From Newsgroup: sci.physics.research

Muscular father trains his son in tug of war https://drive.google.com/file/d/1JHyDVrriwjUnlFoGUaHRFfpIsHBf4gYA/view?usp=sharing

The son pulls to the left with all his strength, while the father pulls
to the right just enough to keep the rope steady: the two opposing
forces balance each other, and the rope stays steady.

The father and son's feet are firmly planted on the ground, where
there's no stupid ice that could favor one or the other.

When the father wants his son to win, he decreases his strength and the
rope moves to the left; when he wants to pull his son toward him, he
increases his strength and the rope moves to the right.

How can the son's reaction be the same as the father's action when he
pulls him towards him (and the rope accelerates to the right) or when he pretends to let the son win (and the rope accelerates to the left)?

Luigi Fortunati

[[Mod. note --
Several points:

First, the picture in the google drive shows the rope NOT being
horizontal, i.e., there are vertical components to the forces.
I'm going to simplify things by instead pretending that the rope
is horizontal (i.e., that the father lowers his hands to be at the
same height as the son's hands) so that we only need to consider
horizontal forces.

Second, we can also greatly simplifies things if we neglect any
stretching of the rope. This lets us refer to "the acceleration
of the rope" without having to worry about different parts of the
rope having different accelerations.

With these simplifications, we can analyse the system relatively
easiy: Let's start by listing the forces of interest:
F_father_on_rope = rightward pull of father on rope (> 0)
F_rope_on_father = leftward pull of rope on father (< 0)
F_son_on_rope = leftward pull of son on rope (< 0)
F_rope_on_son = rightward pull of rope on son (> 0)

The net force on the rope is just the difference between the
father's rightward pull on the rope and the son's leftward pull,
i.e.,
F_net_on_rope = F_father_on_rope + F_son_on_rope

Newton's 2nd law says that the rope accelerates (horizontally)
if and only if F_net_on_rope is nonzero, i.e., if and only if
F_father_on_rope and F_son_on_rope are different in magnitude.
More precisely, Newton's 2nd law says that
F_net_on_rope = F_father_on_rope + F_son_on_rope = m_rope a
where m_rope is the mass of the rope and /a/ is the (horizontal)
acceleration of the rope.

Newton's 3rd law only applies to bodies that directly exert forces
on each other. In this case, the father directly exerts a force on
the rope, and vice versa, so Newton's 3rd law says that
F_rope_on_father = - F_father_on_rope
And, the son directly exerts a force on the rope, and vice versa, so
Newton's 3rd law says that
F_rope_on_son = - F_son_on_rope

Because the father and son don't directly exert forces on each other,
Newton's 3rd law says nothing about the relationship between
F_father_on_rope and F_son_on_rope.

Now, consider Luigi's question

How can the son's reaction be the same as the father's action when he
pulls him towards him (and the rope accelerates to the right) or when he pretends to let the son win (and the rope accelerates to the left)?

We need to be precise about which forces we're considering.
As I noted, Newton's 3rd law says nothing about the relationship
between F_father_on_rope and F_son_on_rope. However, Newton's 2nd
law does relate those, as described above.
-- jt]]
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From Newsgroup: sci.physics.research

Luigi Fortunati wrote:

So, it's the two opposing forces on the rope that determine who wins and
who loses in a tug of war.

It is not at all true what is said, that is, that the two opposing
forces (in tug-of-war) are always equal and opposite!

In the end, the father wins if the force F_father_on_the_rope is greater than the force F_son_on_the_rope. The son wins if it's the opposite, and they tie if the two forces are equal.

That's exactly what I was getting at.

Can you confirm that this is indeed the case?

Luigi Fortunati.

[[Mod. note --
Yes, that's correct.

Newton's 3rd law tells us that the two forces
F_father_on_rope (pulling right on the rope)
and
F_rope_on_father (pulling left on the father)
*are* always of the same magnitude.

Newton's 3rd law also tells us that the two forces
F_son_on_rope (pulling left on the rope)
and
F_rope_on_son (pulling right on the son)
*are* always of the same magnitude.

But the two forces
F_father_on_rope (pulling right on the rope)
and
F_son_on_rope (pulling left on the rope),
are in general NOT of the same magnitude. None of Newton's laws
specifies which of these is larger in magnitude.
-- jt]]
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From Newsgroup: sci.physics.research

Il 20/12/2025 08:36, Luigi Fortunati ha scritto:

Luigi Fortunati wrote:

So, it's the two opposing forces on the rope that determine who wins and
who loses in a tug of war.

It is not at all true what is said, that is, that the two opposing
forces (in tug-of-war) are always equal and opposite!

In the end, the father wins if the force F_father_on_the_rope is greater
than the force F_son_on_the_rope. The son wins if it's the opposite, and
they tie if the two forces are equal.

That's exactly what I was getting at.

Can you confirm that this is indeed the case?

Luigi Fortunati.

[[Mod. note --
Yes, that's correct.

Newton's 3rd law tells us that the two forces
F_father_on_rope (pulling right on the rope)
and
F_rope_on_father (pulling left on the father)
*are* always of the same magnitude.

Newton's 3rd law also tells us that the two forces
F_son_on_rope (pulling left on the rope)
and
F_rope_on_son (pulling right on the son)
*are* always of the same magnitude.

But the two forces
F_father_on_rope (pulling right on the rope)
and
F_son_on_rope (pulling left on the rope),
are in general NOT of the same magnitude. None of Newton's laws
specifies which of these is larger in magnitude.
-- jt]]

Very clear explanation, thank you.

If there's a rope between the father and the son, the third law applies between the father and the rope, and also between the son and the rope,
but it doesn't apply to the father and son because they aren't in direct contact.

So, let's eliminate the rope and put the father and son in direct
contact, as in the image https://ibb.co/rRJ5mvMC where they push each
other directly with their own hands.

The father says to the son: "Push toward me with your hands with all
your strength," and the son pushes.

The father uses only part of his force, just enough to balance the
son's, and thus the two opposing forces are equal, and both remain in
place (even though their masses are different).

Then the father wants the son to win, and reduces his force with his
hands, so that the son's force prevails and the father retreats a little.

Afterward, the father increases his strength to surpass that of his son
and advances, while the son is forced to retreat, his strength no longer
able to counteract that of his father.

How can all this alternating opposing forces be reconciled with the
third law that dictates their eternal equality?

Luigi Fortunati
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From Newsgroup: sci.physics.research

In article <10i5vnm$1t2p1$1@dont-email.me>, Luigi Fortunati asked
about the forces in a tug-of-war where there's no rope, i.e., where
two people push or pull directly on each other. This "push-of-war"
has a father (the stronger of the two people) on the right and a son
(the weaker of the two people) on the left, each pushing on the other
with outstreatched arms (the father pushes left on the son, and the
son pushes right on the father). Luigi specified that both people's
feet are planted solidly on the ground, and don't slip.

