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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.
 2 years ago
 2 years 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.
 2 years ago
 2 years ago

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TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Here is what I did so far.
 2 years 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\[\tan^{1}(\frac{dy}{dx})=\theta\] \[\theta=\tan^{1}(2x)\] \[F=mgsin(\theta)\]
 2 years ago

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

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

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

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

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

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