M1 June 2013 Q3
3.
A box of mass 2 kg is held in equilibrium on a fixed rough inclined plane by a rope. The rope lies in a vertical plane containing a line of greatest slope of the inclined plane. The rope is inclined to the plane at an angle \(\alpha\), where \(\tan\alpha = \tfrac{3}{4}\), and the plane is at an angle of 30\(^\circ\) to the horizontal, as shown in Figure 1. The coefficient of friction between the box and the inclined plane is \(\tfrac{1}{3}\) and the box is on the point of slipping up the plane. By modelling the box as a particle and the rope as a light inextensible string, find the tension in the rope. (8)
| Scheme | Marks |
|---|---|
| \(T\cos\alpha - F = 2g\cos 60^\circ\) | M1 A1 |
| \(T\sin\alpha + R = 2g\cos 30^\circ\) | M1 A1 |
| \(F = \tfrac{1}{3}R\) | B1 |
| eliminating \(F\) and \(R\) | DM1 |
| \(T = g\left(1 + \dfrac{1}{\sqrt{3}}\right)\), 1.6g (or better), 15.5, 15 (N) | DM1 A1 |
| (8) | |
| (8 marks) |
Notes
First M1 for resolving parallel to the plane with correct no. of terms and both \(T\) and \(2g\) terms resolved.
First A1 for a correct equation. (use of \(\alpha\) instead of 30\(^\circ\) or 60\(^\circ\) or vice versa is an A error not M error; similarly if they use sin(3/5) or cos(4/5) when resolving, this can score M1A0)
Second M1 for resolving perpendicular to the plane with correct no. of terms and both \(T\) and \(2g\) terms resolved.
Second A1 for a correct equation (use of \(\alpha\) instead of 30\(^\circ\) or 60\(^\circ\) or vice versa is an A error not M error; similarly if they use sin(3/5) or cos(4/5) when resolving, this can score M1A0)
B1 for \(F = 1/3\ R\) seen or implied.
Third M1, dependent on first two M marks and appropriate angles used when resolving in both equations, for eliminating \(F\) and \(R\).
Fourth M1 dependent on third M1, for solving for \(T\)
Third A1 for 15(N) or 15.5 (N).
N.B. The first two M marks can be for two resolutions in any directions.
Use of \(\tan\alpha = 4/3\) leads to an answer of 17.83…and can score max 7/8.