To properly understand this system, we need to consider *all* the
forces acting on the father and the son, including the forces their
legs/feet exert on the ground, and the Newton's-3rd-law reaction
of the ground on their legs/feet.

To keep things simple, let's treat the ground as a Newtonian
inertial reference frame. Let's only consider horizontal forces
and motions, and let's ignore the changes in shape of the people's
bodies, i.e., let's treat father and son as having the *same*
(horizontal) acceleration with respect to the ground.



Let's list all the (horizontal) forces acting in this system:

The son is trying to push the father's body to the right. To resist
this, the father's feet must push *right* on the ground, with with a
force (applied by the father's leg and hip muscles) of magnitude /F_father_on_ground/.

By Newton's 3rd law, this means that the ground must react on the
father's feet with a force of equal magnitude but opposite direction,
i.e., the ground exerts a force on the father's feet of magnitude /F_father_on_ground/ pushing *left*.

Similarly, the father is trying to pull the son's body to the left.
To try to resist this, the son's feet must push *left* on the ground,
with with a force (applied by the son's leg and hip muscles) of
magnitude /F_son_on_ground/.

By Newton's 3rd law, this means that the ground must react on the
son's feet with a force of equal magnitude but opposite direction,
i.e., the ground exerts a force on the son's feet of magnitude /F_son_on_ground/ pushing *right*.

And finally, the father and son also push directly on each other:
The father pushes left on the son with a force of magnitude
/F_father_vs_son/. The *son* pushes right on the father with a
force which, by Newton's 3rd law, must be of equal magnitude (/F_father_vs_son/) but opposite direction (pulling right).



Now let's collect all the (horizontal) forces acting on each person:

Forces acting on the *father*:
* The son pushes right on the father with a force of magnitude
/F_father_vs_son/.
* The ground has a reaction force on the father, of magnitude
/F_father_on_ground/ pushing left.
So, the net force to the right acting on the father is
F_net_on_father = F_father_vs_son - F_father_on_ground (1)

Forces acting on the *son*:
* The father pushes left on the son with a force of magnitude
/F_father_vs_son/.
* The ground has a reaction force on the son, of magnitude
/F_son_on_ground/ pushing right.
So, the net force to the right acting on the son is
F_net_on_son = F_son_on_ground - F_father_vs_son (2)

Finally, the net force to the right acting on the two people is just
F_net_on_father_and_son
= F_net_on_father + F_net_on_son
= (F_father_vs_son - F_father_on_ground)
+ (F_son_on_ground - F_father_vs_son)
= F_son_on_ground - F_father_on_ground (3)



Let's first consider the case where the pull-of-war is a tie, with
both father and son stationary (and hence unaccelerated horizontally).
We'll apply Newton's 2nd law three times:

Applying Newton's 2nd law to the father, we see that the fact that
the father is unaccelerated (horizontally) means /F_net_on_father = 0/,
so by (1) we have
F_father_vs_son = F_father_on_ground (4)

Applying Newton's 2nd law to the son, we see that the fact that
the son is unaccelerated (horizontally) means /F_net_on_son = 0/,
so by (2) we have
F_father_vs_son = F_son_on_ground (5)

Applying Newton's 2nd law to the combined father+son system,
we see that the fact that the combined system is unaccelerated
(horizontally) means /F_net_on_father_and_son = 0/, so by (3) we
have
F_father_on_ground = F_son_on_ground (6)

Combining (4), (5), and (6), we have (still for the case where the
pull-of-war is a tie)
F_father_on_ground = F_father_vs_son = F_son_on_ground (7)



Now let's say the father wants to win for a while. Since the father
is stronger than the son, the father can increase F_father_on_ground
so that
F_father_on_ground > F_son_on_ground (8)
and hence by (3)
F_net_on_father_and_son < 0 (9)
and thus (by Newton's 2nd law applied to the combined father+son system)
both people accelerate to the left.

Since both people are accelerating to the left, we know (by Newton's
2nd law applied to the father) that
F_net_on_father < 0 (10)
and thus by (1) we must have
F_father_on_ground > F_father_vs_son (11)

Similarly, we know (by Newton's 2nd law applied to the son) that
F_net_on_son < 0 (12)
and thus by (2) we must have
F_father_vs_son > F_son_on_ground (13)

Combining (11) and (13), we have
F_father_on_ground > F_father_vs_son > F_son_on_ground (14)
which is also consistent with (8).

In fact, we can quantify the differences between the three forces in (14).

Suppose that the father and son are both accelerating to the right
with an acceleration /a/ (/a = 0/ means the push-of-war is a tie,
/a < 0/ means the father is winning, /a/ > 0 means the son is winning)

Newton's 2nd law applied to the combined father+son system says
F_net_on_father_and_son = (m_father+m_son) a (15)
so that by (3)
F_son_on_ground - F_father_on_ground = (m_father+m_son) a (16)
and hence
a = (F_son_on_ground - F_father_on_ground) / (m_father+m_son) (17)

Applying Newton's 2nd law to the father, we have that
F_net_on_father = m_father a (18)
so by (1) we have
F_father_vs_son - F_father_on_ground = m_father a (19)
i.e.
F_father_vs_son = F_father_on_ground + m_father a (20)
so that by (17)
F_father_vs_son
= F_father_on_ground + (m_father/(m_father+m_son))
(F_son_on_ground - F_father_on_ground) (21)

Similarly, applying Newton's 2nd law to the son, we have that
F_net_on_son = m_son a (22)
so by (2) we have
F_son_on_ground - F_father_vs_son = m_son a (23)
i.e.
F_father_vs_son = F_son_on_ground - m_son a (24)
so that by (17)
F_father_vs_son
= F_son_on_ground - (m_son/(m_father+m_son))
(F_son_on_ground - F_father_on_ground) (25)

As a check, (21) and (25) should give the same result: (21) says
F_father_vs_son
= (1 - (m_father/(m_father+m_son)) F_father_on_ground
+ (m_father/(m_father+m_son)) F_son_on_ground (26)
= (m_son /(m_father+m_son)) F_father_on_ground
+ (m_father/(m_father+m_son)) F_son_on_ground (27)
while (25) says that
F_father_vs_son
= (1 - m_son/(m_father+m_son)) F_son_on_ground
+ ( m_son/(m_father+m_son)) F_father_on_ground
= ( m_son/(m_father+m_son)) F_father_on_ground
+ (m_father/(m_father+m_son)) F_son_on_ground (29)
and we see that (27) and (29) are in fact identical.



