 Solution

A numerical answer needs an assumption about fault orientation. For a cohesionless fault optimally oriented for Mohr-Coulomb failure criterion, with vertical ${\sigma }_1$, write $\phi$ for the internal friction angle and $\delta$ for the fault dip. The angle between the plane and ${\sigma }_1$ is $90^{\circ }-\delta$, so the optimal normal-fault dip relation givesHence ${\mu }_f=\tan 40^{\circ }=0.839\simeq 0.84$ under that model.

For the Mohr circle, plot effective compression ${\sigma }_n^{\prime }$ horizontally and shear stress $\tau$ vertically. Draw a circle with diameter $[{\sigma }_3^{\prime },{\sigma }_1^{\prime }]$ tangent to $\tau ={\mu }_f{\sigma }_n^{\prime }$. Its centre and radius obey $R/C=\sin \phi$. One illustrative normalization is ${\sigma }_3^{\prime }=1$, ${\sigma }_1^{\prime }=4.59891$, with tangency at the resolved stresses on the $65^{\circ }$ plane. Pore fluid pressure shifts the total-stress circle horizontally; the friction law uses effective stress.

The dip does not determine the actual sliding coefficient of friction without the optimal-orientation assumption. A reactivated inherited plane may be nonoptimal, cohesion may matter, and the stress ratio and pore fluid pressure are not measured. For an arbitrary compression-positive stress state,at cohesionless sliding. Thus the requested single value is a conditional inference, not a friction measurement.