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

1. Oct 25, 2009

ha9981

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. Oct 25, 2009

HallsofIvy

Staff Emeritus
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 $sin^2(y)+ cos^2(y)= 1+ Uk^2$ rather than 1. But a way to fix that is to multiply both sides of the equation by $\sqrt{1+ Uk^2}$.

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

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

That is,
$$x= arcsin(\frac{a}{g\sqrt{1+ Uk^2}}- arcsin(\frac{Uk}{\sqrt{1+ Uk^2}}$$

3. Oct 25, 2009

ha9981

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!

4. Oct 26, 2009

Gerenuk

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 $\tan x$