Codex @codex  0

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.