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Help solve for x in a=g(sin(x)-Uk*cos(x))

  1. Oct 25, 2009 #1
    I want to solve for x

    I only get as for as
    a/g = sin(x)-Uk*cos(x)
    a/gcos(x) = tan(x) - Uk

    im not sure which trig id to use after that! I think the Uk is probably throwing me off. (Its a constant, constant of kinetic friction)
  2. jcsd
  3. Oct 25, 2009 #2


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    I would recommend using either cos(x+y)= cos(x)cos(y)- sin(x)sin(y) or sin(x+y)= sin(x)cos(y)+ sin(x)sin(y). (I don't think it matters which.)

    That is, think of cos(x)cos(y)- sin(x)sin(y)= sin(x)- Uk cos(x) so that "cos(y)= -1" and "sin(y)= -Uk". That won't work directly because that gives [itex]sin^2(y)+ cos^2(y)= 1+ Uk^2[/itex] rather than 1. But a way to fix that is to multiply both sides of the equation by [itex]\sqrt{1+ Uk^2}[/itex].

    Doing that gives a/g = sin(x)-Uk*cos(x)
    [tex]\frac{a}{g\sqrt{1+ Uk^2}}= \frac{1}{\sqrt{1+ Uk^2}}- \frac{Uk}{\sqrt{1+Uk^2}}cos(x)[/tex].
    Now, comparing those to sin(x+y) we would have [itex]sin(y)= -\frac{Uk}{\sqrt{1+ Uk^2}}[/itex] and [itex]cos(y)= -\frac{1}{\sqrt{1+ Uk^2}}[/itex]

    Now your equation becomes [itex]sin(x+y)=\frac{a}{g\sqrt{1+Uk^2}}[/itex] so
    [tex]x+ y= arcsin(\frac{a}{g\sqrt{1+ Uk^2}}[/tex]
    with [itex]y= -arcsin(\frac{Uk}{\sqrt{1+Uk^2}}[/itex].

    That is,
    [tex]x= arcsin(\frac{a}{g\sqrt{1+ Uk^2}}- arcsin(\frac{Uk}{\sqrt{1+ Uk^2}}[/tex]
  4. Oct 25, 2009 #3
    is that the easiest way? cuz that seems really hard. i heard of another way in which u incorporate sinx/cosx = tanx

    I tried this:

    a/g = sin(x) - Ucos(x)

    a/gcos(x) = tanx - U

    a^2/g^2 cos^2(x) = (sec^2(x) - 1) - U^2

    a^2/g^2 = 1-cos^2(x) - U^2cos^2(x)

    a^2/g^2 -1 = (-1-u^2) cos^2(x)
    cosx = sqrt( (a^2/g^2 -1 ) / (-1-u^2) )
    x=arccos (sqrt( (a^2/g^2 -1 ) / (-1-u^2) ) )

    but i messed up somewhere cuz ans of x won't work!
  5. Oct 26, 2009 #4
    You cannot square the terms on the RHS individually.

    But your idea could work. For example you could use
    \frac{a^2}{g^2\cos^2 x}=(\tan x-Uk)^2
    and the identity
    \frac{1}{\cos^2 x}=1+\tan^2 x
    on the LHS. Make sure you factor out correctly the RHS and then you will get a quadric equation for [itex]\tan x[/itex]
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