A book is resting in equilibrium on a horizontal shelf and against a vertical wall, and makes an angle $\theta$ with the horizontal. A horizontal force of magnitude $P$ newtons is applied at the bottom right edge of the book (half way between the front and back cover). The force $P$ acts in a vertical plane which passes through the centre of mass $G$ of the book, and is perpendicular to the wall.
The book is modelled by a uniform rectangular block of height $10$ cm and width $4$cm and a weight of $20$ N. The coefficients of friction between the book and shelf, and the book and wall are each $0.4$.
Show that the least value of $P$ needed to move the book is given by:
$$P=\frac{84\cos{\theta} + 33.6\sin{\theta}}{10\sin{\theta}-8.64\cos{\theta}}$$
Find the interval in which the value of $\theta$ must lie.
Working's out so far:
Let $S$ be the normal reaction on the wall acting in line with the centre of mass $G$ of the book. Let $F_2$ be the frictional force acting on the wall perpendicular to $S$.
Also, on the shelf, let $R$ be the normal reaction acting in line with $G$ of the book and $F_1$ be the frictional force perpendicular to $R$. I will call the weight $W$ (for now) and the horizontal force is acting on the bottom edge of the book is $P$.
Could you check/confirm the following, firstly?
Resolving horizontally:
$$Scosθ + F_1sinθ = F_2sinθ + Rcosθ + P$$
Resolving vertically:
$$ Ssinθ + F_1cosθ + W = F_2cosθ + Rsinθ$$
Taking moments about $G$ ($S$ and $R$ act along the line of action of $G$, so do not appear in the moment equation):
$$5Scosθ + 5Pcosθ = 5F_1sinθ + 5F_2sinθ$$
Clockwise moments = anti-clockwise moments.
Also, $F_1=μR$ and $F_2=μS$, where $μ=0.4$.
