1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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
    [tex]
    \frac{a^2}{g^2\cos^2 x}=(\tan x-Uk)^2
    [/tex]
    and the identity
    [tex]
    \frac{1}{\cos^2 x}=1+\tan^2 x
    [/tex]
    on the LHS. Make sure you factor out correctly the RHS and then you will get a quadric equation for [itex]\tan x[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help solve for x in a=g(sin(x)-Uk*cos(x))
  1. Solve x^x=x (Replies: 7)

  2. Sin x =cosh x (Replies: 12)

  3. Inverse of sin(x)+x (Replies: 10)

Loading...