| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Lamina hinged at point with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments problem requiring students to find the center of mass of an isosceles triangle, then take moments about the hinge point A. While it involves multiple steps (finding CoM coordinates, resolving forces, taking moments), these are routine techniques for M2 students with no novel insight required. The setup is clearly defined and the method is straightforward, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance Notes |
| Centre of mass of triangle is at \(G\), where \(AG=8a\) | B1 | Or equivalent. Seen or implied. e.g. \(\frac{2}{3}\times 5a\cos\theta\left(=\frac{40a}{13}\right)\) from \(AB\) |
| \(\sin\theta=\dfrac{5}{13}\) | B1 | Or equivalent. Any correct trig ratio for an angle in the triangle. Seen or implied |
| \(\text{M}(A)\): \(13aF=W\times 8a\times\dfrac{5}{13}\) | M1 | Dimensionally correct with resolved component of their \(8a\). Condone sin/cos confusion. If \(g\) appears, mark as an accuracy error |
| A1 | Correct substituted equation (any form) | |
| \(F=\dfrac{40W}{169}\) (N) | A1 | \(0.24W\) or better |
# Question 2:
| Working/Answer | Mark | Guidance Notes |
|---|---|---|
| Centre of mass of triangle is at $G$, where $AG=8a$ | B1 | Or equivalent. Seen or implied. e.g. $\frac{2}{3}\times 5a\cos\theta\left(=\frac{40a}{13}\right)$ from $AB$ |
| $\sin\theta=\dfrac{5}{13}$ | B1 | Or equivalent. Any correct trig ratio for an angle in the triangle. Seen or implied |
| $\text{M}(A)$: $13aF=W\times 8a\times\dfrac{5}{13}$ | M1 | Dimensionally correct with resolved component of their $8a$. Condone sin/cos confusion. If $g$ appears, mark as an accuracy error |
| | A1 | Correct substituted equation (any form) |
| $F=\dfrac{40W}{169}$ (N) | A1 | $0.24W$ or better |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3eb71ecb-fa88-4cca-a2b6-bcf11f1d689b-04_760_669_118_641}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The uniform lamina $A B C$ has sides $A B = A C = 13 a$ and $B C = 10 a$. The lamina is freely suspended from $A$. A horizontal force of magnitude $F$ is applied to the lamina at $B$, as shown in Figure 1. The line of action of the force lies in the vertical plane containing the lamina. The lamina is in equilibrium with $A B$ vertical. The weight of the lamina is $W$.
Find $F$ in terms of $W$.\\
VILM SIHI NI JAIUM ION OC\\
VANV SIHI NI I III M LON OO\\
VI4V SIHI NI JAIUM ION OC
\hfill \mbox{\textit{Edexcel M2 2021 Q2 [5]}}