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TomLikesPhysics
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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.
 one year ago
 one year ago
TomLikesPhysics Group Title
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.
 one year ago
 one year ago

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TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Here is what I did so far.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
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\[\tan^{1}(\frac{dy}{dx})=\theta\] \[\theta=\tan^{1}(2x)\] \[F=mgsin(\theta)\]
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
So you seem in the right area
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Then \[\theta=\arctan(a2x)\]
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Yes, it just looks kind of ugly if we plug that into the equation for the force. F=mgsin(arctan(2ax)
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1351340440158:dwNo, its not
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1351340462854:dw
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
F=mgsqrt(1+(2ax)^2)
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Ah... Pythagoras... I see...Now it looks better. So this is the best we can do?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Simplify sqrt(1+(2ax)^2) maybe You could try to x/y decompose the force, but that's probably uglier
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Although I think it's unsimplifiable
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc3FydCgxKyh4KV4yKSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi04IiwiNSIsIi02LjQyIiwiMS41OCJdfV0 It's probably correct the force graph looks about right
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1351349861448:dw The normal (restrictive) force is always at a normal to the curve
 one year ago
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