Coplanar Forces and Equilibrium

Perpendicular forces: $|R| = \sqrt{3^2+4^2} = 5$ N, at $\tan^{-1}(4/3) = 53.1°$ north of east.

$F_x = 20\cos 30° = 17.3$ N and $F_y = 20\sin 30° = 10$ N.

On a horizontal surface $R = mg = 98.1$ N, so $\mu R = 0.4(98.1) = 39.2$ N.

Smooth means no friction, so $a = g\sin 20° = 9.81(0.3420) = 3.36\ \text{m s}^{-2}$. The mass is irrelevant.

Moments about $A$: $4R_B = 250(2) + 150(1) = 650$, so $R_B = 162.5$ N. Then $R_A = 400 - 162.5 = 237.5$ N.

At limiting friction $\mu mg\cos\alpha = mg\sin\alpha$, so $\mu = \tan\alpha = \tan 30° = 0.577$.