\includegraphics{figure_3} A uniform square lamina of side \(2a\) 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 \(\tan\theta\). [8]

Question 3:
3 | Let N be normal reaction at B and F the frictional force acting downwards
Tcos=F+W | B1
→Tsin=N | B1
Moments about B: Ta=Wsina+Wcosa | M1A1 | A moments equation with all relevant forces.
1
F = N used
2 | M1
 1 
(cos+sin) cos− sin =1 oe
 2  | M1 | Combine to obtain equation in .
Equation in trigonometric functions only.
1 3
cossin= (sin)2
2 2
sin(cos−3sin)=0 | M1 | Solve trigonometric equation.
1
tan=
3 | A1
8
Question | Answer | Marks | Guidance

\includegraphics{figure_3}

A uniform square lamina of side $2a$ 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 $\tan\theta$. [8]

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q3 [8]}}