| Topic: |
Science > Physics |
| User: |
"Svilen" |
| Date: |
27 Jul 2003 12:58:13 PM |
| Object: |
question about "work" |
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
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| User: "Paul Cardinale" |
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| Title: Re: question about "work" |
28 Jul 2003 10:27:38 AM |
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(Svilen) wrote in message news:<73a35da8.0307270958.7a380edc@posting.google.com>...
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
The basic physics explanation is a simplification.
Initially, the force must be greater than mg in order to provide
vertical acceleration. Once the object is in motion, the force can be
reduced to mg.
As the desired height is approached, the force must be reduced so the
unit will decellerate. To get the total energy expended, integrate F
dy (where y is your vertical axis). You can take it as a math
exercise to show that regardless of how F varies during the lift, that
as long as the object is at rest at the beginnging and end, then the
result will always be mgH.
Paul Cardinale
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| User: "Svilen" |
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| Title: Re: question about "work" |
29 Jul 2003 01:01:54 AM |
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Paul,
I completely agree with you. That is what others have mentioned in
this post and that's what I meant when I suggested that I can't accept
describing the lift as a sequence of equilibrium states. One has to
make the ***** body move with a velocity of V, starting from a
state with V=0.
pcardinale@volcanomail.com (Paul Cardinale) wrote in message news:<64050551.0307280727.401933ca@posting.google.com>...
sminchev@yahoo.com (Svilen) wrote in message news:<73a35da8.0307270958.7a380edc@posting.google.com>...
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
The basic physics explanation is a simplification.
Initially, the force must be greater than mg in order to provide
vertical acceleration. Once the object is in motion, the force can be
reduced to mg.
As the desired height is approached, the force must be reduced so the
unit will decellerate. To get the total energy expended, integrate F
dy (where y is your vertical axis). You can take it as a math
exercise to show that regardless of how F varies during the lift, that
as long as the object is at rest at the beginnging and end, then the
result will always be mgH.
Paul Cardinale
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| User: "Edward Green" |
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| Title: Re: question about "work" |
29 Jul 2003 11:26:24 AM |
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(Svilen) wrote in message news:<73a35da8.0307282201.67d6c2ae@posting.google.com>...
Paul, I completely agree with you.
Which shows what a vile little kiss-up you are -- no offense intended
to the innocent object of your up-kissing.
pcardinale@volcanomail.com (Paul Cardinale) wrote in message news:<64050551.0307280727.401933ca@posting.google.com>...
The basic physics explanation is a simplification.
Initially, the force must be greater than mg in order to provide
vertical acceleration. Once the object is in motion, the force can be
reduced to mg.
As the desired height is approached, the force must be reduced so the
unit will decellerate. To get the total energy expended, integrate F
dy (where y is your vertical axis). You can take it as a math
exercise to show that regardless of how F varies during the lift, that
as long as the object is at rest at the beginnging and end, then the
result will always be mgH.
nulldev00@aol.com (Edward Green) wrote in message news:<2a0cceff.0307271720.66e26a00@posting.google.com>...
So how do we distinguish the two cases? The answer is we can't, given
only that the applied force is mg: we know at least the body's not in
free-fall, but whether its in static equilibrium, worked upon or
itself doing work (as in a constantly descending lift) is additional
information. If you want to know how nature "decides" which situation
we're in, it must be that to start the lift going up we momentarily
increased f > mg, and to stop the lift at the to, we momentarily
decreased f < mg -- we ignore these transients in calculating work
done and anyway, they should cancel.
I said exactly the same thing as Paul concerning the necessity of
accelerating and deaccelerating the body from rest and the resulting
cancelling hiccups in the work. And he you fawn on, and me you dump
on, and why? Probably because your wheedling little mind perceives
some advantage to both actions. Maybe you think you can curry more
favor with your betters by licking and dumping with alternate ends?
nulldev00@aol.com (Edward Green) wrote in message news:<2a0cceff.0307281850.5a0eea68@posting.google.com>...
Well ... it is perfectly damn evident that ...
OK. That's pretty damn unambiguous.
