Action and reaction

[Luigi Fortunati]

In my animation
https://www.geogebra.org/m/jcd2m2xf
there is a spring between a trolley and a wall.

At point P we apply a blue force to the right with our hand, so that
the trolley starts moving at a constant speed v.

On the other side of point P, there is the spring that contracts and
reacts with a red opposing force, which has the same direction and
intensity as the blue force but the opposite way.

Is it correct to say that the blue and red forces are the action and
reaction at the point P (between the hand and the spring) of which the
third principle speaks?

[[Mod. note --

First, two preliminary points:

1. Your description suggests that the trolley begins at rest and
instantaneously starts moving when the animation starts. That
(an instantaneous jump in velocity) can't happen. Since we're trying
to focus on the various forces acting at the (moving) point A as the
trolley moves, let's instead say that the trolley was already in motion
before your animation starts, and that during the duration of your
animation the (blue) applied force is adjusted so as to maintain the
trolley in uniform (constant-velocity) motion to the right.

2. The the coloring of the red and blue force arrows in the animation
seems to be interchanged from the coloring in your description. I'll
follow the usage of your description. That is, your description says
that we apply a (blue) force on the trolley in the *rightward* direction
with our hand (so as to maintain the trolley in uniform motion to the
right. But your animation shows the rightward force in red. Your
description correctly says that the compressed spring exerts a (red)
force on the trolley in the *leftward* direction. But the animation
shows the leftward force in blue. For the rest of my discussion I'll
follow the coloring of your description.

Now to your question: are the blue and red forces a Newton's-3rd-law
action-reaction pair?

No, they are not, because they act on the same object (the trolley),
whereas Newton's-3rd-law action-reaction forces always act on
*different* objects. [Recall that for any two bodies X and Y, if X
applies a force to Y, then the Newton's-3rd-law "reaction" force is Y
applying a force (of equal magnitude but opposite direction) to X.]

In your animation, during the period of the animation there are
actually 3 "interesting" objects in motion (the hand, the trolley,
and the left end of the spring), each of which has forces acting on it.

Let's list the various Newton's-3rd-law action-reaction forces here:
(A) The hand applies a "blue" rightward force to the trolley
(B) the Newton's-3rd-law reaction to force (A) is that the trolley
applies a leftward force to the hand
(C) the left end of the compressed spring applies a "red" leftward force
to the trolley
(D) the Newton's-3rd-law reaction to force (C) is that the trolley
applies a rightward force to the end of the spring

It's useful to draw free-body diagrams of each of the 3 objects here:

HAND
<----------*---------->
The leftward force is (B), applied by the trolley to the hand.
The rightward force is the force which is (i.e., which must be)
applied by the person's arm to the hand to keep the hand moving with
uniform speed to the right.


TROLLEY
<----------*---------->
The leftward force is the "red" force (C) applied by
the end of the spring to the trolley.
The rightward force is the "blue" force (A) applied by the hand
to the trolley.


LEFT END OF SPRING
<----------*---------->
The leftward force is the force applied to the left end of the spring
by the rest of the spring as the spring is compressed.
The rightward force is (D), applied by the trolley to the left end
of the spring.


Notice that for each of these 3 objects, there are two forces applied
to the object, which are of the same magnitude but opposite in direction,
so the net force acting on the object is zero. Hence each of these 3
objects moves at a uniform velocity.
-- jt]]

[Luigi Fortunati]

Post by Luigi Fortunati
[[Mod. note --
Now to your question: are the blue and red forces a Newton's-3rd-law
action-reaction pair?
No, they are not, because they act on the same object (the trolley),
whereas Newton's-3rd-law action-reaction forces always act on
*different* objects.

You are right, the presence of the trolley implies the presence of 3
bodies instead of 2.

So, the animation I made is not good and I changed it like that
<https://www.geogebra.org/m/an3veznf>
eliminating the trolley.

Now, the tip of the finger acts directly against the tip of the spring
(which reacts).

Post by Luigi Fortunati
Your description suggests that the trolley begins at rest and
instantaneously starts moving when the animation starts. That
(an instantaneous jump in velocity) can't happen.

You're right.

The change of speed from zero to v cannot be instantaneous and, then,
let's say that it happens in a hundredth of a second (any other time
interval is fine too).

I would like to make the animation (necessarily in slow motion) with
this initial acceleration but I need confirmation: in this time interval
is it correct that the blue force is greater than the red one?

[Jonathan Thornburg [remove -color to reply]]

Luigi Fortunati <***@gmail.com> wrote:
[[previous animation of finger pushing trolley, which pushes on
a spring, compressing the spring]]

Post by Luigi Fortunati
So, the animation I made is not good and I changed it like that
<https://www.geogebra.org/m/an3veznf>
eliminating the trolley.
Now, the tip of the finger acts directly against the tip of the spring
(which reacts).

Post by Luigi Fortunati
Your description suggests that the trolley begins at rest and
instantaneously starts moving when the animation starts. That
(an instantaneous jump in velocity) can't happen.

You're right.
The change of speed from zero to v cannot be instantaneous and, then,
let's say that it happens in a hundredth of a second (any other time
interval is fine too).
I would like to make the animation (necessarily in slow motion) with
this initial acceleration but I need confirmation: in this time interval
is it correct that the blue force is greater than the red one?

There are 3 forces of particular interest here:

A: The finger exerts a rightward force on the point P which is at the
end of the spring.

B: The Newton's-3rd-law reaction to force A is that the end of the
spring exerts a leftward force on the hand. By Newton's 3rd law,
this force B has the same magnitude as force A.

