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TomLikesPhysics
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.
Here is what I did so far.
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|\[\tan^{-1}(\frac{dy}{dx})=\theta\] \[\theta=\tan^{-1}(2x)\] \[F=-mgsin(\theta)\]
So you seem in the right area
Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.
Then \[\theta=\arctan(a2x)\]
Yes, it just looks kind of ugly if we plug that into the equation for the force. F=-mgsin(arctan(2ax)
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?
|dw:1351340440158:dw|No, its not
Ah... Pythagoras... I see...Now it looks better. So this is the best we can do?
Simplify sqrt(1+(2ax)^2) maybe You could try to x/y decompose the force, but that's probably uglier
Although I think it's unsimplifiable
Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc3FydCgxKyh4KV4yKSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi04IiwiNSIsIi02LjQyIiwiMS41OCJdfV0- It's probably correct- the force graph looks about right
The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?
|dw:1351349861448:dw| The normal (restrictive) force is always at a normal to the curve