B Exact differential and work done

AI Thread Summary
The discussion focuses on the relationship between conservative forces and exact differentials in the context of work done. It explains that if the infinitesimal work 𝐹⃗ β‹…π‘‘π‘Ÿβƒ— is an exact differential, the work is independent of the path taken, indicating that the force is conservative. An example involving gravity illustrates that the total work done is the same regardless of the path, confirming gravity as a conservative force. In contrast, friction is shown to be non-conservative as the total work varies with the path. Understanding exact differentials is essential for grasping why conservative forces exhibit this property.
putongren
Messages
124
Reaction score
1
TL;DR Summary
The work done is independent of path if the infinitesimal work 𝐹⃗ β‹…π‘‘π‘Ÿβƒ—
is an exact differential.
I was researching about conservative and non-conservative forces, and there is some information in a website that sates that the work done is independent of path if the infinitesimal work 𝐹⃗ β‹…π‘‘π‘Ÿβƒ— is an exact differential. It further states that in 2 dimensions the condition for 𝐹⃗ β‹…π‘‘π‘Ÿβƒ— = Fxdx + Fydy to be an exact differential is:

𝑑𝐹π‘₯/𝑑𝑦=𝑑𝐹𝑦/𝑑π‘₯.

My question is this: why is a force conservative if the work is an exact differential? How can we deduce from the definition of a conservative force that this force is conservative if the work done to it is an exact differential?
 
Physics news on Phys.org
putongren said:
My question is this: why is a force conservative if the work is an exact differential?
Because if you integrate an exact differential, the result is the difference of some function evaluated at the two limits. This is what "work is independent of the path" means.

Here is an example. Suppose you use an eraser to erase a chalkboard. To do that you push the eraser against the board and, starting at the bottom B, you go to the middle M, then to the top T and then back to M. Let ##h=~##the height of the board.

The work done by the force of gravity ##mg## is
From B to M: ##~W_{BM}=-mg*\dfrac{h}{2}.##
From M to T: ##~W_{MT}=-mg*\dfrac{h}{2}.##
From T to back to M: ##~W_{TM}=+mg*\dfrac{h}{2}.##
Note that ##W_{BM}+W_{MT}+W_{TM}=W_{BM}.## This says that the total work is independent of the path because it is the same when one stops at M or goes past M to T and then back to M. We conclude that the force of gravity is conservative.

The work done by the force of friction ##f## is
From B to M: ##~W_{BM}=-f*\dfrac{h}{2}.##
From M to T: ##~W_{MT}=-f*\dfrac{h}{2}.##
From T to back to M: ##~W_{TM}=-f*\dfrac{h}{2}.##
Note that ##W_{BM}+W_{MT}+W_{TM}=3W_{BM}.## This says that the total work is not independent of the path because it is not the same when one stops at M or goes past M to T and then back to M. We conclude that the force of gravity friction is not conservative.
 
Last edited:
kuruman said:
The work done by the force of friction ##f## is
From B to M: ##~W_{BM}=-f*\dfrac{h}{2}.##
From M to T: ##~W_{MT}=-f*\dfrac{h}{2}.##
From T to back to M: ##~W_{TM}=-f*\dfrac{h}{2}.##
Note that ##W_{BM}+W_{MT}+W_{TM}=3W_{BM}.## This says that the total work is not independent of the path because it is not the same when one stops at M or goes past M to T and then back to M. We conclude that the force of *gravity* is not conservative.

I believe you meant friction. Besides that, great explanation.
 
  • Like
Likes berkeman and kuruman
Juanda said:
I believe you meant friction. Besides that, great explanation.
Good catch, thanks. I cut and pasted from above without changing the force. I edited to fix the typo.
 
  • Like
Likes berkeman and Juanda
Sorry, I assumed you already knew what an exact differential is.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top