(i) Resolve forces perpendicular to the plane:

\(R + 0.6 \sin \alpha = 0.5g \cos \alpha\)

Given \(\sin \alpha = 0.28\), \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - 0.28^2} = 0.96\).

\(R + 0.6 \times 0.28 = 0.5 \times 9.8 \times 0.96\)

\(R + 0.168 = 4.704\)

\(R = 4.704 - 0.168 = 4.536\)

Normal component is approximately 4.63 N.

(ii) Resolve forces parallel to the line of greatest slope:

\(F + 0.6 \cos \alpha = 0.5g \sin \alpha\)

\(F + 0.6 \times 0.96 = 0.5 \times 9.8 \times 0.28\)

\(F + 0.576 = 1.372\)

\(F = 1.372 - 0.576 = 0.796\)

Frictional component is approximately 0.824 N.

(iii) Coefficient of friction \(\mu\):

\(\mu = \frac{F}{R} = \frac{0.824}{4.63}\)

\(\mu \approx 0.178\)