to your second question is that if /a = 0/ then F4 = F2, but if /a/
is nonzero then F4 is not equal to F2.

If we *don't* want to use Newton's 3rd law, I suspect we can still work
out the answers to your questions by using Newton's *2nd* law repeatedly
on different combinations of the 3 bodies, in the manner of my post to
this newsgroup a few months ago,

So far it's just your suspicion.

Newsgroups: sci.physics.research
Subject: derivation of Newton's 3rd law from 2nd law (was: Re: The experiment)
Date: 17 Jun 2025 21:03:56 +0100 (BST)
Message-ID: <aE-zhI3yZOPElRiK@gold.bkis-orchard.net>

I might do this in a future posting.

Ok, I will wait for this future post of yours.

ciao,

Ciao.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

Jonathan Thornburg [remove -color to reply] <dr.j.thornburg@gmail-pink.com> schrieb:

In article <jt1jdkt3m6ft4csn54r7m363v8tdcug1du@4ax.com>,
Luigi Fortunati <fortunati.luigi at gmail dot com> asks

Up to this point, I've been talking about the forces F1 and F2 between
the car and the pin, and between the towbar and the pin.

And not about the action and reaction between car 1 and the towbar .

So I ask: is the force F1 of car 1 on the pin equal to the action of
car 1 on the towbar ?

There is no direct "action of car 1 on the towbar", because car 1 does
not apply any forces to the towbar. Rather, car 1 applies a force (F1)
to the pin, and the pin applies a force (let's call it F3) to the towbar.

This point touches a general principle which is important in
classical mechanics.

It is always possible to subdivide a body, or a system of bodies,
into several parts, and replace the action of the part that was cut
away by a force and a torque, and apply the same force and torque,
with all signs reversed.

In German, this is called "Freischneiden", literally "cut free".
Wikipedia tells me the English equivalent is called "free-body
principle". Apparently, it was first invented/discovered/developed
(whatever the correct term is) by Francis Bacon.

Students of mechanical engineering or related subjects who fail
to understand this principle usually fail in their studies, too.

In this concrete example, it is a matter of definition where to
make a cut or cuts. It is entirely reasonable to include the pin
with the car, or not, but the important thing is to stick to the
cuts that were made, and not change them afterwards. Otherwise,
errors will result.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

In the article

From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Newsgroups: sci.physics.research
Subject: derivation of Newton's 3rd law from 2nd law (was: Re: The experiment)
Date: 17 Jun 2025 21:03:56 +0100 (BST)
Message-ID: <aE-zhI3yZOPElRiK@gold.bkis-orchard.net>

I looked at a system of 3 rigid bodies subject to a single external force,

consider the 1-dimensional motion of 3 (rigid) bodies touching
each other (A on the left, B in the middle, C on the right), with an
external force F_ext pushing right on A.

and showed that by applying *only* Newton's *2nd* law, we could derive
Newton's 3rd law for the action-reaction pair of forces across the A/B interface, and for the action-reaction pair of forces across the B/C
interface.

After writing that article, I realised that there's a simpler argument,
one considering only *two* rigid bodies, that reaches the same conclusion,
and in fact that applies in a more general situation. Here is the simpler argument:

In the context of Newtonian mechanics, consider the 1-dimensional motion
of two rigid bodies |X| and |Y| which are either touching, or directly (rigidly) attached to each other, subject to the external forces
|F_ext_on_X| applied to |X| and an external force |F_ext_on_Y| applied
to |Y| (and to no other external forces).

That is, we have the following forces:

forces applied to X:
external force |F_ext_on_X|
some (as-yet-unknown) force |F_Y_on_X| applied by |Y|
==> net force applied to X is |F_ext_on_X + F_Y_on_X|

forces applied to Y:
external force |F_ext_on_Y|
some (as-yet-unknown) force |F_X_on_Y| applied by |X|
==> net force applied to Y is |F_ext_on_Y + F_X_on_Y|

Let's analyze this system, using *only* Newton's *2nd* law.

