Re: The rope

In article <100ip4l$2dm9s$1@dont-email.me>, Luigi Fortunati wrote:

In the diagram https://www.geogebra.org/classic/pnbsvfuk there is the
hand holding end A of the 2 kg rope and there is end B of the rope
holding the 15 kg bucket full of water.
>
The hand exerts the blue force +17 (upward) on the rope, and the rope
reacts with the red force -17 (downward).
>
And the rope exerts the blue force +15 (upward) on the bucket, and the
bucket reacts with the red force -15 (downward).
>
So, contrary to what Newton says, the rope does NOT exert the same
force on the hand (-17) and on the bucket (+15)!

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

[[Mod. note --
Newton's 3rd law applies separately at each location where forces are
applied:
* At the top of the top, Newton's 3rd law says the hand force on the
  rope (17 upward) is equal in magnitude and opposite in direction from
  the rope force on the hand (17 downard).
* At the bottom of the top, Newton's 3rd law says the hand force on the
  bucket (15 upward) is equal in magnitude and opposite in direction from
  the bucket force on the hand (15 downard).

In article <100qjvv$8ala$1@dont-email.me>, Luigi Fortunati then wrote:

You probably meant here that the forces +15 and -15 at the bottom of
the top are between the rope and the bucket and not between the hand
and the bucket.
 
In any case, this is all your interpretation and not that of Newton who
writes: "[the rope] will impede the progress of the stone as much as it
will promote the progress of the horse".

The key distinction is that Newton was referring to a situation where
the rope is treated as massless, with no forces acting on it except for
the pulls (tension) at each end.  In particular, Newton was considering
the case where (among other assumptions) there are no gravitational
forces acting on the rope in the direction of motion (horizontal),
whereas your system has a rope with nonzero mass, with nonzero
gravitational forces acting on the rope in the (vertical) direction
of motion.

NOTE: Everything that I'm writing here assumes that the rope has a fixed
      length, i.e., that it doesn't stretch.  If you want to replace the
      rope with a string-that-can-stretch, things get a bit more
      complicated, and I won't go there in this article.

So, for Newton the rope *always* pulls equally at both ends and never
with force 17 on one end and 15 on the other.

If there are no other (i.e., gravitational or inertial) forces acting
on the rope, that's true.  Notably, if the rope is idealized as being
massless, then there are no gravitational or inertial forces acting on
the rope, and the rope tension is indeed the same at both ends.

But, if there are gravitational and/or inertial forces acting on the rope,
then no, Newton's laws do not say that the rope tension is the same at
both ends.  More precisely, in this case Newton's *3rd* law doesn't
directly relate the tension at one end of the rope with the tension
at the other end of the rope, but Newton's *2nd* law does relate these.

There are two common cases where the tension can differ between the two
ends of the rope.  Both cases require the rope to have nonzero mass (i.e.,
if we idealize the rope as massless, then neither of these cases applies):
* The case you described, where (the rope has nonzero mass and) we are
  considering vertical forces in a gravitational field.
[It's instructive to think about what your system would
look like if your entire diagram were in a free-falling
inertial reference frame, where there are effectively no
gravitational forces, and the bucket's weight were
replaced with a spring pulling downwards with force 15.
In this case the rope tension would be the *same* (= 15)
at the top of the rope and at the bottom of the rope.]
* The other common case is where the rope (has nonzero mass and) the
  whole system is accelerating in the direction of the rope's length.
  In this case Newton's *2nd* law applied to the rope tells us that
  the tensions at the two ends of the rope must differ by m_rope * a.

And in the tug of war, are the forces on the side of team A (which are
at different points) related to those on the side of team B?

Let's assume the tug-of-war rope to be horizontal, so we can neglect
gravitational forces.  And let's treat the Earth's surface as an inertial
reference frame for horizontal motion.  Since (we are assuming that)
the rope doesn't stretch, team A, the rope, and team B, all have the
same horizontal acceleration.

If this horizontal acceleration is zero (i.e., if team A, the rope,
and team B are all stationary, or more generally if they are all in
uniform motion in the horizontal direction), then Newton's 2nd law
applied to the rope tells us that the net (horizontal) force acting
on the rope must be zero.  This force is precisely the difference
between the total force applied by team A to the rope and the total
force applied by team B to the rope.  Thus we infer that (IF team A,
the rope, and team B, are all unaccelerated in the horizontal direction)
the total force applied by team A to the rope must equal the total
force applied by team B to the rope.

If, on the other hand, the horizontal acceleration is nozero (let's
say the acceleration is in the direction away from team A and towards
team B), then Newton's 2nd law applied to the rope says that the total
force applied by team B to the rope must be larger than the total force
applied by team A to the rope.

Notice that in each of these cases, we used Newton's *2nd* law, not
Newton's 3rd law.


In general, if the rope's mass is nonzero, we have to consider *all*
the forces acting on the rope.  That is, we apply Newton's *2nd* law
to the rope:
(a) there's a force pulling on one end of the rope
(b) there's a force pulling on the other end of the rope
(c) if the rope is vertical (or more generally, non-horizontal) in
    a gravitational field, then there's a gravitational force applied
    to the rope (the fact that the gravitational force is distributed
    along the rope's length doesn't matter here)
Newton's 2nd law then says that
   F_a + F_b + F_c = m_rope*a                                     (1)

Newton's *3rd* law doesn't relate (a) and (b).  But equation (1) (which
is just Newton's *2nd* law applied on the rope) lets us relate (a) and
(b).

What Newton's 3rd law does tell us is that
* At one end of the rope, the force applied to the rope by whatever
  object is pulling on the rope (call this object X, e.g., in your
  vertical-rope example X might be the hand at the top of the rope),
  is equal in magnitude, and opposite in direction, to the force the
  rope applies to object X.
* And, at the other end of the rope, the force applied to the rope by
  whatever object is pulling on the rope there (call this object Y,
  e.g., in your vertical-rope example Y might be the bucket at the
  bottom of the rope), is equal in magnitude, and opposite in direction,
  to the force the rope applies to object Y.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
   (he/him; currently on the west coast of Canada)
   "The crisis takes a much longer time coming than you think, and then
    it happens much faster than you would have thought. -- Rudiger Dornbusch