Rotation 4

Relevant wiki: Torque - Equilibrium

Since both sides are symmetric w.r.t. a vertical axis passing through the center of the string, we'll focus on only one of the planks.

Consider plank to be of negligible thickness.

The forces operating on this plank are showing in the diagram above where $T$ is the tension in the string, $mg$ is the gravitational force acting upon the center of mass of the plank, $N$ is the normal reaction force and $f$ is the force of friction between the plank and the surface of the ground.

Balancing forces in vertical direction, we have

$N = mg.$

Since $T$ is unknown, it's best to balance torque about point $O$ . Thus we have

$mg \left( \dfrac{\ell}{2} \cos \theta \right) - N \left( \ell \cos \theta \right) + f \left( \ell \sin \theta \right) = 0 \\ \implies mg \left( \dfrac{\ell}{2} \cos \theta \right) - mg \left( \ell \cos \theta \right) + f \left( \ell \sin \theta \right) = 0 \\ \implies \tan \theta = \dfrac{mg}{2f}$

For minimum $\tan \theta$ , $f$ should be maximum.

From the condition of no slipping, we know that

$f \le f_{\text{max possible}} \implies f \le \mu mg$

Thus maximum value of $f$ is $\mu mg$ . Hence minimum value of $\tan \theta$ is

$\tan \theta = \dfrac{mg}{2 \mu mg} = \dfrac{1}{2 \mu} = \dfrac{1}{2 \times 0.5} = 1.$

As we know that for $\theta \in \left[ 0 , \dfrac{\pi}{2} \right]$ , $\theta$ decreases as $\tan \theta$ decreases, so

$\theta_{\text{min}} = \arctan (1) = \boxed{45^{\circ}}.$