Why is work done expressed as F.dx instead of x.dF?

[Mr Real]

Okay, I think further discussions will not prove to be much fruitful at this point. But this discussion helped in clearing up some of my doubts. So thanks to everyone for helping me.
And I just hope that if someone at some point of time reads all this and knows what we all have been missing, they'll just post or start a conversation
Thank you all!

Warm regards
Mr R

 

[Dale]

someone at some point of time reads all this and knows what we all have been missing

I don't think we have been missing anything. You may not like the answer, but that doesn't mean that the answer is missing.

 

[Mr Real]

I don't think we have been missing anything. You may not like the answer, but that doesn't mean that the answer is missing.

It's not that I didn't like the answers. As I said some of my doubts were cleared (e.g. when someone if force is constant then x.dF is equal to zero which is not so for F.dx that proved why when force is constant x.dF cannot have anything to do with work). Like this I think there is a simple answer to the question: does x.dF mean anything or has anything to do with work? and that'll be pointed out to us in due time.

 

[Dale]

I think there is a simple answer to the original question too

See post 4 and post 64

I think there is a simple answer to the question: does x.dF mean anything or has anything to do with work?

The answer is indeed simple: "no"

 

[jack action]

does x.dF mean anything or has anything to do with work?

Let me show you what you are trying to do with a simpler example. Here is a prism with known side length $a$, $b$ and $c$:

Just like $Fdx$ represents work, $ab$ represents the area of one side of this prism. $a$ is a length, $b$ is a length and length times a length is an area.

Asking what is the meaning of $xdF$ is like asking what is the meaning of $ac$. $ac$ is meaningless in this case. Even if $a$ is a length and $c$ is a length, $ac$ is not an area.

So the answer to your question is $xdF$ means nothing, just like $ac$ means nothing. And I doubt someone will ever find meaning to them, even if it is mathematically possible to multiply both variables together. If it is meaningless, then it certainly doesn't have anything to do with work.

 

[Pedro Zanotta]

I´m not a physics, so, sorry for my comentary (if it turns out to be too naive). But there is a fundamental problem, in my opinion, with your doubt. As I understand it, you presume a fixed distance (x) and the force is varying, right? Then you integrate it to get the work done (in terms of dF). Ok? But if x is constant, there is no deslocation. So, although the force had been applied, there were no work done (as you had said elsewhere). I think this is a definition, physiclly speaking. Now, if force has been applied, then could work has been done? That depends: suppose you have an ideal elastic ribbon. You tied it to a stone and to a tree. Depending of the stone and the tree, the system could stand still. So, the ribbon is "tense" (there are forces) but the stone and the tree stay there. Now, if you have a machine (suppose a caterpillar) then you could try to displace the stone. But the stone is too heavy for the catterpillar. So, the motor works, but the stone endures. The motor has done work (it rotates, burn oil, etc) that could be measured. But, as the stone didn´t move, no mechanical work (in terms of the displacement of the stone) has been done.
All of this make sense?
My two coins
Pedro
(Sorry for my english)

 

[sophiecentaur]

@ Pedro Zanotta I think that the right message is in there somewhere Basic thing is that no work is done on a system if there is no movement, however great the force is. Caterpillar is too weak to make any change so no work. Machine is strong enough to get over the stiction of the stone and can do work.
Remember that this definition of work is 'pure' and takes no account of sweat and groaning whilst pushing at something that just won't move or in supporting a dead weight that stays in the same place. This is why scenarios such as the one you propose tend to get in the way of the actual Physics. Physics tries to reduce things to the very basics.

 

[Mr Real]

As I understand it, you presume a fixed distance (x) and the force is varying, right? Then you integrate it to get the work done (in terms of dF). Ok? But if x is constant, there is no deslocation.

Actually, no x.dF does not mean that x is considered fixed, it is constant only for the small force dF but it can vary for the total force F (just like in F.dx we assume F constant for small displacement dx but it can vary for the total displacement, x).

