#### Answer

$0.577$

#### Work Step by Step

We can find the required coefficient of static friction as follows:
$\Sigma M_A=0$
$\implies 490.5\times 0.6-T cos60^{\circ}(0.3 cos60^{\circ}+0.6)$
This simplifies to:
$T=490.5N$
Now $\Sigma F_x=0$
$T\times sin 60-R_A=0$
$\implies 490.5 sin60^{\circ}-R_A=0$
$\implies R_A=424.8N$
and $\Sigma F_y=0$
$\implies \mu R_A+Tcos60^{\circ}-W=0$
$\implies \mu (424.8)+490.5cos60^{\circ}-490.5=0$
$\implies \mu=0.577$