(Svilen) wrote in message news:<73a35da8.0307282201.67d6c2ae@posting.google.com>...
One has to
make the ***** body move with a velocity of V, starting from a
state with V=0.
I suppose "damn" is in the public domain, but we know about imitation
and flattery, don't we? Seems like a borrowing in your style --
probably a glitch on your "ape" and "disdain" classifying system.
I have no doubt you are equally vile in person, which possibility I am
kindly spared by an after all not entirely non-beneficent diety.
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| User: "Edward Green" |
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| Title: Re: question about "work" |
29 Jul 2003 02:02:12 PM |
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(Svilen) wrote in message news:<73a35da8.0307282201.67d6c2ae@posting.google.com>...
Oops ... please disregard former rant: I seem to have conflated you
with "undeniable". You only ignored my earlier identical answer, you
didn't dis it. My apologies.
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| User: "Svilen" |
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| Title: Re: question about "work" |
29 Jul 2003 06:03:14 PM |
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No offence, Ed. Sorry for not mentioning your contribution to this
discussion. If you read back my reply to Paul you'll see that I admit
there that similar things were suggested also by other people in the
post - well i didn't mentioned you explicitely, sorry.
My thanks also to Sam Wormley for giving me different looks at a
force.
As to Undeniable - well sir, I'm not interested in "what" the formulas
are in the books. I am interested in "why" the formulas are the way
they are. It is that simple. And all you're offering is the "what" not
the "why".
nulldev00@aol.com (Edward Green) wrote in message news:<2a0cceff.0307291102.546ae27b@posting.google.com>...
sminchev@yahoo.com (Svilen) wrote in message news:<73a35da8.0307282201.67d6c2ae@posting.google.com>...
Oops ... please disregard former rant: I seem to have conflated you
with "undeniable". You only ignored my earlier identical answer, you
didn't dis it. My apologies.
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| User: "Undeniable" |
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| Title: Re: question about "work" |
01 Aug 2003 06:32:21 AM |
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(Edward Green) wrote in message news:<2a0cceff.0307311203.62f5d696@posting.google.com>...
undeniablefact@yahoo.com (Undeniable) wrote in message news:<1f366ae.0307300502.7c089f57@posting.google.com>...
sminchev@yahoo.com (Svilen) wrote in message news:<73a35da8.0307291503.64cda8ee@posting.google.com>...
As to Undeniable - well sir, I'm not interested in "what" the formulas
are in the books. I am interested in "why" the formulas are the way
they are. It is that simple. And all you're offering is the "what" not
the "why".
Before you reach the level required in investigating "why" things are
this way, you must first go through the painfull process of
understanding "what" are those things.
Not bad. So far so good.
In you postings you have exhibited a high level of confusion and
misunderstanding of simple mechanics concepts developed by people with
intelligence many orders of magnitude higher than yours and understood
by myriads of others.
You on the other hand have demonstrated a high level of personal
buffoonery, and continue to reinforce this demonstration with every
fresh post.
You have failed to show any contradictions
present, save presenting an alternative you agree with. The burden of
proof lies with you in showing both the alleged contradictions and
presenting an alternative. In both cases. you have only succeded in
demonstrating that you lack formal training because if you had any you
would know that all the concepts you wonder about all very well
understood.
You have a paranoid ability to confuse simple questions with crankdom.
Specifically, you confuse the work done by a force F in moving a body
from point A to point B against a gravitation potential with the work
converted into petential energy.
I think you meant "penitential energy".
Anyway -- what the hell is the alleged confusion here? "The work done
by a force F in moving a body from point A to point B against a
gravitational potential" and "the work converted into potential
energy" sound like the samething to me.
My prior of you as a rude savant is being challenged: you may simply
be rude.
Several others that posted answers
evidently have the same confussion as you do. Those that are confussed
as you are will always be nice to you. Those that are trying to help
you will be rough with you. This is because they can sense an
arrogance in your posts and they would like to prevent you from
attaining the irreversible state of crankiness. I have nothing
against you as a person but whenever someone assaults science with
arrogance I respond and so do others that have gone through the
painfull process of understanding "what is it" before "why is it".
