| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | January |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question with routine composite lamina calculations and a straightforward optimization. Part (a) involves standard techniques (moments about axes, removing a section), while part (b) is basic framework statics with moment equilibrium and resolution of forces. All methods are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(80\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 48\begin{pmatrix}6\\2\end{pmatrix} + 12\begin{pmatrix}1\\-3\end{pmatrix} + 20\begin{pmatrix}11\\9\end{pmatrix}\) | M1 | Correct method for c.m. |
| \(80\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = \begin{pmatrix}520\\240\end{pmatrix}\) | B1 | Total mass correct |
| B1 | One c.m. on RHS correct [If separate components considered, B1 for 2 correct] | |
| \(\bar{x} = 6.5\) | E1 | |
| \(\bar{y} = 3\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider \(x\) coordinate: \(520 = 76 \times 6.4 + 4x\) | M1 | Using additive principle o.e. on \(x\) components |
| B1 | Areas correct. Allow FT from masses from (i) | |
| so \(x = 8.4\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y\) coordinate is 1 so we need \(240 = 76\bar{y} + 4 \times 1\) and \(\bar{y} = 3.10526...\) | B1 | Position of centre of square |
| M1 | ||
| so 3.11 (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Moments about C: \(4R = 120 \times 3 + 120 \times 2\) | M1 | Moments equation. All terms present |
| so \(4R = 600\) and \(R = 150\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| [Diagram showing framework with forces] | B1 | |
| A\(\uparrow\): \(150 + T_{AE}\cos 30 = 0\) | M1 | Equilibrium at a pin-joint |
| \(T_{AE} = -100\sqrt{3}\) so \(100\sqrt{3}\) N (C) | A1 | Any form. Sign correct. Neglect (C) |
| E\(\downarrow\): \(120 + T_{AE}\cos 30 + T_{EB}\cos 30 = 0\) | M1 | Equilibrium at E, all terms present |
| \(T_{EB} = 20\sqrt{3}\) so \(20\sqrt{3}\) N (T) | F1 | Any form. Sign follows working. Neglect (T). |
| F1 | T/C consistent with answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider \(\rightarrow\) at E, using (ii) gives ED as thrust | E1 | Clearly explained. Accept 'thrust' correctly deduced from wrong answers to (ii). |
# Question 3:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $80\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 48\begin{pmatrix}6\\2\end{pmatrix} + 12\begin{pmatrix}1\\-3\end{pmatrix} + 20\begin{pmatrix}11\\9\end{pmatrix}$ | M1 | Correct method for c.m. |
| $80\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = \begin{pmatrix}520\\240\end{pmatrix}$ | B1 | Total mass correct |
| | B1 | One c.m. on RHS correct [If separate components considered, B1 for 2 correct] |
| $\bar{x} = 6.5$ | E1 | |
| $\bar{y} = 3$ | A1 | cao |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider $x$ coordinate: $520 = 76 \times 6.4 + 4x$ | M1 | Using additive principle o.e. on $x$ components |
| | B1 | Areas correct. Allow FT from masses from (i) |
| so $x = 8.4$ | A1 | cao |
## Part (a)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y$ coordinate is 1 so we need $240 = 76\bar{y} + 4 \times 1$ and $\bar{y} = 3.10526...$ | B1 | Position of centre of square |
| | M1 | |
| so 3.11 (3 s.f.) | A1 | cao |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Moments about C: $4R = 120 \times 3 + 120 \times 2$ | M1 | Moments equation. All terms present |
| so $4R = 600$ and $R = 150$ | E1 | |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| [Diagram showing framework with forces] | B1 | |
| A$\uparrow$: $150 + T_{AE}\cos 30 = 0$ | M1 | Equilibrium at a pin-joint |
| $T_{AE} = -100\sqrt{3}$ so $100\sqrt{3}$ N (C) | A1 | Any form. Sign correct. Neglect (C) |
| E$\downarrow$: $120 + T_{AE}\cos 30 + T_{EB}\cos 30 = 0$ | M1 | Equilibrium at E, all terms present |
| $T_{EB} = 20\sqrt{3}$ so $20\sqrt{3}$ N (T) | F1 | Any form. Sign follows working. Neglect (T). |
| | F1 | T/C consistent with answers |
## Part (b)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider $\rightarrow$ at E, using (ii) gives ED as thrust | E1 | Clearly explained. Accept 'thrust' correctly deduced from wrong answers to (ii). |
---3
\begin{enumerate}[label=(\alph*)]
\item A uniform lamina made from rectangular parts is shown in Fig. 3.1. All the dimensions are centimetres. All coordinates are referred to the axes shown in Fig. 3.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-4_691_529_427_762}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of the centre of mass of the lamina is 6.5 and find the $y$-coordinate.
A square of side 2 cm is to be cut from the lamina. The sides of the square are to be parallel to the coordinate axes and the centre of the square is to be chosen so that the $x$-coordinate of the centre of mass of the new shape is 6.4
\item Calculate the $x$-coordinate of the centre of the square to be removed.
The $y$-coordinate of the centre of the square to be removed is now chosen so that the $y$-coordinate of the centre of mass of the final shape is as large as possible.
\item Calculate the $y$-coordinate of the centre of mass of the lamina with the square removed, giving your answer correct to three significant figures.
\end{enumerate}\item Fig. 3.2 shows a framework made from light rods of length 2 m freely pin-jointed at $\mathrm { A } , \mathrm { B } , \mathrm { C }$, D and E. The framework is in a vertical plane and is supported at A and C. There are loads of 120 N at B and at E . The force on the framework due to the support at A is $R \mathrm {~N}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1785fde-a6ce-4f8b-9948-4b4dd973ce84-5_448_741_459_662}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
each rod is 2 m long
\begin{enumerate}[label=(\roman*)]
\item Show that $R = 150$.
\item Draw a diagram showing all the forces acting at the points $\mathrm { A } , \mathrm { B } , \mathrm { D }$ and E , including the forces internal to the rods.
Calculate the internal forces in rods AE and EB , and determine whether each is a tension or a thrust. [You may leave your answers in surd form.]
\item Without any further calculation of the forces in the rods, explain briefly how you can tell that rod ED is in thrust.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2006 Q3 [20]}}