Neither Slipping Nor Tipping

Let us denote the tension is the string by $T$ . The friction force is $F$ , the normal force is $N$ and the weight is $mg$ . Using similar triangles we conclude that the $x$ (horizontal) component of the tension is $T_x=\frac{2}{\sqrt{13}} T$ to the left, and the $y$ component is $T_y=\frac{3}{\sqrt{13}} T$ downwards. From the balance of forces we get

$T_x-F=0$

$T_y+mg-N=0$

We also need the torque equation. We will use the bottom of the stick as the pivot point. We get

$3T_x-1T_y-\frac{1}{2} mg=0$

or

$3\frac{2}{\sqrt{13}} T-\frac{3}{\sqrt{13}} T-\frac{1}{2} mg=0$

From this we get the tension:

$T=\frac{\sqrt{13}}{6} mg$ and $T_x=mg/3$ , $T_y=mg/2$

The coefficient of friction is

$\mu=F/N=\frac{T_x}{T_y+mg}= \frac{1/2}{1+1/2}=2/9=0.222$

This system is in equilibrium, meaning the net force and net torque acting on the rod is equal to $0$ . First, let's set up torque equilibrium. There will be twisting on the rod due to the gravitational force and the the force exerted on the rod by the rope. We will resolve the torque around the pivot point in contact with the ground. We know that the height of the point where the rope is attached to the rod is $3$ , and we also know the distance between the anchor point of the rope and the pivot point of the rod, as well as the length of the bottom of the right triangle formed by the rod and the line perpendicular to the ground. This means we can form two right triangles, one with the rope as the hypotenuse, and one with the rod as the hypotenuse. From the dimensions of these right triangles, we can setup torque equilibrium:

$mg \sin (\tan^{-1} \frac{1}{3}) \ \frac{\sqrt{10}}{2} \ = \ F_t \sin(\tan^{-1} \frac{2}{3} \ - \ \tan^{-1} \frac{1}{3}) \sqrt{10} \ \Rightarrow \ F_t \ = \ \frac{mg \sin (\tan^{-1} \frac{1}{3})}{2\sin(\tan^{-1} \frac{2}{3} \ - \ \tan^{-1} \frac{1}{3})} \ = \ 5.8951m$

Notice how we multiply the gravitational torque by half the radius of the rod (as this is the centre of mass), and the torque due to tension is multiplied by the full length of the rod. We had to resolve the forces into their components perpendicular to the rod. Now, the total downward force acting on the rod will be equal to the normal force, so we have:

$F_t \cos(\tan^{-1} \frac{2}{3}) \ + \ mg \ = \ N$

Finally, the frictional force has to be equal to the component of the force exerted by the rope on the rod parallel to the ground:

$F_t \sin(\tan^{-1} \frac{2}{3}) \ = \ N \mu_k$

So we get:

$\mu_k \ = \ \frac{F_t \sin(\tan^{-1} \frac{2}{3})}{N} \ = \ \frac{F_t \sin(\tan^{-1} \frac{2}{3})}{F_t \cos(\tan^{-1} \frac{2}{3}) \ + \ mg} \ = \ \frac{5.8951 \sin(\tan^{-1} \frac{2}{3})}{5.8951 \cos(\tan^{-1} \frac{2}{3}) \ + \ g} \ \approx \ 0.2222$