| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics question requiring composite centre of mass calculation (splitting trapezium into rectangle and triangles) and equilibrium with moments. The 'show that' part guides students through the geometry, and part (b) is straightforward application of moments about a point. Slightly easier than average A-level due to the structured guidance and routine techniques. |
| 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 |
| Areas: \(ABF = \frac{3}{2}a^2\), \(BCEF = 9a^2\), \(CDE = \frac{3}{2}a^2\), lamina \(= 12a^2\) | B1 | Any equivalent ratios e.g. \(3:18:3:24\) |
| Distances from \(AD\): \(ABF = a\), \(BCEF = \frac{3}{2}a\), \(CDE = a\), lamina \(= \bar{y}\) | B1 | Or correct distances from a parallel axis |
| Moments about \(AD\) | M1 | Or moments about a parallel axis |
| \(\left(\frac{3}{2}a^2 \times a\right) + \left(9a^2 \times \frac{3}{2}a\right) + \left(\frac{3}{2}a^2 \times a\right) = 12a^2\bar{y}\) | A1 | Correct unsimplified equation for their axis |
| \(\bar{y} = \frac{11a}{8}\) | A1* | Correct given answer correctly obtained. If centre of mass at \((xa, ya)\) then \(a\) might not be seen in working. Otherwise with no \(a\) in working maximum score is B1B0M1A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(F\): \(Mg \times \left(3a - \frac{11a}{8}\right) = 3aT\) | M1 | A complete method to obtain an equation in \(T\) only |
| \(T = \frac{13Mg}{24}\) \((0.54166666\ldots Mg)\) | A1 | \(0.54Mg\) or better |
## Question 1:
**Part 1(a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Areas: $ABF = \frac{3}{2}a^2$, $BCEF = 9a^2$, $CDE = \frac{3}{2}a^2$, lamina $= 12a^2$ | B1 | Any equivalent ratios e.g. $3:18:3:24$ |
| Distances from $AD$: $ABF = a$, $BCEF = \frac{3}{2}a$, $CDE = a$, lamina $= \bar{y}$ | B1 | Or correct distances from a parallel axis |
| Moments about $AD$ | M1 | Or moments about a parallel axis |
| $\left(\frac{3}{2}a^2 \times a\right) + \left(9a^2 \times \frac{3}{2}a\right) + \left(\frac{3}{2}a^2 \times a\right) = 12a^2\bar{y}$ | A1 | Correct unsimplified equation for their axis |
| $\bar{y} = \frac{11a}{8}$ | A1* | Correct given answer correctly obtained. If centre of mass at $(xa, ya)$ then $a$ might not be seen in working. Otherwise with no $a$ in working maximum score is B1B0M1A0A0 |
**Part 1(b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $F$: $Mg \times \left(3a - \frac{11a}{8}\right) = 3aT$ | M1 | A complete method to obtain an equation in $T$ only |
| $T = \frac{13Mg}{24}$ $(0.54166666\ldots Mg)$ | A1 | $0.54Mg$ or better |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{99e1d643-7408-4793-9ebc-b33c91bc5fab-02_474_716_246_676}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform plane lamina is in the shape of an isosceles trapezium $A B C D E F$, as shown shaded in Figure 1.
\begin{itemize}
\item $B C E F$ is a square
\item $A B = C D = a$
\item $B C = 3 a$
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the lamina from $A D$ is $\frac { 11 a } { 8 }$
\end{itemize}
The mass of the lamina is $M$\\
The lamina is suspended by two light vertical strings, one attached to the lamina at $A$ and the other attached to the lamina at $F$
The lamina hangs freely in equilibrium, with $B F$ horizontal.
\item Find, in terms of $M$ and $g$, the tension in the string attached at $A$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2022 Q1 [7]}}