| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.8 This is a multi-step M2 centre of mass problem requiring: (a) geometric reasoning about the centroid of an equilateral triangle, (b) composite body centre of mass calculation using the removal method with verification, and (c) equilibrium analysis of a suspended lamina. While the techniques are standard M2 content, the problem requires careful geometric setup, algebraic manipulation, and integration of multiple concepts, making it moderately challenging but within typical M2 scope. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| from triangle properties, area of \(BCD = \frac{1}{3}\) area of \(ABD\) | B1 |
| \(\therefore\) area of \(BCD = \frac{1}{3}(\frac{1}{2} \times 2d \times \sqrt{3}d) = \frac{1}{3}\sqrt{3}d^2\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| portion | mass | y |
| \(ABD\) | \(\sqrt{3}d^2\rho\) | \(\frac{1}{3}\sqrt{3}d\) |
| \(BCD\) | \(\frac{1}{3}\sqrt{3}d^2\rho\) | \(\frac{1}{9}\sqrt{3}d\) |
| \(ABCD\) | \(\frac{4}{3}\sqrt{3}d^2\rho\) | \(\bar{y}\) |
| \(\rho =\) mass per unit area; \(y\) coords. taken vert. from \(BD\) | M3 A3 | |
| \(\bar{y} = \frac{\frac{8}{9}d^3\rho}{\frac{4}{3}\sqrt{3}d^2\rho} = \frac{4d\rho}{3\sqrt{3}} = \frac{4}{9}\sqrt{3}d\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = \tan^{-1}\frac{\frac{4}{9}\sqrt{3}d}{d} = \tan^{-1}\frac{4\sqrt{3}}{9}\) | M1 A1 | |
| req'd angle \(= 60 - \theta = 22.4°\) (1dp) | M1 A1 | (15) |
**Part (a):**
from triangle properties, area of $BCD = \frac{1}{3}$ area of $ABD$ | B1 |
$\therefore$ area of $BCD = \frac{1}{3}(\frac{1}{2} \times 2d \times \sqrt{3}d) = \frac{1}{3}\sqrt{3}d^2$ | M1 A1 |
**Part (b):**
| portion | mass | y | my |
|---------|------|---------|---------|
| $ABD$ | $\sqrt{3}d^2\rho$ | $\frac{1}{3}\sqrt{3}d$ | $d^3\rho$ |
| $BCD$ | $\frac{1}{3}\sqrt{3}d^2\rho$ | $\frac{1}{9}\sqrt{3}d$ | $\frac{1}{9}d^3\rho$ |
| $ABCD$ | $\frac{4}{3}\sqrt{3}d^2\rho$ | $\bar{y}$ | $\frac{8}{9}d^3\rho$ |
$\rho =$ mass per unit area; $y$ coords. taken vert. from $BD$ | M3 A3 |
$\bar{y} = \frac{\frac{8}{9}d^3\rho}{\frac{4}{3}\sqrt{3}d^2\rho} = \frac{4d\rho}{3\sqrt{3}} = \frac{4}{9}\sqrt{3}d$ | M1 A1 |
**Part (c):**
$\theta = \tan^{-1}\frac{\frac{4}{9}\sqrt{3}d}{d} = \tan^{-1}\frac{4\sqrt{3}}{9}$ | M1 A1 |
req'd angle $= 60 - \theta = 22.4°$ (1dp) | M1 A1 | (15)
---
**Total: (75)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f8ece90a-5042-4db1-9855-ffe74333a899-4_542_625_959_589}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Figure 2 shows a uniform lamina $A B C D$ formed by removing an isosceles triangle $B C D$ from an equilateral triangle $A B D$ of side $2 d$. The point $C$ is the centroid of triangle $A B D$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of triangle $B C D$ in terms of $d$.
\item Show that the distance of the centre of mass of the lamina from $B D$ is $\frac { 4 } { 9 } \sqrt { 3 } d$.\\
(8 marks)\\
The lamina is freely suspended from the point $B$ and hangs at rest.
\item Find in degrees, correct to 1 decimal place, the acute angle that the side $A B$ makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [15]}}