| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| 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 requiring systematic application of the centre of mass formula for composite shapes and moments in equilibrium. Part (a) involves setting up coordinates, finding centres of mass of triangles, and solving an equation—routine but multi-step. Part (b) is a straightforward moments problem once the geometry is established. While requiring careful work, it follows standard M2 techniques without novel insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
\includegraphics{figure_1}
A uniform lamina $ABCD$ is formed by removing the isosceles triangle $ADC$ of height $h$ metres, where $h < 2\sqrt{3}$, from a uniform lamina $ABC$ in the shape of an equilateral triangle of side 4 m, as shown in Figure 1. The centre of mass of $ABCD$ is at $D$.
\begin{enumerate}[label=(\alph*)]
\item Show that $h = \sqrt{3}$ [7]
\end{enumerate}
The weight of the lamina $ABCD$ is $W$ newtons. The lamina is freely suspended from $A$. A horizontal force of magnitude $F$ newtons is applied at $B$ so that the lamina is in equilibrium with $AB$ vertical. The horizontal force acts in the vertical plane containing the lamina.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $F$ in terms of $W$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q4 [11]}}