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### Luigi Fortunati

Feb 18, 2023, 3:37:53â€¯AMFeb 18
to
What is the difference between the force accelerating the mass (F=ma)
and the force deforming the mass (Hooke)?

Can a force accelerate mass without deforming it?

[[Mod. note --
1. Deformation can be quasi-static or dynamic, whereas acceleration
is necessarily dynamic.
2. That depends on the force and the body-being-accelerated. If the
force is somehow applied equally to each part of the body (e.g.,
a uniform gravitational field in the Newtonian perspective), then
the body can be accelerated without any deformation. Or, if the
body is either very small or very stiff, and/or the acceleration
is very small, then the deformation may be negligibly small.
An important limiting case of this is the acceleration of a point
mass, which we define as a mass with no internal structure; a point
mass doesn't deform under acceleration. Electrons are a well-known
example. But if a force is applied to a macroscopic body, and is
*not* applied equally to each part of the body, then yes, the body
will deform.
-- jt]]

### Stefan Ram

Feb 19, 2023, 10:01:26â€¯AMFeb 19
to
Luigi Fortunati <fortuna...@gmail.com> writes:
>What is the difference between the force accelerating the mass (F=ma)
>and the force deforming the mass (Hooke)?

It's the same force, but applied to two different systems.

One system is a point of mass "m" that is not held in position.
When we call "F" the force acting on it, then the acceleration
of the point is "a".

When applying one version of Hooke's law, the force F is now
applied to one end of an elastic wire the other end of which
is fixed, and then the relative elongation e is proportional
to the force F (if the force if not too strong.).

So, when applied to one system, the force is proportional
to the acceleration, when applied to another system,
to the elongation.

>Can a force accelerate mass without deforming it?

The two laws just discussed had a simplified concept of a force:
they talked about a force applied at just one point: at a point
mass or at one end of a wire. When we continue to use this simplified
concept we can now apply a force to the end of an elastic wire which
is not held in position.

In this case, there should be some kind of deformation as
the parts of the wire near to the point upon which the force
acts start to move earlier than the remote parts of the
wire, when one imagines the wire as a grid of points
(atoms) connected by small springs (van der Waals forces).

Here both aspects of the force come into play: the points
have inertia (F=ma) and they are connected by springs which
deform (Hooke's law) when a force is applied. A mass point
at the end of an elastic wire can act somewhat like an
anchoring of the wire because of its (the point's) inertia.

For simplification, one might think of a one-dimensional
model: a long chain of links each of which is a mass point
and a spring. O////O////O////O////O////O///O////O////O.
Then, when one starts to push or pull the mass point at one
end of the chain, the effects start to travel through the
chain.

But when you have a small and rigid body and a small force, you
can accelerate the body without deforming it in good approximation.

### Luigi Fortunati

Feb 20, 2023, 4:09:12â€¯AMFeb 20
to
Luigi Fortunati il 17/02/2023 17:37:48 ha scritto:
> What is the difference between the force accelerating the mass (F=ma)
> and the force deforming the mass (Hooke)?
>
> Can a force accelerate mass without deforming it?
>
>
> [[Mod. note --
> 1. Deformation can be quasi-static or dynamic, whereas acceleration
> is necessarily dynamic.

Exact.

And this means that in one case it depends on the reference and in the other it doesn't.

In one case it is relative, in the other it is absolute.

But how can force be relative and also absolute?

> 2. That depends on the force and the body-being-accelerated. If the
> force is somehow applied equally to each part of the body (e.g.,
> a uniform gravitational field in the Newtonian perspective), then
> the body can be accelerated without any deformation. Or, if the
> body is either very small or very stiff, and/or the acceleration
> is very small, then the deformation may be negligibly small.
> An important limiting case of this is the acceleration of a point
> mass, which we define as a mass with no internal structure; a point
> mass doesn't deform under acceleration. Electrons are a well-known
> example. But if a force is applied to a macroscopic body, and is
> *not* applied equally to each part of the body, then yes, the body
> will deform.
> -- jt]]

No one is able to know what happens to the point mass, whether it deforms or not.

