| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 8 |
| 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 composite shapes (square + triangle), finding the centre of mass using moments, then applying equilibrium conditions. The calculation is straightforward with given answer to verify in part (i), and part (ii) is routine application of moments and vertical equilibrium. Slightly above average due to composite shape handling but well within typical M2 scope. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) CG of triangle is \(\frac{2}{3}a\) horizontally from A. Moments: \(\frac{1}{8}W \times \frac{5}{3}a + \frac{2}{3}W \times \frac{3}{4}a = W \times \bar{x}\). Hence \(\bar{x} = \frac{11}{12}a\) | M1, A1, A1, A1 (4 marks) | For equating moments about A, or equivalent; For a correct unsimplified equation; For a correct unsimplified equation; Given answer correctly shown |
| (ii) \(R_A \times 2a = W \times \frac{7}{3}a \Rightarrow R_A = \frac{7}{6}W\). \(R_A + R_D = W \Rightarrow R_D = \frac{1}{6}W\) | M1, A1, M1, A1 (4 marks) | For one moments equation; For one correct answer; For resolving, or a second moments equation; For a second correct answer |
(i) CG of triangle is $\frac{2}{3}a$ horizontally from A. Moments: $\frac{1}{8}W \times \frac{5}{3}a + \frac{2}{3}W \times \frac{3}{4}a = W \times \bar{x}$. Hence $\bar{x} = \frac{11}{12}a$ | M1, A1, A1, A1 (4 marks) | For equating moments about A, or equivalent; For a correct unsimplified equation; For a correct unsimplified equation; Given answer correctly shown
(ii) $R_A \times 2a = W \times \frac{7}{3}a \Rightarrow R_A = \frac{7}{6}W$. $R_A + R_D = W \Rightarrow R_D = \frac{1}{6}W$ | M1, A1, M1, A1 (4 marks) | For one moments equation; For one correct answer; For resolving, or a second moments equation; For a second correct answer
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\includegraphics[max width=\textwidth, alt={}, center]{b96a99a6-3df4-4000-9bf1-aab7ab954b4a-2_389_698_1706_694}
A uniform lamina $A B C D$ has the shape of a square of side $a$ adjoining a right-angled isosceles triangle whose equal sides are also of length $a$. The weight of the lamina is $W$. The lamina rests, in a vertical plane, on smooth supports at $A$ and $D$, with $A D$ horizontal (see diagram).\\
(i) Show that the centre of mass of the lamina is at a horizontal distance of $\frac { 11 } { 9 } a$ from $A$.\\
(ii) Find, in terms of $W$, the magnitudes of the forces on the supports at $A$ and $D$.
\hfill \mbox{\textit{OCR M2 Q3 [8]}}