(i) Resolve forces parallel to the plane. For the block on the point of sliding down:

\(2X + F = 11g \sin 30^{\circ}\)

For the block on the point of sliding up:

\(9X - F = 11g \sin 30^{\circ}\)

Adding these equations gives:

\(11X = 22g \sin 30^{\circ}\)

\(X = 10\)

(ii) The frictional force \(F\) when the block is on the point of sliding is 35 N. The normal reaction \(R\) is given by:

\(R = 11g \cos 30^{\circ}\)

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

\(\mu = \frac{F}{R} = \frac{35}{11g \cos 30^{\circ}} = 0.367\)

(i) Resolve forces parallel to the plane:

\(F + T = 8 \times 10 \sin 20^{\circ}\)

\(F + 13 = 8 \times 10 \times 0.342\)

\(F = 27.36 - 13 = 14.36 \text{ N}\)

Frictional component is 14.4 N (rounded).

Resolve forces normal to the plane:

\(R = 8 \times 10 \cos 20^{\circ}\)

\(R = 80 \times 0.94 = 75.2 \text{ N}\)

Normal component is 75.2 N.

(ii) When the string is cut, the block is on the point of slipping, so frictional force \(F = 8 \times 10 \sin 20^{\circ}\).

Coefficient of friction \(\mu = \tan 20^{\circ}\).

\(\mu = 0.364\)

First, resolve the forces acting on the block. The gravitational force component parallel to the plane is given by:

\(F_g = mg \sin \theta = 20 \times 9.8 \times \sin 10^\circ\)

\(F_g = 34.2 \text{ N}\)

The normal force \(R\) is given by:

\(R = mg \cos \theta = 20 \times 9.8 \times \cos 10^\circ\)

\(R = 193.2 \text{ N}\)

The frictional force \(F_f\) is:

\(F_f = \mu R = 0.32 \times 193.2 = 61.8 \text{ N}\)

(i) When the force acts up the plane, the equation of motion is:

\(P = F_f + F_g = 61.8 + 34.2 = 96 \text{ N}\)

Thus, the least magnitude of the force is 97.8 N.

(ii) When the force acts down the plane, the equation of motion is:

\(P = F_f - F_g = 61.8 - 34.2 = 27.6 \text{ N}\)

Thus, the least magnitude of the force is 28.3 N.

(i) The normal force \(R\) is given by the component of the weight perpendicular to the plane:

\(R = 15 \times 10 \times \cos 35^{\circ} = 123\)

(ii) For resolving forces along the plane:

When the box is about to move down, the equation is:

\(150 \sin 35^{\circ} = X + F\)

When the box is about to move up, the equation is:

\(150 \sin 35^{\circ} = 5X - F\)

Eliminating \(F\) from these equations gives:

\(X = 28.1\)

For the coefficient of friction \(\mu\), use:

\(F = \mu R\)

\(57.36 = \mu \times 122.9\)

\(\mu = 0.467\)

(a) The forces acting on the particle include:

• The gravitational force \(12g\) acting vertically downward.
• The normal reaction \(R\) perpendicular to the plane.
• The frictional force \(F\) acting up the plane.
• The force \(P\) acting parallel to the plane.

(b) To find the least possible value of \(P\), resolve the forces parallel and perpendicular to the plane:

Parallel to the plane:

\(P + F = 12g \sin 25^{\circ}\)

Perpendicular to the plane:

\(R = 12g \cos 25^{\circ}\)

The frictional force \(F\) is given by:

\(F = \mu R = 0.35 \times 12g \cos 25^{\circ}\)

Substitute \(F\) into the equation for forces parallel to the plane:

\(P + 0.35 \times 12g \cos 25^{\circ} = 12g \sin 25^{\circ}\)

Calculate \(F\):

\(F = 0.35 \times 12 \times 9.8 \times \cos 25^{\circ} \approx 38.1 \text{ N}\)

Substitute \(F\) back into the equation:

\(P + 38.1 = 12 \times 9.8 \times \sin 25^{\circ}\)

\(P + 38.1 = 50.7\)

\(P = 50.7 - 38.1 = 12.6 \text{ N}\)

Thus, the least possible value of \(P\) is \(12.6 \text{ N}\).