| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Standard +0.8 This is a challenging M2 equilibrium problem requiring resolution of forces in two directions, taking moments about a strategic point, and using limiting friction at two contact points simultaneously. The geometry with the peg partway along the rod adds complexity, and students must carefully handle the normal and friction forces at both contacts to find the two unknowns systematically. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Take moments about \(A\) | M1 | Must be dimensionally correct. Condone sin/cos confusion |
| \(5N = 4\cos\theta W\) | A1 | |
| \(N = \frac{12}{25}W = 0.48W\) Given Answer | A1 | |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(G = \frac{1}{4}N = 0.12W\) | B1 | Seen or implied |
| Resolve vertically: \(R + N\cos\theta + G\sin\theta = W\) | M1, A1 | Needs all terms. Condone sin/cos confusion and sign errors. \((R = 0.616W)\) |
| Resolve horizontally: \(F + G\cos\theta = N\sin\theta\) | M1, A1 | Needs all terms. Condone sin/cos confusion and sign errors. \((F = 0.312W)\) |
| \(\mu = \frac{N\sin\theta - G\cos\theta}{W - N\cos\theta - G\sin\theta}\) | DM1 | Use \(F = \mu R\) to find \(\mu\). Dependent on 2 preceding M marks |
| \(= \frac{0.48W \times 0.8 - 0.12W \times 0.6}{W - 0.48W \times 0.6 - 0.12W \times 0.8} = \frac{0.312}{0.616}\) | ||
| \(= 0.51\) \((0.50649\ldots)\) \(\left(\frac{39}{77}\right)\) | A1 | |
| (7 marks, 10 marks total for Q5) |
# Question 5:
## Part 5a:
| Working | Mark | Notes |
|---------|------|-------|
| Take moments about $A$ | M1 | Must be dimensionally correct. Condone sin/cos confusion |
| $5N = 4\cos\theta W$ | A1 | |
| $N = \frac{12}{25}W = 0.48W$ **Given Answer** | A1 | |
| **(3 marks)** | | |
## Part 5b:
| Working | Mark | Notes |
|---------|------|-------|
| $G = \frac{1}{4}N = 0.12W$ | B1 | Seen or implied |
| Resolve vertically: $R + N\cos\theta + G\sin\theta = W$ | M1, A1 | Needs all terms. Condone sin/cos confusion and sign errors. $(R = 0.616W)$ |
| Resolve horizontally: $F + G\cos\theta = N\sin\theta$ | M1, A1 | Needs all terms. Condone sin/cos confusion and sign errors. $(F = 0.312W)$ |
| $\mu = \frac{N\sin\theta - G\cos\theta}{W - N\cos\theta - G\sin\theta}$ | DM1 | Use $F = \mu R$ to find $\mu$. Dependent on 2 preceding M marks |
| $= \frac{0.48W \times 0.8 - 0.12W \times 0.6}{W - 0.48W \times 0.6 - 0.12W \times 0.8} = \frac{0.312}{0.616}$ | | |
| $= 0.51$ $(0.50649\ldots)$ $\left(\frac{39}{77}\right)$ | A1 | |
| **(7 marks, 10 marks total for Q5)** | | |
**NB:** One of the two equations for part (b) could be a moments equation:
- M$(P)$: $1 \times W\cos\theta + 5F\sin\theta = 5R\cos\theta$
- M$(B)$: $3N + 8R\cos\theta = 4W\cos\theta + 8F\sin\theta$
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be16c17a-c4db-4f0c-9f32-8d5614b4f2f3-12_440_1047_246_447}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform rod $A B$ of length 8 m and weight $W$ newtons rests in equilibrium against a rough horizontal peg $P$. The end $A$ is on rough horizontal ground. The friction is limiting at both $A$ and $P$. The distance $A P$ is 5 m , as shown in Figure 1. The rod rests at angle $\theta$ to the horizontal, where $\tan \theta = \frac { 4 } { 3 }$. The rod is in a vertical plane which is perpendicular to $P$. The coefficient of friction between the rod and $P$ is $\frac { 1 } { 4 }$ and the coefficient of friction between the rod and the ground is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the normal reaction between the rod and $P$ is $0.48 W$ newtons.
\item Find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q5 [10]}}