M1 June 2016 Q5

Figure 2: particle P of mass 2 kg on a plane inclined at 20 degrees, pushed by a 40 N force at 30 degrees to the plane — Figure 2

A particle \(P\) of mass 2 kg is held at rest in equilibrium on a rough plane by a constant force of magnitude 40 N. The direction of the force is inclined to the plane at an angle of 30\(^\circ\). The plane is inclined to the horizontal at an angle of 20\(^\circ\), as shown in Figure 2. 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 up the plane, find the value of \(\mu\) (10)

Mark scheme

Scheme	Marks
\(\mu R\)	B1
\(R = 2g\cos 20^\circ + 40\cos 60^\circ\)	M1 A2
\(F = 40\cos 30^\circ - 2g\cos 70^\circ\)	M1 A2
\(\mu = \dfrac{40\cos 30^\circ - 2g\cos 70^\circ}{2g\cos 20^\circ + 40\cos 60^\circ}\)	M1 M1
\(= 0.73\) or 0.727	A1
	(10 marks)

Notes

B1 for \(\mu R\) seen or implied.

First M1 for resolving perpendicular to the plane with usual rules (must be using \(2(g)\) with 20\(^\circ\) or 70\(^\circ\) and 40 with 30\(^\circ\) or 60\(^\circ\))

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

Second M1 for resolving parallel to the plane with usual rules (must be using \(2(g)\) with 20\(^\circ\) or 70\(^\circ\) and 40 with 30\(^\circ\) or 60\(^\circ\))

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

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

Fourth M1 independent for solving for \(\mu\)

Fifth A1 for 0.727 or 0.73

N.B. They may choose to resolve in 2 other directions e.g. horizontally and vertically.

N.B. If \(F\) is replaced by \(-F\) in the second equⁿ, treat this as an error unless they subsequently explain that they have their \(F\) acting in the wrong direction, in which case they could score full marks for the question.