I prefer to talk about macroscopic bodies where what is happening is clear and evident, as in my simulation
https://www.geogebra.org/m/zjbrrcet

### Luigi Fortunati

Feb 20, 2023, 2:25:07â€¯PMFeb 20
to
Stefan Ram il 19/02/2023 16:01:21 ha scritto:
> Luigi Fortunati <fortuna...@gmail.com> writes:
>> What is the difference between the force accelerating the mass (F=ma)
>> and the force deforming the mass (Hooke)?
>
> It's the same force, but applied to two different systems.
>
> One system is a point of mass "m" that is not held in position.
> When we call "F" the force acting on it, then the acceleration
> of the point is "a".

I had never heard this one: does F=ma only apply to material points and
not to bodies?

> When applying one version of Hooke's law, the force F is now
> applied to one end of an elastic wire the other end of which
> is fixed, and then the relative elongation e is proportional
> to the force F (if the force if not too strong.).

Right, Hooke's law only holds in these cases.

But I spoke of the force that (generally) deforms the mass and not only
of that which stretches the elastic threads.

Force does infinite things: it crushes even the hardest metals (the
press), breaks the shell of the egg, breaks the shopping bag when we
load it too much, all things that deform the masses without necessarily
accelerating or stretching them.

> So, when applied to one system, the force is proportional
> to the acceleration, when applied to another system,
> to the elongation.

This is precisely the profound meaning of my question: are they two
different types of force, how different are acceleration and
lengthening?

I note that the acceleration is relative to the reference, while the
lengthening is not.

Elongation is absolute, acceleration is not.

And how is the strength? Is it absolute or is it relative?

https://www.geogebra.org/m/zjbrrcet
to clarify what I mean.

There is the rigid chord AB and there is the elastic and spherical mass
m.

When we start the rotation in the inertial frame of the laboratory, two
things happen: the mass accelerates and, at the same time, it stretches
outwards.

Acceleration is centripetal only, elongation is centrifugal only.

How can centripetal force alone generate the two opposite effects?

### Stefan Ram

Feb 21, 2023, 2:59:23â€¯AMFeb 21
to
Luigi Fortunati <fortuna...@gmail.com> writes:
>Stefan Ram il 19/02/2023 16:01:21 ha scritto:
>I had never heard this one: does F=ma only apply to material points and
>not to bodies?

The point is a simplification. An extended body has a mass density
at each of its points, and the force has to be replaced by
a force field, which assigns a force to each point in space.

>This is precisely the profound meaning of my question: are they two
>different types of force, how different are acceleration and lengthening?

In "F=ma", "F" means "/the sum/ of all forces acting on the point".
When you pull one end of a spring, your pull is one force, but the
spring also exerts another force. When the end of the spring stops
moving, the sum of these two forces is zero, so, F=0 and a=0. The
lenght of the spring actually describes the force that the spring
exerts on its ends.

### Luigi Fortunati

Feb 22, 2023, 3:10:14â€¯AMFeb 22
to
Stefan Ram il 21/02/2023 08:59:19 ha scritto:
> In "F=ma", "F" means "/the sum/ of all forces acting on the point".
> When you pull one end of a spring, your pull is one force, but the
> spring also exerts another force. When the end of the spring stops
> moving, the sum of these two forces is zero, so, F=0 and a=0. The
> lenght of the spring actually describes the force that the spring
> exerts on its ends.

Ok, this description of yours fits perfectly into point B of my animation
https://www.geogebra.org/m/zjbrrcet
because B is the end of the spring (the elastic body of mass m).

First question: we are in the inertial frame of the laboratory and on
point B (during rotation at constant angular speed) two equal and
opposite forces act: are they the centripetal and centrifugal forces?

Second question (we are still in the inertial reference of the
laboratory): on which point does only the centripetal force act and not
the centrifugal one?

I hope that these two questions are answered, and are not forgotten as
has been the case in several previous cases.

