| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring standard techniques: finding COM of composite shapes using moments, equilibrium angles using trigonometry, and resolving forces. While it involves several steps and careful bookkeeping, the methods are all standard FM2 procedures with no novel insights required. The 'show that' in part (a) provides a target to verify calculations. |
| 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 |
|---|---|---|
| \(M(AC)\) | M1 | 2.1 |
| \((27-6)a^2d = 27a^2 \times 3a - 6a^2 \times 2a(= 69a^3)\) | A1 | 1.1b |
| \(\Rightarrow d = \frac{69}{21}a = \frac{23}{7}a\) * | A1* | 2.2a |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(AB)\) | M1 | 3.1a |
| \((27-6)a^2\bar{y} = 27a^2 \times 2a - 6a^2 \times 1.5a(= 45a^3)\) | A1 | 1.1b |
| \(\bar{y} = \frac{45}{21}a = \frac{15}{7}a\) | A1 | 1.1b |
| \(\tan\theta° = \frac{15}{23}\) | M1 | 3.1a |
| \(\theta = 33 \quad (\theta = 33.111...)\) | A1 | 1.1b |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(A)\) | M1 | 3.1a |
| \(6aX + \frac{23}{7}a \times W = \frac{23}{7}a \times 4W\) | A1 | 1.1b |
| A1 | 1.1b | |
| \(X = \frac{23}{14}W\) | A1 | 1.1b |
| (4) | ||
| (12 marks) |
**4a**
| $M(AC)$ | M1 | 2.1 |
| $(27-6)a^2d = 27a^2 \times 3a - 6a^2 \times 2a(= 69a^3)$ | A1 | 1.1b |
| $\Rightarrow d = \frac{69}{21}a = \frac{23}{7}a$ * | A1* | 2.2a |
| | (3) | |
**4b**
| $M(AB)$ | M1 | 3.1a |
| $(27-6)a^2\bar{y} = 27a^2 \times 2a - 6a^2 \times 1.5a(= 45a^3)$ | A1 | 1.1b |
| $\bar{y} = \frac{45}{21}a = \frac{15}{7}a$ | A1 | 1.1b |
| $\tan\theta° = \frac{15}{23}$ | M1 | 3.1a |
| $\theta = 33 \quad (\theta = 33.111...)$ | A1 | 1.1b |
| | (5) | |
**4c**
| $M(A)$ | M1 | 3.1a |
| $6aX + \frac{23}{7}a \times W = \frac{23}{7}a \times 4W$ | A1 | 1.1b |
| | A1 | 1.1b |
| $X = \frac{23}{14}W$ | A1 | 1.1b |
| | (4) | |
| | (12 marks) | |
**Notes for 4a:**
- M1: Take moments about $AC$ or a parallel axis. Dimensionally correct terms. Correct number of terms.
- A1: Correct unsimplified equation
- A1*: Obtain the given answer including "$d =$" from correct working
**Notes for 4b:**
- M1: Take moments about $AB$ or a parallel axis. Dimensionally correct terms. Correct number of terms.
- A1: Correct unsimplified equation
- A1: Correct vertical distance. Any equivalent form
- M1: Correct use of trig to find a relevant angle
- A1: 2 sf or better ($33.111\ldots$)
**Notes for 4c:**
- M1: Complete method to form an equation in $X$ e.g by taking moments about $A$. $5W$ must be split correctly and equation must be dimensionally correct.
- A1: Unsimplified equation with at most one error
- A1: Correct unsimplified equation
- A1: $3.1W$ or better ($3.1428\ldots W$)4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fd8bc7b5-adee-4d67-b15d-571255b00b83-12_351_597_246_735}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The uniform triangular lamina $A B C$ has $A B$ perpendicular to $A C$, $A B = 9 a$ and $A C = 6 a$. The point $D$ on $A B$ is such that $A D = a$.
The rectangle $D E F G$, with $D E = 2 a$ and $E F = 3 a$, is removed from the lamina to form the template shown shaded in Figure 3.
The distance of the centre of mass of the template from $A C$ is $d$.
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 23 } { 7 } a$
The template is freely suspended from $A$ and hangs in equilibrium with $A B$ at an angle $\theta ^ { \circ }$ to the downward vertical through $A$.
\item Find the value of $\theta$
A new piece, of exactly the same size and shape as the template, is cut from a lamina of a different uniform material. The template and the new piece are joined together to form the model shown in Figure 4. Both parts of the model lie in the same plane.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fd8bc7b5-adee-4d67-b15d-571255b00b83-12_369_1185_1667_440}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
The weight of $C P Q R S T A$ is $W$\\
The weight of $A D G F E B C$ is $4 W$\\
The model is freely suspended from $A$.\\
A horizontal force of magnitude $X$, acting in the same vertical plane as the model, is now applied to the model at $T$ so that $A C$ is vertical, as shown in Figure 4.
\item Find $X$ in terms of $W$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2024 Q4 [12]}}