Given \(\tan \alpha = 2.4\), we find \(\sin \alpha = \frac{12}{13}\) and \(\cos \alpha = \frac{5}{13}\) using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

Resolving forces parallel to the plane:

\(0.6g \sin \alpha = F + P \cos \alpha\)

Resolving forces perpendicular to the plane:

\(R = 0.6g \cos \alpha + P \sin \alpha\)

Using \(F = \mu R\):

\(0.6g \sin \alpha - P \cos \alpha = 0.4(0.6g \cos \alpha + P \sin \alpha)\)

Substituting \(g = 9.8\), \(\sin \alpha = \frac{12}{13}\), and \(\cos \alpha = \frac{5}{13}\):

\(6 \times \frac{12}{13} - P \times \frac{5}{13} = 2.4 \times \frac{5}{13} + 0.4P \times \frac{12}{13}\)

Simplifying gives:

\(6(12/13) - P(5/13) = 2.4(5/13) + 0.4P(12/13)\)

Solving for \(P\):

\(P = 6.12\)