| Exam Board | SPS |
|---|---|
| Module | SPS FM Mechanics (SPS FM Mechanics) |
| Year | 2022 |
| Session | January |
| Marks | 14 |
| Topic | Centre of Mass 1 |
| Type | Centre of mass with variable parameter |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths mechanics question requiring centre of mass calculation for a trapezium (algebraic manipulation with given answer), toppling condition analysis (standard technique), and inclined plane toppling (applying equilibrium conditions). While it involves several steps and FM content, the techniques are standard textbook applications without requiring novel geometric insight or complex problem-solving strategies. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
The diagram shows the cross-section through the centre of mass of a uniform solid prism. The cross-section is a trapezium ABCD with AB and CD perpendicular to AD. The lengths of AB and AD are each 5 cm and the length of CD is $(a + 5)$ cm.
\includegraphics{figure_7}
\begin{enumerate}[label=(\roman*)]
\item Show the distance of the centre of mass of the prism from AD is
$$\frac{a^2 + 15a + 75}{3(a + 10)} \text{ cm.}$$
[5]
\end{enumerate}
The prism is placed with the face containing AB in contact with a horizontal surface.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the greatest value of $a$ for which the prism does not topple.
[3]
\end{enumerate}
The prism is now placed on an inclined plane which makes an angle $\theta^o$ with the horizontal. AB lies along a line of greatest slope with B higher than A.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Using the value for $a$ found in part (ii), and assuming the prism does not slip down the plane, find the great value of $\theta$ for which the prism does not topple.
[6]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Mechanics 2022 Q7 [14]}}