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anonymous
 4 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.
anonymous
 4 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.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here is what I did so far.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0So you seem in the right area

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then \[\theta=\arctan(a2x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, it just looks kind of ugly if we plug that into the equation for the force. F=mgsin(arctan(2ax)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351340440158:dwNo, its not

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351340462854:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah... Pythagoras... I see...Now it looks better. So this is the best we can do?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Simplify sqrt(1+(2ax)^2) maybe You could try to x/y decompose the force, but that's probably uglier

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Although I think it's unsimplifiable

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItc3FydCgxKyh4KV4yKSIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi04IiwiNSIsIi02LjQyIiwiMS41OCJdfV0 It's probably correct the force graph looks about right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351349861448:dw The normal (restrictive) force is always at a normal to the curve
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