Apparently, you haven't gone through that painfull process yourself
and I suggest to you if you are interested in mechanics to seek formal
training rather than wondering "why" thigs are the way they are.
Questions involving "why" fall withinn the subject of philosophy
called Metaphysics. Unless you understand well Physics first, going
into Metaphysics is a dangerous endevour and can even cost you your
mental stability, one proof is all these cranks posting at sci.* NGs.
Your thoughts overall aren't bad out of context, merely wildly
inappropriate. Noting that on the one hand we have a weight sitting
on a table, doing nothing, and on another hand we have a weight being
lifted, and on either hand we have the same force mg, raises the
obvious question: so what is different if it's not the force? It's a
good question, a thoughtful question, and not a particularly cranky
question. You're the nut who delivers a tirade about meat-eating to
the poor Joe who just ordered a burger at the diner. Take it outside,
please.
Edward Green,
I suppose you fell a victim of a crank (that is Svilen) who dragged
you to the lowest level possible.
These USENET postings Edward are read by many people, students,
layman, etc., and cranks like Svilen generate amazing levels of
confusion in the minds of those people who come here to learn
something. I will spend some of my valuable time to explain to you why
Svilen is a crank in my opinion, not merely a confused person who is
asking a physics question about work, acutally he didn't. Please
recall his original post with my comments:
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body.
He is using a logical fallacy known as "an appeal to majority" to
support his confusion. "Everyone knows" is a fallacious premise. Only
he knows that way. This is because as I explained the work done by the
force is NOT mgh. That is the work converted into potential energy.
Svilen is trying to reinforce a confusion on other's minds using the
appeal to majority and hidding his false statement in it.
But now suppose that the body lays on the floor and experiences a
gravity >force mg. You want to lift it up and according to the formula
you apply an >equal and opposite force of mg.
This is a statement to further reinforce his fallacy on his audience
by assuming that the force to lift the body is only mg. You can apply
as much force as you like to lift a body. Part of the work done by
that force will be converted into potenial energy and the rest into
kinetic and possibly other forms of energy, depending on how the task
is accomplished and the conditions present (friction etc.). There is
no formula saying that the force you apply on a body must be mg as he
alleges to support his fallacy. He is confusing the work converted
into potential energy with the work done by a force F.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the
work
done on a positive charge to move it against the force of the field."
Here the fallacy is extended in order to refute not just mechanics but
other fields. This is not the type of question, Edward, asked by a
confused person. A person who is willing to learn asks the question in
the following manner:
"I do not understand the formula for work. What does mgh represent?
Any help will be appreciated", or something along these lines.
I am spending the time to answer your ad hominen attacks not because I
care what you think of me. You are a crank just like Svilen and this
is proven by your postings. This is done for helping others in quickly
identifying the intentions of a poster and responding properly. These
threads have become a playground for cranks, against the original
purpose of serving as a medium for the exchange of valuable knowledge.
Your entropy has reached a maximum level Edward, the highest level of
disorder possible.
Anyway -- what the hell is the alleged confusion here? "The work done
by a force F in moving a body from point A to point B against a
gravitational potential" and "the work converted into potential
energy" sound like the samething to me.
It sounds like the same thing to you because you are at the same level
of confusion as Svilen. Cranks are supposed to support each other I
heard, nothing new about that.
I am tempted to quote Uncle AL, I hope he won't get mad at me:
"Christ, join your own civilization."
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| User: "Edward Green" |
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| Title: Re: question about "work" |
01 Aug 2003 02:10:25 PM |
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(Undeniable) wrote in message news:<1f366ae.0308010332.3e26154a@posting.google.com>...
<...>
I am spending the time to answer your ad hominen attacks ...
Do you think just possibly that opening a dialogue by pleasantly
calling somebody an "idiot" invites that kind of thing? I know it's a
crazy idea, but you might want to consider it.
No doubt we have both spent too much of our valuable time on this.
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| User: "Edward Green" |
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| Title: Re: question about "work" |
31 Jul 2003 02:37:42 PM |
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(Svilen) wrote in message news:<73a35da8.0307291503.64cda8ee@posting.google.com>...