 

[jack action]

Actually, no x.dF does not mean that x is considered fixed, it is constant only for the small force dF but it can vary for the total force F (just like in F.dx we assume F constant for small displacement dx but it can vary for the total displacement, x).

$xdF$ means $x(F_2 - F_1)$ and $Fdx$ means $F(x_2 - x_1)$. Do you see that $dF$ is not a force, it is a force increment and $dx$ is not a position, it is a position increment?

$F$ is not equivalent to $dF$ and $x$ is not equivalent to $dx$.

 

[sophiecentaur]

Actually, no x.dF does not mean that x is considered fixed,

Yes it does. If you want x to vary then you need to state what it varies with thus:- x(F)dF
X can be any function of F that fits a physical situation. And that is where you are getting things wrong. Maths follows the Science. If it is a model of a Physical situation then you can progress from there. You cannot invent a mathematical expression and assume that is has any relevance at all to Science.
At this stage I have a feeling that you just don't want to be wrong, rather than use this thread as a learning experience.

 

[jack action]

If you want x to vary then you need to state what it varies with thus:- x(F)dF

But it still wouldn't define work as x(F) is not a displacement but just a definition of a fixed position x with respect to a force F. The dx have meaning that x(F) doesn't have.

 

[Mr Real]

If you want x to vary then you need to state what it varies with thus:- x(F)dF.

Yes, that's correct but what I said there is that x.dF doesn't mean that x is constant, because it can vary (see I didn't say there that x.dF definitely means that x is variable, rather I pointed out to the user that he was wrongly assuming that x.dF means x is constant, the point is: x can vary)

At this stage I have a feeling that you just don't want to be wrong, rather than use this thread as a learning experience.

Actually, no I have no problem in being wrong, I just want a satisfactory answer, that's why I've spent so much time posting messages here, replying to them, pondering over other people's replies,etc.

 

[Dale]

I pointed out to the user that he was wrongly assuming that x.dF means x is constant, the point is: x can vary

I agree. The integrand is not assumed to be constant. Even when doing numerical integration it is uncommon to approximate it as piecewise constant.

I just want a satisfactory answer

You have had several correct answers. Have you been satisfied with them? If not, why not; shouldn't a correct answer be satisfactory?

 

[Mr Real]

You have had several correct answers. Have you been satisfied with them? If not, why not; shouldn't a correct answer be satisfactory?

Yes, I have got several good answers and they have partially answered my questions, but not completely.

 

[sophiecentaur]

, I just want a satisfactory answer,

You have been given the perfectly satisfactory answer that your arbitrary bit of maths can't be assumed to have a physical interpretation. It's just not satisfactory for you.

 

[Dale]

Yes, I have got several good answers and they have partially answered my questions, but not completely.

I don't see any question of yours which has been partially answered.

1) You wanted to know why the integral of x.dF did not also give work just like F.dx. The complete answer was that it's a calculus error to switch the quantities like that. The area under a curve is not the same as the area to the left of the curve. So whatever x.dF might represent it is not generally work.

2. You wanted to know why textbooks don't spend time discussing this. The complete answer is that there are an infinite number of mistakes you could make, and this is a very rare one. Authors focus primarily on the correct ways, not one of the infinite incorrect ways. When they do address mistakes, it is only the common ones.

3. You wanted to know why the integral of x.dF is not useful in its own right. The complete answer is twofold, first, it is just an odd expression. It is not common to express x as a function of F, and it is often not even possible to do so, and the quantity dF is a weird quantity without physical use elsewhere. Second, there is no known relationship of that quantity to other known physical quantities of interest, nor does it posses known properties that make it a primary quantity of interest.

What could possibly be considered incomplete about this discussion? Not only has this community correctly and completely answered your question, we have been exceptionally patient with you. This discussion should have ended on the first page as soon as it was pointed out that it is a mathematical mistake to switch the terms that way. So your answer has been correct, complete, and exceptionally supportive.

It is highly frustrating to have such thoughtful and helpful replies from the community regarded as insufficient.

At this point I am going to close this thread. I only hope that in the future you can recognize when you have received such high quality responses, and be a little more encouraging to the community that provided them.