Static equilibirum or rotation 
I am unable to get horizontal force in equilibrium and i get $$\mu_s$$ a negative value and greater than 1 also which is absurd please let me know how to do.
I got $$f_s  = -T\sin 15$$
I used moment as 0 at bottom to get T then using horizontal equilibrium to get friction and then using vertical equilibrium to get N and then using mu times normal reaction concept to get mu
 A: We have
$$
\cases{
\vec T = -T_0(\sin\beta, \cos\beta),\ \ \ \text{rope tension}\\
\vec W = m g(0,-1),\ \ \ \ \ \ \ \ \ \ \ \text{weight}\\
\vec R = V(-\mu,1),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{soil bar tip reaction}\\
A = L(-\cos\alpha, \sin\alpha),\ \ \ \ \text{where the rope is fixed}\\
O = \ \ \ \ \text{tip rod support location}
}
$$
The equilibrium conditions are
$$
\cases{\vec T + \vec W +  \vec R = \vec 0\\
\vec T\times (A-O) + \vec W \times \frac 12 (A-O) = \vec 0
}
$$
and after solving for $T_0,V,\mu$ we obtain
$$
\mu = \frac{1}{2 \tan (\alpha )+\cot (\beta )}\approx 0.181
$$ 
A: I said in a comment that $f_s$ looked OK. But that may not have been exactly true.
Assuming you follow a convention where horizontal forces to the right are positive and forces to the left are negative,
the horizontal component of the string's force on the bar is
$-T\sin 15^\circ.$
In order for the bar to be in horizontal equilibrium, the floor must exert a horizontal force $T\sin 15^\circ$ on it.
Therefore the bar exerts an equal and opposite horizontal force on the floor:
$-T\sin 15^\circ.$
Both of these forces are frictional forces and could be called $f_s,$ though I suppose you probably meant $f_s$ to be the force exerted by the floor on the bar.
If so, I was wrong; you had reversed the sign on that force.
The convention that I'm familiar with is that a coefficient of friction $\mu$ is positive, and that if $N$ is the normal force and is positive, then
$\mu N$ gives the magnitude (or maximum magnitude) of the frictional force.
You have to use other knowledge to figure out the direction in which the frictional force acts, namely in whichever direction would either prevent the relative motion of the two objects or reduce the relative motion.
So if you were given a problem in which a certain frictional force was to the left, and you had a positive normal force $N$, the signed representation of the frictional force would be $-\mu N.$
So what you should have written is simply
$$ T\sin 15^\circ =  \mu_s (mg - T\cos 15^\circ).$$
But you can't show that by math alone. You have to use physics.
A: Use the contact point with the floor as the pivot to establish the rotational equilibrium below
$$(T\cos15)(L\cos42) = (T\sin15)(L\sin42)+ W(\frac12L\cos42)$$
Solve for the tension $T$ in terms of $W$, the weight of the bar,
$$T = \frac{\cos42}{2\cos57}W$$
Then, plug it into the horizontal equilibrium equation 
$$T\sin15 = \mu_s(W-T\cos15)$$
to obtain the static coefficient
$$\mu_s = \frac{T\sin15}{W - T\cos15}=\frac{\cos42\sin15}{2\cos57-\cos42\cos15}=\frac1{\cot15-2\tan42}=0.518$$
