| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite shapes (square minus triangle) and particle attachment. Part (a) is routine calculation using the composite body formula, part (b) is similar by symmetry, and part (c) requires taking moments about the suspension point with the lamina hanging in equilibrium. All techniques are standard 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 |
6.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{266c4f52-f35f-459c-9184-836b0f3baf5b-20_570_608_287_669}
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\caption{Figure 3}
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\end{figure}
The uniform lamina $A B C D$ is a square with sides of length $6 a$. The point $E$ is the midpoint of side $A D$. The triangle $C D E$ is removed from the square to form the uniform lamina $L$, shown in Figure 3. The centre of mass of $L$ is at the point $G$.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of $G$ from the side $A B$ is $\frac { 7 } { 3 } a$.
\item Find the distance of $G$ from the side $A E$.
The mass of $L$ is $M$. A particle of mass $k M$ is attached to $L$ at the point $E$. The lamina, with the particle attached, is freely suspended from $A$ and hangs in equilibrium with the diagonal $A C$ vertical.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q6 [13]}}