| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics question requiring: (a) centre of mass calculation using composite bodies with coordinate geometry, (b) equilibrium with two vertical strings (straightforward moment equation), and (c) suspended equilibrium requiring moments about the suspension point. While systematic, it demands careful coordinate work, understanding of composite COM, and multiple equilibrium applications—moderately challenging for FM2 level. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(CM)\) | M1 | Form moments equation about \(CM\) or parallel axis. Must be dimensionally correct. Condone sign errors |
| \(36a^{2}\times2a = (162a^{2}-36a^{2})d\quad(=126a^{2}d)\) | A1, A1 | Unsimplified with at most one error; Correct unsimplified equation |
| \(d = \frac{4}{7}a\) * | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(A)\) | M1 | Complete method to find tension e.g. take moments about \(A\). Must be dimensionally correct. Condone sign errors |
| \(T\times18a = W\times\left(9a+\frac{4}{7}a\right)\) | A1 | Correct unsimplified equation |
| \(T = \frac{67}{126}W\) | A1 | Or equivalent (\(0.53W\) or better) (\(0.53174\ldots W\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(AB)\) | M1 | Complete method to find relevant vertical distance for centre of mass of \(L\). Condone sign errors |
| \(126a^{2}\bar{y} = 162a^{2}\times6a - 36a^{2}\times(6a+4a)\) | A1, A1 | Unsimplified with at most one error; Correct unsimplified equation |
| \(\bar{y} = \frac{34}{7}a\) | A1 | Seen or implied |
| Correct use of trig | M1 | Correct use of trig and the given answer to part (a) to find a relevant angle |
| \(\tan\theta = \frac{34}{67}\) | A1ft | Correct unsimplified equation for required angle. Follow their \(\bar{y}\) |
| \(\theta = 27°\) or better | A1 | 27 or better (26.906……) |
## Question 4:
### Part 4a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(CM)$ | M1 | Form moments equation about $CM$ or parallel axis. Must be dimensionally correct. Condone sign errors |
| $36a^{2}\times2a = (162a^{2}-36a^{2})d\quad(=126a^{2}d)$ | A1, A1 | Unsimplified with at most one error; Correct unsimplified equation |
| $d = \frac{4}{7}a$ * | A1* | Obtain **given answer** from correct working |
### Part 4b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A)$ | M1 | Complete method to find tension e.g. take moments about $A$. Must be dimensionally correct. Condone sign errors |
| $T\times18a = W\times\left(9a+\frac{4}{7}a\right)$ | A1 | Correct unsimplified equation |
| $T = \frac{67}{126}W$ | A1 | Or equivalent ($0.53W$ or better) ($0.53174\ldots W$) |
### Part 4c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(AB)$ | M1 | Complete method to find relevant vertical distance for centre of mass of $L$. Condone sign errors |
| $126a^{2}\bar{y} = 162a^{2}\times6a - 36a^{2}\times(6a+4a)$ | A1, A1 | Unsimplified with at most one error; Correct unsimplified equation |
| $\bar{y} = \frac{34}{7}a$ | A1 | Seen or implied |
| Correct use of trig | M1 | Correct use of trig and the given answer to part (a) to find a relevant angle |
| $\tan\theta = \frac{34}{67}$ | A1ft | Correct unsimplified equation for required angle. Follow their $\bar{y}$ |
| $\theta = 27°$ or better | A1 | 27 or better (26.906……) |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fcae18f-6588-4b71-8b7f-c8408de591f4-12_819_853_255_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform triangular lamina $A B C$ is isosceles, with $A C = B C$. The midpoint of $A B$ is $M$. The length of $A B$ is $18 a$ and the length of $C M$ is $18 a$.
The triangular lamina $C D E$, with $D E = 6 a$ and $C D = 12 a$, has $E D$ parallel to $A B$ and $M D C$ is a straight line.
Triangle $C D E$ is removed from triangle $A B C$ to form the lamina $L$, shown shaded in Figure 1.
The distance of the centre of mass of $L$ from $M C$ is $d$.
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 4 } { 7 } a$
The lamina $L$ is suspended by two light inextensible strings. One string is attached to $L$ at $A$ and the other string is attached to $L$ at $B$.\\
The lamina hangs in equilibrium in a vertical plane with the strings vertical and $A B$ horizontal.\\
The weight of $L$ is $W$
\item Find, in terms of $W$, the tension in the string attached to $L$ at $B$
The string attached to $L$ at $B$ breaks, so that $L$ is now suspended from $A$. When $L$ is hanging in equilibrium in a vertical plane, the angle between $A B$ and the downward vertical through $A$ is $\theta ^ { \circ }$
\item Find the value of $\theta$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2023 Q4 [14]}}