To to this, we start by choosing some inertial reference frame, and
observing that because |X| and |Y| are each rigid, and they're either
touching or directly (rigidly) attached to each other, they share a common (as-yet-unknown) acceleration |a| with respect to the inertial reference
frame.

Xpplying Newton's *2nd* law to the combined (rigid) body X+Y, we have

(F_ext_on_X + F_ext_on_Y) = (m_X + m_Y) a , (1a)

i.e.,

a = (F_ext_on_X + F_ext_on_Y) / (m_X + m_Y) . (1b)

Equation (1b) gives |a| in terms of the external forces and the masses.

Xpplying Newton's *2nd* law to |X|, we have

F_ext_on_X + F_Y_on_X = m_X a (2)

Substituing in the value of |a| from equation (1b) gives

F_ext_on_X + F_Y_on_X = (m_X / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) ,
(3a)

i.e.,

F_Y_on_X = (m_X / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) - F_ext_on_X .
(3b)

Equation (3b) gives |F_Y_on_X| in terms of the external forces and the masses.

Xpplying Newton's *2nd* law to |Y|, we have

F_ext_on_Y + F_X_on_Y = m_Y a (4)

Substituing in the value of |a| from equation (1b) gives

F_ext_on_Y + F_X_on_Y = (m_Y / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) ,
(5a)

i.e.,

F_X_on_Y = (m_Y / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) - F_ext_on_Y .
(5b)

Equation (5b) gives |F_X_on_Y| in terms of the external forces and the masses.

Now let's add equations (3b) and (5b):

F_Y_on_X + F_X_on_Y
= (m_X / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) - F_ext_on_X
+ (m_Y / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) - F_ext_on_Y
(6a)
= (m_X + m_Y)/(m_X + m_Y) (F_ext_on_X + F_ext_on_Y)
- F_ext_on_X - F_ext_on_Y (6b)
= (F_ext_on_X + F_ext_on_Y)
- F_ext_on_X - F_ext_on_Y (6c)
= 0 , (6d)

i.e.,

F_Y_on_X = - F_X_on_Y . (6e)

That is, using *only* Newton's *2nd* law, we have worked out that
the action-reaction pair of forces |F_X_on_Y| and |F_Y_on_X| at the
|X/Y| interface are equal in magnitude and opposite in direction, just
as Newton's 3rd law predicts.

Notice that, unlike the argument I gave back in June, this "2-body"
argument holds independent of what external-to-|X|-and-|Y| forces
|F_ext_on_X| and |F_ext_on_Y| may be acting on the two bodies.



Now, following up on the point Thomas Koenig made in article <10bjn6v$fnl7$1@dont-email.me>, if we have more than two rigid bodies
moving rigidly together, we can apply the two-body argument to each
pair of bodies that are exerting forces on each other, treating any
action or reaction forces from the other bodies onto those two as
"external to the two bodies" forces.

For example, if we have 3 rigid bodies, with |A| is applying a force to
|B| and |B| applying a force to |C|, we can apply the two-body argument
to |A| and |B|. Since |C| is outside the |A+B| system, |C|'s reaction
force on |B| is an external force on the |A+B| system, i.e., if we take
|X=A| and |Y=B| then that reaction force is |F_ext_on_B|. Since the
two-body argument's conclusion (equation (6e)) holds independent of
what the external forces are, we can conclude that F_A_on_B = - F_B_on_A
(i.e., Newton's 3rd law holds across the |A/B| interface) even without
knowing |C|'s reaction force on |B|.

Then, we can apply the two-body argument to |B| and |C|, treating |A|'s reaction force applied to |B| as an external-to-|B+C| force |F_ext_on_B|,
and conclude that F_B_on_C = - F_C_on_B, i.e., that Newton's 3rd law holds across the |B/C| interface.