If the father wants to loose the pull-of-war for a while, this same
analysis still applies with each '>' changed to a '<' and vice versa:
To loose, the father decreases /F_father_on_ground/ so that
F_father_on_ground < F_son_on_ground
and hence by (3) /F_net_on_father_and_son > 0/ and so both people
accelerate to the right (i.e., /a > 0/).



So, inclusion, now that we look at this system more carefully, we see
that the rope was actually a bit of a distraction. *The actual determinant
of who wins is the balance between how hard the father pushes on the
ground (the force /F_father_on_ground/) versus how hard the son pushes
on the ground (the force /F_son_on_ground/).*

If /F_Father_on_ground = F_son_on_ground/ then the pull-of-war is a
tie, and (7) says that the direct forces /F_father_vs_son/ between father
and son are necessarily equal in magnitude to the forces pushing on the
ground.

If /F_father_on_ground > F_son_on_ground/ then the father wins and
(17) gives the father's and son's acceleration to the left (/a < 0/).
In this case the direct forces /F_father_vs_son/ betwen father and
son have magnitude in between those of the father's and son's forces
on the ground, as given by (14), and in more detail by (27) and (29).

Finally, if /F_father_on_ground < F_son_on_ground/ then the son wins
and (17) gives the father's and son's acceleration to the right (a > 0).
In this case the inequalities in (14) are reversed, and we have
F_father_on_ground < F_father_vs_son < F_son_on_ground (30)
In this case the direct forces /F_father_vs_son/ betwen father and
son still have magnitude in between those of the father's and son's
forces on the ground, and are given (27) and (29).
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; on the west coast of Canada)
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From Newsgroup: sci.physics.research

Il 25/12/2025 08:28, Jonathan Thornburg [remove -color to reply] ha scritto:

In article <10i5vnm$1t2p1$1@dont-email.me>, Luigi Fortunati asked
about the forces in a tug-of-war where there's no rope, i.e., where
two people push or pull directly on each other. This "push-of-war"
has a father (the stronger of the two people) on the right and a son
(the weaker of the two people) on the left, each pushing on the other
with outstreatched arms (the father pushes left on the son, and the
son pushes right on the father). Luigi specified that both people's
feet are planted solidly on the ground, and don't slip.

To properly understand this system, we need to consider *all* the
forces acting on the father and the son, including the forces their
legs/feet exert on the ground, and the Newton's-3rd-law reaction
of the ground on their legs/feet.

To keep things simple, let's treat the ground as a Newtonian
inertial reference frame. Let's only consider horizontal forces
and motions, and let's ignore the changes in shape of the people's
bodies, i.e., let's treat father and son as having the *same*
(horizontal) acceleration with respect to the ground.

Let's list all the (horizontal) forces acting in this system:

The son is trying to push the father's body to the right. To resist
this, the father's feet must push *right* on the ground, with with a
force (applied by the father's leg and hip muscles) of magnitude /F_father_on_ground/.

By Newton's 3rd law, this means that the ground must react on the
father's feet with a force of equal magnitude but opposite direction,
i.e., the ground exerts a force on the father's feet of magnitude /F_father_on_ground/ pushing *left*.

Similarly, the father is trying to pull the son's body to the left.
To try to resist this, the son's feet must push *left* on the ground,
with with a force (applied by the son's leg and hip muscles) of
magnitude /F_son_on_ground/.

By Newton's 3rd law, this means that the ground must react on the
son's feet with a force of equal magnitude but opposite direction,
i.e., the ground exerts a force on the son's feet of magnitude /F_son_on_ground/ pushing *right*.

And finally, the father and son also push directly on each other:
The father pushes left on the son with a force of magnitude /F_father_vs_son/. The *son* pushes right on the father with a
force which, by Newton's 3rd law, must be of equal magnitude (/F_father_vs_son/) but opposite direction (pulling right).

Now let's collect all the (horizontal) forces acting on each person:

Forces acting on the *father*:
* The son pushes right on the father with a force of magnitude
/F_father_vs_son/.
* The ground has a reaction force on the father, of magnitude
/F_father_on_ground/ pushing left.
So, the net force to the right acting on the father is
F_net_on_father = F_father_vs_son - F_father_on_ground (1)

Forces acting on the *son*:
* The father pushes left on the son with a force of magnitude
/F_father_vs_son/.
* The ground has a reaction force on the son, of magnitude
/F_son_on_ground/ pushing right.
So, the net force to the right acting on the son is
F_net_on_son = F_son_on_ground - F_father_vs_son (2)

Finally, the net force to the right acting on the two people is just
F_net_on_father_and_son
= F_net_on_father + F_net_on_son
= (F_father_vs_son - F_father_on_ground)
+ (F_son_on_ground - F_father_vs_son)
= F_son_on_ground - F_father_on_ground (3)

Let's first consider the case where the pull-of-war is a tie, with
both father and son stationary (and hence unaccelerated horizontally).
We'll apply Newton's 2nd law three times:

Applying Newton's 2nd law to the father, we see that the fact that
the father is unaccelerated (horizontally) means /F_net_on_father = 0/,
so by (1) we have
F_father_vs_son = F_father_on_ground (4)

Applying Newton's 2nd law to the son, we see that the fact that
the son is unaccelerated (horizontally) means /F_net_on_son = 0/,
so by (2) we have
F_father_vs_son = F_son_on_ground (5)

Applying Newton's 2nd law to the combined father+son system,
we see that the fact that the combined system is unaccelerated
(horizontally) means /F_net_on_father_and_son = 0/, so by (3) we
have
F_father_on_ground = F_son_on_ground (6)

Combining (4), (5), and (6), we have (still for the case where the pull-of-war is a tie)
F_father_on_ground = F_father_vs_son = F_son_on_ground (7)

Now let's say the father wants to win for a while. Since the father
is stronger than the son, the father can increase F_father_on_ground
so that
F_father_on_ground > F_son_on_ground (8)

When the father increases his force F_father_on_ground, not only does
this force increase, but his force F_father_vs_son also increases.

The father cannot increase his push to the right (against the ground)
without also increasing his push to the left (against the son)!

Or not?

But the son can't further increase his force F_son_vs_father because it
was already at its maximum!

And so, the force F_father_vs_son becomes greater and no longer equal to F_son_vs_father, contrary to what Newton's third law states.

Luigi Fortunati
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From Newsgroup: sci.physics.research

We're considering a tug-of-war where (a) there's no rope, i.e., where
two people push/pull directly on each other, and (b) we've switched
the direction of forces, so each person is now *pushing* on the other.
This "push-of-war" has a father (the stronger of the two people) on
the right and a son (the weaker of the two people) on the left, each
pushing on the other, so the the father pushes *left* on the son, and
the son pushes *right* on the father. Luigi specified that both people's
feet are planted solidly on the ground, and don't slip, and we're
assuming the ground to be a Newtonian inertial reference frame.

