Further Math ASs 3.2.6 Toppling and sliding

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Figure for Question CAIE A-Level Further Math AS 3.2.6 Toppling and Sliding Questions question 1 — CAIE A-Level Further Math AS

A uniform square lamina of side 2 a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle $\theta$ with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is $\frac{1}{2}$ .

Given that the vertex B is about to slip up the wall, find the value of $\tan \theta$ .

Let N be normal reaction at B and F the frictional force acting downwards
$\uparrow T \cos \theta=F+W$ $\rightarrow T \sin \theta=N$
Moments about $B: T a=W \sin \theta \times a+W \cos \theta \times a$
A moments equation with all relevant forces.

$F=\frac{1}{2} N$ used
$(\cos \theta+\sin \theta)\left(\cos \theta-\frac{1}{2} \sin \theta\right)=1$ ое
Combine to obtain equation in $\theta$ .
Equation in trigonometric functions only.

$\begin{aligned} \frac{1}{2} \cos \theta \sin \theta=\frac{3}{2}(\sin \theta)^{2} \sin \theta(\cos \theta-3 \sin \theta)=0 \end{aligned}$
Solve trigonometric equation.

$\tan \theta=\frac{1}{3}$
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CAIE A-Level Further Math AS 3.2.6 Toppling and Sliding Questions question 1