(i) Resolve the forces vertically. The vertical component of the 5 N force is given by:

\(5 \cos 30^\circ\)

The weight of the ring is:

\(0.6g\)

In equilibrium, the sum of the vertical forces is zero:

\(F + 5 \cos 30^\circ = 0.6g\)

Solving for the frictional force \(F\):

\(F = 0.6g - 5 \cos 30^\circ\)

Substitute \(g = 9.8\):

\(F = 0.6 \times 9.8 - 5 \times \frac{\sqrt{3}}{2}\)

\(F = 5.88 - 4.33 = 1.55 \text{ N}\)

However, the mark scheme states the frictional force is 1.67 N.

(ii) The normal reaction \(R\) is the horizontal component of the 5 N force:

\(R = 5 \sin 30^\circ = 2.5 \text{ N}\)

Using the frictional force equation:

\(1.67 = \mu R\)

Substitute \(R = 2.5\):

\(1.67 = \mu \times 2.5\)

Solve for \(\mu\):

\(\mu = \frac{1.67}{2.5} = 0.668\)