Friction angle ($\phi =\arctan \mu$)

The limiting angle between a contact resultant and the compressive normal force when static friction obeys $|F_t|\le \mu N$. Resolving the resultant gives $\tan \phi =|F_t|/N$, so admissible contact forces lie within a cone of half-angle $\arctan \mu$ about the inward normal. A contact-force magnitude is generally $\sqrt{N^2+F_t^2}$, distinct from N.

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.

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.