| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Standard +0.3 This is a standard M2 moments problem requiring equilibrium conditions (resolving forces and taking moments) with friction at limiting equilibrium. The setup is straightforward with given angle and distances, requiring systematic application of three equilibrium equations to find μ. Slightly easier than average due to clear geometry and standard method. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about A: \(3S = 100 \times 2 \times \cos\alpha\) | M1 A1 | |
| Resolving vertically: \(R + S\cos\alpha = 100\) | M1 A1 | |
| Resolving horizontally: \(S\sin\alpha = F\) | M1 A1 | Most alternative methods need 3 independent equations, each worth M1A1. Can be done in 2 e.g. resolving horizontally and taking moments about \(X\): \(R \times 2 \times \cos\alpha = S \times (3 - 2\cos^2\alpha)\) scores M2A2 |
| Substitute trig values: \(\left(S = \frac{200\sqrt{8}}{9}\right)\), \(R = 100 - \frac{1600}{27} = \frac{1100}{27} \approx 40.74\), \(F = \frac{200\sqrt{8}}{27} \approx 20.95\) | DM1 A1 | Substitute trig values to obtain correct values for \(F\) and \(R\) (exact or decimal equivalent) |
| \(F \leq \mu R\), \(200\sqrt{8} \leq \mu \times 1100\), \(\mu \geq \frac{200\sqrt{8}}{1100} = \frac{2\sqrt{8}}{11}\) | M1 | |
| Least possible \(\mu\) is \(0.514\) (3sf), or exact | A1 |
# Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about A: $3S = 100 \times 2 \times \cos\alpha$ | M1 A1 | |
| Resolving vertically: $R + S\cos\alpha = 100$ | M1 A1 | |
| Resolving horizontally: $S\sin\alpha = F$ | M1 A1 | Most alternative methods need 3 independent equations, each worth M1A1. Can be done in 2 e.g. resolving horizontally and taking moments about $X$: $R \times 2 \times \cos\alpha = S \times (3 - 2\cos^2\alpha)$ scores M2A2 |
| Substitute trig values: $\left(S = \frac{200\sqrt{8}}{9}\right)$, $R = 100 - \frac{1600}{27} = \frac{1100}{27} \approx 40.74$, $F = \frac{200\sqrt{8}}{27} \approx 20.95$ | DM1 A1 | Substitute trig values to obtain correct values for $F$ and $R$ (exact or decimal equivalent) |
| $F \leq \mu R$, $200\sqrt{8} \leq \mu \times 1100$, $\mu \geq \frac{200\sqrt{8}}{1100} = \frac{2\sqrt{8}}{11}$ | M1 | |
| Least possible $\mu$ is $0.514$ (3sf), or exact | A1 | |
---7.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{c5760fa5-3c7f-4e29-87a2-b3b4145b9361-14_442_986_264_479}
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\caption{Figure 4}
\end{center}
\end{figure}
A uniform plank $A B$, of weight 100 N and length 4 m , rests in equilibrium with the end $A$ on rough horizontal ground. The plank rests on a smooth cylindrical drum. The drum is fixed to the ground and cannot move. The point of contact between the plank and the drum is $C$, where $A C = 3 \mathrm {~m}$, as shown in Figure 4. The plank is resting in a vertical plane which is perpendicular to the axis of the drum, at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 3 }$. The coefficient of friction between the plank and the ground is $\mu$. Modelling the plank as a rod, find the least possible value of $\mu$.
\hfill \mbox{\textit{Edexcel M2 2011 Q7 [10]}}