| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed triangle/rectangle/square |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question using composite bodies (rectangle minus triangle) with straightforward coordinate geometry and equilibrium. Part (a) requires routine application of the composite body formula, and part (b) involves basic moment equilibrium about a suspension point. Both parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
\includegraphics{figure_1}
A uniform lamina $ABCD$ is made by taking a uniform sheet of metal in the form of a rectangle $ABED$, with $AB = 3a$ and $AD = 2a$, and removing the triangle $BCE$, where $C$ lies on $DE$ and $CE = a$, as shown in Fig. 1.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina from $AD$.
[5]
\end{enumerate}
The lamina has mass $M$. A particle of mass $m$ is attached to the lamina at $B$. When the loaded lamina is freely suspended from the mid-point of $AB$, it hangs in equilibrium with $AB$ horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $m$ in terms of $M$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q3 [9]}}