| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| 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 M2 centre of mass question requiring composite shapes (rectangle + triangle), followed by equilibrium with applied force. The geometry is straightforward, calculations are routine, and the method is well-practiced in M2 courses. Slightly easier than average due to the simple isosceles triangle and clear setup. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| Mass ratios: \(18a^2\), \(9a^2\), \(27a^2\) | B1 |
| Centres of mass: \(\frac{3}{2}a\), \(4a\), \(\bar{x}\) from \(AE\) | B1 |
| Moments about \(AE\): \(18a^2 \times \frac{3}{2}a + 9a^2 \times 4a = 27a^2 \times \bar{x}\) | M1A1 |
| \(\bar{x} = \frac{2a}{3}\) | A1 (5) |
| Notes: First B1 allow any correct ratios i.e. do not need \(a^2\); Second B1 could be scored if distances consistently measured from somewhere else; First M1 for attempt at correct moments equation with correct no. of terms and condone sign errors; First A1 for a correct equation; Second A1 for 7a/3, 2.3a, or better |
| Answer | Marks |
|---|---|
| Moments about \(A\): \(4g \times \bar{x} = 6aF\) | M1 A1ft |
| Vertical component (\(Y\)) of force at \(A = 4g\) | B1 |
| Force at \(A\): \(\sqrt{X^2 + Y^2}\) | M1 |
| \(= 42.1\) (N) or 42 (N) | A1 (5) [10] |
| Notes: First M1 for a moments equation (could be about \(A\), \(B\), \(C\), \(D\) or \(E\)), dim correct with correct no. of terms etc and allow sign errors; First A1 ft for a correct equation with their \(\bar{x}\); First B1 for vert component (\(Y\)) = 4g seen; Second M1, independent but must have found an \(X\) and a \(Y\), for attempt at magnitude \(\sqrt{X^2 + Y^2}\), using their \(X\) and \(Y\); Second A1 \(\frac{2g\sqrt{373}}{9}\) or 42 (N) or 42.1 (N); N.B. Note that \(X = F\); Possible moments equations: \(M(A) : 4g\bar{x} = 6aF\); \(M(B) : 4g(3a - \bar{x}) + 6aF = 3aY\); \(M(C) : 4g(6a - \bar{x}) + 3aF + 3aX = 6aY\); \(M(D) : 4g(3a - \bar{x}) + 6aX = 3aY\); \(M(E) : 4g\bar{x} = 6aX\) |
## 4a
| Mass ratios: $18a^2$, $9a^2$, $27a^2$ | B1 |
| Centres of mass: $\frac{3}{2}a$, $4a$, $\bar{x}$ from $AE$ | B1 |
| Moments about $AE$: $18a^2 \times \frac{3}{2}a + 9a^2 \times 4a = 27a^2 \times \bar{x}$ | M1A1 |
| $\bar{x} = \frac{2a}{3}$ | A1 (5) |
| **Notes:** First B1 allow any correct ratios i.e. do not need $a^2$; Second B1 could be scored if distances consistently measured from somewhere else; First M1 for attempt at correct moments equation with correct no. of terms and condone sign errors; First A1 for a correct equation; Second A1 for 7a/3, 2.3a, or better |
## 4b
| Moments about $A$: $4g \times \bar{x} = 6aF$ | M1 A1ft |
| Vertical component ($Y$) of force at $A = 4g$ | B1 |
| Force at $A$: $\sqrt{X^2 + Y^2}$ | M1 |
| $= 42.1$ (N) or 42 (N) | A1 (5) [10] |
| **Notes:** First M1 for a moments equation (could be about $A$, $B$, $C$, $D$ or $E$), dim correct with correct no. of terms etc and allow sign errors; First A1 ft for a correct equation with their $\bar{x}$; First B1 for vert component ($Y$) = 4g seen; Second M1, independent but must have found an $X$ and a $Y$, for attempt at magnitude $\sqrt{X^2 + Y^2}$, using their $X$ and $Y$; Second A1 $\frac{2g\sqrt{373}}{9}$ or 42 (N) or 42.1 (N); N.B. Note that $X = F$; **Possible moments equations:** $M(A) : 4g\bar{x} = 6aF$; $M(B) : 4g(3a - \bar{x}) + 6aF = 3aY$; $M(C) : 4g(6a - \bar{x}) + 3aF + 3aX = 6aY$; $M(D) : 4g(3a - \bar{x}) + 6aX = 3aY$; $M(E) : 4g\bar{x} = 6aX$ |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{790546bf-38a4-4eb7-876e-941fe58f4a48-07_671_661_239_635}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform lamina $A B C D E$ is made by joining a rectangular lamina $A B D E$ to a triangular lamina $B C D$ along the edge $B D$. The rectangle has length $6 a$ and width $3 a$. The triangle is isosceles, with $B C = C D$, and the distance from $C$ to $B D$ is $3 a$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina, $A B C D E$, from $A E$.
The lamina $A B C D E$ is freely suspended from $A$. A horizontal force of magnitude $F$ newtons is applied to the lamina at $D$. The line of action of the force lies in the vertical plane containing the lamina. The lamina is in equilibrium with $A E$ vertical. The mass of the lamina is 4 kg .
\item Find the magnitude of the force exerted on the lamina at $A$.\\
DO NOT WIRITE IN THIS AREA
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q4 [10]}}