| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| 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 M2 centre of mass problem requiring decomposition of a trapezium into simpler shapes (rectangle and triangle), finding the centroid using standard formulas, then applying moment equilibrium about a pivot with a given angle. All techniques are routine for M2 students with no novel insight required, making it slightly easier than average. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| com of \(\Delta\) 3 cm right of C | B1 | |
| \((48+27)\bar{x} = 48x4 + 27x11\) | M1, A1 | |
| \(\bar{x} = 6.52\) | A1 | |
| com of \(\Delta\) 2 cm above AD | B1 | |
| \((48+27)\bar{y} = 48x3 + 27x2\) | M1, A1 | |
| \(\bar{y} = 2.64\) | A1 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(14F\) | B1 | can be implied e.g. \(7/\sin30°\). F |
| \(3g\cos30° \times 6.52\) | B1 | 7.034 (AG) or \((6.52 - 2.64\tan30°)\) |
| \(3g\sin30° \times 2.64\) | B1 | \(52.0°\) (GAH) or (above)\(\times\cos30°\); \((5.00)\times\cos30°\) (4.33) |
| \(14F = 3g\cos30°\times6.52 - 3g\sin30°\times2.64\) | M1 | \(14F = 3\times9.8\times7.034\times\cos52.0°\) |
| \(F = 9.09\) N | A1 | 5 |
# Question 8:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| com of $\Delta$ 3 cm right of C | B1 | |
| $(48+27)\bar{x} = 48x4 + 27x11$ | M1, A1 | |
| $\bar{x} = 6.52$ | A1 | |
| com of $\Delta$ 2 cm above AD | B1 | |
| $(48+27)\bar{y} = 48x3 + 27x2$ | M1, A1 | |
| $\bar{y} = 2.64$ | A1 | **8** |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $14F$ | B1 | can be implied e.g. $7/\sin30°$. F |
| $3g\cos30° \times 6.52$ | B1 | 7.034 (AG) or $(6.52 - 2.64\tan30°)$ |
| $3g\sin30° \times 2.64$ | B1 | $52.0°$ (GAH) or (above)$\times\cos30°$; $(5.00)\times\cos30°$ (4.33) |
| $14F = 3g\cos30°\times6.52 - 3g\sin30°\times2.64$ | M1 | $14F = 3\times9.8\times7.034\times\cos52.0°$ |
| $F = 9.09$ N | A1 | **5** | |8 (i)
Fig. 1
A uniform lamina $A B C D$ is in the form of a right-angled trapezium. $A B = 6 \mathrm {~cm} , B C = 8 \mathrm {~cm}$ and $A D = 17 \mathrm {~cm}$ (see Fig. 1). Taking $x$ - and $y$-axes along $A D$ and $A B$ respectively, find the coordinates of the centre of mass of the lamina.\\
(ii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-5_481_1079_991_575}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The lamina is smoothly pivoted at $A$ and it rests in a vertical plane in equilibrium against a fixed smooth block of height 7 cm . The mass of the lamina is 3 kg . $A D$ makes an angle of $30 ^ { \circ }$ with the horizontal (see Fig. 2). Calculate the magnitude of the force which the block exerts on the lamina.
\hfill \mbox{\textit{OCR M2 2008 Q8 [13]}}