How to derive coefficient of friction?

AI Thread Summary
The discussion focuses on deriving the coefficient of static friction (μ_s) for a box at rest on an inclined plane at the critical angle. The correct formula is μ_s = tan(θ), as μ_s is dimensionless and cannot equal mgtan(θ). Participants emphasize the importance of drawing a free body diagram to visualize the forces at play, particularly in the limiting case where the box remains stationary. The gravitational force components are analyzed, leading to the conclusion that the teacher's reference to mgtan(θ) is incorrect. Overall, the key takeaway is that μ_s should be derived as tan(θ) for the scenario described.
Meteo
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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?
 
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Meteo said:
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
 
\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
 
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.
 
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