| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Standard +0.8 This is a challenging M2 statics problem requiring geometric analysis to find contact point position, then resolving forces in two directions plus taking moments with friction limiting at two surfaces simultaneously. The geometry is non-trivial (involving perpendicular from cylinder center to rod), and the friction conditions at both contacts create a system requiring careful force resolution and moment calculation. Significantly harder than routine equilibrium problems but standard for harder M2 questions. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC = 4a \tan 60° = 4a\sqrt{3}\) | M1 A1 | Or \(\frac{4a}{\tan 30°}\) or \(\sqrt{(8a)^2 - (4a)^2}\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| use of \(F = \mu R\) at either A or C | M1 | |
| 3 independent equations required. Award M1A1 for each in the order seen. If more than 3 relevant equations seen, award the marks for the best 3. | ||
| \(M(A), \quad R_C.4a\sqrt{3} = W.3a\sqrt{3}\cos60°\) | M1 A1 | \(R_C = \frac{3W}{8}\) |
| \((\uparrow), \quad R_A + R_C\cos60° + F_C\cos30° = W\) | M1 A1 | \(R_A = \frac{5W}{8}\) |
| \((\rightarrow), \quad F_A - R_C\cos30° - F_C\cos60° = 0\) | M1 A1 | \(F_A = R_C \frac{\sqrt{3}}{3}\) |
| \(M(C)\) \(a\sqrt{3}\cos60° + F_A.4a\sqrt{3}\sin60° = R_A.4a\sqrt{3}\cos60°\) | ||
| Parallel: \(F_A \cos 60 + R_A \cos 30 + F_C = W \cos 30\) | ||
| Perpendicular: \(R_C + R_A \cos 60 = F_A \cos 30 + W \cos 60\) | ||
| solving to give \(\mu = \frac{\sqrt{3}}{5}\); 0.346 or 0.35. | DM1 A1 | Equation in \(\mu\) only. Dependent on 4 M marks for their equations. |
| Reactions in the wrong direction(s) – check carefully | ||
| (9) | ||
| [11] |
**Part (a):**
| $AC = 4a \tan 60° = 4a\sqrt{3}$ | M1 A1 | Or $\frac{4a}{\tan 30°}$ or $\sqrt{(8a)^2 - (4a)^2}$ |
| | (2) | |
**Part (b):**
| use of $F = \mu R$ at either A or C | M1 | |
| 3 independent equations required. Award M1A1 for each in the order seen. If more than 3 relevant equations seen, award the marks for the best 3. | | |
| $M(A), \quad R_C.4a\sqrt{3} = W.3a\sqrt{3}\cos60°$ | M1 A1 | $R_C = \frac{3W}{8}$ |
| $(\uparrow), \quad R_A + R_C\cos60° + F_C\cos30° = W$ | M1 A1 | $R_A = \frac{5W}{8}$ |
| $(\rightarrow), \quad F_A - R_C\cos30° - F_C\cos60° = 0$ | M1 A1 | $F_A = R_C \frac{\sqrt{3}}{3}$ |
| $M(C)$ $a\sqrt{3}\cos60° + F_A.4a\sqrt{3}\sin60° = R_A.4a\sqrt{3}\cos60°$ | | |
| Parallel: $F_A \cos 60 + R_A \cos 30 + F_C = W \cos 30$ | | |
| Perpendicular: $R_C + R_A \cos 60 = F_A \cos 30 + W \cos 60$ | | |
| solving to give $\mu = \frac{\sqrt{3}}{5}$; 0.346 or 0.35. | DM1 A1 | Equation in $\mu$ only. Dependent on 4 M marks for their equations. |
| Reactions in the wrong direction(s) – check carefully | | |
| | (9) |
| | [11] | |
---A rough circular cylinder of radius $4a$ is fixed to a rough horizontal plane with its axis horizontal. A uniform rod $AB$, of weight $W$ and length $6a\sqrt{3}$, rests with its lower end $A$ on the plane and a point $C$ of the rod against the cylinder. The vertical plane through the rod is perpendicular to the axis of the cylinder. The rod is inclined at 60° to the horizontal, as shown in Figure 1.
\includegraphics{figure_1}
\begin{enumerate}[label=(\alph*)]
\item Show that $AC = 4a\sqrt{3}$ [2]
\end{enumerate}
The coefficient of friction between the rod and the cylinder is $\frac{\sqrt{3}}{3}$ and the coefficient of friction between the rod and the plane is $\mu$. Given that friction is limiting at both $A$ and $C$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $\mu$. [9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q4 [11]}}