| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Toppling and sliding of solids |
| Difficulty | Standard +0.3 This is a standard M2 moments question requiring identification of toppling condition (vertical through pivot point) and comparison with sliding condition (F = μR). Part (i) uses basic geometry (tan θ = r/h) and part (ii) compares tan θ values. While it requires understanding of equilibrium concepts, the mathematical execution is straightforward with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| (i) CM is vertically above lowest point of base. Hence \(\tan\alpha = \frac{6}{7.5} \Rightarrow \alpha = 38.7°\) | B1, M1, A1 (3 marks) | For stating or implying correct geometry; For appropriate trig calculation; For correct answer 38.7 |
| (ii) Cylinder slides when \(\tan\theta = \frac{3}{4}\). But \(\frac{3}{4} < 0.8\), so \(\theta < \alpha\). Hence it slides first (at inclination \(36.9°\)) | B1, M1, A1, A1 (4 marks) | For stating or implying limiting friction case; For comparing \(\tan\alpha\) to \(\tan\theta\), or equivalent; For correct comparison of the angles; For correct conclusion of sliding first |
(i) CM is vertically above lowest point of base. Hence $\tan\alpha = \frac{6}{7.5} \Rightarrow \alpha = 38.7°$ | B1, M1, A1 (3 marks) | For stating or implying correct geometry; For appropriate trig calculation; For correct answer 38.7
(ii) Cylinder slides when $\tan\theta = \frac{3}{4}$. But $\frac{3}{4} < 0.8$, so $\theta < \alpha$. Hence it slides first (at inclination $36.9°$) | B1, M1, A1, A1 (4 marks) | For stating or implying limiting friction case; For comparing $\tan\alpha$ to $\tan\theta$, or equivalent; For correct comparison of the angles; For correct conclusion of sliding first
---2 A uniform circular cylinder, of radius 6 cm and height 15 cm , is in equilibrium on a fixed inclined plane with one of its ends in contact with the plane.\\
(i) Given that the cylinder is on the point of toppling, find the angle the plane makes with the horizontal.
The cylinder is now placed on a horizontal board with one of its ends in contact with the board. The board is then tilted so that the angle it makes with the horizontal gradually increases.\\
(ii) Given that the coefficient of friction between the cylinder and the board is $\frac { 3 } { 4 }$, determine whether or not the cylinder will slide before it topples, justifying your answer.
\hfill \mbox{\textit{OCR M2 Q2 [7]}}