Answer
$$0.339$$
Work Step by Step
The forces acting on the uniform plank are
1. The normal force $F_{N1}$ on the plank exerted by the ground
2. The normal force $F_{N2}$ on the plank exerted by the wall
3. The static frictional force $f_s=\mu_sF_{N1}$
4. The weight of the climber $W$
At the static equilibrium, the net force acting on the the climber zero.
Therefore,
at horizontal equilibrium: $f_s-F_{N2}\sin\theta=0\implies f_s=F_{N2}\sin\theta\;.............(1)$
at vertical equilibrium: $F_{N1}+F_{N2}\cos\theta-W=0\implies F_{N1}=W-F_{N2}\cos\theta\;...............(2)$
The net torque about the contact point between the plank and the ground is also zero.
$W\frac{l}{2}\cos\theta-F_{N2}\cos\theta h-F_{N2}\sin\theta h=0$
or, $F_{N2}=\frac{Wl}{2h(\cot\theta+\tan\theta)}\;...............(3)$
Now the coefficient of the static friction is given by
$\mu_s=\frac{f_s}{F_{N1}}$
or, $\mu_s=\frac{F_{N2}\sin\theta}{W-F_{N2}\cos\theta}$ (using Eq. $1$ and Eq. $2$)
or, $\mu_s=\frac{\sin\theta}{\frac{W}{F_{N2}}-\cos\theta}$
or, $\mu_s=\frac{\sin\theta}{\frac{2h(\cot\theta+\tan\theta)}{l}-\cos\theta}$ (using Eq. $3$ )
Substituting the given values
$\mu_s=\frac{\sin70^{\circ}}{\frac{2\times3.05\times(\cot70^{\circ}+\tan70^{\circ})}{6.10}-\cos70^{\circ}}$
$\boxed{\mu_s=0.339}$
Therefore, the coefficient of static friction between the plank and ground is $0.339$
