| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 9 |
| 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 a composite lamina and suspended equilibrium. Part (a) requires routine application of the formula for centre of mass of composite shapes (subtracting the removed triangle), and part (b) uses the standard principle that the centre of mass hangs vertically below the suspension point. The geometry is straightforward with given dimensions, requiring only basic trigonometry and no novel insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
3.
\begin{figure}[h]
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\caption{Fig. 2}
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A uniform plane lamina is in the shape of an isosceles triangle $A B C$, where $A B = A C$. The mid-point of $B C$ is $M , A M = 30 \mathrm {~cm}$ and $B M = 40 \mathrm {~cm}$. The mid-points of $A C$ and $A B$ are $D$ and $E$ respectively. The triangular portion $A D E$ is removed leaving a uniform plane lamina $B C D E$ as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the lamina $B C D E$ is $6 \frac { 2 } { 3 } \mathrm {~cm}$ from $B C$.\\
(6 marks)\\
The lamina $B C D E$ is freely suspended from $D$ and hangs in equilibrium.
\item Find, in degrees to one decimal place, the angle which $D E$ makes with the vertical.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [9]}}