| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Toppling on inclined plane |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring routine application of the formula for a triangular lamina's centre of mass (⅓ from each edge), followed by a straightforward toppling condition where the vertical line through the centre of mass must pass through the base. The geometry is simple (right-angled triangle with given dimensions) and the method is textbook standard, making it slightly easier than average for M2 level. |
| Spec | 6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) c.o.m. = \(\frac{1}{3}\) dist. from B to C = \(\frac{1}{3} \times 9 = 3\) cm from AB | M1 A1 | |
| (ii) \(AB = \sqrt{(15^2 - 9^2)} = 12\) cm | M1 A1 | |
| c.o.m. = \(\frac{1}{3}\) dist. from B to A = \(\frac{1}{3} \times 12 = 4\) cm from BC | M1 A1 | |
| (b) lamina will not topple if vertical through c.o.m. passes between B and C | B1 | |
| max. \(\theta\) when it passes through B | B1 | |
| \(\tan\theta = \frac{3}{4} \therefore \theta = 36.9°\) (1dp) | M1 A1 | (10) |
**(a)**
(i) c.o.m. = $\frac{1}{3}$ dist. from B to C = $\frac{1}{3} \times 9 = 3$ cm from AB | M1 A1 |
(ii) $AB = \sqrt{(15^2 - 9^2)} = 12$ cm | M1 A1 |
c.o.m. = $\frac{1}{3}$ dist. from B to A = $\frac{1}{3} \times 12 = 4$ cm from BC | M1 A1 |
**(b)** lamina will not topple if vertical through c.o.m. passes between B and C | B1 |
max. $\theta$ when it passes through B | B1 |
| | |
$\tan\theta = \frac{3}{4} \therefore \theta = 36.9°$ (1dp) | M1 A1 | (10)3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{086ace58-0aa9-4f36-95c3-5698d14f511e-2_369_684_1356_555}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Figure 1 shows a uniform triangular lamina $A B C$ placed with edge $B C$ along the line of greatest slope of a plane inclined at an angle $\theta$ to the horizontal. The lengths $A C$ and $B C$ are 15 cm and 9 cm respectively and $\angle A B C$ is a right angle.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina from
\begin{enumerate}[label=(\roman*)]
\item $\quad A B$,
\item $B C$.
Assuming that the plane is rough enough to prevent the lamina from slipping,
\end{enumerate}\item find in degrees, correct to 1 decimal place, the maximum value of $\theta$ for which the lamina remains in equilibrium.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [10]}}