From: dr.j.thornburg@gmail-pink.com
In the article
From: "Jonathan Thornburg [remove -color to reply]"
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:
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
,
> 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
[continued in next message]
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