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- anonymous

I have a parabola slide and I am looking for a function of the downhill force which only depends on the x and/or y coordinate.
I came up with a function but I am not sure if this is correct.
You can see everything on the attached picture.

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- anonymous

- jamiebookeater

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- anonymous

Here is what I did so far.

- anonymous

I assume that it is safe to assume that the parabola is y=x^2, with the tip at (0,0)
dy/dx=2x
|dw:1351340058220:dw|\[\tan^{-1}(\frac{dy}{dx})=\theta\]
\[\theta=\tan^{-1}(2x)\]
\[F=-mgsin(\theta)\]

- anonymous

So you seem in the right area

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- anonymous

Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.

- anonymous

Then \[\theta=\arctan(a2x)\]

- anonymous

Yes, it just looks kind of ugly if we plug that into the equation for the force.
F=-mgsin(arctan(2ax)

- anonymous

Is this really the finaly answer?
CanĀ“t we do something to make it simpler or look better? Is there perhaps a different way to determine the force in terms of x and/or y?

- anonymous

|dw:1351340440158:dw|No, its not

- anonymous

|dw:1351340462854:dw|

- anonymous

F=-mgsqrt(1+(2ax)^2)

- anonymous

Ah... Pythagoras... I see...Now it looks better.
So this is the best we can do?

- anonymous

Simplify sqrt(1+(2ax)^2) maybe
You could try to x/y decompose the force, but that's probably uglier

- anonymous

Although I think it's unsimplifiable

- anonymous

Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.

- anonymous

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc3FydCgxKyh4KV4yKSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi04IiwiNSIsIi02LjQyIiwiMS41OCJdfV0-
It's probably correct- the force graph looks about right

- anonymous

The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?

- anonymous

|dw:1351349861448:dw| The normal (restrictive) force is always at a normal to the curve

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