(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 \(X \approx 43.4\).