C: If the spring is compressed, the body of the spring exerts a leftward
force on the end of the spring (the point P). Assuming that the spring
obeys Hooke's law, this force C is proportional to the distance the
spring is compressed. Notice that this means that force C depends on
the distance the spring is compressed, but NOT directly on the force A.

The net force on the end of the spring (point P) is the signed sum of A
and C. That is, since A acts to the right and C acts to the left, the
net force to the right acting on P is A - C. This means that by Newton's
2nd law,

(rightward acceleration of P) = (A - C)/M

where M is the "effective mass" of the spring (not all of the spring's
mass moves with the same acceleration, so the "effective mass" is going
to be something like 1/2 the actual mass of the spring).

Initially the spring isn't compressed, so C = 0. Therefore, to
initially accelerate the end of the spring (point P) to the right, A
must be positive. As the spring begins to compress, C will increase,
and in order to continue accelerating point P to the right, A - C must
remain positive.

Once the end of the spring (point P) has reached a suitable velocity,
A should be decreased to equal C, so that A - C = 0. To maintain P at
a constant velocity, as the spring compresses and C increases, A should
be increased so that A continues to equal C.

Thus a free-body diagram of the end of the spring (point P) at the
initial time (just after the force A is "turned on" would look like this:

P -----------> (right arrow = A, force C = 0)

Somewhat later, but still during the initial acceleration phase, the
free-body diagram might look like this (I've chosen A to increase so
that A - C is a constant during the acceleration phase, giving a constant
acceleration of point P):

<---- P ---------------> (right arrow = A, left arrow = C)

During the later constant-velocity phase, the free-body diagram might
look like this:

<------- P -------> (right arrow = A, left arrow = C)

For your animation, the motion of the end of the spring (point P) should
be governed by Newton's 2nd law, i.e., at each time step in your animation,

C = determined by how far the spring is compressed
A = you get to choose this to give whatever acceleration profile
for P you want
(rightward acceleration of P) = (A - C)/M

Note that B doesn't appear in the dynamics of point P because it doesn't
act on point P. Rather, B acts on the finger, and we can assume that
the arm/hand exerts whatever forces on the finger are needed so that the
finger's motion is whatever we desire.

As to your question (about equality or inequality of the red & blue
forces in your animation), I'm a bit confused by the arrows in your
animation: Your animation shows a red arrow labelled "Finger" pointing
to the right, but the legend below the animation shows "Finger action"
with a blue arrow pointing to the right. And the animation shows a
blue arrow labelled "Spring" pointing to the left, but the legend shows
"Spring reaction" with a red arrow pointing to the left.

What we can say for sure is that B = A (these are a Newton's-3rd-law
action-reaction pair), while C varies with the compression of the spring.
Remember, A and C both act on the end of the spring (point P), whereas
B acts on the finger.

For a future version of your animation, it might be useful to use 3
different colors for the 3 forces I've described as A, B, and C. It
might also be useful to show a free-body diagram of the end of the spring
(point P) at each time step.

--
-- "Jonathan Thornburg [remove -color to reply]" <***@gmail-pink.com>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently on the west coast of Canada
"Why would we install sewers in London? Everyone keeps getting cholera
again and again so there's obviously no reason to install sewers. We
just need to get used to this as the new normal."
-- 2022-Jul-25 tweet by "Neoliberal John Snow"

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] domenica 04/09/2022 alle

Post by Jonathan Thornburg [remove -color to reply]
[[previous animation of finger pushing trolley, which pushes on
a spring, compressing the spring]]

Post by Luigi Fortunati
So, the animation I made is not good and I changed it like that
<https://www.geogebra.org/m/an3veznf>
eliminating the trolley.

As to your question (about equality or inequality of the red & blue
forces in your animation), I'm a bit confused by the arrows in your
animation: Your animation shows a red arrow labelled "Finger" pointing
to the right, but the legend below the animation shows "Finger action"
with a blue arrow pointing to the right. And the animation shows a
blue arrow labelled "Spring" pointing to the left, but the legend shows
"Spring reaction" with a red arrow pointing to the left.

There is some bug in the Geogebra program that inverts the colors.

If you reload the program with the right button, the colors are reset
with the blue arrow going to the right and the red arrow to the left.

I will answer all the rest of your very interesting post after careful
consideration.

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] domenica 04/09/2022 alle

Post by Jonathan Thornburg [remove -color to reply]
[[previous animation of finger pushing trolley, which pushes on
a spring, compressing the spring]]

Post by Luigi Fortunati
So, the animation I made is not good and I changed it like that
<https://www.geogebra.org/m/an3veznf>
eliminating the trolley.

A: The finger exerts a rightward force on the point P which is at the
end of the spring.
B: The Newton's-3rd-law reaction to force A is that the end of the
spring exerts a leftward force on the hand. By Newton's 3rd law,
this force B has the same magnitude as force A.
C: If the spring is compressed, the body of the spring exerts a leftward
force on the end of the spring (the point P). Assuming that the spring
obeys Hooke's law, this force C is proportional to the distance the
spring is compressed. Notice that this means that force C depends on
the distance the spring is compressed, but NOT directly on the force A.

Reflecting on what you wrote to me, I believe that my mistake is to
have believed that the forces acted on a point, in this case on point
P.

Instead the forces act on the bodies.

I made the new animation
https://www.geogebra.org/m/mxkeks7j
in which I enlarged the ends of the finger and the spring until you can
see the extreme atom of the finger acting against the extreme atom of
the spring.

The spring atom accelerates to the right because the force it receives
from the left is greater than that it receives from the right.

But the forces are only two (atom against atom) and not 3.

Is the animation (which concerns only the initial slow motion phase)
correct?