| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Practical friction scenarios |
| Difficulty | Standard +0.3 This is a standard two-part mechanics question on friction and equilibrium on slopes. Part (a) is straightforward application of limiting equilibrium (μ = tan θ). Part (b) requires resolving forces on a steeper slope and finding the deficit between friction and weight component. Both parts follow textbook methods with no novel problem-solving required, making it slightly easier than average for M1. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks |
|---|---|
| resolve perp. to plane: \(R - 4g\cos 10 = 0 \Rightarrow R = 4g\cos 10\) | M1 A1 |
| resolve // to plane: \(F - 4g\sin 10 = 0 \Rightarrow F = 4g\sin 10\) | M1 A1 |
| \(\mu = \frac{F}{R} = \tan 10 = 0.176\) (3sf) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| let extra force be \(P\) | M1 | |
| resolve perp. to plane: \(R - 4g\cos 30 = 0 \Rightarrow R = 4g\cos 30\) | M1 A1 | |
| \(F = \mu R = 5.986\) | A1 | |
| resolve // to plane: \(F + P - 4g\sin 30 = 0\) | M1 A1 | |
| \(P = 2g - F = 13.6\) N | A1 | (11) |
**Part (a):**
resolve perp. to plane: $R - 4g\cos 10 = 0 \Rightarrow R = 4g\cos 10$ | M1 A1 |
resolve // to plane: $F - 4g\sin 10 = 0 \Rightarrow F = 4g\sin 10$ | M1 A1 |
$\mu = \frac{F}{R} = \tan 10 = 0.176$ (3sf) | M1 A1 |
**Part (b):**
let extra force be $P$ | M1 |
resolve perp. to plane: $R - 4g\cos 30 = 0 \Rightarrow R = 4g\cos 30$ | M1 A1 |
$F = \mu R = 5.986$ | A1 |
resolve // to plane: $F + P - 4g\sin 30 = 0$ | M1 A1 |
$P = 2g - F = 13.6$ N | A1 | (11)
---A sledge of mass 4 kg rests in limiting equilibrium on a rough slope inclined at an angle 10° to the horizontal. By modelling the sledge as a particle,
\begin{enumerate}[label=(\alph*)]
\item show that the coefficient of friction, $\mu$, between the sledge and the ground is 0.176 correct to 3 significant figures. [6 marks]
\end{enumerate}
The sledge is placed on a steeper part of the slope which is inclined at an angle 30° to the horizontal. The value of $\mu$ remains unchanged.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the minimum extra force required along the line of greatest slope to prevent the sledge from slipping down the hill. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [11]}}