| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite shapes and suspended equilibrium. Part (a) requires symmetry recognition, part (b) uses the standard formula for composite rectangles (straightforward calculation), and part (c) applies the principle that the centre of mass hangs directly below the suspension point (tan θ = horizontal distance / vertical distance). All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{3072\rho}{384\rho} = 8\) so must lie on \(BF\) | M1 A1 | |
| \(\bar{y} = \frac{4608\rho}{384\rho} = 12\) \(\therefore\) dist. from \(AB = 20 \text{ cm}\) | M1 A1 | |
| (c) \(\tan\theta = \frac{8}{20}\) \(\therefore \theta = 21.8°\) (1dp) | M1 A1 | (12 marks) |
**(a), (b)** Table completed with mass per unit area, $x, y$ coordinates taken horiz./vert. from $G$
$\bar{x} = \frac{3072\rho}{384\rho} = 8$ so must lie on $BF$ | M1 A1 |
$\bar{y} = \frac{4608\rho}{384\rho} = 12$ $\therefore$ dist. from $AB = 20 \text{ cm}$ | M1 A1 |
**(c)** $\tan\theta = \frac{8}{20}$ $\therefore \theta = 21.8°$ (1dp) | M1 A1 | (12 marks)5.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{f0e751be-f095-4a56-8ee9-8433cc4873e9-3_591_609_785_623}
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\caption{Fig. 2}
\end{center}
\end{figure}
Figure 2 shows a uniform plane lamina $A B C D E G$ in the shape of a letter ' $L$ ' consisting of a rectangle $A B F G$ joined to another rectangle $C D E F$. The sides $A B$ and $D E$ are both 8 cm long and the sides $E G$ and $G A$ are of length 24 cm and 32 cm respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the lamina lies on the line $B F$.
\item Find the distance of the centre of mass from the line $A B$.
The uniform lamina in Figure 2 is a model of the letter ' $L$ ' in a sign above a shop. The letter is normally suspended from a wall at $A$ and $B$ so that $A B$ is horizontal but the fixing at $B$ has broken and the letter hangs in equilibrium from the point $A$.
\item Find, in degrees to one decimal place, the acute angle $A G$ makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [12]}}