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DoITPoMS Teaching & Learning Packages Granular Materials Proof of the formula for the angle of repose as a function of θ and φ

# Proof of the formula for the angle of repose as a function of θ and φ

Four smooth identical spheres pack together in a regular tetrahedron. Suppose that the centres of three spheres are at A, B and C on a flat surface in the above diagram for a general φ' = 60° - φ , with the fourth sphere at D above the plane on which A, B and C lie. It is expedient to choose C to be at the origin (0, 0, 0) in an orthonormal $$\hat x$$ - $$\hat y$$ - $$\hat z$$  space. Friction can be assumed to hold these particles together in a stable configuration.

When φ' = 0°, CB is parallel to the $$\hat x$$-axis, whereas when φ' = 60°, AB is parallel to the $$\hat x$$-axis.

If a rotation is now made about the $$\hat x$$ axis by an angle θ, then A, B and D rotate to new positions in $$\hat x$$ - $$\hat y$$ - $$\hat z$$  space. Projected onto the $$\hat x$$ - $$\hat y$$  plane, these positions will be Ap, Bp and Dp, respectively. The condition for stability of the tetrahedral arrangement (assumed still to be made stable by a sufficient level of friction) is that Dp must lie in the new projected triangle whose corners are Ap, Bp and C.

For simplicity, we can choose the sides of the regular tetrahedron to be of unit length.

B and A are initially at (cos φ', sin φ', 0) and (cos (φ' + 60°), sin (φ' + 60°), 0) respectively, while D is initially at

$\left( {\frac{1}{{\sqrt 3 }}\cos (\varphi ' + 30^\circ ),\frac{1}{{\sqrt 3 }}\sin (\varphi ' + 30^\circ ),\frac{{\sqrt 2 }}{{\sqrt 3 }}} \right){\rm{ }}$

A rotation R of θ about the $$\hat x$$-axis rotates vectors R to new positions R' in $$\hat x$$ - $$\hat y$$ - $$\hat z$$  space so that R' = Rr where

$R = \left( {\begin{array}{*{20}{c}} 1&0&0\\ 0&{\cos \theta }&{ - \sin \theta }\\ 0&{\sin \theta }&{\cos \theta } \end{array}} \right){\rm{ }}$

Hence, for example, the ‘old’ vector [010] rotates to become the ‘new’ vector [0 , cos θ, sin θ]. Similarly, the vector CB, CA and CD rotate to the new positions

$\left( {\cos \varphi ',\cos \theta \sin \varphi ',\sin \theta \sin \varphi '} \right),$

$\left( {\cos {\rm{ }}(\varphi ' + 60^\circ ),\cos \theta \sin {\rm{ }}(\varphi ' + 60^\circ ),\sin \theta \sin {\rm{ }}(\varphi ' + 60^\circ )} \right),$

and

$\left( {\frac{1}{{\sqrt 3 }}\cos (\varphi ' + 30^\circ ),\frac{1}{{\sqrt 3 }}\cos \theta \sin (\varphi ' + 30^\circ ) - \frac{{\sqrt 2 }}{{\sqrt 3 }}\sin \theta ,\frac{1}{{\sqrt 3 }}\sin \theta \sin (\varphi ' + 30^\circ ) + \frac{{\sqrt 2 }}{{\sqrt 3 }}\cos \theta } \right){\rm{ }}$

respectively.

The projection of D onto the $$\hat x$$ - $$\hat y$$  plane, Dp, will lie in the triangle, CBpAp until the vector CDp is parallel to the vector CBp, i.e., the limiting condition is that

$\left[ {\cos \varphi ',\cos \theta \sin \varphi ',0} \right]$

is parallel to

$\left[ {\frac{1}{{\sqrt 3 }}\cos (\varphi ' + 30^\circ ),\frac{1}{{\sqrt 3 }}\cos \theta \sin (\varphi ' + 30^\circ ) - \frac{{\sqrt 2 }}{{\sqrt 3 }}\sin \theta ,0} \right]{\rm{ }}$

i.e., a critical θ, θc, is defined by the equality

$\frac{{\cos \varphi '}}{{\cos {\theta _{\rm{c}}}\sin \varphi '}} = \frac{{\cos (\varphi ' + 30^\circ )}}{{\cos {\theta _{\rm{c}}}\sin (\varphi ' + 30^\circ ) - \sqrt 2 \sin {\theta _{\rm{c}}}}}$

Now, φ' = 60° - φ. Substituting for φ' in this equation, the condition becomes

$\frac{{\cos (60^\circ - \varphi )}}{{\cos {\theta _{\rm{c}}}\sin (60^\circ - \varphi )}} = \frac{{\sin \varphi }}{{\cos {\theta _{\rm{c}}}\cos \varphi - \sqrt 2 \sin {\theta _{\rm{c}}}}}$

i.e., the condition

$2\sqrt 2 \sin {\theta _{\rm{c}}}\cos (60^\circ - \varphi ) = \cos {\theta _{\rm{c}}}$

using some elementary trigonometry, or better still the condition

${\theta _{\rm{c}}} = \arctan \frac{1}{{2\sqrt 2 \cos (60^\circ - \varphi )}}{\rm{ }}$