## Engineering Mechanics: Statics & Dynamics (14th Edition)

$0.577$
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 $\Sigma F_y=0$ $\implies \mu R_A+Tcos60^{\circ}-W=0$ $\implies \mu (424.8)+490.5cos60^{\circ}-490.5=0$ $\implies \mu=0.577$