Standard +0.3 This is a standard M1 equilibrium problem on a slope with friction. It requires resolving forces in two directions and using the limiting friction condition, but follows a well-practiced method with straightforward trigonometry (given tan α = 3/4). The 8 marks reflect multiple steps rather than conceptual difficulty—slightly easier than average for M1.
\includegraphics{figure_1}
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 \(\tan \alpha = \frac{3}{4}\), and the plane is at an angle of \(30°\) to the horizontal, as shown in Figure 1. The coefficient of friction between the box and the inclined plane is \(\frac{1}{2}\) 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]
- 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° or 60° 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° or 60° 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.
$T\cos\alpha - F = 2g\cos 60°$ | M1 A1
$T\sin\alpha + R = 2g\cos 30°$ | M1 A1
$F = \frac{1}{3}R$ | B1
eliminating $F$ and $R$ | DM1
$T = g(1 + \frac{1}{\sqrt{3}}), 1.6g$ (or better), 15.5, 15 (N) | DM1 A1
| (8)
| [8]
**Notes for Question 3:**
- 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° or 60° 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° or 60° 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.
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\includegraphics{figure_1}
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 $\tan \alpha = \frac{3}{4}$, and the plane is at an angle of $30°$ to the horizontal, as shown in Figure 1. The coefficient of friction between the box and the inclined plane is $\frac{1}{2}$ 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]
\hfill \mbox{\textit{Edexcel M1 2013 Q3 [8]}}