How to derive coefficient of friction?

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Homework Help Overview

The discussion revolves around deriving the coefficient of static friction (\mu_s) for a box at rest on an inclined plane at the critical angle. Participants are examining the relationship between the forces acting on the box and the angle of inclination.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of the formula \mu_s = mgtan(\theta) and question the dimensional consistency of this expression. There are references to free body diagrams and the balance of forces acting on the box.

Discussion Status

Some participants have pointed out potential errors in the original formula proposed, suggesting that \mu_s should equal tan(\theta) instead. There is an ongoing exploration of the conditions under which these relationships hold, particularly in the context of static friction and the critical angle.

Contextual Notes

There is a mention of the dimensionality of \mu_s, which is dimensionless, raising questions about the validity of the proposed formula involving mass and gravitational components.

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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