(i) Resolve forces vertically and horizontally at B:

For vertical equilibrium: \(T_C \times \frac{2}{2.5} - T_A \times \frac{1.5}{2.5} = 0\)

For horizontal equilibrium: \(T_C \times \frac{1.5}{2.5} + T_A \times \frac{2}{2.5} = 8\)

Solving these equations, we find:

\(T_A = 6.4 \text{ N}\)

\(T_C = 4.8 \text{ N}\)

(ii) Resolve forces vertically:

\(F + 0.2g = T_A \times \frac{1.5}{2.5}\)

Normal force: \(N = T_A \times \frac{2}{2.5}\)

Using \(\mu = \frac{F}{N}\), we find:

\(\mu = \frac{3.84 - 2}{5.12} = 0.359\)

(i) Resolve forces vertically on R: \(2T \cos \alpha = 0.6g\), where \(\alpha = \frac{1}{2} \angle ARB\). Solving gives tension \(T = 5 \text{ N}\).

(ii) Resolve forces horizontally on B: Frictional component \(F = T \sin \alpha = 4 \text{ N}\). Resolve forces vertically on B: Normal component \(N = 0.4g + T \cos \alpha = 7 \text{ N}\).

(iii) Using \(\mu = \frac{F}{N}\), the coefficient of friction \(\mu = \frac{4}{7}\) or 0.571.

(i) To find the tension in the longer string, resolve the forces acting on P in the direction of PS. The angle between PS and the vertical is 36.9° (since \\sin^{-1}(0.8) = 36.9°\\). The tension in the longer string T_{PS} is given by:

\(T_{PS} = 5 \sin 36.9^{\circ}\)

Calculating gives \(T_{PS} = 3 \text{ N}\).

(ii) The frictional force F acting on S is found by resolving horizontally. The horizontal component of the tension is:

\(F = T \cos(\sin^{-1}(0.6))\)

\(F = 2.4 \text{ N}\).

(iii) Given the coefficient of friction \(\mu = 0.75\), and S is in limiting equilibrium, the normal reaction R is:

\(R = \frac{2.4}{0.75}\)

Resolving vertically for S gives:

\(W + T \sin(\sin^{-1}(0.6)) = R\)

\(W = 1.4 \text{ N}\).

(i) To show that the tension in BM is 5 N, resolve forces at M horizontally. Using the sine rule in the triangle of forces, we have:

\(T \cdot \sin 60^{\circ} = 5 \cdot \sin 60^{\circ}\)

or using Lami's theorem:

\(\frac{T}{\sin 120^{\circ}} = \frac{5}{\sin 120^{\circ}}\)

Thus, the tension is 5 N.

(ii) For resolving forces on B horizontally, we have:

\(N = T \cdot \sin 30^{\circ}\)

or from symmetry:

\(N = \frac{5}{2}\)

For resolving forces on B vertically:

\(0.2 \times 10 + F = T \cdot \cos 30^{\circ}\)

Using \(F = \mu R\), we find:

\(2.33 = 2.5 \mu\)

Thus, the coefficient of friction is 0.932.

(iii) For the ring on the point of sliding down the wire:

\((0.2 + m)g - 2.33 = 5 \cdot \cos 30^{\circ}\)

or:

\(mg = 2(2.33)\)

Solving gives \(m = 0.466\).

In limiting equilibrium, the frictional force \(F\) is given by \(F = \mu R\), where \(\mu = 0.25\) is the coefficient of friction and \(R\) is the normal reaction.

The horizontal component of the tension \(T\) is \(T \cos \alpha = 0.96T\), which equals the frictional force \(F\). Thus, \(F = 0.96T\).

The vertical component of the tension is \(T \sin \alpha\), and the weight of the ring is \(0.2g\). The normal reaction \(R\) is given by:

\(R = 0.2g - T\sin \alpha\)

Given \(\cos \alpha = 0.96\), we find \(\sin \alpha\) using \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\(\sin \alpha = \sqrt{1 - 0.96^2} = 0.28\)

Thus, \(R = 0.2 \times 9.8 - T \times 0.28 = 1.96 - 0.28T\).

Using \(F = \mu R\), we have:

\(0.96T = 0.25(1.96 - 0.28T)\)

Solving for \(T\):

\(0.96T = 0.49 - 0.07T\)

\(0.96T + 0.07T = 0.49\)

\(1.03T = 0.49\)

\(T = \frac{0.49}{1.03} \approx 0.4757\)

Rounding to three significant figures, \(T = 0.485\).