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tech / sci.physics.research / Gravitational mass and inertial mass

SubjectAuthor
* Gravitational mass and inertial massLuigi Fortunati
`* Re: Gravitational mass and inertial massLuigi Fortunati
 `* Re: Gravitational mass and inertial massLuigi Fortunati
  `- Re: Gravitational mass and inertial massLuigi Fortunati

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Gravitational mass and inertial mass

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From: fortunati.luigi@gmail.com (Luigi Fortunati)
Newsgroups: sci.physics.research
Subject: Gravitational mass and inertial mass
Date: Thu, 22 Feb 2024 09:49:45 PST
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 by: Luigi Fortunati - Thu, 22 Feb 2024 17:49 UTC

In my animation https://www.geogebra.org/m/kqjzk5gt there are the two
bodies mA and mB (of equal mass) connected via an inextensible wire,
ideally massless.

The wire experiences two opposite forces: the force FA from the body mA
and the opposing force FB from the body mB.

The action starts from the body mA which exerts its gravitational force
mg on the wire (where m is the mass of the body mA).

And what does the massless wire do? If it were not connected to the
body mB, it would not oppose any resistance to the force FA received by
the body mA, so the body mA would only be a gravitational mass and
that's it.

And instead, the wire (tied to the body mB) transmits to the body mA
the opposite reaction FB from the body mB and, thus, the body mA must
react inertially to this opposing braking force and consequently also
becomes an inertial mass.

This is what determines the prevalence of the force FA (generated by
the gravitational and inertial mass mA) compared to the force FB
(generated by the mass mB which is only inertial).

That's why the blue strength of my animation is always greater than the
red strength.

And what is the ratio between the blue force FA (gravitational +
inertial) and the red force FB (inertial only)?

I will try to answer this question but, first, I would like to know if
my animation and what I wrote contain errors.

Luigi Fortunati

[[Mod. note -- The animationn appears to show FA larger in magnitude
than FB. This is incorrect: assuming that we idealize the string as
massless, then FA and FB have equal magnitudes.

This system is known as "Atwood's machine"; see
https://en.wikipedia.org/wiki/Atwood_machine
for more information.
-- jt]]

Re: Gravitational mass and inertial mass

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From: fortunati.luigi@gmail.com (Luigi Fortunati)
Newsgroups: sci.physics.research
Subject: Re: Gravitational mass and inertial mass
Date: 3 Apr 2024 06:58:33 GMT
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 by: Luigi Fortunati - Wed, 3 Apr 2024 06:58 UTC

Luigi Fortunati il 22/02/2024 02:49:45 ha scritto:
> In my animation https://www.geogebra.org/m/kqjzk5gt there are the two bodies A and B (of equal mass m) connected via an inextensible wire, ideally massless.
> [[Mod. note -- This system is known as "Atwood's machine"; see
> https://en.wikipedia.org/wiki/Atwood_machine
> for more information.
> -- jt]]

My animation is not Atwood_machine because only body A moves vertically, while body B slides horizontally on a frictionless plane.

> And what is the ratio between the blue force FA=mg (gravitational + inertial) and the red force FB (inertial only)?

It turns out to me that FB=FA(mB/(mA+mB))

In our case where mA=mB=m, the force FB is FB=1/2 of FA.

It's correct?

Luigi Fortunati

Re: Gravitational mass and inertial mass

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From: fortunati.luigi@gmail.com (Luigi Fortunati)
Newsgroups: sci.physics.research
Subject: Re: Gravitational mass and inertial mass
Date: 19 Apr 2024 10:52:59 GMT
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 by: Luigi Fortunati - Fri, 19 Apr 2024 10:52 UTC

Il 03/04/2024 08:58, Luigi Fortunati ha scritto:
> Luigi Fortunati il 22/02/2024 02:49:45 ha scritto:
>> In my animation https://www.geogebra.org/m/kqjzk5gt there are the two bodies A and B (of equal mass m) connected via an inextensible wire, ideally massless.
>> [[Mod. note -- This system is known as "Atwood's machine"; see
>> https://en.wikipedia.org/wiki/Atwood_machine
>> for more information.
>> -- jt]]
>
> My animation is not Atwood_machine because only body A moves vertically, while body B slides horizontally on a frictionless plane.
>
>> And what is the ratio between the blue force FA=mg (gravitational + inertial) and the red force FB (inertial only)?
>
> It turns out to me that FB=FA(mB/(mA+mB))
>
> In our case where mA=mB=m, the force FB is FB=1/2 of FA.

In my animation, the string AB receives the blue force FA=mg from body A
and transmits it to point B of body B.

However, the black force -FB that the string transmits to body B is only
half the force it received at point A.

What happened to the force that was lost? Where is the error in the
animation or in the reasoning I did?

Luigi Fortunati

Re: Gravitational mass and inertial mass

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From: fortunati.luigi@gmail.com (Luigi Fortunati)
Newsgroups: sci.physics.research
Subject: Re: Gravitational mass and inertial mass
Date: 4 May 2024 09:50:12 GMT
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 by: Luigi Fortunati - Sat, 4 May 2024 09:50 UTC

Il 19/04/2024 12:52, Luigi Fortunati ha scritto:
>>> In my animation https://www.geogebra.org/m/kqjzk5gt there are the two bodies A and B (of equal mass m) connected via an inextensible wire, ideally massless.
> The string AB receives the blue force FA=mg from body A
> and transmits it to point B of body B.
>
> However, the black force -FB that the string transmits to body B is only
> half the force it received at point A.
>
> What happened to the force that was lost? Where is the error in the
> animation or in the reasoning I did?

Since no one wanted (or knew) to answer me, I thought on my own and
found the error in my animation https://www.geogebra.org/m/kqjzk5gt

I thought like this...

The system composed of the two bodies A and B and the string AB behaves
like a single body because it accelerates in the same way at all its
points, without exception.

There cannot be a point that accelerates and another point that does not
accelerate.

And therefore, in the direction of acceleration the opposing forces (at
each point) cannot be balanced, otherwise there would be no net force
and there would be no acceleration.

Instead, the acceleration is there and the prevailing force must be there.

Take, for example, point B accelerating to the right.

From the side of the string AB the black force comes to him which pulls
him to the right, from the side of the body B the red force comes to him
which pulls him to the left.

If the two opposing forces (black and red) were equal, point B would
accelerate neither to the right nor to the left.

And since point B accelerates to the right, it means that the black
force cannot be equal to the red force but must be *greater*!

And then, I created the new animation
https://www.geogebra.org/m/zxt8ysv6 where the black and blue forces are
equal to mg, and the red and brown forces are equal to mg/2).

How to reconcile my last animation (which is correct) with the third
law, that of action and reaction?

Luigi Fortunati


tech / sci.physics.research / Gravitational mass and inertial mass

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