| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Friction inequality derivation |
| Difficulty | Standard +0.3 This is a standard M1 friction problem requiring resolution of forces in two directions, use of F ≤ μR, and algebraic manipulation with a given angle. The tan α = 4/3 hint makes finding sin α and cos α straightforward (3-4-5 triangle). All steps follow a well-practiced routine with no novel insight required, making it slightly easier than average. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vertical: \(R - P\sin\alpha = W\) | M1, A1 | Equation for vertical equilibrium. Correct number of terms, forces resolved. Condone sign errors and sin/cos confusion. M0 for inequality |
| Horizontal: \(F = P\cos\alpha\) OR \(F_{MAX} \geq P\cos\alpha\) | M1, A1 | Equation for horizontal equilibrium. N.B. Allow \(F \geq P\cos\alpha\) |
| \(F \leq \frac{1}{4}R\) or \(F = \frac{1}{4}R\) seen or implied | M1 | M0 for \(F < \frac{1}{4}R\) or \(F > \frac{1}{4}R\) or \(F \geq \frac{1}{4}R\) |
| Produce dimensionally correct inequality or equation in \(P\) and \(W\) only (trig need not be substituted). e.g. \(\frac{1}{4}(W+P\sin\alpha) \geq P\cos\alpha\) | M1 | Eliminate \(F\) and \(R\) to form inequality/equation in \(P\) and \(W\) only |
| \(P \leq \frac{5W}{8}\) or \(\frac{5W}{8} \geq P\) | A1* cso | Reach given answer with exact working and correct use of inequality |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical: $R - P\sin\alpha = W$ | M1, A1 | Equation for vertical equilibrium. Correct number of terms, forces resolved. Condone sign errors and sin/cos confusion. M0 for inequality |
| Horizontal: $F = P\cos\alpha$ **OR** $F_{MAX} \geq P\cos\alpha$ | M1, A1 | Equation for horizontal equilibrium. N.B. Allow $F \geq P\cos\alpha$ |
| $F \leq \frac{1}{4}R$ **or** $F = \frac{1}{4}R$ seen or implied | M1 | M0 for $F < \frac{1}{4}R$ or $F > \frac{1}{4}R$ or $F \geq \frac{1}{4}R$ |
| Produce dimensionally correct inequality or equation in $P$ and $W$ only (trig need not be substituted). e.g. $\frac{1}{4}(W+P\sin\alpha) \geq P\cos\alpha$ | M1 | Eliminate $F$ and $R$ to form inequality/equation in $P$ and $W$ only |
| $P \leq \frac{5W}{8}$ **or** $\frac{5W}{8} \geq P$ | A1* cso | Reach given answer with exact working and correct use of inequality |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2737a11-4a15-41e9-9f87-31a705a8948b-16_314_815_246_625}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A particle of weight $W$ newtons lies at rest on a rough horizontal surface, as shown in Figure 3.\\
A force of magnitude $P$ newtons is applied to the particle.\\
The force acts at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 4 } { 3 }$\\
The coefficient of friction between the particle and the surface is $\frac { 1 } { 4 }$\\
Given that the particle does not move, show that
$$P \leqslant \frac { 5 W } { 8 }$$
\hfill \mbox{\textit{Edexcel M1 2023 Q6 [7]}}