I analyzed this system in the article

Newsgroups: sci.physics.research
Subject: Re: Tug of War
From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Date: Wed, 24 Dec 2025 23:28:13 PST
Message-ID: <mt2.1.4-66957-1766647692@gold.bkis-orchard.net>

and here (below) I'll refer to some of the numbered equations from that article.

In my analysis, I made the simplifying assumptions that
(a) we're only considering horizontal forces & motions (in particular,
the forces between father and and son are purely horizontal, with
no vertical components), and
(b) both people's bodies are approximated as rigid, with no relative
motion between different body parts (so we can unambiguously refer
to "the (horizontal) acceleration" of father and son, with no
ambiguity about different body parts having different accelerations).

In article <10imfhe$2q8c6$1@dont-email.me>, Luigi Fortunati asked

When the father increases his force F_father_on_ground, not only does
this force increase, but his force F_father_vs_son also increases.

The father cannot increase his push to the right (against the ground) without also increasing his push to the left (against the son)!

Or not?

But the son can't further increase his force F_son_vs_father because it
was already at its maximum!

And so, the force F_father_vs_son becomes greater and no longer equal to F_son_vs_father, contrary to what Newton's third law states.

Let's look at this situation a bit more:

We're starting with both people stationary (so the push-of-war is a tie).
Then my previous analysis leading up to the previous article's equation (7) applies:
F_father_on_ground = F_father_vs_son = F_son_on_ground (7)

To make this concrete, I'll consider the case
m_father = 100 kg
m_son = 50 kg
F_father_on_ground = F_father_vs_son = F_son_on_ground = 600 Newtons

Now suppose the father increases /F_father_on_ground/ (pushing right
on the ground), say to
F_father_on_ground = 630 Newtons.

Assuming that the son can't increase /F_son_on_ground/ (pushing left
on the ground) above the stationary value of 600 Newtons, then by
equation (3) of my previous article, the net force acting on the entire father+son system is
F_net_on_father_and_son
= F_son_on_ground - F_father_on_ground (3)
= 600 N - 630 N = -30 N
i.e., 30 Newtons pushing to the left.

By Newton's 2nd law, this means that the father+son center of mass will
now accelerate to the left with an acceleration of
a = F_net_on_father_and_son / (m_father+m_son)
= -30N / 150 kg
= -0.2 m/s^2
In other words (still assuming both bodies stay rigid), the son would
be pushed backwards, and his feet would start skidding backwards (left)
on the ground. (Actually, the father's body would have to distort a bit
(e.g., bending at knees or hip joints) in order to push his hands left
while keeping his feet fixed on the ground, but that's a relatively
small effect that we can reasonably neglect here.)

Applying Newton's 2nd law to the father, equation (21) of my previous
article gives
F_father_vs_son
= F_father_on_ground + (m_father/(m_father+m_son))
(F_son_on_ground - F_father_on_ground) (21)
= 630N + (100/150)(-30N)
= 610N
or equivalently (applying Newton's 2nd law to the son) equation (25) of my previous article,
F_father_vs_son
= F_son_on_ground - (m_son/(m_father+m_son))
(F_son_on_ground - F_father_on_ground) (25)
= 600N - (50/150)(-30N)
= 610N
We see that the direct force /F_father_vs_son/ that each person exerts
on the other has in fact increased increased a bit (from 600N to 610N).
Notice, however, that it hasn't increased as much as /F_father_on_ground/ increased (from 600N to 630N), so there are still net forces on each of
father and son
F_net_on_father = F_father_vs_son - F_father_on_ground
= 610N - 630N
= -20N (this is of course equal to /m_father a/)
F_net_on_son = F_son_on_ground - F_father_vs_son
= 600N - 610N
= -10N (this is of course equal to /m_son a/)



So far we've assumed that the son is able to keep his body rigid despite
the increased force from the father. But what if that's not true? E.g.,
what if the son's feet have enough friction with the ground to stay fixed,
but the son isn't able to resist the father's increased /F_father_vs_son/, i.e., the son's arms can't apply more than 600N pushing right on the son's hands? In this case there will be a net force to the left on the son's
hands, so the son's hands will move left, and the son's arms will be
forced to retract.



Finally, it should be noted that I made use of both Newton's 2nd law
and Newton's 3rd law in my analysis. In particular, I used Newton's 3rd
law to infer that /F_father_on_son/ and /F_son_on_father/ are equal in

However, I didn't actually need to use Newton's 3rd law to infer this.
In an article back in October,
Newsgroups: sci.physics.research
Subject: derivation of Newton's 3rd law from 2nd law (was: Re: The experiment)
From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Date: Wed, 01 Oct 2025 22:45:04 PDT
Message-ID: <mt2.1.4-3531-1759383904@gold.bkis-orchard.net>
I showed that if we have *any* pair of rigid bodies applying forces
to each other, we can use Newton's 2nd law to prove that derive Newton's
3rd law must apply to the forces the bodies exert on each other:

In that October article I wrote:
# 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).
#
# [[...]]
#
# 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.

Applying this result to our present push-of-war context, this shows
that assuming *only* Newton's 2nd law (and *not* assuming that Newton's
3rd law holds), we can infer that /F_father_on_son/ and /F_son_on_father/
must be equal in magnitude and opposite in direction.

So, all that's really needed for my analysis to be valid is Newton's
2nd law.
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"addiction produces tremendous profit potential for those who can exploit
it, so, in a society where money is basically God, enormous amounts of
talent and energy go into figuring out ways of creating more addicts,
and then exploiting their addictions. (This is what smart phones are
becoming). -- Paul Campos, /Lawyers, Guns, and Money/ blog, 2024-09-23
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From Newsgroup: sci.physics.research

Il 31/12/2025 03:31, Jonathan Thornburg [remove -color to reply] ha scritto:

We're considering a tug-of-war where (a) there's no rope, i.e., where
two people push/pull directly on each other, and (b) we've switched
the direction of forces, so each person is now *pushing* on the other.
This "push-of-war" has a father (the stronger of the two people) on
the right and a son (the weaker of the two people) on the left, each
pushing on the other, so the the father pushes *left* on the son, and
the son pushes *right* on the father. Luigi specified that both people's feet are planted solidly on the ground, and don't slip, and we're
assuming the ground to be a Newtonian inertial reference frame.

I analyzed this system in the article

Newsgroups: sci.physics.research
Subject: Re: Tug of War
From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Date: Wed, 24 Dec 2025 23:28:13 PST
Message-ID: <mt2.1.4-66957-1766647692@gold.bkis-orchard.net>

and here (below) I'll refer to some of the numbered equations from that article.

