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One of a pair of fault orientations that can form under a common principal stress field with opposite senses of resolved shear stress. For newly formed, optimally oriented faults obeying the same Mohr-Coulomb failure criterion, their acute intersection angle is $90^{\circ }-\varphi$, where $\varphi$ is the friction angle. The maximum compressive principal stress bisects that acute angle. This permits an estimate $\mu =\tan \varphi$ only under the common-formation and optimal-orientation assumptions, rather than for arbitrary inherited faults.

In the symmetric pinch geometry of normal versus resultant force in a frictional pinch, each inward normal force $N$ has downward vertical component $N\sin \alpha$, while limiting upward static friction contributes $\mu N\cos \alpha$. The maximum net upward support is thereforeFor $0\le \alpha <\pi /2$, lifting a positive weight requires $\mu >\tan \alpha$. At equality the two vertical contributions cancel for every grip magnitude; below it, squeezing harder increases a downward resultant rather than supplying upward support.

The condition compares geometry with the friction angle $\arctan \mu$. It explains why a sufficiently steep outward-widening object cannot be raised by this particular symmetric squeeze alone. The conclusion assumes the stated face orientation and ideal contact law; a different surface orientation, an adhesive contact or a ledge changes the force balance.

For symmetric upper faces inclined by alpha from the vertical, inward normals push partly downward. Limiting upward friction gives vertical support $2N(\mu \cos \alpha -\sin \alpha )$. Balancing weight yields the displayed minimum normal grip. The corresponding minimum total contact magnitude per face is $N_{\min }\sqrt{1+{\mu }^2}$. It is achieved at the limiting friction angle. The formula assumes symmetric contacts with cancelling horizontal force and torque and positive denominator.

The PDF's apex angle is $30^{\circ }$, so each contacted face makes $\alpha =15^{\circ }$ with the vertical. Let N be the inward normal force per face. Its vertical component is downward, $-N\sin \alpha$; limiting upward static friction contributes $\mu N\cos \alpha$. Symmetric contacts cancel horizontal forces and torque. Incipient lift consequently requiresThis is the usual normal gripping force convention, using $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. Friction can lift the prism only if $\mu >\tan \alpha$, which is satisfied here.

If “force applied” means the magnitude of the entire finger-contact resultant, including its tangential component, the normal versus resultant force in a frictional pinch givesThe optimal resultant lies at the limiting friction angle $\arctan \mu$ from the inward normal. Stating which force is being measured avoids confusing these two legitimate quantities.

Treat the two strands as newly formed conjugate faults obeying the same Mohr-Coulomb failure criterion, rather than merely inherited faults reactivated in a later stress field. A fault strike is an unoriented line, so $296^{\circ }$ is equivalent to $116^{\circ }$. The acute angle between the fault strikes is therefore $116^{\circ }-60^{\circ }=56^{\circ }$. The formal compressive acute bisector puts ${\sigma }_1$ at azimuth $88^{\circ }$ and ${\sigma }_3$ at $178^{\circ }$. This orientation must also be checked against the mapped senses of strike-slip faulting, as discussed below.

Let $\beta =28^{\circ }$ be the angle between either fault plane and ${\sigma }_1$. Tangency of Mohr's circle to $|\tau |=c+\mu {\sigma }_n^{\prime }$, where ${\sigma }_n^{\prime }$ is effective stress normal to the plane and $c$ is cohesion, gives $\beta =45^{\circ }-\varphi /2$ and $\mu =\tan \varphi$. ThusIn the Mohr circle, the fault planes are $112^{\circ }$ apart around the circle, because a physical plane rotation is doubled. The figure uses $c=0$ to display both tangencies; a common nonzero cohesion changes the intercept, not the inferred friction angle. The inference of $\mu$ requires optimally oriented conjugate faults at formation: the two fault strikes alone cannot measure a friction coefficient for arbitrarily reactivated faults.

Consistency limitation: the mapped westward motion of the Anatolian block corresponds to left-lateral strike-slip faulting on the East Anatolian strand and right-lateral strike-slip faulting on the North Anatolian strand. East–west maximum compression for the acute-bisector construction instead predicts the opposite pair. The stress orientation compatible with the mapped senses lies near the north–south, obtuse bisector; the planes are then about $62^{\circ }$ from ${\sigma }_1$, which would give $\varphi =90^{\circ }-2(62^{\circ })=-34^{\circ }$ in the optimal-formation formula. A negative friction angle is inadmissible for ordinary frictional strength. Thus $0.675$ is the formal magnitude inferred from unoriented angles, while the full map does not support a self-consistent positive-friction, optimally formed conjugate pair. The common-stress assumption can still describe inherited or anisotropically controlled faults, but then their orientations no longer determine a unique coefficient.