M1 June 2017 Q4

Figure 1: particle on a plane inclined at angle alpha, pushed by a horizontal force of 10 N — Figure 1

A particle \(P\) of mass 5 kg is held at rest in equilibrium on a rough inclined plane by a horizontal force of magnitude 10 N. The plane is inclined to the horizontal at an angle \(\alpha\) where \(\tan\alpha = \dfrac{3}{4}\), as shown in Figure 1. The line of action of the force lies in the vertical plane containing \(P\) and a line of greatest slope of the plane. The coefficient of friction between \(P\) and the plane is \(\mu\). Given that \(P\) is on the point of sliding down the plane, find the value of \(\mu\). (9)

Mark scheme

Scheme	Marks
\(F = \mu R\)	B1
\((\nwarrow),\ \ R = 10\sin\alpha + 5g\cos\alpha\ \ (45.2)\)	M1 A2
\((\nearrow),\ \ F = 5g\sin\alpha - 10\cos\alpha\ \ (21.4)\)	M1 A2
\(\mu = \dfrac{g\sin\alpha - 2\cos\alpha}{2\sin\alpha + g\cos\alpha} = 0.47\) or 0.473	M1 A1
	(9 marks)

Notes

B1 for \(F = \mu R\) seen or implied

First M1 for resolving perpendicular to the plane with usual rules

First and second A1’s for a correct equation. A1A0 if one error.

Second M1 for resolving parallel to the plane with usual rules

Third and fourth A1’s for a correct equation. A1A0 if one error.

If \(m\) is used instead of 5, penalise once in each equation.

Third M1 independent for eliminating \(R\) to produce an equation in \(\mu\) only. Does not need to be \(\mu = \ldots\).

Fifth A1 for 0.47 or 0.473.