In my analysis, I made the simplifying assumptions that
(a) we're only considering horizontal forces & motions (in particular,
the forces between father and and son are purely horizontal, with
no vertical components), and
(b) both people's bodies are approximated as rigid, with no relative
motion between different body parts (so we can unambiguously refer
to "the (horizontal) acceleration" of father and son, with no
ambiguity about different body parts having different accelerations).

In article <10imfhe$2q8c6$1@dont-email.me>, Luigi Fortunati asked

When the father increases his force F_father_on_ground, not only does
this force increase, but his force F_father_vs_son also increases.

The father cannot increase his push to the right (against the ground)
without also increasing his push to the left (against the son)!

Or not?

But the son can't further increase his force F_son_vs_father because it
was already at its maximum!

And so, the force F_father_vs_son becomes greater and no longer equal to
F_son_vs_father, contrary to what Newton's third law states.

Let's look at this situation a bit more:

We're starting with both people stationary (so the push-of-war is a tie). Then my previous analysis leading up to the previous article's equation (7) applies:
F_father_on_ground = F_father_vs_son = F_son_on_ground (7)

To make this concrete, I'll consider the case
m_father = 100 kg
m_son = 50 kg
F_father_on_ground = F_father_vs_son = F_son_on_ground = 600 Newtons

Now suppose the father increases /F_father_on_ground/ (pushing right
on the ground), say to
F_father_on_ground = 630 Newtons.

Assuming that the son can't increase /F_son_on_ground/ (pushing left
on the ground) above the stationary value of 600 Newtons, then by
equation (3) of my previous article, the net force acting on the entire father+son system is
F_net_on_father_and_son
= F_son_on_ground - F_father_on_ground (3)
= 600 N - 630 N = -30 N
i.e., 30 Newtons pushing to the left.

By Newton's 2nd law, this means that the father+son center of mass will
now accelerate to the left with an acceleration of
a = F_net_on_father_and_son / (m_father+m_son)
= -30N / 150 kg
= -0.2 m/s^2
In other words (still assuming both bodies stay rigid), the son would
be pushed backwards, and his feet would start skidding backwards (left)
on the ground. (Actually, the father's body would have to distort a bit (e.g., bending at knees or hip joints) in order to push his hands left
while keeping his feet fixed on the ground, but that's a relatively
small effect that we can reasonably neglect here.)

Applying Newton's 2nd law to the father, equation (21) of my previous
article gives
F_father_vs_son
= F_father_on_ground + (m_father/(m_father+m_son))
(F_son_on_ground - F_father_on_ground) (21)
= 630N + (100/150)(-30N)
= 610N
or equivalently (applying Newton's 2nd law to the son) equation (25) of my previous article,
F_father_vs_son
= F_son_on_ground - (m_son/(m_father+m_son))
(F_son_on_ground - F_father_on_ground) (25)
= 600N - (50/150)(-30N)
= 610N

By the 3rd law, if the father's force on the son is equal to 610N, the
son's force on the father should also be equal to 610N.


But if the son's maximum force is 600N, who will help him increase it to
610N?


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

From Newsgroup: sci.physics.research

In article <mt2.1.4-99339-1767148287@gold.bkis-orchard.net> I wrote

We're considering a tug-of-war where (a) there's no rope, i.e., where
two people push/pull directly on each other, and (b) we've switched
the direction of forces, so each person is now *pushing* on the other.
This "push-of-war" has a father (the stronger of the two people) on
the right and a son (the weaker of the two people) on the left, each
pushing on the other, so the the father pushes *left* on the son, and
the son pushes *right* on the father. Luigi specified that both people's feet are planted solidly on the ground, and don't slip, and we're
assuming the ground to be a Newtonian inertial reference frame.

[[...]]

We're starting with both people stationary (so the push-of-war is a tie). Then my previous analysis leading up to the previous article's equation (7) applies:
F_father_on_ground = F_father_vs_son = F_son_on_ground (7)

To make this concrete, I'll consider the case
m_father = 100 kg
m_son = 50 kg
F_father_on_ground = F_father_vs_son = F_son_on_ground = 600 Newtons

Now suppose the father increases /F_father_on_ground/ (pushing right
on the ground), say to
F_father_on_ground = 630 Newtons.

[[...]]

In other words (still assuming both bodies stay rigid), the son would
be pushed backwards, and his feet would start skidding backwards (left)
on the ground. (Actually, the father's body would have to distort a bit (e.g., bending at knees or hip joints) in order to push his hands left
while keeping his feet fixed on the ground, but that's a relatively
small effect that we can reasonably neglect here.)

Applying Newton's 2nd law to the father, equation (21) of my previous
article gives
F_father_vs_son

[[...]]

= 610N

In article <10j2nl2$2egs7$1@dont-email.me>, Luigi Fortunati asks

By the 3rd law, if the father's force on the son is equal to 610N, the
son's force on the father should also be equal to 610N.

That's right.

But if the son's maximum force is 600N, who will help him increase it to 610N?

To answer this we need a finer-scale analysis. Let's idealize the son's
body as rigid torso/legs, with arms pushing (with maximum force 600N) on
hands. Since the father is pushing left on the hands with a force 610N, Newton's 3rd law does indeed stay that the son's hands push on the father
with a force equal in magnitude (610N) and opposite in direction (pushing right).

But what is the net force acting on the son's hands? The father is pushing left with a force 610N, but (we're assuming) the son's arm muscles can only push right on the son's hands with a maximum force of 600N, so there's a
net force on the son's hands of 10N pushing left. By Newton's 2nd law,
that means the son's hands must accelerate to the left.

In other words (assuming the son's feet stay stationary with respect to
the ground, and the son's legs/torso stay rigid), the son's arms must
retract, allowing the son's hands to move left (closer to the son's body).
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
"In the whole of human history across the entire planet not one deity
has volunteered Novocain. It is a telling omission."
-- Uncle Al, 11.11.2008, comment on
http://scienceblogs.com/goodmath/2008/11/evolution_produces_better_ante.php --- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

Il 02/01/2026 09:12, Jonathan Thornburg [remove -color to reply] ha scritto:

In article <mt2.1.4-99339-1767148287@gold.bkis-orchard.net> I wrote

We're considering a tug-of-war where (a) there's no rope, i.e., where
two people push/pull directly on each other, and (b) we've switched
the direction of forces, so each person is now *pushing* on the other.
This "push-of-war" has a father (the stronger of the two people) on
the right and a son (the weaker of the two people) on the left, each
pushing on the other, so the the father pushes *left* on the son, and
the son pushes *right* on the father. Luigi specified that both people's
feet are planted solidly on the ground, and don't slip, and we're
assuming the ground to be a Newtonian inertial reference frame.

