| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with particle - suspended equilibrium |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring finding the centroid of a uniform triangular frame, then applying equilibrium conditions when a particle is attached and the system is suspended. The geometry is straightforward (isosceles triangle), and the methods are routine applications of taught techniques with no novel insight required. Slightly easier than average due to the symmetric geometry and clear structure. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Mass ratio: \(AB=10\), \(AC=10\), \(BC=12\), frame \(=32\) | B1 | |
| Distance from \(BC\): \(4, 4, 0, \bar{x}\) | B1 | |
| Moments about \(BC\): \(10\times4+10\times4+0=32\bar{x}\) | M1 A1 | |
| \(\bar{x} = \frac{80}{32} = 2\frac{1}{2}\) (2.5) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Moments about \(B\): \(Mg \times 6\sin\theta = Mg \times (\bar{x}\cos\theta - 6\sin\theta)\) | M1 A1 A1 | |
| \(12\sin\theta = \bar{x}\cos\theta\) | ||
| \(\tan\theta = \frac{\bar{x}}{12}\) | ||
| \(\theta = 11.768... = 11.8°\) | A1 | (4 marks) |
| Alternative: C of M of loaded frame at distance \(\frac{1}{2}\bar{x}\) from \(D\) along \(DA\) | B1 | |
| \(\tan\theta = \frac{\frac{1}{2}\bar{x}}{6}\) | M1 A1 | |
| \(\theta = 11.8°\) | A1 |
# Question 3:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| Mass ratio: $AB=10$, $AC=10$, $BC=12$, frame $=32$ | B1 | |
| Distance from $BC$: $4, 4, 0, \bar{x}$ | B1 | |
| Moments about $BC$: $10\times4+10\times4+0=32\bar{x}$ | M1 A1 | |
| $\bar{x} = \frac{80}{32} = 2\frac{1}{2}$ (2.5) | A1 | (5 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| Moments about $B$: $Mg \times 6\sin\theta = Mg \times (\bar{x}\cos\theta - 6\sin\theta)$ | M1 A1 A1 | |
| $12\sin\theta = \bar{x}\cos\theta$ | | |
| $\tan\theta = \frac{\bar{x}}{12}$ | | |
| $\theta = 11.768... = 11.8°$ | A1 | (4 marks) |
| **Alternative:** C of M of loaded frame at distance $\frac{1}{2}\bar{x}$ from $D$ along $DA$ | B1 | |
| $\tan\theta = \frac{\frac{1}{2}\bar{x}}{6}$ | M1 A1 | |
| $\theta = 11.8°$ | A1 | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0a4e4cdd-bec4-4059-b9f7-9ce00bc34b71-08_613_629_125_660}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle ABC , where $\mathrm { AB } = \mathrm { AC } = 10 \mathrm {~cm}$ and $\mathrm { BC } = 12 \mathrm {~cm}$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the frame from $B C$.
The frame has total mass M . A particle of mass M is attached to the frame at the mid-point of BC . The frame is then freely suspended from B and hangs in equilibrium.
\item Find the size of the angle between BC and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [9]}}