The internal force

[Luigi Fortunati]

In my animation
https://www.geogebra.org/m/sywszege
when the stone hits the wall, it experiences the blue force directed to
the left.

During the collision, the internal particles of the stone (in tension)
exert their force towards the right on the particle in front and
towards the left on the particle behind.

Is the resultant of all internal forces a zero force or is it a force
equal and opposite to the blue force exerted by the wall on the stone?

[Luigi Fortunati]

Post by Luigi Fortunati
In my animation
https://www.geogebra.org/m/sywszege
when the stone hits the wall, it experiences the blue force directed to=
the left.
During the collision, the internal particles of the stone (in tension)
exert their force towards the right on the particle in front and
towards the left on the particle behind.
Is the resultant of all internal forces a zero force or is it a force
equal and opposite to the blue force exerted by the wall on the stone?

Regarding my animation, I based myself on the following considerations
which seem correct to me.

The stone is a set of billions of billions of particles which, in my
animation, for simplicity, I have reduced to 5 (from A to E).

It is not the stone (in its entirety) that hits the wall but only its
point E.

By the third law, the force of the stone on point F of the wall is
equal and opposite to the force of the wall on point E of the stone
(let's say it is equal to 5 in some unit).

As a result of the collision between E and F, also the points D and E
exchange action and reaction forces of magnitude 4.

Between C and D the forces are of magnitude 3, between B and C of
magnitude 2 and between A and B of magnitude 1.

Always for the third principle.

[Richard Livingston]

On Saturday, April 1, 2023 at 3:40:45=E2=80=AFPM UTC-5, Luigi Fortunati wro=

Post by Luigi Fortunati

Post by Luigi Fortunati
In my animation
https://www.geogebra.org/m/sywszege
when the stone hits the wall, it experiences the blue force directed to=

=

Post by Luigi Fortunati

Post by Luigi Fortunati
the left.
During the collision, the internal particles of the stone (in tension)=
exert their force towards the right on the particle in front and
towards the left on the particle behind.
Is the resultant of all internal forces a zero force or is it a force=
equal and opposite to the blue force exerted by the wall on the stone?

Regarding my animation, I based myself on the following considerations
which seem correct to me.
The stone is a set of billions of billions of particles which, in my
animation, for simplicity, I have reduced to 5 (from A to E).
...
Between C and D the forces are of magnitude 3, between B and C of
magnitude 2 and between A and B of magnitude 1.
Always for the third principle.

A more precise way to model this, the way it is done in computer codes
of elastic collisions, is to model the material as a fine array of masses
connected to adjacent masses by short stiff springs. When the point
E contacts F there is actually a short spring that reduces the theoretical
instantaneous stop of E to a rapid deceleration.

As E comes to a stop the spring to D compresses and causes D to slow
down, then C then B, etc.

The impulse from E colliding with F propagates back through the mass
(D to C to B to A etc) at the speed of sound in that material.

Rich L.

[Luigi Fortunati]

Post by Richard Livingston

Post by Luigi Fortunati
In my animation
https://www.geogebra.org/m/sywszege
...
Regarding my animation, I based myself on the following considerations=
which seem correct to me.
The stone is a set of billions of billions of particles which, in my
animation, for simplicity, I have reduced to 5 (from A to E).
...
Between C and D the forces are of magnitude 3, between B and C of
magnitude 2 and between A and B of magnitude 1.
Always for the third principle.

A more precise way to model this, the way it is done in computer codes
of elastic collisions, is to model the material as a fine array of mass=

es

Post by Richard Livingston
connected to adjacent masses by short stiff springs. When the point
E contacts F there is actually a short spring that reduces the theoreti=

cal

Post by Richard Livingston
instantaneous stop of E to a rapid deceleration.
As E comes to a stop the spring to D compresses and causes D to slow
down, then C then B, etc.
The impulse from E colliding with F propagates back through the mass
(D to C to B to A etc) at the speed of sound in that material.
Rich L.

I agree with you: the impulse of the collision goes leftward from
particle E to A but not with the same intensity.

Is it correct to say that the exchange of forces decreases in the
proportion of 5 between F and E, 4 between E and D, 3 between D and C,
2 between C and B and 1 between B and A?