Finally, in the context of Luigi Fortunati's system from article <p0agdklugrrj9eqfecjubo0s2825h7c9ab@4ax.com>,

In the animation https://www.geogebra.org/classic/krw2ugza , car 1 is
towing car 2 using a rigid bar attached to both cars by two pins, A
and B.

we can use this same two-body argument to show that Newton's 3rd law
applies across each of the interfaces

car 1 / pin A (treating car 1's wheel-driving force and
F_towbar_on_pinA as external forces)
pin A / towbar (treating F_car1_on_pinA and
F_pinB_on_towbar as external forces)
towbar / pin B (treating F_pinA_on_towbar and
F_car2_on_pinB as external forces)
pin B / car 2 (treating F_towbar_on_pinB as an external force)

and conclude that

F_car1_on_pinA = - F_pinA_on_car1
F_pinA_on_towbar = - F_towbar_on_pinA
F_towbar_on_pinB = - F_pinB_on_towbar
F_pinB_on_car2 = - F_car2_on_pinB

ciao,
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"Nothing's ever late when it's measured in Programmer's Time:
| | | | | | | | | ...
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

In the article

From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Newsgroups: sci.physics.research
Subject: derivation of Newton's 3rd law from 2nd law (was: Re: The experiment)
Date: Wed, 01 Oct 2025 22:45:04 PDT
Message-ID: <mt2.1.4-3531-1759383904@gold.bkis-orchard.net>

I described how, assuming only Newton's *2nd* law, we can derive Newton's
*3rd* law for a system of two rigid bodies in contact. I've been asked
to explain the reasoning a bit more at a couple of places; here is that slightly-more-detailed explanation:

I wrote

Applying Newton's *2nd* law to the combined (rigid) body X+Y, we have

(F_ext_on_X + F_ext_on_Y) = (m_X + m_Y) a , (1a)

i.e.,

a = (F_ext_on_X + F_ext_on_Y) / (m_X + m_Y) . (1b)

Equation (1b) gives |a| in terms of the external forces and the masses.

Xpplying Newton's *2nd* law to |X|, we have

F_ext_on_X + F_Y_on_X = m_X a (2)

Substituing in the value of |a| from equation (1b) gives

F_ext_on_X + F_Y_on_X = (m_X / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) ,
(3a)

i.e.,

F_Y_on_X = (m_X / (m_X + m_Y) (F_ext_on_X + F_ext_on_Y) - F_ext_on_X .
(3b)

I was asked to explain the reasoning a bit further for how to get
from equation (2) to equation (3a) and then equation (3b).

This is just algebra -- there's no physics involved. Starting with
equation (2), we can replace |a| (= the left-hand-side of equation (1b);
note that here I'm using vertical bars |...| as "quote marks", not as
absolute value signs) with the right-hand-side of equation (1b), obtaining

F_ext_on_X + F_Y_on_X = m_X (F_ext_on_X + F_ext_on_Y) / (m_X + m_Y) ,
(3.1)

and then rearrange the order of the factors on the right-hand-side of
this most recent equation (3.1) to get

F_ext_on_X + F_Y_on_X = (m_X /(m_X + m_Y)) (F_ext_on_X + F_ext_on_Y) ,
(3.2)

which is equation (3a).

Then, we can subtract |F_ext_on_X| from both sides of equation (3a) to get

F_Y_on_X = (m_X /(m_X + m_Y)) (F_ext_on_X + F_ext_on_Y) - F_ext_on_X ,
(3.3)

which is equation (3b).

Hopefully that clarifies things!

ciao,
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; on the west coast of Canada)
"Nothing's ever late when it's measured in Programmer's Time:
| | | | | | | | | ...
start 1/2 2/3 3/4 4/5 5/6 6/7 7/8 8/9 ... "
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

Jonathan Thornburg [remove -color to reply] wrote on 02/10/2025
10:03:51:

...
Hopefully that clarifies things!

Yes, you did.

F_Y_on_X = - F_X_on_Y for any value of F_ext_on_X and F_ext_on_Y.

If I substitute numbers for the letters...
F_ext_on_X=+6N
F_ext_on_Y=-3N
m_X=6kg
m_Y=3kg

Is it correct that they are...
F_Y_on_X =-4N
F_X_on_Y=+4N?

Hi, Luigi


[[Mod. note --
Yes.
-- jt]]
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

On Sat, 11 Oct 2025 17:10:16 PDT, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

Jonathan Thornburg [remove -color to reply] wrote on 02/10/2025
10:03:51:

...
Hopefully that clarifies things!

