| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring composite shapes (rectangle + triangle), finding the centroid using moments, then applying equilibrium conditions for a suspended lamina. All steps are routine textbook methods with straightforward geometry and trigonometry, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| CoM of triangle \(= \frac{1}{3} \times cv(12)\) from \(BD\) | B1 | OR \(\frac{2}{3}\times cv(12)\) from C. CoM of triangle |
| M1 | Table of values idea | |
| \((80 + 60)x_G\) | A1 | |
| \(= 4(80) + 12(60)\) | A1 | |
| \(x_G = 7.43\) cm | A1 | \(7.42857\ldots\) or \(\frac{52}{7}\) cm |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan\theta = (8 - x_G)/5\) | M1 | Using tan to find a relevant angle |
| \(\tan\theta = 0.5714\ldots/5\) | A1ft | ft their \(x_G\) to target angle with the vertical |
| \(\theta = 6.52°\) | A1 | \(6.5198\ldots\) Allow \(6.5(0)\) from \(x_G = 7.43\) |
| [3] |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| CoM of triangle $= \frac{1}{3} \times cv(12)$ from $BD$ | B1 | OR $\frac{2}{3}\times cv(12)$ from C. CoM of triangle |
| | M1 | Table of values idea |
| $(80 + 60)x_G$ | A1 | |
| $= 4(80) + 12(60)$ | A1 | |
| $x_G = 7.43$ cm | A1 | $7.42857\ldots$ or $\frac{52}{7}$ cm |
| **[5]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta = (8 - x_G)/5$ | M1 | Using tan to find a relevant angle |
| $\tan\theta = 0.5714\ldots/5$ | A1ft | ft their $x_G$ to target angle with the vertical |
| $\theta = 6.52°$ | A1 | $6.5198\ldots$ Allow $6.5(0)$ from $x_G = 7.43$ |
| **[3]** | | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{5bfd0285-71cb-4dcb-8545-a379653f9a3e-2_778_579_1304_744}
A uniform lamina $A B C D E$ consists of a rectangle $A B D E$ and an isosceles triangle $B C D$ joined along their common edge. $A B = D E = 8 \mathrm {~cm} , A E = B D = 10 \mathrm {~cm}$ and $B C = C D = 13 \mathrm {~cm}$ (see diagram).\\
(i) Find the distance of the centre of mass of the lamina from $A E$.\\
(ii) The lamina is freely suspended from $B$ and hangs in equilibrium. Calculate the angle that $B D$ makes with the vertical.
\hfill \mbox{\textit{OCR M2 2014 Q3 [8]}}