No offence, Ed. Sorry for not mentioning your contribution to this
discussion. If you read back my reply to Paul you'll see that I admit
there that similar things were suggested also by other people in the
post - well i didn't mentioned you explicitely, sorry.
My thanks also to Sam Wormley for giving me different looks at a
force.
As to Undeniable - well sir, I'm not interested in "what" the formulas
are in the books. I am interested in "why" the formulas are the way
they are. It is that simple. And all you're offering is the "what" not
the "why".
A remarkably civil answer, all things considered. Thanks.
I agree with you -- even though others may like to harp on physics
only offering "what" not "why". I think at least a deeper explanation
is possible than the simple "you give it a kick, it starts, you give
it another kick, it stops". That covers the F = ma aspects, but
doesn't explain how it is that the mg in one case is just sitting
there doing nothing, in the other doing work.
I suggested a starting point: that all possible physical processes
generate entropy. The way the system knows which way the lift is to
go, whether worked on or itself doing work, is by entropy generation
-- which of course we have set up with external motors and brakes.
I have a semi-coherent idea that force itself is _always_ identified
with entropy creation -- which is a little stronger than the standard
statements. We have for example the elastomer which generate force by
a purely entropic mechanism: the entropy of the long chains is reduced
when they are stretched out, while there energy is unchanged. My
hunch is all force can be thought of this way ... not just the
transparent elastomeric case ... but that's how you say ... only a
nascent idea.
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| User: "Edward Green" |
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| Title: Re: question about "work" |
29 Jul 2003 05:14:46 PM |
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(Undeniable) wrote in message news:<1f366ae.0307290624.638d9199@posting.google.com>...
sminchev@yahoo.com (Svilen) wrote in message news:<73a35da8.0307282201.67d6c2ae@posting.google.com>...
Paul,
I completely agree with you. ...
And speak of the devil ...
You completely agree with something that you do not seem to
understand. What you think you understand is not what Paul is taking
about. You confuse the calculation of potential and work converted
into it with motion dynamics. The potential is defined as the work
done is moving to height h without any acceleration. There is no
physics textbook around describing an actual lift as a sequence of
equilibrium states. You probably read that in a cranky book, like the
one pusblished recently by a pathetic crank who thinks the equation
for work is a trick and a conspiracy. I know this crank is going
around NGs trying to convince poeple there is a major contradiction in
Newtonian mechanics work principle. I hope that is not you and you are
simply a confused person. One can overcome confusion by getting help
and stydying but a crank is in a terminal condition, no hope of coming
back to reality.
There are more forms of crankiness in heaven and earth than are known
in your philosophy, Horatio.
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| User: "Undeniable" |
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| Title: Re: question about "work" |
28 Jul 2003 05:18:12 AM |
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(Svilen) wrote in message news:<73a35da8.0307270958.7a380edc@posting.google.com>...
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
You confuse statics and dynamics. The Work required to lift a mass
against a gravitational potential is found by integrating the Virtual
Work deltaW necessary to move the body for an infinitesimal distance
deltaR over the length of the path. You seem to confuse the notion of
work with the dynamics of motion. The same work mgh is required
whether this is done in 1 second or in infinite time. The time it
takes for the body to reach the height h will depend on the force
applied and any excess work that is not required for the lift will be
converted into kinetic energy. The work required will be converted to
potential energy and it can be recovered if the mass is set to fall in
a theoretical situation, where some kind of device will convert the
kinetic energy back to potential stored energy, like a spring.
When students or layman start thinking in terms of work alone for
solving problems they get really confussed and several contradictions
can arise in their minds. This is because Work has the same units with
Energy and it is a part of a bigger equation, the Lagrangian. In the
latter, only the kinetic and potential energy get to play a role and
used determine the dynamics of motion.
