anonymous
  • 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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Here is what I did so far.
1 Attachment
anonymous
  • 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
  • anonymous
So you seem in the right area

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anonymous
  • anonymous
Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.
anonymous
  • anonymous
Then \[\theta=\arctan(a2x)\]
anonymous
  • anonymous
Yes, it just looks kind of ugly if we plug that into the equation for the force. F=-mgsin(arctan(2ax)
anonymous
  • 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
  • anonymous
|dw:1351340440158:dw|No, its not
anonymous
  • anonymous
|dw:1351340462854:dw|
anonymous
  • anonymous
F=-mgsqrt(1+(2ax)^2)
anonymous
  • anonymous
Ah... Pythagoras... I see...Now it looks better. So this is the best we can do?
anonymous
  • anonymous
Simplify sqrt(1+(2ax)^2) maybe You could try to x/y decompose the force, but that's probably uglier
anonymous
  • anonymous
Although I think it's unsimplifiable
anonymous
  • anonymous
Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.
anonymous
  • anonymous
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc3FydCgxKyh4KV4yKSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi04IiwiNSIsIi02LjQyIiwiMS41OCJdfV0- It's probably correct- the force graph looks about right
anonymous
  • anonymous
The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?
anonymous
  • anonymous
|dw:1351349861448:dw| The normal (restrictive) force is always at a normal to the curve

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