Luigi Fortunati

[[Mod. note --
1. You seem to be using the word "impulse" in a way different from how
it's used in physics ( https://en.wikipedia.org/wiki/Impulse_(physics) ).
2. I don't quite know what you mean by the term "exchange of forces".
But if we interpret your question as asking if the forces
F_FE = the force F exerts on E
F_ED = the force E exerts on D
F_DC = the force D exerts on C
F_CB = the force C exerts on B
F_BA = the force B exerts on A
are in the ratio 5:4:3:2:1, then we can apply Newton's 2nd law to try
to find out. To do this, let's introduce a bit of terminology: Let's
* we'll call the entire ball (the combined system A+B+C+D+E) "ABCDE"
* we'll call the combined system A+B+C+D "ABCD"
* we'll call the combined system A+B+C "ABC"
* we'll call the combined system A+B "AB"
Now we observe that
* the net (horizontal) force acting on ABCDE is F_FE, i.e.,
F_FE is the only (horizontal) external force acting on ABCDE
* the net (horizontal) force acting on ABCD is F_ED, i.e.,
F_ED is the only (horizontal) external force acting on ABCD
* the net (horizontal) force acting on ABC is F_DC, i.e.,
F_DC is the only (horizontal) external force acting on ABC
* the net (horizontal) force acting on AB is F_CB, i.e.,
F_CB is the only (horizontal) external force acting on AB
* the net (horizontal) force acting on A is F_BA, i.e.,
F_BA is the only (horizontal) external force acting on A
* the net (horizontal) force on A is F_BA
If we assume A, B, C, D, and E to all have the same mass m, then
* the mass of ABCDE is m_ABCDE = 5m
* the mass of ABCD is m_ABCD = 4m
* the mass of ABC is m_ABC = 3m
* the mass of AB is m_AB = 2m
* the mass of A is m_A = m
Since the masses m_ABCDE:m_ABCD:m_ABC:m_AB:m_A are in the ratio
5:4:3:2:1, by F_net=ma the net forces will be in the same ratio
if and only if the accelerations of the center-of-masses of
ABCDE, ABCD, ABC, AB, and A are all the same. But in reality these
accelerations *won't* all be the same -- the springs will compress
by different amounts under the different forces, so the balls won't
all accelerate at the same rate, and the center-of-mass accelerations
of ABCDE, ABCD, ABC, AB, and A *won't* all be the same. So, we
conclude that the forces F_FE:F_ED:F_DC:F_CB:F_AB *aren't* in the
ratio 5:4:3:2:1. Figuring out the actual (time-dependent) forces
requires a more careful analysis of the dynamics of the collision.
-- jt]]

[Luigi Fortunati]

Post by Luigi Fortunati

Post by Luigi Fortunati
In my animation
https://www.geogebra.org/m/sywszege
...

[[Mod. note --
1. You seem to be using the word "impulse" in a way different from how
it's used in physics ( https://en.wikipedia.org/wiki/Impulse_(physics) ).
2. I don't quite know what you mean by the term "exchange of forces".
But if we interpret your question as asking if the forces
F_FE = the force F exerts on E
F_ED = the force E exerts on D
F_DC = the force D exerts on C
F_CB = the force C exerts on B
F_BA = the force B exerts on A
are in the ratio 5:4:3:2:1, then we can apply Newton's 2nd law to try
to find out. To do this, let's introduce a bit of terminology: Let's
* we'll call the entire ball (the combined system A+B+C+D+E) "ABCDE"
* we'll call the combined system A+B+C+D "ABCD"
* we'll call the combined system A+B+C "ABC"
* we'll call the combined system A+B "AB"
Now we observe that
* the net (horizontal) force acting on ABCDE is F_FE, i.e.,
F_FE is the only (horizontal) external force acting on ABCDE
* the net (horizontal) force acting on ABCD is F_ED, i.e.,
F_ED is the only (horizontal) external force acting on ABCD
* the net (horizontal) force acting on ABC is F_DC, i.e.,
F_DC is the only (horizontal) external force acting on ABC
* the net (horizontal) force acting on AB is F_CB, i.e.,
F_CB is the only (horizontal) external force acting on AB
* the net (horizontal) force acting on A is F_BA, i.e.,
F_BA is the only (horizontal) external force acting on A
* the net (horizontal) force on A is F_BA
If we assume A, B, C, D, and E to all have the same mass m, then
* the mass of ABCDE is m_ABCDE = 5m
* the mass of ABCD is m_ABCD = 4m
* the mass of ABC is m_ABC = 3m
* the mass of AB is m_AB = 2m
* the mass of A is m_A = m
Since the masses m_ABCDE:m_ABCD:m_ABC:m_AB:m_A are in the ratio
5:4:3:2:1, by F_net=ma the net forces will be in the same ratio
if and only if the accelerations of the center-of-masses of
ABCDE, ABCD, ABC, AB, and A are all the same. But in reality these
accelerations *won't* all be the same -- the springs will compress
by different amounts under the different forces, so the balls won't
all accelerate at the same rate, and the center-of-mass accelerations
of ABCDE, ABCD, ABC, AB, and A *won't* all be the same. So, we
conclude that the forces F_FE:F_ED:F_DC:F_CB:F_AB *aren't* in the
ratio 5:4:3:2:1. Figuring out the actual (time-dependent) forces
requires a more careful analysis of the dynamics of the collision.
-- jt]]

The stone is not a point and it is not even an elementary particle: it
is a set of tiny adjacent masses connected by short stiff springs, as
Richard Livingston rightly pointed out.

