(i) Resolve the forces horizontally: \(F = 0.195 \cos \theta = 0.195 \times \frac{12}{13} = 0.18 = \frac{9}{50}\).

Resolve the forces vertically: \(R = 0.24 + 0.195 \sin \theta = 0.24 + 0.195 \times \frac{5}{13} = 0.315 = \frac{63}{200}\).

Using \(\mu = \frac{F}{R}\), we find \(\mu = \frac{9}{50} \div \frac{63}{200} = \frac{4}{7}\) or 0.571.

(ii) When \(\theta\) is above the horizontal, resolve vertically: \(R = 0.24 - 0.195 \sin \theta = 0.24 - 0.195 \times \frac{5}{13} = 0.165 = \frac{33}{200}\).

Using Newton's second law for motion along the rod: \(0.195 \times \frac{12}{13} - \left( \frac{4}{7} \right) \times 0.165 = 0.024a\).

Solve for \(a\): \(a = 3.57 \text{ m/s}^2\).

(i) Resolve the forces vertically:

\(12 + 15 \, \sin 30^\circ = R\)

Resolve the forces horizontally for friction:

\(F = 15 \, \cos 30^\circ\)

Use the formula for the coefficient of friction:

\(\mu = \frac{F}{R} = \frac{15 \, \cos 30^\circ}{12 + 15 \, \sin 30^\circ}\)

Calculate \(\mu\):

\(\mu = 0.666\)

(ii) Calculate the frictional force:

\(F = 0.666(12 - 15 \, \sin 30^\circ)\)

Apply Newton's second law horizontally:

\(15 \, \cos 30^\circ - F = 1.2a\)

Solve for \(a\):

\(a = 8.33 \, \text{m s}^{-2}\)

First, resolve the tension in the string into horizontal and vertical components. The horizontal component is \(T \cos 15^{\circ} = 2.5 \cos 15^{\circ}\) and the vertical component is \(T \sin 15^{\circ} = 2.5 \sin 15^{\circ}\).

The normal reaction \(R\) is equal to the horizontal component of the tension: \(R = 2.5 \cos 15^{\circ}\).

The frictional force \(F\) is given by \(F = \mu R\). Therefore, \(F = \mu \times 2.5 \cos 15^{\circ}\).

In limiting equilibrium, the vertical forces balance: \(2.5 \sin 15^{\circ} = 0.03g + F\).

Substitute \(F = \mu \times 2.5 \cos 15^{\circ}\) into the equation: \(2.5 \sin 15^{\circ} = 0.03g + \mu \times 2.5 \cos 15^{\circ}\).

Rearrange to solve for \(\mu\):

\(\mu = \frac{2.5 \sin 15^{\circ} - 0.03g}{2.5 \cos 15^{\circ}}\).

Calculate \(\mu\) using \(g = 9.8\):

\(\mu = \frac{2.5 \times 0.2588 - 0.03 \times 9.8}{2.5 \times 0.9659} \approx 0.144\).

To solve this problem, we need to resolve the forces acting on the ring both horizontally and vertically.

1. **Horizontal Resolution:**
The horizontal component of the tension \(T\) is \(T \cos 30^{\circ}\). This is balanced by the normal reaction \(R\) from the rod, so:

\(R = T \cos 30^{\circ}\)

2. **Vertical Resolution:**
The vertical component of the tension \(T\) is \(T \sin 30^{\circ}\). The weight of the ring is \(2g\) N (where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\)). The frictional force \(F\) can act either upwards or downwards to prevent motion, so we consider both cases:

**Case 1:** Friction prevents upwards motion:
\(F = T \sin 30^{\circ} - 2g\)

**Case 2:** Friction prevents downwards motion:
\(-F = T \sin 30^{\circ} - 2g\)

3. **Frictional Force:**
The frictional force \(F\) is given by \(F = \mu R\), where \(\mu = 0.24\). Substituting for \(R\) from the horizontal resolution:

\(F = 0.24 \times T \cos 30^{\circ}\)

4. **Solving for \(T\):**
Using the equations for \(F\) in both cases, we have:

\(T \sin 30^{\circ} - 2g = \pm 0.24 \times T \cos 30^{\circ}\)

Rearranging gives:

\(T = \frac{2g}{\sin 30^{\circ} \pm 0.24 \cos 30^{\circ}}\)

5. **Calculating Values:**
Substituting \(g = 9.8 \text{ m/s}^2\), \(\sin 30^{\circ} = 0.5\), and \(\cos 30^{\circ} = \sqrt{3}/2 \approx 0.866\):

\(T = \frac{2 \times 9.8}{0.5 \pm 0.24 \times 0.866}\)

Calculating for both cases:

**Case 1:**
\(T = \frac{19.6}{0.5 + 0.24 \times 0.866} \approx 28.3 \text{ N}\)

**Case 2:**
\(T = \frac{19.6}{0.5 - 0.24 \times 0.866} \approx 68.5 \text{ N}\)

Thus, the two values of \(T\) are \(28.3 \text{ N}\) and \(68.5 \text{ N}\).

(i) Resolve horizontally to find the normal force \(R\):

\(R = T \sin 60^\circ\)

Resolve vertically to find the frictional force \(F\):

\(F = W + T \cos 60^\circ\)

Given \(W = 4 \times 10 = 40\) N (weight of the ring),

\(F = 40 + T \cos 60^\circ\)

(ii) Use the equation for limiting friction:

\(F = \mu R\)

Substitute \(\mu = 0.7\), \(R = T \sin 60^\circ\), and \(F = 40 + T \cos 60^\circ\):

\(40 + 0.5T = 0.7 \times 0.866T\)

Solve for \(T\):

\(40 + 0.5T = 0.6062T\)

\(40 = 0.1062T\)

\(T = \frac{40}{0.1062} \approx 377\)