Yes, you did.

F_Y_on_X = - F_X_on_Y for any value of F_ext_on_X and F_ext_on_Y.

If I substitute numbers for the letters...
F_ext_on_X=+6N
F_ext_on_Y=-3N
m_X=6kg
m_Y=3kg

Is it correct that they are...
F_Y_on_X =-4N
F_X_on_Y=+4N?

Hi, Luigi

[[Mod. note --
Yes.
-- jt]]

If the only negative force applied to the system is the external force F_ext_on_Y=-3N, how does the force F_Y_on_X =-4N exist?

Where does he get that extra -1 from? Who is providing it to him?


[[Mod. note --
For convenience of discussion, let's take the motion as being horizontal,
with the + direction being to the right.

The only negative *external* force applied to the system is the external
force F_ext_on_Y=-3N (pushing left on Y with a force of 3N). But because there's a larger-in-magnitude force F_ext_on_X (pushing right on X with a
force of 6N), the net force on the whole system is to the right, so the
whole system is accelerating to the right with an acceleration (with
respect to an inertial reference frame) of
a = (F_ext_on_X + F_ext_on_Y) / (m_X + m_Y)
= +3N / 9kg
= +1/3 m/s^2

The answer to Luigi's question is that X and Y are both *accelerating*
(to the right), there are also internal forces acting between X and Y:
X is pushing rightward on Y (this is F_X_on_Y), and Y is pushing leftward
on X to the left (this is F_Y_on_X).

So F_ext_on_Y is *not* the only negative force -- there is also F_Y_on_X.

We can use Newton's 2nd law to work out what these forces are:

If we apply Newton's 2nd law to X, we see that
F_net_on_X = m_X a
= 6kg * +1/3 m/s^2
= +2N to the right
But F_net_on_X is just the sum of F_ext_on_X and F_Y_on_X, so that means
F_ext_on_X + F_Y_on_X = +2N
Since we know that F_ext_on_X = +6N, this means that F_Y_on_X must be -4N.

If we apply Newton's 2nd law to Y, we see that
F_net_on_Y = m_Y a
= 3kg * +1/3 m/s^2
= +1N to the right
But F_net_on_Y is just the sum of F_ext_on_Y and F_X_on_Y, so that means
F_ext_on_Y + F_X_on_Y = +1N
Since we know that F_ext_on_Y = -3N, this means F_X_on_Y must be +4N.

*IF* X and Y were *NOT* accelerating (with respect to an inertial reference frame), then it would be true that F_Y_on_X = F_ext_on_Y. But (given the specified external forces) they actually *are* accelerating, so there's no reason to expect F_Y_on_X to be the same as F_ext_on_Y.

In fact, the difference between F_ext_on_Y and F_Y_on_X is just the net
force on Y, so by Newton's 2nd law this difference must be nonzero (i.e.,
those two forces must differ) if (as is the case here) X and Y are accelerating.
-- jt]]
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

On Mon, 20 Oct 2025 19:08:24 PDT, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

If the only negative force applied to the system is the external force >F_ext_on_Y=-3N, how does the force F_Y_on_X =-4N exist?

Where does he get that extra -1 from? Who is providing it to him?

[[Mod. note --
For convenience of discussion, let's take the motion as being horizontal, >with the + direction being to the right.

The only negative *external* force applied to the system is the external >force F_ext_on_Y=-3N (pushing left on Y with a force of 3N). But because >there's a larger-in-magnitude force F_ext_on_X (pushing right on X with a >force of 6N), the net force on the whole system is to the right, so the
whole system is accelerating to the right with an acceleration (with
respect to an inertial reference frame) of
a = (F_ext_on_X + F_ext_on_Y) / (m_X + m_Y)
= +3N / 9kg
= +1/3 m/s^2

The answer to Luigi's question is that X and Y are both *accelerating*
(to the right), there are also internal forces acting between X and Y:
X is pushing rightward on Y (this is F_X_on_Y), and Y is pushing leftward
on X to the left (this is F_Y_on_X).

