| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane - force parallel to slope |
| Difficulty | Standard +0.3 This is a standard M1 inclined plane friction problem requiring resolution of forces parallel and perpendicular to the plane, application of F=μR at limiting equilibrium, and basic algebraic manipulation to show a given ratio. Part (c) tests conceptual understanding of static friction. While multi-part with 11 marks total, it follows a routine template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | [Diagram showing forces on inclined plane] | B2 |
| (b) | resolve perp. to plane: \(R - 2g\cos 30° = 0 \therefore R = g\sqrt{3}\) | M1 A1 |
| \(F = \mu R = \frac{1}{\sqrt{3}} \times g\sqrt{3} = g\) | M1 A1 | |
| resolve // to plane: \(H - F - 2g\sin 30° = 0\) | M1 | |
| \(H - g - g = 0 \therefore H = 2g\) | A1 | |
| \(F : H = g : 2g = 1 : 2\) | A1 | |
| (c) | friction varies between \(\mu R\) up plane (to prevent movement down plane) and \(\mu R\) down plane (to prevent movement up plane) | B2 |
(a) | [Diagram showing forces on inclined plane] | B2 |
(b) | resolve perp. to plane: $R - 2g\cos 30° = 0 \therefore R = g\sqrt{3}$ | M1 A1 |
| $F = \mu R = \frac{1}{\sqrt{3}} \times g\sqrt{3} = g$ | M1 A1 |
| resolve // to plane: $H - F - 2g\sin 30° = 0$ | M1 |
| $H - g - g = 0 \therefore H = 2g$ | A1 |
| $F : H = g : 2g = 1 : 2$ | A1 |
(c) | friction varies between $\mu R$ up plane (to prevent movement down plane) and $\mu R$ down plane (to prevent movement up plane) | B2 | (11) |\includegraphics{figure_2}
A particle $P$, of mass 2 kg, lies on a rough plane inclined at an angle of 30° to the horizontal. A force $H$, whose line of action is parallel to the line of greatest slope of the plane, is applied to the particle as shown in Figure 2. The coefficient of friction between the particle and the plane is $\frac{1}{\sqrt{3}}$.
Given that the particle is on the point of moving up the plane,
\begin{enumerate}[label=(\alph*)]
\item draw a diagram showing all the forces acting on the particle, [2 marks]
\item show that the ratio of the magnitude of the frictional force to the magnitude of $H$ is equal to $1 : 2$ [7 marks]
\end{enumerate}
The force $H$ is now removed but $P$ remains at rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use the principle of friction to explain how this is possible. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q5 [11]}}