### Richard Livingston

Feb 22, 2023, 3:10:15â€¯AMFeb 22
to
On Monday, February 20, 2023 at 1:25:07 PM UTC-6, Luigi Fortunati wrote:
> Stefan Ram il 19/02/2023 16:01:21 ha scritto:
> > Luigi Fortunati <fortuna...@gmail.com> writes:
> >> What is the difference between the force accelerating the mass (F=ma)
> >> and the force deforming the mass (Hooke)?
> >
> > It's the same force, but applied to two different systems.
> >
> > One system is a point of mass "m" that is not held in position.
> > When we call "F" the force acting on it, then the acceleration
> > of the point is "a".
> I had never heard this one: does F=ma only apply to material points and
> not to bodies?
...

You are making this more complex than it needs to be. Rather than thinking of
two different kinds of force, you need to model the objects (masses). F=ma
is best thought of as applying to point or rigid masses. If the situation
requires modeling elasticity of the object, then it is modeled as an array of
point masses with springs (and maybe dampers) between the point masses.

Many situations do not require such detailed models of the objects, such as
planets orbiting outside the Roche limit. However if you are impacting such
a planet with a rock, then you need the elastic model I just described.

One place where a slightly different model is required is a planet inside the
Roche limit, or when calculating tidal forces. Then the gravitational
"force" must be applied to each point mass in the elastic model. As this
"force" will be different for each point, those differences in gravitational
"force" will result is elastic deformation of the body.

Rich L.

### Stefan Ram

Feb 22, 2023, 5:26:58â€¯PMFeb 22
to
Luigi Fortunati <fortuna...@gmail.com> writes:
>I hope that these two questions are answered, and are not forgotten as
>has been the case in several previous cases.

But I don't visit web pages. So I leave it to
someone else to answer questions related to web
pages.

### Luigi Fortunati

Feb 22, 2023, 5:31:23â€¯PMFeb 22
to
Ok fine.

But all this has nothing to do with the question of whether force is
relative or absolute.

I would like to know, for example, why on earth I should specify the
"frame of reference" of the force of my weight on the floor if I am
always 80 kg-weight whatever the reference.

This is what interests me to clarify: if the force is not a motion, why
should it vary if I observe it from another reference?

And then, why should I "observe" the force if the force is not an
observable quantity?

There is no "speed" of force, only the "strength" of force.

I put a dynamometer on it, measure it and that's it!

Luigi.

[[Mod. note --
It's important not to confuse *weight* and *mass*. Your *mass* is always
the same, but your *weight* is different between [for example] (a) you're
standing on a floor attached to the Earth's surface), (b) you're standing
on the floor of an Einstein elevator which is current accelerating upwards
with respect to the Earth's surface, and (c) you're standing on the floor
of a space station in a free-fall orbit.
-- jt]]

### Sylvia Else

Feb 23, 2023, 3:18:15â€¯AMFeb 23
to
Your example is not so mysterious. The acceleration towards the centre
is proportional to the square of the angular velocity and proportional
to the radius of curvature. The angular velocity is the same for the
entire body, but the radius of curvature varies. The parts of the body
that are further away from the centre are thus accelerating more, and
require a greater force per unit mass. This has to be provided by
stretching the body, so that it provides the extra force needed.

For the part of the body closer to the centre of radius, it is
accelerating less, so some of the centripetal force has to be cancelled
by an outward force resulting from stretching the body.

These extra forces cancel out where part of the body is accelerating
towards the centre at a rate consistent with the centripetal force
provided by the tether.

Sylvia.

### Luigi Fortunati

Feb 23, 2023, 4:41:26â€¯PMFeb 23
to
I wasn't referring specifically to you but to the audience of newsgroup
goers for my previous discussions left to die because no one was able
or willing to respond.

[[Mod. note --
Newsgroup participants may "drop out" for any number of reasons,
including work committments, medical or other personal/family issues,
and various other things unrelated to physics. I don't think one can
reliably infer much of anything from "no one responded to my article".
Now let's get back to discussing physics.
-- jt]]

### Luigi Fortunati

Feb 23, 2023, 4:41:40â€¯PMFeb 23
to
Sylvia Else il 23/02/2023 09:18:10 ha scritto:
> Your example is not so mysterious. The acceleration towards the centre
> is proportional to the square of the angular velocity and proportional
> to the radius of curvature. The angular velocity is the same for the
> entire body, but the radius of curvature varies. The parts of the body
> that are further away from the centre are thus accelerating more, and
> require a greater force per unit mass. This has to be provided by
> stretching the body, so that it provides the extra force needed.
>
> For the part of the body closer to the centre of radius, it is
> accelerating less, so some of the centripetal force has to be cancelled=

> by an outward force resulting from stretching the body.
>
> These extra forces cancel out where part of the body is accelerating
> towards the centre at a rate consistent with the centripetal force
> provided by the tether.
>
> Sylvia.