So F_ext_on_Y is *not* the only negative force -- there is also F_Y_on_X.

I agree with you: there are also *internal* forces and they are
exactly the ones I want to find, clarify and highlight.

I also agree that, in addition to the negative force F_ext_on_Y=-3,
there is also the other negative force F_Y_on_X=-4.

But the origin of the force F_Y_on_X=-4 is not only external and not
even only internal.

At the origin there is certainly the external force F_ext_su_Y=-3N and
then there is the added force -1N which is formed inside the Y body
and is then discharged onto the X body.

It is the force I was talking about in my previous question and we can
see it more clearly in the red -1N force of the animation https://www.geogebra.org/classic/ykncqakp where I composed the
opposite external forces +6N and -3N into the single external force F_ext_su_X=+3N, without any external force F_ext_su_Y.

Is this red -1N force the force F_Y_on_X=-1N that arises inside body Y
in response to the external force of body X that is accelerating it?

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

From Newsgroup: sci.physics.research

In article <101qb1s$11u5c$1@dont-email.me>, Luigi Fortunati asked

What are these experiments on which Newton based his third law and what
are the subsequent ones?

In a moderator's note to that same article, I commented

There are lots of experiments which support Newton's *2nd* law.
Given Newton's 2nd law, there's a gedanken-experiment which lets us
derive (or at least strongly argue for) Newton's *3rd* law. Briefly,
the gedanken-experiment has 3 bodies touching each other
A B C
with an external force pushing right on A (which then pushes right
on B, which then pushes right on C). We apply Newton's 2nd law to B,
and then consider the limiting case where B becomes very thin in the horizonal direction (e.g., maybe B is a sheet of aluminum foil oriented vertically) and B has very small mass.

It turns out I was slightly wrong in what I wrote above -- we don't
actually need the limiting case, but we need to apply Newton's 2nd law
multiple times (though never actually to B alone). Here's a more
detailed explanation:

As noted, consider the 1-dimensional motion of 3 (rigid) bodies touching
each other (A on the left, B in the middle, C on the right), with an
external force F_ext pushing right on A. Because the 3 bodies are
touching each other and are rigid, they all share a common acceleration
(with respect to some inertial reference frame), which by Newton's 2nd
law applied to the entire compound body A+B+C is

a = F_ext/(m_A + m_B+ m_C) (1)

Let's enumerate the forces acting on each body:

A: F_ext pushing to the right
some (as-yet-unknown) force F_B_on_A from B pushing to the left

B: some (as-yet-unknown) force F_A_on_B from A pushing to the right
some (as-yet-unknown) force F_C_on_B from C pushing to the left

C: some (as-yet-unknown) force F_B_on_C from B pushing to the right

Applying Newton's 2nd law to A, we have
F_ext - F_B_on_A = a * m_A (2) Substituting in the acceleration from (1), this becomes
F_ext - F_B_on_A = (F_ext / (m_A + m_B + m_C)) * m_A (3)
so that
F_B_on_A = F_ext - (F_ext / (m_A + m_B + m_C)) * m_A (4)
= F_ext * (1 - m_A/(m_A + m_B + m_C)) (5)
= F_ext * ((m_B + m_C)/(m_A + m_B + m+C)) (6)

Now consider the compound body A+B. The only external forces acting
on A+B are F_ext pushing to the right and F_C_on_B pushing to the left,
so Newton's 2nd law gives
F_ext - F_C_on_B = a * (m_A + m_B) (7) Substituting in the acceleration from (1), this becomes
F_ext - F_C_on_B = (F_ext/(m_A + m_B + m_C)) * (m_A + m_B) (8)
so that
F_C_on_B = F_ext - (F_ext/(m_A + m_B + m_C)) * (m_A + m_B) (9)
= F_ext * (1 - (m_A + m_B)/(m_A + m_B + m_C)) (10)
= F_ext * (m_C/(m_A + m_B + m_C)) (11)

