| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with particle attached |
| Difficulty | Challenging +1.8 This is a complex composite lamina problem requiring multiple centre of mass calculations (rectangle, removed triangle, added quarter circle, and attached particle), careful coordinate geometry with the given constraints, and then a suspended equilibrium calculation. The multi-stage nature, need to track signs correctly when removing mass, and the geometric relationships make this substantially harder than standard centre of mass questions, though the individual techniques are A-level standard. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
2. A metal sign is formed by removing triangle $B C D$ from a rectangular lamina $A C E F$ made of uniform material, and adding a quarter circle XYZ, made of the same uniform material, with a particle attached to its vertex at $Y$. The sign is supported by two light chains fixed at $E$ and $F$.
The quarter circle has radius 24 cm and the particle at $Y$ has a mass equal to half of that of the removed triangle. $X D$ is parallel to $A C$ and $B Z$ is parallel to $A F$. The dimensions, in cm , are as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{3578a810-46da-4d9e-a98f-248be72a517a-3_885_636_712_715}
\begin{enumerate}[label=(\alph*)]
\item Calculate the distance of the centre of mass of the sign from
\begin{enumerate}[label=(\roman*)]
\item $A F$,
\item $A C$.
\end{enumerate}\item The support at $F$ comes loose so that the sign is freely suspended at $E$ by one chain alone. Given that it then hangs in equilibrium, calculate the angle that $E F$ makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 6 2019 Q2 [14]}}