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Feb 18, 2023, 3:37:53â€¯AM2/18/23

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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]]

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]]

Feb 19, 2023, 10:01:26â€¯AM2/19/23

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.
>What is the difference between the force accelerating the mass (F=ma)

>and the force deforming the mass (Hooke)?

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?

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.

Feb 20, 2023, 4:09:12â€¯AM2/20/23

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.
> 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.

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]]

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

Feb 20, 2023, 2:25:07â€¯PM2/20/23

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
> 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".

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.).

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.

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?

I made the simulation

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?

Feb 21, 2023, 2:59:23â€¯AM2/21/23

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
>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?

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?

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.

Feb 22, 2023, 3:10:14â€¯AM2/22/23

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
> 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.

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.

Feb 22, 2023, 3:10:15â€¯AM2/22/23

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?

...
> 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.

Feb 22, 2023, 5:26:58â€¯PM2/22/23

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.

I have not forgotten to answer your questions.
>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.

Feb 22, 2023, 5:31:23â€¯PM2/22/23

to

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]]

Feb 23, 2023, 3:18:15â€¯AM2/23/23

to

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.

Feb 23, 2023, 4:41:26â€¯PM2/23/23

to

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]]

Feb 23, 2023, 4:41:40â€¯PM2/23/23

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=
> 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

> 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.

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.

Feb 24, 2023, 1:45:26â€¯AM2/24/23

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
> 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?

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.
> 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.

combine to produce no acceleration.

>

> The forces generate the rotation which, being a motion, can disappear

> when the reference is changed.

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.

transformation.

Sylvia.

Feb 26, 2023, 2:35:31â€¯AM2/26/23

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.
>> 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.

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.

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,

> 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.

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!
> transformation.

Luigi.

Mar 7, 2023, 12:18:28â€¯PM3/7/23

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''.

---

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''.

---

Mar 7, 2023, 6:16:55â€¯PM3/7/23

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.
>What is the difference between the Newton's force accelerating the mass (F=ma)

>and the Einstein's force E=mc^2?

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

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".)

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