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 il 30/03/2023 15:59:01 ha scritto:=
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).
...
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.
esLuigi Fortunati il 30/03/2023 15:59:01 ha scritto:
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=
connected to adjacent masses by short stiff springs. When the pointcal
E contacts F there is actually a short spring that reduces the theoreti=
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.
Richard Livingston il 02/04/2023 05:55:09 ha scritto:...
Luigi Fortunati il 30/03/2023 15:59:01 ha scritto:
...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
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]]
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]]
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.
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