| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| 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 routine application of the formula for centre of mass of composite bodies (rectangle minus triangle), while part (b) uses basic trigonometry with the equilibrium condition that the centre of mass hangs directly below the suspension point. The calculations are straightforward with no conceptual surprises, making it slightly easier than average for M2 material. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
3.
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\caption{Figure 1}
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The uniform lamina $A B D E$ is in the shape of a rectangle with $A B = 8 a$ and $B D = 6 a$. The triangle $B C D$ is isosceles and has base $6 a$ and perpendicular height $6 a$. The template $A B C D E$, shown shaded in Figure 1, is formed by removing the triangular lamina $B C D$ from the lamina $A B D E$.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the template is $\frac { 14 } { 5 } a$ from $A E$.
The template is freely suspended from $A$ and hangs in equilibrium with $A B$ at an angle of $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$, giving your answer to the nearest whole number.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2023 Q3 [8]}}