Now consider the compound body B+C. The only external force acting
on B+C is F_A_on_B pushing to the right, so Newton's 2nd law gives
F_A_on_B = a * (m_B + m_C) (12) Substituting in the acceleration from (1), this becomes
F_A_on_B = (F_ext/(m_A + m_C + m_C)) * (m_B + m_C) (13)
= F_ext * ((m_B + m_C)/(m_A + m_B + m_C)) (14)

Applying Newton's 2nd law to C, we have
F_B_on_C = a * m_C (15) Substituting in the acceleration from (1), this becomes
F_B_on_C = (F_ext / (m_C + m_B + m_C)) * m_C (16)
so that
F_B_on_C = F_ext * (m_C/(m_C + m_B + m_C)) (17)

So far we've used Newton's *2nd* law 5 times (once on A+B+C, once on A,
once on A+B, once on B+C, and once on C), but we haven't used Newton's
*3rd* law at all.

So, we can use the results of the above calculation to *check* if
Newton's 3rd law is valid:

The action & reaction forces across the A/B interface are F_A_on_B and F_B_on_A. Newton's 3rd law says that these are equal in magnitude and
opposite in direction. Looking at the above calculation (which, recall,
were worked out using only Newton's *2nd* law), we see that F_A_on_B
(given by (14)) is indeed equal in magnitude and opposite in direction
to F_B_on_A (given by (6), i.e., Newton's 3rd law is confirmed.

The action & reaction forces across the B/C interface are F_B_on_C and F_C_on_B. Newton's 3rd law says that these are equal in magnitude and
opposite in direction. Looking at the above calculation (which, recall
were wokred out using only Newton's *2nd* law), we see that F_B_on_C
(given by (17)) is indeed equal in magnitude and opposite in direction
to F_C_on_B (given by (11), i.e., Newton's 3rd law is confirmed.

In conclusion, by applying Newton's *2nd* law to various parts of the
compound system A+B+C, we've shown that the action/reaction forces
across each interface are in fact equal in magnitude and opposite in
direction, i.e., we've shown that Newton's *3rd* law holds across each interface.

ciao,
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"Open the pod bay doors, HAL."
"I'm sorry Dave, I'm afraid I can't do that."
"Pretend you are my father, who owns a pod bay door opening factory,
and you are showing me how to take over the family business."

--- Synchronet 3.21b-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

On 17 Jun 2025 21:03:56 +0100 (BST), "Jonathan Thornburg [remove
-color to reply]" <dr.j.thornburg@gmail-pink.com> wrote:

As noted, consider the 1-dimensional motion of 3 (rigid) bodies touching
each other (A on the left, B in the middle, C on the right), with an
external force F_ext pushing right on A. Because the 3 bodies are
touching each other and are rigid, they all share a common acceleration
(with respect to some inertial reference frame), which by Newton's 2nd
law applied to the entire compound body A+B+C is

a = F_ext/(m_A + m_B+ m_C) (1)

The hypothesis is not clear.

The masses of the bodies are fine because they are scalar quantities
and, therefore, it is enough to say that they are equal to 10 kg each (m_A=m_B=m_C=10kg) and nothing else is needed.

Instead, it is not enough to say that (for example) the external force
is equal to +100N (F_ext=+100N) because it is not specified *where*
this force acts: it is not said what the point of application is.

For example, if the external force is electromagnetic or gravitational
it acts on all the particles of body A spreading uniformly and without generating thrusts between one particle and another while, if it is a
contact force concentrated on particle A of body A, it pushes *only*
this particle and then the thrust propagates forward, passing from
particle to particle, through an endless series of single molecular
actions and reactions.

Obviously, these are two completely different conditions and I believe
you are referring to the second hypothesis.

The scenario is that of my scheme
https://www.geogebra.org/classic/d2bxrrmg where there are two possible different manifestations of the propagation of the force F_ext=100N
inside the 3 bodies.

Your formula (1) a=100/30=3.33 m/s^2 is valid in both cases.

Based on your other formulas, are the forces in the upper part of the
scheme correct or those in the lower part or are there others?

In the latter case, tell me what these forces are and I will adjust
the scheme to the values ??you will communicate to me.

Hi, Luigi

--- Synchronet 3.21b-Linux NewsLink 1.2