| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Standard +0.8 This is a multi-part statics problem involving a rod resting on a peg, requiring force diagrams, resolving forces in two directions, taking moments about a strategic point, and applying friction laws. While the individual techniques are standard M2 content, the combination of steps, the geometry with the peg creating two contact points, and the friction coefficient calculation make this moderately challenging—above average but not requiring exceptional insight. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Diagram showing: Normal reaction at \(A\) (vertical), Weight \(6g\) downward at midpoint, Normal reaction at \(C\) (perpendicular to rod), Friction at \(C\) (along rod) | B1 B1 | B1 for weight + reaction at A; B1 for normal and friction at C |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Moments about \(C\): \(R_A \times 3\cos 20° = 6g \times 0.5\cos 20°\) | M1 A1 | |
| \(R_A = \frac{6g \times 0.5}{3} = g = 9.8\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve vertically: \(R_A + N_C\cos 20° - F_C\sin 20° = 6g\) | M1 A1 | |
| Resolve horizontally: \(N_C\sin 20° + F_C\cos 20° = 0\)... corrected: \(N_C\sin 20° = F_C\cos 20°\) rearranged | M1 | |
| \(N_C = \frac{6g - R_A}{\cos 20° + \sin 20°\tan 20°}\) | ||
| \(N_C \approx 46.6\) N | A1 | |
| \(F_C = N_C\tan 20° \approx 17.0\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mu = \frac{F_C}{N_C} = \frac{17.0}{46.6} \approx 0.364\) | M1 A1 |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Diagram showing: Normal reaction at $A$ (vertical), Weight $6g$ downward at midpoint, Normal reaction at $C$ (perpendicular to rod), Friction at $C$ (along rod) | B1 B1 | B1 for weight + reaction at A; B1 for normal and friction at C |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Moments about $C$: $R_A \times 3\cos 20° = 6g \times 0.5\cos 20°$ | M1 A1 | |
| $R_A = \frac{6g \times 0.5}{3} = g = 9.8$ N | A1 | |
## Part (c)(i) and (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve vertically: $R_A + N_C\cos 20° - F_C\sin 20° = 6g$ | M1 A1 | |
| Resolve horizontally: $N_C\sin 20° + F_C\cos 20° = 0$... corrected: $N_C\sin 20° = F_C\cos 20°$ rearranged | M1 | |
| $N_C = \frac{6g - R_A}{\cos 20° + \sin 20°\tan 20°}$ | | |
| $N_C \approx 46.6$ N | A1 | |
| $F_C = N_C\tan 20° \approx 17.0$ N | A1 | |
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mu = \frac{F_C}{N_C} = \frac{17.0}{46.6} \approx 0.364$ | M1 A1 | |
---\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show the forces acting on the rod.
\item Find the magnitude of the normal reaction force between the rod and the ground.
\item \begin{enumerate}[label=(\roman*)]
\item Find the normal reaction acting on the rod at $C$.
\item Find the friction force acting on the rod at $C$.
\end{enumerate}\item In this position, the rod is on the point of slipping.
Calculate the coefficient of friction between the rod and the peg.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-15_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q7 [12]}}