If you cannot grasp these concepts and there are still doubts in your
mind it is not because there is something wrong with physics. That may
be because your particular talent is more suitable for another
profession, that of a lawyer for instance. A physicist must be able to
see the grand picture and infer the facts whereas a lawyer tries to
manipulate the facts to develop a big picture. This is perfectly ok if
one knows what he's doing. If not, he can be humiliated, confused and
doomed in the contradictions that are only in his mind.
You got the message I hope.
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| User: "Svilen" |
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| Title: Re: question about "work" |
28 Jul 2003 10:15:31 AM |
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(Undeniable) wrote in message news:<1f366ae.0307280218.fb1c136@posting.google.com>...
sminchev@yahoo.com (Svilen) wrote in message news:<73a35da8.0307270958.7a380edc@posting.google.com>...
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
You confuse statics and dynamics. The Work required to lift a mass
against a gravitational potential is found by integrating the Virtual
Work deltaW necessary to move the body for an infinitesimal distance
deltaR over the length of the path. You seem to confuse the notion of
work with the dynamics of motion. The same work mgh is required
whether this is done in 1 second or in infinite time.
Actually I don't confuse motion and work. It is perfectly clear from
my post that I have not mentioned time. Of course the same amount of
work can be done in different periods of time. And yes, I guess
everyone on this forum can figure out that total work is integration
of deltaW.
The time it
takes for the body to reach the height h will depend on the force
applied and any excess work that is not required for the lift will be
converted into kinetic energy. The work required will be converted to
potential energy
But isn't it true that if a body is in equilibrium and you apply force
to move it to a hight h this force will need to change this
equilibrium i.e. bring the body to a motion with a certain speed -
that is you have deltaV in a certain time which is acceleration, hence
force? I don't quite agree that you can represent the transition to a
hight h as a sequence of transitions through equilibrium states.
and it can be recovered if the mass is set to fall in
a theoretical situation, where some kind of device will convert the
kinetic energy back to potential stored energy, like a spring.
When students or layman start thinking in terms of work alone for
solving problems they get really confussed and several contradictions
can arise in their minds. This is because Work has the same units with
Energy
And isn't Energy the ability to do work?
and it is a part of a bigger equation, the Lagrangian. In the
latter, only the kinetic and potential energy get to play a role and
used determine the dynamics of motion.
If you cannot grasp these concepts and there are still doubts in your
mind it is not because there is something wrong with physics. That may
be because your particular talent is more suitable for another
profession, that of a lawyer for instance. A physicist must be able to
see the grand picture and infer the facts whereas a lawyer tries to
manipulate the facts to develop a big picture. This is perfectly ok if
one knows what he's doing. If not, he can be humiliated, confused and
doomed in the contradictions that are only in his mind.
You got the message I hope.
Yes, I think I got the message as everyone else here did. However, if
you thought someone here thought there was something wrong with
physics, you were completely wrong. You better read again the original
post. I believe that people here discuss things even "simple" as this
one and treat each other with respect. No body here needs to see how
great a physisit you are and how well you understand the Lagrangian.
And even if this is really so, the devine nature of it doesn't allow
you to call someone an idiot simply because he expresses his opinion -
this is in connection to your post further down.
I hope you got the message.
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| User: "Svilen" |
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| Title: Re: question about "work" |
27 Jul 2003 04:14:04 PM |
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OK, let's make things simpler by assuming that it is not a
biological/muscle force that's doing the work. Just any force that's
trying to lift the body up. Now, I understand that if the body sits on
the table there is already the force -mg that's holding against the
weight. In order to lift it up 1m for example a force has to be
applied that is mg+F - mg component will compensate for the gravity
and F will overcome it. Coming from the general definition of work it
seems to me that the work that force is doing should be
(mg+F)(deltaH), and not only mg(deltaH) as the books say. So, where's
the difference coming from?
Sam Wormley <swormley1@mchsi.com> wrote in message news:<3F241EB0.E2970BF2@mchsi.com>...
Svilen wrote:
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
For a body sitting on the table there is already an equal and
opposite force from the table holding up against gravity. To
move it higher against gravity requires work, w = mg(delta-h).
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| User: "Edward Green" |
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| Title: Re: question about "work" |
29 Jul 2003 02:10:59 PM |
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(Svilen) wrote in message news:<73a35da8.0307271314.5e0f50d5@posting.google.com>...
