Two uniform boards, tied together with the help of a string, are balanced on a surface as shown in figure.

 

The coefficient of static friction between boards and surface is 0.5. The minimum value of $\theta$, for which this type of arrangement is possible is

For the equilibrium of the entire structure, ${\mathrm{N}}_1+{\mathrm{N}}_2=2 \mathrm{Mg}$

          $\mathrm{Mg}\left(\frac{\mathrm{l}}{2} \cos \mathrm{\theta }\right)+{\mathrm{N}}_2\times 2\mathrm{x}-\mathrm{Mg}\left(\frac{\mathrm{l}}{2} \cos \mathrm{\theta }+2\mathrm{x}\right)=0$

Given  ${\mathrm{N}}_2=\mathrm{Mg}, {\mathrm{N}}_1=\mathrm{Mg}$

For individual boards, ${\mathrm{T}}_1={\mathrm{f}}_1, {\mathrm{N}}_1=\mathrm{Mg}$

$\begin{array}{l}\mathrm{and} {\mathrm{T}}_1\times \mathrm{l} \sin \mathrm{\theta }=\mathrm{Mg} \frac{\mathrm{l}}{2} \cos \mathrm{\theta }\\ {\mathrm{f}}_1={\mathrm{T}}_1=\frac{\mathrm{Mg}}{2 \tan \mathrm{\theta }}\end{array}$

For safe equilibrium, ${\mathrm{f}}_1<{\mathrm{f}}_{\mathrm{L}}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{Mg}$

           $\frac{\mathrm{Mg}}{2 \tan \mathrm{\theta }}={\mathrm{\mu }}_{\mathrm{s}} \mathrm{Mg} \Rightarrow \tan \mathrm{\theta }>\frac{1}{2{\mathrm{\mu }}_{\mathrm{s}}} \Rightarrow \mathrm{\theta }=45$

So the minimum value of $\mathrm{\theta }$ for this type of arrangement is 45^o.