Line integral and path dependence question

[Old Guy]

Given F = iy - jx (this is my first post; not sure how you do vector notation here but I'm showing vectors in bold - hope that works). The problem is to show that this is a non-conservative force by integrating from the origin to (1,1) (ie, the path is y=x), and then do it again from the origin to (0,1) and from there to (1,1).

I just don't get how to set up the integrals - can someone help? I understand the concept behind the line integrals, but hit a brickwall trying to apply it. Thanks.

 

[neutrino]

_C\int_{A}^{B}\vec{F}.d\vec{r}

The first part is quite simple - direct substitution. For the second part, what are x and dx =? when you go from (0,0) to (0,1)?

 

[Old Guy]

Sorry, I don't know how to show integrals in here, but what I believe you do is that F remains the same, and you dot it with dr, which will be (dx,0) for the path from (0,0) to (0,1), and (0,dy) for the path from (0,1) to (1,1). Right so far?

 

[neutrino]

Sorry, I don't know how to show integrals in here

Refer to this https://www.physicsforums.com/showthread.php?t=8997

but what I believe you do is that F remains the same, and you dot it with dr, which will be (dx,0) for the path from (0,0) to (0,1), and (0,dy) for the path from (0,1) to (1,1). Right so far?

You're right about the differentials, but what about x and y along those paths?

 

[Old Guy]

Thanks for the reference. And I think you've hit on what I don't get - what about x and y? F is the same regardless of path, isn't it? So what's the point (and how is it calculated)? I think that's what I"m missing.

 

[neutrino]

F = yi - xj, but what's y equal to when you're on the x-axis; in other words, when moving from (0,0) to (0,1)?

 

[Old Guy]

y=0 along the x-axis

 

[neutrino]

EDIT: Sorry, I've been confusing coordinates here. When you're moving from origin to (0,1) dx = 0 and x =0...it's along the y-axis. I apologise.

 

[Old Guy]

I'm not clear on what the question is now . . .

 

[neutrino]

Okay, here's how you do it for the path from (0,0) to (0,1)

\int_{(0,0)}^{(0,1)}(y\hat{i} - x\hat{j}).(dx\hat{i} + dy\hat{i})

But since you're moving along a path in which x is a constant, in this case, zero, the above reduces to...

\int_{y=0}^{y=1}(ydx - xdy) = 0 (x = 0 and dx = 0)

Is that clear?

 

[Old Guy]

Yes, I got that.

 

[Old Guy]

And I also don't understand how this could come to 0, because the particle was moved over a finite distance by a force - SOME amount of work should have been done!

 

[neutrino]

Not necessarily. The work can be zero even in simple circumstances such as carrying a box ("horizontally"). There is a force, the box was moved through a finite distance, yet no (physical) work was done.

When you consider line integrals such as the above, which is a generalisation of the formula W = F.d, you evaluate the work at every point along the path, and then add it all up. It can be positive, negative, or zero.

 

[Old Guy]

Right, I do recall that. On this problem, however, the work seems to calculate to 0 for either path, which would imply the force is conservative, which my book says it is not (and I also calculated curl F to confirm this). What am I missing here?

 

[neutrino]

Did you calculate the integral from (0,1) to (1,1)?

 

[Old Guy]

Yes; it was also 0.

 

[neutrino]

Can you show your work?

 

[Old Guy]

Yes, this also came to 0.

 

[neutrino]

The path from (0,1) to (1,1) is line y = 1, dy=0

 

[Old Guy]

I still need a little time to figure out the LaTex stuff, but what I did was for the first step (y,-x) dot (dx,dy). dx=0, so I get -xdy, integrate to get -xy, and evaluat from y=0 to y=1 gives -x.

For the second step, (y,-x) dot (dx,dy) has dy=0, so I get ydx, integrate to get xy, and evaluate from x=0 to x=1 gives y. Summing, I get -x+y. Is the point being made that -x+y does not =0 because the values of x and y along the path do not =0? I'm looking now at my direct integration, which resulted in an integration of (xdx + ydy) from 0 to 1, for which I got 1. Is the force non-conservative because -x+y does not =1?