Does the book hold on the wall?

Let us first look at what happens if the weight is zero, $F_g=0$ . The normal force (that pushes the book to the wall) is $N=F \cos 30= 0.866 F$ . The friction force $F_f \le \mu N= (0.5)(0.866) F= 0.433 F$ has to balance the force parallel to the wall, $F \sin 30 =0.5 F$ . Since the maximum value of the friction force is $0.433F$ , there is no way to have stable equilibrium. Adding the weight destroys the stabilty even more.

The force F can be divided into tangential and normal components: $\vec F = \vec F_\parallel + \vec F_\perp = - F \sin(30^\circ) \vec e_z + F \cos(30^\circ) \vec e_x$ The normal component of the force determines the maximal frictional force: $F_f \leq \mu |\vec F_\perp|$ Together with gravitation, a net force results in the vertical direction to $\begin{aligned} (\vec F_g + \vec F + \vec F_f) \cdot \vec e_z &\leq - M g - F \sin(30^\circ) + \mu F \cos(30^\circ) \\ &= - M g - \frac{1}{2} F + \frac{1}{2} \frac{\sqrt{3}}{2} F \\ &= - M g - \frac{2 - \sqrt{3}}{4} F\\ &< 0 \end{aligned}$ Since the total force is always negative, the book fall towards regardless of the strength of the force $\vec F$ .