| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| 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 centre of mass question requiring decomposition into rectangles, calculation of composite centre of mass (part a), equilibrium with two supports using moments (part b), and finding angle when freely suspended (part c). All techniques are routine for FM2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratio table: 16, 6, 4, 26 with distances \(a\), \(3.5a\), \(3a\), \(\bar{x}\) from \(AJ\) and \(4a\), \(0\), \(5a\), \(\bar{y}\) from \(AB\) | B1 | Correct mass ratios seen or implied |
| Correct distances from vertical axis | B1 | Correct distances from their vertical axis |
| M(\(AJ\)) | M1 | Correct strategy including appropriate division of lamina, moments about axis parallel to \(AJ\). Terms dimensionally correct. Condone sign errors. |
| \(16a + 21a + 12a (= 49a) = 26\bar{x}\) | A1 | Correct unsimplified moments equation for correct division of template |
| \(26\bar{x} = 49a \Rightarrow \bar{x} = \frac{49}{26}a\) | A1* | Obtain given answer from complete and correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M(\(A\)) | M1 | Complete method to find the tension. Dimensionally correct equations. |
| \(5a \times T = \frac{49}{26}a \times W\) | A1 | Correct unsimplified equation for the tension |
| \(T = \frac{49}{130}W\) | A1 | Correct simplified (\(0.38W\) or better) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distances from \(AB\) as in table above | B1 | Distances from a horizontal axis for complete correct division (could be found in (a) but need to be used here) |
| M(\(AB\)): \(64a + 6a + 20a = 26\bar{y}\) | M1 | Moments about a horizontal axis. Terms dimensionally correct. Condone sign errors. |
| \(\bar{y} = \frac{90}{26}a\) | A1 | Correct vertical distance (any equivalent form) \(\left(\frac{59}{13}a \text{ from } JI\right)\) seen or implied |
| \(\tan\theta = \frac{49}{90}\) | M1 | Use trig. with \(\frac{49}{26}a\) and their \(\bar{y}\), or equivalent, to find a relevant angle |
| \(\theta = 28.6°\ (29°)\) | A1 | \(29°\) or better. \(28.5658...°\), \(0.499\) rads |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratio table: 16, 6, 4, 26 with distances $a$, $3.5a$, $3a$, $\bar{x}$ from $AJ$ and $4a$, $0$, $5a$, $\bar{y}$ from $AB$ | B1 | Correct mass ratios seen or implied |
| Correct distances from vertical axis | B1 | Correct distances from their vertical axis |
| M($AJ$) | M1 | Correct strategy including appropriate division of lamina, moments about axis parallel to $AJ$. Terms dimensionally correct. Condone sign errors. |
| $16a + 21a + 12a (= 49a) = 26\bar{x}$ | A1 | Correct unsimplified moments equation for correct division of template |
| $26\bar{x} = 49a \Rightarrow \bar{x} = \frac{49}{26}a$ | A1* | Obtain **given answer** from complete and correct working |
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| M($A$) | M1 | Complete method to find the tension. Dimensionally correct equations. |
| $5a \times T = \frac{49}{26}a \times W$ | A1 | Correct unsimplified equation for the tension |
| $T = \frac{49}{130}W$ | A1 | Correct simplified ($0.38W$ or better) |
## Question 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distances from $AB$ as in table above | B1 | Distances from a horizontal axis for complete correct division (could be found in (a) but need to be used here) |
| M($AB$): $64a + 6a + 20a = 26\bar{y}$ | M1 | Moments about a horizontal axis. Terms dimensionally correct. Condone sign errors. |
| $\bar{y} = \frac{90}{26}a$ | A1 | Correct vertical distance (any equivalent form) $\left(\frac{59}{13}a \text{ from } JI\right)$ seen or implied |
| $\tan\theta = \frac{49}{90}$ | M1 | Use trig. with $\frac{49}{26}a$ and their $\bar{y}$, or equivalent, to find a relevant angle |
| $\theta = 28.6°\ (29°)$ | A1 | $29°$ or better. $28.5658...°$, $0.499$ rads |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a7901165-1679-4d30-9444-0c27020e32ea-08_547_410_246_829}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The uniform lamina $A B C D E F G H I J$ is shown in Figure 3.\\
The lamina has $A J = 8 a , A B = 5 a$ and $B C = D E = E F = F G = G H = H I = I J = 2 a$.\\
All the corners are right angles.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the lamina from $A J$ is $\frac { 49 } { 26 } a$
A light inextensible rope is attached to the lamina at $A$ and another light inextensible rope is attached to the lamina at $B$. The lamina hangs in equilibrium with both ropes vertical and $A B$ horizontal. The weight of the lamina is $W$.
\item Find, in terms of $W$, the tension in the rope attached to the lamina at $B$.
The rope attached to $B$ breaks and subsequently the lamina hangs freely in equilibrium, suspended from $A$.
\item Find the size of the angle between $A J$ and the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2021 Q3 [13]}}