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?
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]]
Luigi Fortunati <fortunati.luigi@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.
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?
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.
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.I had never heard this one: does F=ma only apply to material points and
When we call "F" the force acting on it, then the acceleration
of the point is "a".
not to bodies?
I hope that these two questions are answered, and are not forgotten as
has been the case in several previous cases.
...I had never heard this one: does F=ma only apply to material points andWhat 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?
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.
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 <fortunati.luigi@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.
But I don't visit web pages. So I leave it to
someone else to answer questions related to web
pages.
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.
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?
The shape in a different reference frame in described by the Lorentz transformation.
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''.
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