| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths mechanics question requiring centre of mass calculations for composite shapes, equilibrium with moments, and geometric reasoning. While it involves several steps and FM2 content (inherently harder), the techniques are standard: subtract areas/masses for the removed semicircle, take moments about a point for equilibrium, and use geometry when suspended freely. The calculations are methodical rather than requiring novel insight, making it moderately above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: Rectangle \(8a^2\), Semicircle \(\frac{1}{2}\pi a^2\), Sign \(a^2\!\left(8 - \frac{\pi}{2}\right)\); distances from \(AD\): \(a\), \(\frac{4a}{3\pi}\), \(h\) | B1 | Correct mass ratios |
| Moments about \(AD\) | M1 | |
| \(a^2\!\left(8-\frac{\pi}{2}\right)h = 8a^2 \times a - \frac{1}{2}\pi a^2 \times \frac{4a}{3\pi} = 8a^3 - \frac{2}{3}a^3 = \frac{22}{3}a^3\) | A1 | Correct unsimplified equation |
| \(h = \dfrac{22}{3}a \div \left(8 - \dfrac{\pi}{2}\right) = \dfrac{44a}{3(16-\pi)}\) | A1* | Deduce given answer clearly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(A\): \(2aT = \dfrac{44a}{3(16-\pi)}W\) | M1 | |
| \(T = \dfrac{hW}{2a} = \dfrac{22W}{3(16-\pi)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Take moments about \(AB\) to find distance of com from \(AB\) | M1 | |
| \(8a^2 \times 2a - \frac{1}{2}\pi a^2 \times d = \left(8 - \frac{1}{2}\pi\right)a^2 \times v\) | A1 | |
| \(v = \dfrac{32a - \pi d}{16 - \pi}\) | A1 | |
| Correct trig for the given angle | M1 | |
| \(\tan\alpha = \dfrac{11}{18} = \dfrac{h}{v} = \dfrac{44a}{3(32a - \pi d)}\) | A1ft | |
| \((24a = 32a - \pi d,\ 8a = \pi d)\ \Rightarrow d = \dfrac{8a}{\pi}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct mass ratios | B1 | |
| Need all three terms, must be dimensionally correct | M1 | |
| Correct unsimplified equation | A1 | |
| Show sufficient working to justify the given answer and a 'statement' that the required result has been achieved | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Could also take moments about B or about the c.o.m. and use | M1 | |
| cso | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| All terms and dimensionally correct | M1 | |
| Correct unsimplified equation | A1 | |
| Or equivalent | A1 | |
| Condone tan the wrong way up | M1 | |
| Equation in \(a\) and \(d\); follow through on their \(v\) | A1 | |
| cao | A1 |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: Rectangle $8a^2$, Semicircle $\frac{1}{2}\pi a^2$, Sign $a^2\!\left(8 - \frac{\pi}{2}\right)$; distances from $AD$: $a$, $\frac{4a}{3\pi}$, $h$ | B1 | Correct mass ratios |
| Moments about $AD$ | M1 | |
| $a^2\!\left(8-\frac{\pi}{2}\right)h = 8a^2 \times a - \frac{1}{2}\pi a^2 \times \frac{4a}{3\pi} = 8a^3 - \frac{2}{3}a^3 = \frac{22}{3}a^3$ | A1 | Correct unsimplified equation |
| $h = \dfrac{22}{3}a \div \left(8 - \dfrac{\pi}{2}\right) = \dfrac{44a}{3(16-\pi)}$ | A1* | Deduce given answer clearly |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $A$: $2aT = \dfrac{44a}{3(16-\pi)}W$ | M1 | |
| $T = \dfrac{hW}{2a} = \dfrac{22W}{3(16-\pi)}$ | A1 | |
---
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about $AB$ to find distance of com from $AB$ | M1 | |
| $8a^2 \times 2a - \frac{1}{2}\pi a^2 \times d = \left(8 - \frac{1}{2}\pi\right)a^2 \times v$ | A1 | |
| $v = \dfrac{32a - \pi d}{16 - \pi}$ | A1 | |
| Correct trig for the given angle | M1 | |
| $\tan\alpha = \dfrac{11}{18} = \dfrac{h}{v} = \dfrac{44a}{3(32a - \pi d)}$ | A1ft | |
| $(24a = 32a - \pi d,\ 8a = \pi d)\ \Rightarrow d = \dfrac{8a}{\pi}$ | A1 | |
# Question 5:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct mass ratios | B1 | |
| Need all three terms, must be dimensionally correct | M1 | |
| Correct unsimplified equation | A1 | |
| Show sufficient working to justify the given answer and a 'statement' that the required result has been achieved | A1* | |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Could also take moments about B **or** about the c.o.m. and use | M1 | |
| cso | A1 | |
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| All terms and dimensionally correct | M1 | |
| Correct unsimplified equation | A1 | |
| Or equivalent | A1 | |
| Condone tan the wrong way up | M1 | |
| Equation in $a$ and $d$; follow through on their $v$ | A1 | |
| cao | A1 | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f06704e5-454c-41c1-9577-b1210f60480d-12_693_515_210_781}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A shop sign is modelled as a uniform rectangular lamina $A B C D$ with a semicircular lamina removed.
The semicircle has radius $a , B C = 4 a$ and $C D = 2 a$.\\
The centre of the semicircle is at the point $E$ on $A D$ such that $A E = d$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the sign is $\frac { 44 a } { 3 ( 16 - \pi ) }$ from $A D$.
The sign is suspended using vertical ropes attached to the sign at $A$ and at $B$ and hangs in equilibrium with $A B$ horizontal.
The weight of the sign is $W$ and the ropes are modelled as light inextensible strings.
\item Find, in terms of $W$ and $\pi$, the tension in the rope attached at $B$.
The rope attached at $B$ breaks and the sign hangs freely in equilibrium suspended from $A$, with $A D$ at an angle $\alpha$ to the downward vertical.
Given that $\tan \alpha = \frac { 11 } { 18 }$
\item find $d$ in terms of $a$ and $\pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 Q5 [12]}}