It is not the stone that arrives on the wall but it is the E particle
that arrives on the F particle.

Particles from A to D do not interact at all on the wall, they merely
exert their force on another particle of the stone itself, which is why
I spoke of *internal forces*.

[Luigi Fortunati]

In the previous animation there was a collision between the stone and
the wall.

In my new animation
https://www.geogebra.org/m/eg5gbtua
the body AE is stationary because it is placed on the floor.

By the third law, the *internal* forces of action and reaction are as
given: the sum of the blue forces (directed downwards) is equal to 15
(1+2+3+4+5) and the sum of the red forces (directed up) is equal to 10
(1+2+3+4).

The resultant of the internal forces is, therefore, a blue force 5,
directed downwards.

The red force 5 between the floor and particle A is not a force
internal to the body AE.

It's correct?

[Richard Livingston]

...

Post by Luigi Fortunati

Post by Richard Livingston
A more precise way to model this, the way it is done in computer codes
of elastic collisions, is to model the material as a fine array of mass

...

Post by Luigi Fortunati

Post by Richard Livingston
Rich L.

I agree with you: the impulse of the collision goes leftward from
particle E to A but not with the same intensity.
Is it correct to say that the exchange of forces decreases in the
proportion of 5 between F and E, 4 between E and D, 3 between D and C,
2 between C and B and 1 between B and A?
Luigi Fortunati
[[Mod. note --
...
2. I don't quite know what you mean by the term "exchange of forces".
But if we interpret your question as asking if the forces
F_FE = the force F exerts on E
F_ED = the force E exerts on D
F_DC = the force D exerts on C
F_CB = the force C exerts on B
F_BA = the force B exerts on A
are in the ratio 5:4:3:2:1, then we can apply Newton's 2nd law to try
to find out. To do this, let's introduce a bit of terminology: Let's
...
ratio 5:4:3:2:1. Figuring out the actual (time-dependent) forces
requires a more careful analysis of the dynamics of the collision.
-- jt]]

As Luigi has modeled it, the ball is a one dimensional transmission
line. Object F applies an impulse on E causing the "spring" between
them to compress and bringing E to a stop. Then E applies an
impulse on D compressing that spring and bringing D to a stop.
etc. etc. When the impulse reaches A there is nothing beyond
A to apply an impulse to. As a result A not only stops but reverses
direction. This stretches the "spring" between A and B causing B
to reverse direction, pulling on C, etc. The final result is that the
ball bounces off the wall and flies off to the left.

In answer to Luigi's question, all the impulses are the same
magnitude, just applied at different times per the speed of
sound.

Read up on transmission lines for a more detailed understanding.

Rich L.

[Luigi Fortunati]

Post by Richard Livingston
As Luigi has modeled it, the ball is a one dimensional transmission
line. Object F applies an impulse on E causing the "spring" between
them to compress and bringing E to a stop. Then E applies an
impulse on D compressing that spring and bringing D to a stop.
etc. etc. When the impulse reaches A there is nothing beyond
A to apply an impulse to. As a result A not only stops but reverses
direction. This stretches the "spring" between A and B causing B
to reverse direction, pulling on C, etc. The final result is that the
ball bounces off the wall and flies off to the left.
In answer to Luigi's question, all the impulses are the same
magnitude, just applied at different times per the speed of
sound.
Read up on transmission lines for a more detailed understanding.
Rich L.

You turned the rock of my animation
https://www.geogebra.org/m/sywszege
in a ball and that's okay.

And you're right, pushes to the right are equal to pushes to the left.

Particle E is the first to be stopped by particle F, while particle D
continues to advance to the right, slowing down and compressing the
spring DE which acquires a force it did not have before.

Then it is the turn of the particle C which advances by slowing down and
compressing the spring CD.

Then particle B does the same and lastly particle A which compresses the
spring AB.

When particle A stops, the compression of the springs is at its maximum:
there is a series of internal forces which did not exist before and
which are now ready to push from one side to the other.

The spring force DE pushes E to the right and pushes D to the left, the
force CD pushes D to the right and pushes C to the left, and so on.

They are internal forces of the ball that (globally) push it to the
right and to the left (what you call "stretch of the spring").

The pushto the right has opposition, push to the left does not.

But the forces must have an outlet and, therefore, the forces to the
right are exhausted with compression (without acceleration) and those to
the left with acceleration (without compression).

Luigi Fortunati