Solving Mechanics Question: Find Norm of F

[Ed Aboud]

Homework Statement

Let $$r = \sqrt{x^2 +y^2 +z^2} = \lVert x \rVert _\mathbb{R} _^3$$
be the Euclidean distance of the point $$x = (x,y,z) \in \mathbb{R} ^3$$ from the origin.

And $$e_r := \nabla r$$

Let F be a central force , i.e.,


$$\underline{F} = - \nabla U(r)$$

for some function $U : \mathbb{R} \rightarrow \mathbb{R}$

Show that

$$\underline{F} = \pm \lVert \underline{F} \rVert e_r$$

What is $$\lVert F \rVert$$ ?

Homework Equations

The Attempt at a Solution

I just don't really understand what is being asked here.
Any help would be appreciated.
Thanks.

 

[Delphi51]

Why don't you write up the gradient for tau in partial derivative notation.
Also do it for the gradient of function U. Take a good look at the "show that" line in this notation and perhaps inspiration will strike!

 

[Ed Aboud]

Ok so you get $$\nabla U(r) = ( \frac{\partial }{\partial x}U(r) + \frac{\partial }{\partial y}U(r) + \frac{\partial }{\partial z}U(r))$$

I'm still not really sure what the question is asking me for.

Thanks again.

 

[Delphi51]

Okay, you've got the left side of the equation you are supposed to find.
Now work on the right side, the gradient of that tau = square root(x^2 + y^2 + z^2)

 

[Ed Aboud]

So wait you're saying that $$\nabla U(r) = ( \frac{\partial }{\partial x}U(r) + \frac{\partial }{\partial y}U(r) + \frac{\partial }{\partial z}U(r)) = \nabla r$$

 

[Delphi51]

Show that
$$\underline{F} = \pm \lVert \underline{F} \rVert e_r$$

I better confess that I don't know how to do this problem.
I just thought you should try writing expressions for the two sides of this thing they are asking you to show, before giving up. Who knows, once you see it written out in partial derivative form, you may see how to show the two sides are equal.

That expression in your last post isn't finished - you have to do the gradient partial derivatives on the right side and also multiply by that magnitude of F (or whatever the double vertical lines around F mean).

 

[Ed Aboud]

Oh ok, well thanks for trying anyway.

Can anyone else help please?