Answer
$\mu_s = 0.29$
Work Step by Step
We can find the tension $F_T$ in the wire:
$F_T~cos~\theta = m_B~g$
$F_T = \frac{m_B~g}{cos~\theta}$
$F_T = \frac{(5.0~kg)~(9.8~m/s^2)}{cos~30^{\circ}}$
$F_T = 56.58~N$
We can find the coefficient of static friction:
$m_A~g~\mu_s = F_T~sin~\theta$
$\mu_s = \frac{F_T~sin~\theta}{m_A~g}$
$\mu_s = \frac{(56.58~N)~sin~30^{\circ}}{(10~kg)~(9.8~m/s^2)}$
$\mu_s = 0.29$
