| Exam Board | SPS |
|---|---|
| Module | SPS FM Mechanics (SPS FM Mechanics) |
| Year | 2026 |
| Session | January |
| Marks | 12 |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of composite lamina |
| Difficulty | Standard +0.3 This is a standard centre of mass problem for a composite lamina requiring systematic application of the formula for combined centres of mass, followed by equilibrium geometry. The L-shape from two identical rectangles is a common setup, and the suspension equilibrium condition is straightforward once the centre of mass is found. While it requires careful coordinate work and algebraic manipulation, it follows a well-established method with no novel insights needed. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
\includegraphics{figure_2}
The uniform L-shaped lamina $OABCDE$, shown in Figure 2, is made from two identical rectangles. Each rectangle is 4 metres long and $a$ metres wide. Giving each answer in terms of $a$, find the distance of the centre of mass of the lamina from
\begin{enumerate}[label=(\alph*)]
\item $OE$. [4]
\item $OA$. [4]
\end{enumerate}
The lamina is freely suspended from $O$ and hangs in equilibrium with $OE$ at an angle $\theta$ to the downward vertical through $O$, where $\tan \theta = \frac{4}{3}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $a$. [4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Mechanics 2026 Q2 [12]}}