[[...]]

We're starting with both people stationary (so the push-of-war is a tie).
Then my previous analysis leading up to the previous article's equation (7) >> applies:
F_father_on_ground = F_father_vs_son = F_son_on_ground (7) >>
To make this concrete, I'll consider the case
m_father = 100 kg
m_son = 50 kg
F_father_on_ground = F_father_vs_son = F_son_on_ground = 600 Newtons

Now suppose the father increases /F_father_on_ground/ (pushing right
on the ground), say to
F_father_on_ground = 630 Newtons.

[[...]]

In other words (still assuming both bodies stay rigid), the son would
be pushed backwards, and his feet would start skidding backwards (left)
on the ground. (Actually, the father's body would have to distort a bit
(e.g., bending at knees or hip joints) in order to push his hands left
while keeping his feet fixed on the ground, but that's a relatively
small effect that we can reasonably neglect here.)

Applying Newton's 2nd law to the father, equation (21) of my previous
article gives
F_father_vs_son

[[...]]

= 610N

In article <10j2nl2$2egs7$1@dont-email.me>, Luigi Fortunati asks

By the 3rd law, if the father's force on the son is equal to 610N, the
son's force on the father should also be equal to 610N.

That's right.

But if the son's maximum force is 600N, who will help him increase it to
610N?

To answer this we need a finer-scale analysis. Let's idealize the son's
body as rigid torso/legs, with arms pushing (with maximum force 600N) on hands. Since the father is pushing left on the hands with a force 610N, Newton's 3rd law does indeed stay that the son's hands push on the father with a force equal in magnitude (610N) and opposite in direction (pushing right).

But what is the net force acting on the son's hands? The father is pushing left with a force 610N, but (we're assuming) the son's arm muscles can only push right on the son's hands with a maximum force of 600N, so there's a
net force on the son's hands of 10N pushing left. By Newton's 2nd law,
that means the son's hands must accelerate to the left.

Exactly.

In fact, the resultant force acting on the son's hands is -10 N,
resulting from the opposing forces F_father_vs_son (-610 N) and
F_son_muscles (+600 N).

And the same resultant force of -10 N also acts on the father's hands, resulting from the external force F_son_vs_father (+600 N) and the
internal force F_father_muscles (-610 N).

This shows that the force F_father_vs_son (-610 N) is greater than, and
not equal to, the force F_son_vs_father (+600 N).

If there's a mistake in all this, where is it?

In other words (assuming the son's feet stay stationary with respect to
the ground, and the son's legs/torso stay rigid), the son's arms must retract, allowing the son's hands to move left (closer to the son's body).

Yes, the son's hands (and the father's) accelerate to the left because
they are both subject to the net force of -10N.

Luigi.
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From Newsgroup: sci.physics.research

In article <10jageu$12sh6$1@dont-email.me> (Sat, 03 Jan 2026 22:22:58 PST) Luigi Fortunati wrote:

And the same resultant force of -10 N also acts on the father's hands, resulting from the external force F_son_vs_father (+600 N) and the
internal force F_father_muscles (-610 N).

This shows that the force F_father_vs_son (-610 N) is greater than, and
not equal to, the force F_son_vs_father (+600 N).

If there's a mistake in all this, where is it?

Sorry for the delayed reply -- I've been sick. Catching up now.....

To respond to Luigi's question properly, we need to analyze the
biomechanics a bit more carefully. In particular, we need to drop my
previous assumption that the father's body stays rigid, and go back and
redo the analysis without that assumption.

Let's now model the father's body the same way we're already modelling
the son's body, namely, as rigid legs/torso with arms pushing on (i.e., applying a force on) on hands.

That is, we now have
* father's feet are assumed to be fixed on the ground
* father's legs/torso are assumed to be rigid
* father's arms push left on father's hands
with a force of magnitude /F_father_arms_on_father_hands/
* father's hands push left on son's hands
with a force of magnitude /F_father_hands_on_son_hands/
* son's feet are assumed to be fixed on the ground
* son's legs/torso are assumed to be rigid
* son's arms push right on son's hands
with a force of magnitude /F_son_arms_on_son_hands/
* son's hands push right on father's hands
with a force of magnitude /F_son_hands_on_father_hands/

Notice that I am *not* assuming that /F_son_hands_on_father_hands/
must necessarily be the same as /F_father_hands_on_son_hands/ (which
is what Newton's 3rd law would say).

One complication in this analysis is that if a person's hands move,
then necessarily their arms must also be in motion, but not all of
their arms have the same acceleration. The easiest way to model this
is to treat the hand-arm system as having an "effective mass" which
includes all of the hands but only a fraction of the arms, and say that
that the effective mass is the only part of their body that accelerates.
I'll do this from now on.

Applying this to the son and father, let's take
* effective mass of son's hands + arms = 5kg
* effective mass of father's hands + arms = 10kg

When the push-of-war is tied, we have
F_son_arms_on_son_hands = 600N
F_son_hands_on_father_hands = 600N
F_father_arms_on_father_hands = 600N
F_father_hands_on_son_hands = 600N
and clearly the net force on the hands is zero.

Now if the father increases /F_father_arms_on_father_hands/ to 630N,
but the son doesn't (can't) increase /F_son_arms_on_son_hands/ above 600N,
what happens? (This is basically the situation Luigi was asking about.)
We have
F_son_arms_on_son_hands = 600N
F_son_hands_on_father_hands = don't know yet
F_father_arms_on_father_hands = 630N
F_father_hands_on_son_hands = don't know yet

If we look at the son's and father's hands (and the moving parts of their arms), their combined effective mass is 15kg, and the net force acting on
them is
F_net_on_combined_hands
= F_son_arms_on_son_hands - F_father_arms_on_father_hands
= -30N
Applying Newton's 2nd law to the two hands together, we see that they accelerate with an acceleration of
a_hands = F_net/m
= -30N / 15kg = -2 m/s^2
(This is to the left, which is what we expect since the father is winning
the push-of-war.)

Now let's apply Newton's 2nd law to the son's hands:
F_net_on_son_hands = m_son_hands a_hands = 5kg (-2 m/s^2) = -10N
= F_son_arms_on_son_hands - F_father_hands_on_son_hands
= 600N - F_father_hands_on_son_hands
so we must have F_father_hands_on_son_hands = 610N.

Now let's apply Newton's 2nd law to the father's hands:
F_net_on_father_hands = m_father_hands a_hands = 10kg (-2 m/s^2) = -20N
= F_son_hands_on_F_father_hands - F_father_arms_on_father_hands
= F_son_hands_on_F_father_hands - 630N
so we must have F_son_hands_on_F_father_hands = 610N.

