| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane - force at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 friction problem requiring resolution of forces in two directions and application of F=μR at limiting equilibrium. While it involves multiple steps (resolving parallel and perpendicular to plane, applying friction law), the techniques are routine for M1 students and the setup is clearly defined. Part (c) requires showing motion occurs by comparing friction with component of weight, which is a standard check. Slightly above average difficulty due to the horizontal force complicating the geometry, but still a textbook-style question. |
| Spec | 3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| \(F = \frac{2}{3}R\) | B1 |
| \(R(\uparrow): R \cos 30° - F \cos 60° = 6g\) | M1 A1 |
| \(R\frac{\sqrt{3}}{2} - \frac{2}{3}R - \frac{1}{2} = 6g\) | A1 |
| \(R = 88.3 \text{ N (or 88 N)}\) | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R(\leftarrow): P = R \cos 60° + F \cos 30°\) | M1 A1 | |
| \(= 74.7 \text{ N (or 75 N)}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Component of weight \((\checkmark) = 6g \cos 60° = 29.4 \text{ N}\) | B1 | |
| \(R' = 6g \cos 30° = 50.9 \text{ N}\) | M1 A1 | |
| \(F_{\max} = 0.4 R' = 20.36 \text{ N}\) | M1 | |
| Since \(29.4 > 20.36\), the box moves | A1 cso | (5) |
## (a)
$F = \frac{2}{3}R$ | B1 |
$R(\uparrow): R \cos 30° - F \cos 60° = 6g$ | M1 A1 |
$R\frac{\sqrt{3}}{2} - \frac{2}{3}R - \frac{1}{2} = 6g$ | A1 |
$R = 88.3 \text{ N (or 88 N)}$ | (4) |
## (b)
$R(\leftarrow): P = R \cos 60° + F \cos 30°$ | M1 A1 |
$= 74.7 \text{ N (or 75 N)}$ | A1 | (3) |
## (c)
Component of weight $(\checkmark) = 6g \cos 60° = 29.4 \text{ N}$ | B1 |
$R' = 6g \cos 30° = 50.9 \text{ N}$ | M1 A1 |
$F_{\max} = 0.4 R' = 20.36 \text{ N}$ | M1 |
Since $29.4 > 20.36$, the box moves | A1 cso | (5) |
**Total: (12 marks)**
---\includegraphics{figure_2}
A box of mass $6 \text{ kg}$ lies on a rough plane inclined at an angle of $30°$ to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude $P$ newtons, as shown in Fig. 2. The line of action of the force is in the same vertical plane as a line of greatest slope of the plane. The coefficient of friction between the box and the plane is $0.4$. The box is modelled as a particle.
Given that the box is in limiting equilibrium and on the point of moving up the plane, find,
\begin{enumerate}[label=(\alph*)]
\item the normal reaction exerted on the box by the plane, [4]
\item the value of $P$. [3]
\end{enumerate}
The horizontal force is removed.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that the box will now start to move down the plane. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q4 [12]}}