| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| 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 taking moments about a point, resolving forces in two directions, and applying friction formula. The geometry is straightforward with given distances, and part (a) guides students through to the key result needed for part (b). While it requires multiple steps and careful algebra, it follows a well-practiced method with no novel insights required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(R(\uparrow)\): \(R + P\cos\alpha = W\) | M1 A1 | |
| \(M(A)\): \(P \times 2a = W \times 1.5a\cos\alpha\) | M1 A1 | |
| \(\left(p = \frac{3}{4}W\cos\alpha\right)\) | ||
| \(R = W - P\cos\alpha = W - \frac{3}{4}W\cos^2\alpha\) | DM1 | |
| \(= \frac{1}{4}(4 - 3\cos^2\alpha) | W\) ★ | cso A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(\cos\alpha = \frac{2}{3}\): \(R = \frac{2}{3}W\) | B1 | |
| \(R(\rightarrow)\): \(\mu R = P\sin\alpha\) | M1 A1 | |
| Leading to \(\mu = \frac{3}{4}\sin\alpha\) | ||
| \(\left(\sin\alpha = \sqrt{\left(1 - \frac{4}{9}\right) = \frac{\sqrt{5}}{5}}\right)\) | ||
| \(\mu = \frac{\sqrt{5}}{4}\) | awrt 0.56 DM1 A1 | [5] [11] |
**Part (a):**
$R(\uparrow)$: $R + P\cos\alpha = W$ | M1 A1 |
$M(A)$: $P \times 2a = W \times 1.5a\cos\alpha$ | M1 A1 |
$\left(p = \frac{3}{4}W\cos\alpha\right)$ | |
$R = W - P\cos\alpha = W - \frac{3}{4}W\cos^2\alpha$ | DM1 |
$= \frac{1}{4}(4 - 3\cos^2\alpha)|W$ ★ | cso A1 | [6]
**Part (b):**
Using $\cos\alpha = \frac{2}{3}$: $R = \frac{2}{3}W$ | B1 |
$R(\rightarrow)$: $\mu R = P\sin\alpha$ | M1 A1 |
Leading to $\mu = \frac{3}{4}\sin\alpha$ | |
$\left(\sin\alpha = \sqrt{\left(1 - \frac{4}{9}\right) = \frac{\sqrt{5}}{5}}\right)$ | |
$\mu = \frac{\sqrt{5}}{4}$ | awrt 0.56 DM1 A1 | [5] [11]5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2738ce4-4dc5-4cd1-ac3d-0c3fcf21ea71-07_501_918_274_502}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A plank rests in equilibrium against a fixed horizontal pole. The plank is modelled as a uniform rod $A B$ and the pole as a smooth horizontal peg perpendicular to the vertical plane containing $A B$. The rod has length $3 a$ and weight $W$ and rests on the peg at $C$, where $A C = 2 a$. The end $A$ of the rod rests on rough horizontal ground and $A B$ makes an angle $\alpha$ with the ground, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal reaction on the rod at $A$ is $\frac { 1 } { 4 } \left( 4 - 3 \cos ^ { 2 } \alpha \right) W$.
Given that the rod is in limiting equilibrium and that $\cos \alpha = \frac { 2 } { 3 }$,
\item find the coefficient of friction between the rod and the ground.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2008 Q5 [11]}}