So, using *only* Newton's 2nd law, we conclude that in fact
F_son_hands_on_F_father_hands = F_father_hands_on_son_hands = 610N
i.e., Newton's 3rd law does indeed hold here.

ciao,
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; currently on the west coast of Canada)
Alberto Moreira (comp.arch, Jan 1999):
"Cache is, basically, a kluge generated by technological restrictions."
Donald C. Lindsay (also comp.arch, Jan 1999):
"No. It's a kluge generated by deeply fundamental restrictions,
like the speed of light."
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From Newsgroup: sci.physics.research

Il 12/01/2026 04:51, Jonathan Thornburg [remove -color to reply] ha scritto:

In article <10jageu$12sh6$1@dont-email.me> (Sat, 03 Jan 2026 22:22:58 PST) Luigi Fortunati wrote:

And the same resultant force of -10 N also acts on the father's hands,
resulting from the external force F_son_vs_father (+600 N) and the
internal force F_father_muscles (-610 N).

This shows that the force F_father_vs_son (-610 N) is greater than, and
not equal to, the force F_son_vs_father (+600 N).

If there's a mistake in all this, where is it?

Sorry for the delayed reply -- I've been sick. Catching up now.....

To respond to Luigi's question properly, we need to analyze the
biomechanics a bit more carefully. In particular, we need to drop my previous assumption that the father's body stays rigid, and go back and
redo the analysis without that assumption.

Let's now model the father's body the same way we're already modelling
the son's body, namely, as rigid legs/torso with arms pushing on (i.e., applying a force on) on hands.

That is, we now have
* father's feet are assumed to be fixed on the ground
* father's legs/torso are assumed to be rigid
* father's arms push left on father's hands
with a force of magnitude /F_father_arms_on_father_hands/
* father's hands push left on son's hands
with a force of magnitude /F_father_hands_on_son_hands/
* son's feet are assumed to be fixed on the ground
* son's legs/torso are assumed to be rigid
* son's arms push right on son's hands
with a force of magnitude /F_son_arms_on_son_hands/
* son's hands push right on father's hands
with a force of magnitude /F_son_hands_on_father_hands/

Notice that I am *not* assuming that /F_son_hands_on_father_hands/
must necessarily be the same as /F_father_hands_on_son_hands/ (which
is what Newton's 3rd law would say).

One complication in this analysis is that if a person's hands move,
then necessarily their arms must also be in motion, but not all of
their arms have the same acceleration. The easiest way to model this
is to treat the hand-arm system as having an "effective mass" which
includes all of the hands but only a fraction of the arms, and say that
that the effective mass is the only part of their body that accelerates.
I'll do this from now on.

Applying this to the son and father, let's take
* effective mass of son's hands + arms = 5kg
* effective mass of father's hands + arms = 10kg

When the push-of-war is tied, we have
F_son_arms_on_son_hands = 600N
F_son_hands_on_father_hands = 600N
F_father_arms_on_father_hands = 600N
F_father_hands_on_son_hands = 600N
and clearly the net force on the hands is zero.

Now if the father increases /F_father_arms_on_father_hands/ to 630N,
but the son doesn't (can't) increase /F_son_arms_on_son_hands/ above 600N, what happens? (This is basically the situation Luigi was asking about.)
We have
F_son_arms_on_son_hands = 600N
F_son_hands_on_father_hands = don't know yet
F_father_arms_on_father_hands = 630N
F_father_hands_on_son_hands = don't know yet

If we look at the son's and father's hands (and the moving parts of their arms), their combined effective mass is 15kg, and the net force acting on them is
F_net_on_combined_hands
= F_son_arms_on_son_hands - F_father_arms_on_father_hands
= -30N
Applying Newton's 2nd law to the two hands together, we see that they accelerate with an acceleration of
a_hands = F_net/m
= -30N / 15kg = -2 m/s^2
(This is to the left, which is what we expect since the father is winning
the push-of-war.)

Now let's apply Newton's 2nd law to the son's hands:
F_net_on_son_hands = m_son_hands a_hands = 5kg (-2 m/s^2) = -10N
= F_son_arms_on_son_hands - F_father_hands_on_son_hands
= 600N - F_father_hands_on_son_hands
so we must have F_father_hands_on_son_hands = 610N.

Now let's apply Newton's 2nd law to the father's hands:
F_net_on_father_hands = m_father_hands a_hands = 10kg (-2 m/s^2) = -20N
= F_son_hands_on_F_father_hands - F_father_arms_on_father_hands
= F_son_hands_on_F_father_hands - 630N
so we must have F_son_hands_on_F_father_hands = 610N.

How can the son increase his force on his father to +610N if the force
he's exerting (+600N) is already equal to his maximum capacity?

Ciao, Luigi.

[[Mod. note --
The son can't increase /F_son_arms_on_son_hands/. But the son doesn't
directly control /F_son_hands_on_F_father_hands/: the additional 10N by
which /F_son_hands_on_F_father_hands/ exceeds /F_son_arms_on_son_hands/
is essentially due to the inertia of the son's hands (which are accelerating
to the left with an acceleration of /a_hands/).

This is similar to how /F_father_arms_on_father_hands/ is 630N, but /F_father_hands_on_son_hands/ is only 610N -- the 20N difference is due
to the inerta of the father's hands (which are also accelerating to the
left with an acceleration of /a_hands/).
-- jt]]
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From Newsgroup: sci.physics.research

Il 15/01/2026 07:25, Luigi Fortunati ha scritto:

How can the son increase his force on his father to +610N if the force
he's exerting (+600N) is already equal to his maximum capacity?

[[Mod. note --
The son can't increase /F_son_arms_on_son_hands/. But the son doesn't directly control /F_son_hands_on_F_father_hands/: the additional 10N by
which /F_son_hands_on_F_father_hands/ exceeds /F_son_arms_on_son_hands/
is essentially due to the inertia of the son's hands (which are accelerating to the left with an acceleration of /a_hands/).

This is similar to how /F_father_arms_on_father_hands/ is 630N, but /F_father_hands_on_son_hands/ is only 610N -- the 20N difference is due
to the inerta of the father's hands (which are also accelerating to the
left with an acceleration of /a_hands/).
-- jt]]

Is the additional force +10N included in F_son_hands_on_F_father_hands
of +610N a real or apparent force?

Luigi Fortunati

[[Mod. note -- It's a real force, measured in an inertial reference frame.
-- jt]]
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From Newsgroup: sci.physics.research

Il 18/01/2026 03:00, Luigi Fortunati ha scritto:

How can the son increase his force on his father to +610N if the force
he's exerting (+600N) is already equal to his maximum capacity?

