The Tipping Table

Relevant wiki: Friction

Let $R$ be the normal reaction force and $\alpha$ be the angle the table makes to the horizontal and $M$ be the mass of the book
There are three forces acting on the book,
1) Normal reaction $R$ , perpendicular to the inclined plane.
2) Friction force $f$ , up the inclined plane.
3) Gravitational force $Mg$ , vertically downwards.

Resolving all the forces perpendicular to the inclined plane gives: $R - Mg \cos \alpha = 0$ $R = Mg \cos \alpha.$

Let $f$ be the friction force between the book and the table.

As the book is on the verge of slipping, friction force will be limiting, $f = f_L.$ $f_L = \mu R = 0.5 R.$

Resolving all components parallel to the plane gives: $f - Mg \sin \alpha = 0.$

Putting the value of $f$ , gives us: $0.5 Mg \cos \alpha - Mg \sin \alpha = 0$

Dividing by $Mg$ gives us: $\begin{aligned} 0.5 \cos \alpha - \sin \alpha &= 0 \\ 0.5 \cos \alpha &= \sin \alpha \\ \frac{\sin \alpha}{\cos \alpha } &= 0.5 \\ \tan \alpha &= 0.5 \\ \alpha &= tan^{-1} 0.5 = \boxed{26.5651}. \\ \end{aligned}$

Resolving the weight gives $R=Mg cos(\alpha)$ and $F=Mg sin(\alpha)$ as the book is in equilibrium.

Friction is proportional to the reaction force by $F=\mu R$ and using the stated coefficient of friction $\mu =\frac{1}{2}$ we get $Mg sin(\alpha)=$ $\frac{1}{2}$ $Mg cos(\alpha)$

This can be simplified to $sin(\alpha)=$ $\frac{1}{2}$ $cos(\alpha)$ . Rearranging gives us $tan(\alpha)=$ $\frac{1}{2}$ , which can be solved for $\alpha=26.6$

At the moment the book begins to move , we have balance of forces in the direction of movement, mgsinx=F, where F is the friction force, F= 0.5 N = 0.5 mgcosx. So mgsinx = 0.5 mgcosx, then we find x=26.6 deg.