OK, let's make things simpler
But not top-posting? Ok.
by assuming that it is not a
biological/muscle force that's doing the work. Just any force that's
trying to lift the body up. Now, I understand that if the body sits on
the table there is already the force -mg that's holding against the
weight. In order to lift it up 1m for example a force has to be
applied that is mg+F - mg component will compensate for the gravity
and F will overcome it.
No, that's wrong. _Exactly_ the same force is necessary to lift the
wieght against gravity as to "overcome it" -- except for the starting
transient, which may be vanishingly small -- and transient.
The key to the seemingly magical difference between the same force on
the one hand merely holding the weight statically, and on the other
hand doing work against gravity, really does lie in the creation of
entropy ... whether in your muscle, gas in an expanding piston, or
some distant power plant. It looks like a question in simple
dynamics, but thermo is lurking in the background.
Coming from the general definition of work it
seems to me that the work that force is doing should be
(mg+F)(deltaH), and not only mg(deltaH) as the books say. So, where's
the difference coming from?
Your conceptual mistake. The difference between the two situations is
not in the instantaneously applied force but in your set up. A
fundamental difference between a table top and a lift is the ability
to apply force over a distance: the table top gives up after a few
nanometers, the lift does not.
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| User: "Undeniable" |
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| Title: Re: question about "work" |
28 Jul 2003 02:47:57 PM |
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(Svilen) wrote in message news:<73a35da8.0307271314.5e0f50d5@posting.google.com>...
OK, let's make things simpler by assuming that it is not a
biological/muscle force that's doing the work. Just any force that's
trying to lift the body up. Now, I understand that if the body sits on
the table there is already the force -mg that's holding against the
weight. In order to lift it up 1m for example a force has to be
applied that is mg+F - mg component will compensate for the gravity
and F will overcome it. Coming from the general definition of work it
seems to me that the work that force is doing should be
(mg+F)(deltaH), and not only mg(deltaH) as the books say. So, where's
the difference coming from?
Your are confusing work converted to potential energy and total work
done by a force F in moving a particle form point A to point B. If you
read the books carefully, you will notice that W = mgh is the work
converted to potential energy. If you read even more carefully you
will find some place a proper definition for potential energy:
Potential energy is the work we must do in moving a body from point A
to point B WITHOUT acceleration.
No textbook says that if a force F = mg+f is applied to a body in a
state of equilibrium under own weight in a gravitational potential
then the work done by that force is W = mgh. What your books say is
the the work converted to potential energy is equal to mgh.
Look, this is basic stuff. Whenever asking a question one should be
careful and avoid words you're using such as: "I don't quite agree
that you can represent the transition to a hight h as a sequence of
transitions through equilibrium states." because such words will
provoke reactions you won't like.
No one cares if you agree or not. That's the definition and it works.
A better word for you to use in the place of "agree" is "understand".
If you want to learn basic mechanics, USENET is not the proper place.
You will have to either go to a college near you or get a good tutor.
But you must willing to learn and admit you are confused. If you have
the proper attitude you will learn and everything will come clear in
your mind. Otherwise you will remain in a state of confusion and maybe
become another crank going around NGs claiming he found a
contradiction of some sort.
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| User: "Edward Green" |
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| Title: Re: question about "work" |
28 Jul 2003 05:40:58 PM |
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(Undeniable) wrote in message news:<1f366ae.0307280231.49b2e265@posting.google.com>...
sminchev@yahoo.com (Svilen) wrote in message news:<73a35da8.0307271314.5e0f50d5@posting.google.com>...
OK, let's make things simpler by assuming that it is not a
biological/muscle force that's doing the work. Just any force that's
trying to lift the body up. Now, I understand that if the body sits on
the table there is already the force -mg that's holding against the
weight. In order to lift it up 1m for example a force has to be
applied that is mg+F - mg component will compensate for the gravity
and F will overcome it.
You are very confused about this as I can see. Read my post below.
