M1 June 2012 Q3

Figure 2: box on a plane inclined at 30 degrees, held by a horizontal force of 20 N — Figure 2

A box of mass 5 kg lies on a rough plane inclined at 30\(^\circ\) to the horizontal. The box is held in equilibrium by a horizontal force of magnitude 20 N, as shown in Figure 2. The force acts in a vertical plane containing a line of greatest slope of the inclined plane.

The box is in equilibrium and on the point of moving down the plane. The box is modelled as a particle.

Find

(a) the magnitude of the normal reaction of the plane on the box, (4)

(b) the coefficient of friction between the box and the plane. (5)

Force diagram: normal reaction R, friction mu R up the plane, horizontal force 20 and weight 5g on the box, plane at 30 degrees — Mark scheme (a)

Scheme	Marks
\(\perp\) plane \(R = 20\cos 60^\circ + 5g\cos 30^\circ\)	M1 A2(1,0)
\(= 52.4\) (N) or 52	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(F_r = \mu R\)	B1
\(\parallel\) plane \(F + 20\cos 30^\circ = 5g\cos 60^\circ\)	M1 A2(1, 0)
Leading to \(\mu = 0.137\) or 0.14	A1
	(5)
	(9 marks)

Notes

B1 for use of \(F = \mu R\) (could just be on diagram)

First M1 (allow if \(F\) is used rather than \(\mu R\)) for resolving parallel to the plane with usual criteria

First A2 for a correct equation (A1A0 one error, A0A0 for two or more errors)

Second A1 for either 0.14 or 0.137

N.B. If they resolve vertically AND horizontally, there are max 6 marks available (M1A2, M1A2) for the TWO equations, but if they only have one equation, there are no marks available for that equation.

The marks for the horizontal resolution should be entered first on ePen.