Editing Angle of repose (section)

==Formulation==

If the [[coefficient of static friction]] ''μ''<sub>s</sub> is known of a material, then a good approximation of the angle of repose can be made with the following function. This function is somewhat accurate for piles where individual objects in the pile are minuscule and piled in random order.<ref>
{{cite book
 |last1=Nichols |first1=E. L.
 |last2=Franklin |first2=W. S.
 |year=1898
 |title=The Elements of Physics
 |url=https://books.google.com/books?id=8IlCAAAAIAAJ
 |volume=1 |page=101
 |publisher=[[Macmillan Publishers (United States)|Macmillan]]
 |lccn=03027633
}}</ref>

:<math>\tan{(\theta)} \approx \mu_\mathrm{s}\,</math>

where <math>\theta</math> is the angle of repose.

[[File:Free Body Diagram (Angle of Repose).png|thumb|This free body diagram demonstrates the relationship between angle of repose and material on the slope.]]

A simple [[free body diagram]] can be used to understand the relationship between the angle of repose and the stability of the material on the [[slope]]. For the heaped material to resist collapse, the frictional forces must be equivalent to the horizontal component of the [[Gravity|gravitational force]] <math>m g \sin\theta</math>, where <math>m</math> is the mass of the material, <math>g</math> is the gravitational acceleration and <math>\theta</math> is the slope angle:

:<math>m g \sin\theta = f</math>

The frictional force <math>f</math> is equivalent to the multiplication product of the coefficient of static friction <math>\mu</math> and the Normal Force <math>N</math> or <math>mg\cos\theta</math>:

:<math>m g \sin\theta = N \mu </math>

:<math>m g \sin\theta = \mu m g \cos\theta </math>

:<math>\left ( \frac{\sin \theta}{cos\theta} \right ) = \mu </math>

:<math>\theta_R = \arctan(\mu) </math>

Where <math>\theta_R  </math> is the angle of repose, or the angle at which the slope fails under regular conditions, and <math> \mu </math> is the coefficient of static friction of the material on the slope.