Solution

By hypothesis, the coefficient of sliding friction between the pencil and the inclined plane satisfies the condition $$\mu \ge \tan \alpha$$. Indeed, the pencil put at right angles to the generatrix is in equilibrium, which means that $$mg \sin \alpha =F_{fr}$$, where $$mg$$ is the force of gravity, and $$F_{fr}$$ is the force of friction. But $$F_{fr}\le \mu mg\cos \alpha$$. Consequently , $$mg \sin \alpha \le \mu mg\cos \alpha $$, where $$\mu \ge \tan \alpha$$.Thus, the pencil will not slide down the inclined plane for any value of the angle $$\phi$$.The pencil may start rolling down at an angle $$\phi_0$$ such that vector of the force of gravity "leaves" the region of contacts between the pencil and the inclined plane (hatched region in Fig $$183$$).In order to find this angle, we project the center of mass of the pencil (Point $$A$$) on the inclined plane and mark the point of intersection of the vertical passing through the center of mass and the onclined plane (Point $$B$$). Obviously, points $$A$$ and $$B$$ will be at rest for different orientations of the pencil its center of mass remains stationary. In this case, $$AB=2l \cos 30^o\tan \alpha,$$ where $$2l$$ is the side of hexagonal cross section of the pencil, and $$2l\cos 30^o$$ is the radius of the circle inscribed in the hexagonal cross section.As long as point $$B$$ lies in the hatched region, the pencil will not roll down the plane.Let us write the condition for the beginning of rolling down$$\dfrac{AD}{\cos \phi_0}=AB$$ or $$\dfrac{l}{\cos \phi_0}=l\sqrt{3}\tan \alpha$$,whence$$\phi_0=arc\cos \left(\dfrac{1}{\sqrt{3}\tan \alpha}\right)$$.Thus, if the angle $$\phi$$ satisfies the condition$$arc\cos \left(\dfrac{1}{\sqrt{3}\tan \alpha}\right)\le \phi \le \dfrac{\pi}{2}$$,The pencil remains in equilibrium. The expression for the angle $$phi_0$$ is meaningful provided that $$tan \alpha >1/\sqrt{3}$$. The fact that the pencil put parallel to the generatrix rolls down indicates that $$\tan \alpha > 1/\sqrt{3}$$ (proce this).