Proving F(tang) = -dU/ds

1. Jan 22, 2010

Oijl

1. The problem statement, all variables and given/known data
[[F(tang) means the tangential force]]
Consider a bead constrained to move on a tract curved in three-dimensional space, with the bead's position specified by its distance s, measured along the wire, from the origin.

One force on the bead is the normal force N of the wire (which constrains the bead to stay on the wire). If we assume that all other forces (gravity, etc.) are conservative, then their resultant can be derived from a potential energy U. Prove that F(tang) = -dU/ds. This shows that one-dimensional systems of this type can be treated just like linear systems, with x replaced by s and Fx by F(tang).

2. Relevant equations

3. The attempt at a solution
I have before me a solution someone wrote that looks like it could be right, but I can't follow one of the steps. Here is what they have written:

F = N - $$\nabla$$U
Where N is the normal force, and F is the net force.
The tangental force will be F(tang) = $$\hat{}v$$(-$$\nabla$$U)

Now consider a small displacement along the wire. Then we should have
ds = ds $$\hat{}v$$
***
Then dU = ds$$\hat{}v$$$$\nabla$$U
***
= -F(tang)ds

Therefore you can write
-dU/ds = F(tang)

The part I don't follow is enclosed with asterisks.

2. Jan 22, 2010

RoyalCat

I don't quite follow that proof either, but I offer an alternative one:

Since $$\vec F_{conservative_{net}}=-\nabla U$$ (Where we've ignored the normal force completely since it is radial to the wire at all times.)
Then it is an immediate consequence that the tangential component of the RHS is the same as the tangential component of the LHS (Since this is a vector equation)

Find a way to express the components of the RHS along the direction of the wire (In the direction of $$d\vec s$$ using geometry and then use the definition of $$\nabla U$$ and some algebra to get the desired expression.

Hint on the geometry: $$d\vec s = d\vec x+d\vec y+d\vec z$$
Think about right-angled triangles. :) And remember that the magnitudes of these differentials are rarely the same! Do not assume $$|{d\vec x}|=|{d\vec y}|=|{d\vec z}|$$

Last edited: Jan 22, 2010
3. Jan 22, 2010

Oijl

I'm sorry, I'm just not getting it. I'm bad at proofs.

So -{grad}U = F∂x, F∂y, F∂z
and
ds = sqr(dx^2 + dy^2 + dz^2)

And doesn't -U = integral(Fdr)?
So....

At any rate, I've slept from sometime after 6:15am to sometime before 7:15am, and that was my full night's sleep. I'm going to put myself down for another hour and a half, and then I'll be back on here to ask more questions in case there's been a response.

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