I agree with everything you wrote.

We just need to understand the meaning of the forces that cancel each
other out.

For you, are two forces that cancel each other like two forces that are
not there?

Check out my simulation
> https://www.geogebra.org/m/zjbrrcet

In the initial position there are no forces, during the rotation there
are and they cancel each other out.

They are the same thing? No!

Because without rotation (initial position) the body is spherical and
the string is not under tension.

Instead, during the rotation the body is stretched and the string is
under tension.

So, in the second case, the forces exist and both act even when they
cancel each other out.

If we put a dynamometer between point B of the string and point C of
the body of mass m when there is no rotation, it tells us that there
are no forces (and, consequently, there are no tensions and no
elongations) .

If we put it during a rotation, there are forces, tensions and
elongations.

So the forces are there even when they cancel each other out.

They don't disappear.

The forces generate the rotation which, being a motion, can disappear
when the reference is changed.

But do you agree with me that they also generate tension and elongation
which, not being motions, do not disappear when the reference is
changed?

Luigi.

### Sylvia Else

Feb 24, 2023, 1:45:26â€¯AMFeb 24
to
On 24-Feb-23 8:41 am, Luigi Fortunati wrote:
> Sylvia Else il 23/02/2023 09:18:10 ha scritto:
>> Your example is not so mysterious. The acceleration towards the centre
>> is proportional to the square of the angular velocity and proportional
>> to the radius of curvature. The angular velocity is the same for the
>> entire body, but the radius of curvature varies. The parts of the body
>> that are further away from the centre are thus accelerating more, and
>> require a greater force per unit mass. This has to be provided by
>> stretching the body, so that it provides the extra force needed.
>>
>> For the part of the body closer to the centre of radius, it is
>> accelerating less, so some of the centripetal force has to be cancelled=
>
>> by an outward force resulting from stretching the body.
>>
>> These extra forces cancel out where part of the body is accelerating
>> towards the centre at a rate consistent with the centripetal force
>> provided by the tether.
>>
>> Sylvia.
>
> I agree with everything you wrote.
>
> We just need to understand the meaning of the forces that cancel each
> other out.
>
> For you, are two forces that cancel each other like two forces that are
> not there?

If a particle is being pulled in opposite directions by two forces with
the same magnitude, then the particle will not accelerate. The effect of
the two forces together is the same as no force.
>
> Check out my simulation
>> https://www.geogebra.org/m/zjbrrcet
>
> In the initial position there are no forces, during the rotation there
> are and they cancel each other out.
>
> They are the same thing? No!
>
> Because without rotation (initial position) the body is spherical and
> the string is not under tension.
>
> Instead, during the rotation the body is stretched and the string is
> under tension.
>
> So, in the second case, the forces exist and both act even when they
> cancel each other out.

It the forces originating from the stretching of the body that cancel out.

>
> If we put a dynamometer between point B of the string and point C of
> the body of mass m when there is no rotation, it tells us that there
> are no forces (and, consequently, there are no tensions and no
> elongations) .
>
> If we put it during a rotation, there are forces, tensions and
> elongations.
>
> So the forces are there even when they cancel each other out.
>
> They don't disappear.

Cancelling out doesn't mean that forces disappear, just that they
combine to produce no acceleration.

>
> The forces generate the rotation which, being a motion, can disappear
> when the reference is changed.

The acceleration does not disappear with any change of reference frame,
provided that the reference frame is inertial.

>
> But do you agree with me that they also generate tension and elongation
> which, not being motions, do not disappear when the reference is
> changed?
>
> Luigi.

The shape in a different reference frame in described by the Lorentz
transformation.

Sylvia.

### Luigi Fortunati

Feb 26, 2023, 2:35:31â€¯AMFeb 26
to
Sylvia Else il 24/02/2023 07:44:14 ha scritto:
>> For you, are two forces that cancel each other like two forces that are
>> not there?
>
> If a particle is being pulled in opposite directions by two forces with
> the same magnitude, then the particle will not accelerate.

