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June 2023 p42 q5
3997
A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of 35° to the horizontal. The particle is kept in equilibrium by a horizontal force of magnitude \(P\) N acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and plane is 0.4.
Find the least possible value of \(P\).
Solution
Resolve forces perpendicular to the plane:
\(R = P \sin 35 + 0.6g \cos 35\)
Resolve forces parallel to the plane:
\(F + P \cos 35 = 0.6g \sin 35\)
Using \(F = 0.4R\), substitute into the parallel resolution equation:
\(0.4R + P \cos 35 = 0.6g \sin 35\)
Substitute \(R = \frac{0.6g}{\cos 35 + 0.4 \sin 35}\) into the equation: