How to derive coefficient of friction?

[Meteo]

I have a box at rest on a plane raised to the critical angle.

I need to derive the formula \mu_s=mgtan(\theta)
I know f_s/n=\mu
n=f_y=mgcos(\theta)
f_s=f_x=mgsin(\theta)

This leaves me with tan(\theta)=\mu cause the mg cancel out... what did I do wrong?

 

[Andrew Mason]

I have a box at rest on a plane raised to the critical angle.

I need to derive the formula \mu_s=mgtan(\theta)
I know f_s/n=\mu
n=f_y=mgcos(\theta)
f_s=f_x=mgsin(\theta)

This leaves me with tan(\theta)=\mu cause the mg cancel out... what did I do wrong?

\mu_s is dimensionless so it cannot be equal to mgtan\theta. Your answer is correct.

AM

 

[daveed]

\mu_s=tan(\theta)
is what you are looking for

Draw a free body diagram.
This is only in the limiting case that it is the maximum \mu_x which will let the body rest on the incline of such an angle without sliding. So in this case you can use Friction=F_n*\mu
And set that equal to the component of the gravitational force pointing down the incline

 

[Meteo]

Yes, that's what I managed to get. I think the mgtan(\theta) must be a mistake the teacher made since \mu cannot equal mgtan(\theta) because its dimensionless as AM said.