| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Resultant force on lamina |
| Difficulty | Standard +0.3 This is a standard mechanics question involving centre of mass of a composite shape (trapezium decomposed into rectangle and triangle) and toppling equilibrium. Parts (i) and (ii) require routine application of the centre of mass formula for composite bodies, while part (iii) applies the standard toppling condition (line of action through pivot point). All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
\includegraphics{figure_4}
The diagram shows the cross-section $ABCD$ through the centre of mass of a uniform solid prism. $AB = 0.9$ m, $BC = 2a$ m, $AD = a$ m and angle $ABC =$ angle $BAD = 90°$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the distance of the centre of mass of the prism from $AD$. [2]
\item Express the distance of the centre of mass of the prism from $AB$ in terms of $a$. [2]
\end{enumerate}
The prism has weight 18 N and rests in equilibrium on a rough horizontal surface, with $AD$ in contact with the surface. A horizontal force of magnitude 6 N is applied to the prism. This force acts through the centre of mass in the direction $BC$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Given that the prism is on the point of toppling, calculate $a$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2016 Q4 [7]}}