This is certainly true.

> The effect of the two forces together is the same as no force.

However, this is only partially true.

It is true for the "acceleration" effect but not for the other effects:
compression, stretching, deformation or tension.

Obviously I am referring to the effects on bodies and not to the
effects on particles because nobody knows if the particle compresses or
doesn't compress, if it stretches or doesn't stretch.

>> Check out my simulation
>>> https://www.geogebra.org/m/zjbrrcet
>>
>> In the initial position there are no forces, during the rotation there
>> are and they cancel each other out.
>>
>> They are the same thing? No!
>>
>> Because without rotation (initial position) the body is spherical and
>> the string is not under tension.
>>
>> Instead, during the rotation the body is stretched and the string is
>> under tension.
>>
>> So, in the second case, the forces exist and both act even when they
>> cancel each other out.
>
> It the forces originating from the stretching of the body that cancel out.

No, it's not the forces that cancel each other out!

I'll give you a clarifying example.

If they push you from the left, they compress where they hit you and
accelerate you to the right.

There is force and there are two effects: compression and acceleration.

If they push you with the same force, both from the right and from the
left, the compression effect is still there, the acceleration effect is
gone.

So, forces don't disappear and compression doesn't disappear either,
the only thing that disappears is acceleration.

>> If we put a dynamometer between point B of the string and point C of
>> the body of mass m when there is no rotation, it tells us that there
>> are no forces (and, consequently, there are no tensions and no
>> elongations) .
>>
>> If we put it during a rotation, there are forces, tensions and
>> elongations.
>>
>> So the forces are there even when they cancel each other out.
>>
>> They don't disappear.
>
> Cancelling out doesn't mean that forces disappear,

Exact! Did you see that you recognize him too?

> just that they combine to produce no acceleration.

That's right: it's the acceleration that vanishes, not the forces!

>> The forces generate the rotation which, being a motion, can disappear
>> when the reference is changed.
>
> The acceleration does not disappear with any change of reference frame,
> provided that the reference frame is inertial.

Obvious.

Acceleration (which is motion) disappears in accelerated references,
not in inertial ones.

But force does not disappear because it is not motion and tension does
not disappear because it is not motion.

>> But do you agree with me that they also generate tension and elongation
>> which, not being motions, do not disappear when the reference is
>> changed?
>
> The shape in a different reference frame in described by the Lorentz
> transformation.

We are not talking about relativistic speeds here!

Luigi.

### israel socratus

Mar 7, 2023, 12:18:28â€¯PMMar 7
to
What is the difference between the Newton's force accelerating the mass (F=ma)
and the Einstein's force E=mc^2?
1905. Einstein wrote: "On the Electrodynamics of Moving Bodies"
1905. Einstein asked: "Does the Inertia of a Body Depend Upon its Energy"
The answer was "Yes, the Inertia of a Body depends upon its Energy: E=mc^2''.
---

### Stefan Ram

Mar 7, 2023, 6:16:55â€¯PMMar 7
to
israel socratus <socrat...@gmail.com> writes:
>What is the difference between the Newton's force accelerating the mass (F=ma)
>and the Einstein's force E=mc^2?

In "E=mc^2", there is no force. "E" is an energy, not a force.

>The answer was "Yes, the Inertia of a Body depends upon its Energy: E=mc^2''.

This is true if the body has no momentum. Then, the mass
is indeed E/c^2. If a body gains momentum, its energy will
increase, but not its mass.

(We can identify "inertia" with mass.)

If a body has a momentum p, then

E^2 = (pc)^2 + (mc^2)^2.

You can substitute p=0 in this formula, and get "E=mc^2".

I sometimes like to call "pc" the "momentum energy" of a system
and "mc^2" the "mass energy" of a system. So, both the momentum
and the mass contribute to the total energy of a system.

The total energy of a system, however, is not the plain sum of the
momentum energy and the mass energy, but given by the above formula.
(One might say, that the energy E is the "pythagorean sum" of
the momentum energy "pc" and the mass energy "mc^2".)