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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.

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.