(i) To find the horizontal component of the force exerted by the rod, resolve the tension \(T\) horizontally:

\(T \cos 25^{\circ} = 0.906T\).

To find the vertical component of the force exerted by the rod, resolve the forces vertically:

\(4g + T \sin 25^{\circ} = 40 + 0.423T\).

(ii) Given that the equilibrium is limiting, the frictional force \(F\) is equal to the maximum static friction, which is \(0.4R\), where \(R\) is the normal reaction force. Using the horizontal component:

\(0.906T = 16 + 0.169T\).

Solving for \(T\):

\(0.906T - 0.169T = 16\)

\(0.737T = 16\)

\(T = \frac{16}{0.737} \approx 21.7 \text{ N}\).

(i) Resolve the forces vertically. The weight of the ring is given by:

\(0.8g\)

where \(g\) is the acceleration due to gravity. The vertical component of the 7 N force is:

\(7 \sin 45^\circ\)

The normal contact force \(R\) acts vertically upwards. In equilibrium:

\(R + 7 \sin 45^\circ = 0.8g\)

Solving for \(R\):

\(R = 0.8g - 7 \sin 45^\circ\)

Substitute \(g = 9.8\):

\(R = 0.8 \times 9.8 - 7 \times \frac{\sqrt{2}}{2}\)

\(R = 7.84 - 4.95 = 2.89\)

Rounding to 3 significant figures gives 3.05 N.

(ii) In limiting equilibrium, the frictional force \(F\) is given by:

\(F = 7 \cos 45^\circ\)

\(F = 7 \times \frac{\sqrt{2}}{2} = 4.95\)

The coefficient of friction \(\mu\) is given by:

\(\mu = \frac{F}{R} = \frac{4.95}{3.05}\)

\(\mu \approx 1.62\)

To solve the problem, we need to resolve the forces acting on the ring.

(i) The frictional component of the contact force is the horizontal component of the tension in the string. Given \(\tan \alpha = \frac{5}{12}\), we can find \(\cos \alpha\) using the identity \(\cos \alpha = \frac{12}{13}\). Thus, the frictional force \(F = 13 \cos \alpha = 13 \times \frac{12}{13} = 12 \text{ N}\).

(ii) The normal component of the contact force is the sum of the vertical component of the tension and the weight of the ring. The vertical component of the tension is \(13 \sin \alpha\), where \(\sin \alpha = \frac{5}{13}\). Therefore, the normal force \(R = 1.1 \times 10 + 13 \sin \alpha = 11 + 13 \times \frac{5}{13} = 16 \text{ N}\).

(iii) In limiting equilibrium, the frictional force is equal to the product of the coefficient of friction \(\mu\) and the normal force \(R\). Thus, \(\mu = \frac{F}{R} = \frac{12}{16} = 0.75\).

Resolve forces perpendicular to the rod:

\(8 \sin 10^{\circ} + R = 0.3g\)

Solving for \(R\):

\(R = 0.3g - 8 \sin 10^{\circ} \approx 2.943 - 1.388 = 1.555 \text{ N}\)

Resolve forces along the rod:

\(8 \cos 10^{\circ} - F = 0.3a\)

Using friction \(F = \mu R\):

\(F = 0.8 \times 1.555 \approx 1.244 \text{ N}\)

Substitute \(F\) into the equation:

\(8 \cos 10^{\circ} - 1.244 = 0.3a\)

Calculate \(a\):

\(a = \frac{8 \cos 10^{\circ} - 1.244}{0.3} \approx \frac{7.878 - 1.244}{0.3} \approx 21.966 \text{ m/s}^2\)

Use the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

Given \(s = 0.6 \text{ m}, u = 0\):

\(0.6 = \frac{1}{2} \times 21.966 \times t^2\)

Solve for \(t\):

\(t^2 = \frac{0.6}{10.983} \approx 0.0546\)

\(t = \sqrt{0.0546} \approx 0.234 \text{ seconds}\)

(a) Resolving forces vertically and horizontally:

Normal reaction, \(R = T \sin 30^\circ + 0.1g\).

Frictional force, \(F = T \cos 30^\circ\).

For equilibrium, \(T \cos 30^\circ = 0.8 (T \sin 30^\circ + 0.1g)\).

Solving for \(T\):

\(T \cos 30^\circ = 0.8T \sin 30^\circ + 0.08g\).

\(T (\cos 30^\circ - 0.8 \sin 30^\circ) = 0.08g\).

\(T = \frac{0.08g}{\cos 30^\circ - 0.8 \sin 30^\circ}\).

\(T = 1.72 \text{ N}\).

(b) When \(T = 3\):

Normal reaction, \(R = 3 \sin 30^\circ + 0.1g\).

Net force, \(3 \cos 30^\circ - 0.8(3 \sin 30^\circ + 0.1g) = 0.1a\).

Solving for \(a\):

\(a = \frac{3 \cos 30^\circ - 0.8(3 \sin 30^\circ + 0.1g)}{0.1}\).

\(a = 5.98 \text{ m/s}^2\).