## TomLikesPhysics 2 years ago 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.

1. TomLikesPhysics

Here is what I did so far.

2. henpen

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|$\tan^{-1}(\frac{dy}{dx})=\theta$ $\theta=\tan^{-1}(2x)$ $F=-mgsin(\theta)$

3. henpen

So you seem in the right area

4. TomLikesPhysics

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

5. henpen

Then $\theta=\arctan(a2x)$

6. TomLikesPhysics

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

7. TomLikesPhysics

Is this really the finaly answer? Can´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?

8. henpen

|dw:1351340440158:dw|No, its not

9. henpen

|dw:1351340462854:dw|

10. henpen

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

11. TomLikesPhysics

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

12. henpen

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

13. henpen

Although I think it's unsimplifiable

14. TomLikesPhysics

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

15. henpen

16. TomLikesPhysics

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

17. henpen

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

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