Coming from the general definition of work it
seems to me that the work that force is doing should be
(mg+F)(deltaH), and not only mg(deltaH) as the books say. So, where's
the difference coming from?
You are confused.
And you are compensating for some ego deficiency.
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| User: "Chuck Simmons" |
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| Title: Re: question about "work" |
27 Jul 2003 04:43:54 PM |
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Svilen wrote:
OK, let's make things simpler by assuming that it is not a
biological/muscle force that's doing the work. Just any force that's
trying to lift the body up. Now, I understand that if the body sits on
the table there is already the force -mg that's holding against the
weight. In order to lift it up 1m for example a force has to be
applied that is mg+F - mg component will compensate for the gravity
and F will overcome it. Coming from the general definition of work it
seems to me that the work that force is doing should be
(mg+F)(deltaH), and not only mg(deltaH) as the books say. So, where's
the difference coming from?
That is slightly different from the idealized problem. As a practical
matter, to raise the mass m to a height of one meter (H), you will apply
a force of mg+F until a desired velocity is achieved. To stop m at H,
you will have to apply a force of mg-F for the same time you applied
mg+F. In this way, you can indeed raise the mass m and stop it at a
height H. What is interesting is that doing it as stated makes the final
work done mgH anyway.
Chuck
--
... The times have been,
That, when the brains were out,
the man would die. ... Macbeth
Chuck Simmons
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| User: "Igor" |
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| Title: Re: question about "work" |
27 Jul 2003 02:34:13 PM |
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On 27 Jul 2003 10:58:13 -0700, (Svilen) wrote:
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
I think you're confusing the situation of no net force with no motion.
Even though there's no net force on the body, it can still be moving
upward in a state of dynamic equilibrium, at a constant speed. In
that case, the work done by the lifter is mgH, since mg is the minimal
amount of force needed to lift the body. If the lifter pauses with
the body at a height H, it becomes potential energy, and then, if
dropped, kinetic energy as gravity performs work upon it. The best
way to picture work is from the point of view of whatever is putting
energy into the system or taking it out.
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| User: "Edward Green" |
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| Title: Re: question about "work" |
28 Jul 2003 07:06:47 AM |
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Igor <bx238@bfn.org> wrote in message news:<cm98iv47084v9o4q7lmmdkgjjmvoc34q81@4ax.com>...
I think you're confusing the situation of no net force with no motion.
Even though there's no net force on the body, it can still be moving
upward in a state of dynamic equilibrium, at a constant speed. In
that case, the work done by the lifter is mgH, since mg is the minimal
amount of force needed to lift the body. If the lifter pauses with
the body at a height H, it becomes potential energy, and then, if
dropped, kinetic energy as gravity performs work upon it. The best
way to picture work is from the point of view of whatever is putting
energy into the system or taking it out.
Just small comment on yours and Jim Greenfield posts:
One _might_ on cursory reading think that if the body is stopped at
height H "it", the kinetic energy of the body, becomes potential
energy. Of course you both know this is not the case, and the
potential at H relative to ground is mgH regardless of the body's
state of motion. Transients at either end of a descent or ascent
adding to kinetic energy of the of the body are just that, transients,
and will also cancel if we return the body to its former state of
motion at the end of the excursion.
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| User: "Uncle Al" |
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| Title: Re: question about "work" |
27 Jul 2003 01:43:13 PM |
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Svilen wrote:
We've been arguing here about something related to "work" in physics
sens. Everyone knows that if you lift a body with mass m from ground
to a hight of H the work you do on the body is mgH, where mg is the
gravity force on the body. But now suppose that the body lays on the
floor and experiences a gravity force mg. You want to lift it up and
according to the formula you apply an equal and opposite force of mg.
However in this case the two forces on the body will cacel each other
and the body should not move from its position, hence no displacement
and no work is done. How come that te formula still says the work is
mgH? The same question would apply to electrostatic force and the work
done on a positive charge to move it against the force of the field.
Biology is not physics. No matter how much your meat overheats, if
the object does not move you have done no work on it. Obviously,
trying to lift an object will put some give into its elastic
deformation at the contact interface.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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