First, resolve the forces vertically to find the normal reaction \(R\):

\(R = 300g - X \sin \alpha\)

Given \(\sin \alpha = 0.28\), we find \(\alpha \approx 16.26^\circ\).

Now, resolve horizontally:

\(0.96X - F = 0\)

where \(F\) is the frictional force, \(F = 0.5R\).

Substitute \(F\) into the equation:

\(0.96X = 0.5(300g - X \sin \alpha)\)

Solving for \(X\):

\(0.96X = 0.5(300 \times 9.8 - X \times 0.28)\)

\(0.96X = 1470 - 0.14X\)

\(1.1X = 1470\)

\(X = \frac{1470}{1.1} \approx 1363.63\)

Rounding down to the nearest whole number, \(X = 1360\).

(a) The diagram should show the following forces acting on the block:

• The weight of the block, \(4g\), acting vertically downwards.
• The normal reaction, \(R\), acting vertically upwards.
• The frictional force, \(F_r\), acting horizontally opposite to the direction of tension.
• The tension in the string, 30 N, acting at an angle of 24° above the horizontal.

(b) To find the coefficient of friction, resolve the forces horizontally and vertically:

Horizontally: \(30 \cos 24^{\circ} = F\)

\(F = 27.406 \ldots\)

Vertically: \(R + 30 \sin 24^{\circ} = 4g\)

\(R = 40 - 30 \sin 24^{\circ}\)

\(R = 27.797 \ldots\)

The coefficient of friction \(\mu\) is given by \(\mu = \frac{F}{R}\).

\(\mu = \frac{30 \cos 24^{\circ}}{40 - 30 \sin 24^{\circ}}\)

\(\mu = 0.986\) (rounded to three decimal places).

(i) Since the surface is smooth, there is no friction. The block is in equilibrium, so the horizontal components of the forces must balance. The horizontal component of the \(40 \text{ N}\) force is \(40 \cos(90^\circ) = 0\). The horizontal component of the \(X \text{ N}\) force is \(X \cos(30^\circ)\). Therefore, \(X \cos(30^\circ) = 0\), which implies \(X = 20 \text{ N}\).

(ii) For the rough surface, the block is in limiting equilibrium, so the frictional force \(F = 10 \text{ N}\) acts towards \(A\). The normal reaction \(R\) is equal to the weight of the block, \(R = 15g\). The frictional force is given by \(F = \mu R\), so \(10 = \mu \times 15g\). Solving for \(\mu\), we get \(\mu = \frac{10}{15g} = 0.5\).

(i) To find the normal component of the force exerted on B by the ground, resolve the forces vertically. The weight of the block is \(7 \times 10 = 70\) N. The vertical component of the force \(X\) is \(X \cos 15^{\circ}\). In equilibrium, the normal force \(N\) plus the vertical component of \(X\) equals the weight of the block:

\(N + X \cos 15^{\circ} = 70\)

Thus, the normal component is:

\(N = 70 - X \cos 15^{\circ}\)

(ii) For limiting equilibrium, the frictional force \(F\) is equal to the maximum friction \(\mu N\), where \(\mu = 0.4\). The horizontal component of the force \(X\) is \(X \sin 15^{\circ}\). Therefore:

\(X \sin 15^{\circ} = 0.4 (70 - X \cos 15^{\circ})\)

Solving for \(X\):

\(X \sin 15^{\circ} = 28 - 0.4X \cos 15^{\circ}\)

\(X (\sin 15^{\circ} + 0.4 \cos 15^{\circ}) = 28\)

\(X = \frac{28}{\sin 15^{\circ} + 0.4 \cos 15^{\circ}}\)

Calculating gives \(X \approx 43.4\).

(i) To find the value of C, we use the triangle of forces. The horizontal component of the applied force is given by:

\(F = 4 \cos 30^{\circ}\)

The vertical component of the applied force is:

\(R = 10 - 4 \sin 30^{\circ}\)

Using the cosine rule for the triangle of forces:

\(C^2 = (4 \cos 30^{\circ})^2 + (10 - 4 \sin 30^{\circ})^2\)

Calculating gives:

\(C = 8.72\)

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

\(\mu = \frac{4 \cos 30^{\circ}}{10 - 4 \sin 30^{\circ}}\)

Calculating gives:

Coefficient is 0.433 (accept 0.43)