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.