[[Mod. note --
The son can't increase /F_son_arms_on_son_hands/. But the son doesn't
directly control /F_son_hands_on_F_father_hands/: the additional 10N by
which /F_son_hands_on_F_father_hands/ exceeds /F_son_arms_on_son_hands/
is essentially due to the inertia of the son's hands (which are accelerating >> to the left with an acceleration of /a_hands/).

This is similar to how /F_father_arms_on_father_hands/ is 630N, but
/F_father_hands_on_son_hands/ is only 610N -- the 20N difference is due
to the inerta of the father's hands (which are also accelerating to the
left with an acceleration of /a_hands/).
-- jt]]

Is the additional force +10N included in F_son_hands_on_F_father_hands
of +610N a real or apparent force?

Luigi Fortunati

[[Mod. note -- It's a real force, measured in an inertial reference frame.
-- jt]]

I agree: that force is certainly real.

But it contradicts what you wrote, namely that this greater force of +10
N is "due to the inertia of the son's hands."

If it is inertia, it should be an apparent force.

In any case, since you mentioned measurements, I tried to imagine how
these opposing forces could be measured.

The most obvious way to do this is to use a dynamometer.

If we place one between the father's feet and the ground, and another
between the father and son's hands while they are still pushing with an
equal force of 600 N, we'll notice that the dynamometers are all
stationary and not accelerating.

This should convince us that they are receiving the same force from the
right and left.

Instead, when the father increases his force against the ground from
+600N to +630N and the force of his hands against those of his son from
-600N to -610N, the dynamometer at his feet remains stationary and the
one at his hands accelerates to the left from zero velocity to a
non-zero velocity of -v.

Why do the two dynamometers behave differently?

Why does the action-reaction between the father and the ground only
cause compression of the dynamometer at his feet, while the
action-reaction between the father and son's hands also causes
acceleration in addition to compression?

Luigi Fortunati
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From Newsgroup: sci.physics.research

In article <10kn92v$10o8l$1@dont-email.me>, Luigi Fortunati asks

Why does the action-reaction between the father and the ground only
cause compression of the dynamometer at his feet, while the
action-reaction between the father and son's hands also causes
acceleration in addition to compression?

The answer is that the father's feet are (we've assumed) firmly
planted on the ground, i.e., there's no relative motion between the
father's feet and the ground. So the only thing that could actually
be accelerated by the force the father is applying to push the ground
to the right is the entire Earth, and the acceleration of the entire
earth (mass ~ 6e24 kg) under a force of 630N is so small (around 1e-22
m/s^2) that we can neglect it.

In contrast, the effective mass that's being accelerated by the
father-son action-reaction is just the father's and son's hands and
part of their arms, with an effective mass of (we assumed) only 15kg,
so the acceleration is non-trivial.



Luigi also wrote (about part of the father-son action-reaction force):

Is the additional force +10N included in F_son_hands_on_F_father_hands
of +610N a real or apparent force?

Luigi Fortunati

[[Mod. note -- It's a real force, measured in an inertial reference frame. -- jt]]

I agree: that force is certainly real.

But it contradicts what you wrote, namely that this greater force of +10
N is "due to the inertia of the son's hands."

If it is inertia, it should be an apparent force.

On the contrary, inertial forces can be real forces.

As a simpler example, consider a block of mass 10kg sitting on a
sheet of ice (so that it can move horizontally with very little friction).
If I stand at the edge of the ice and push on the block with a horizontal
force of 1N, causing the block to accelerate at 0.1 m/s^2, the block's
reaction force on my hand is an inertial force, and it is a real force.
(This may be easier to imagine if we instead make the block be of mass
1000 kg, so that the block's acceleration is very small (0.001 m/s^2).)
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
on the west coast of Canada
"Now back when I worked in banking, if someone went to Barclays,
pretended to be me, borrowed UKP10,000 and legged it, that was
`impersonation', and it was the bank's money that had been stolen,
not my identity. How did things change?" -- Ross Anderson
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From Newsgroup: sci.physics.research

Il 22/01/2026 07:20, Jonathan Thornburg [remove -color to reply] ha scritto:

In article <10kn92v$10o8l$1@dont-email.me>, Luigi Fortunati asks

Why does the action-reaction between the father and the ground only
cause compression of the dynamometer at his feet, while the
action-reaction between the father and son's hands also causes
acceleration in addition to compression?

The answer is that the father's feet are (we've assumed) firmly
planted on the ground, i.e., there's no relative motion between the
father's feet and the ground. So the only thing that could actually
be accelerated by the force the father is applying to push the ground
to the right is the entire Earth, and the acceleration of the entire
earth (mass ~ 6e24 kg) under a force of 630N is so small (around 1e-22
m/s^2) that we can neglect it.

In contrast, the effective mass that's being accelerated by the
father-son action-reaction is just the father's and son's hands and
part of their arms, with an effective mass of (we assumed) only 15kg,
so the acceleration is non-trivial.

All correct, but I was talking about the acceleration of the
dynamometers and not of the feet, hands, and ground.

Why does the dynamometer in their hands remain stationary until the
father pushes it to the left with a force of -600N?

Evidently because, on the other side, the son is pushing it to the right
with the same force of +600N.

And why does the dynamometer accelerate to the left when the father
increases his force to -610N?

Evidently because the son does NOT oppose the same opposing force of
+610N, otherwise the dynamometer would remain stationary and not accelerate!

If the dynamometer accelerates to the left, it means that the force of
the father (who pushes it to the left) is greater than and not equal to,
that of the son (who pushes it to the right).

Luigi also wrote (about part of the father-son action-reaction force):

Is the additional force +10N included in F_son_hands_on_F_father_hands
of +610N a real or apparent force?

Luigi Fortunati

[[Mod. note -- It's a real force, measured in an inertial reference frame. >>> -- jt]]

I agree: that force is certainly real.

But it contradicts what you wrote, namely that this greater force of +10
N is "due to the inertia of the son's hands."

If it is inertia, it should be an apparent force.

On the contrary, inertial forces can be real forces.

As a simpler example, consider a block of mass 10kg sitting on a
sheet of ice (so that it can move horizontally with very little friction).
If I stand at the edge of the ice and push on the block with a horizontal force of 1N, causing the block to accelerate at 0.1 m/s^2, the block's reaction force on my hand is an inertial force, and it is a real force.

Exactly.

The inertia of the block on the ice, which *reacts* to your push, exerts
a real, not an apparent, force.

Therefore, the inertia of bodies is an apparent force only when there
are no external forces attempting to change their state of motion, but
it becomes a real force when the bodies react to someone attempting to accelerate them.

(This may be easier to imagine if we instead make the block be of mass
1000 kg, so that the block's acceleration is very small (0.001 m/s^2).)

This is also correct: the greater the mass of the bodies, the greater
their actual inertial reaction force.

Luigi Fortunati
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