| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.5 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions and basic trigonometry. While it involves multiple steps (resolving horizontally and vertically, then solving simultaneous equations), the method is standard M1 fare with no conceptual challenges—slightly easier than average due to the clear setup and routine application of equilibrium conditions. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10Cor S\alpha = 8\) \(10\cos\alpha = 8\) \(\alpha = 36.9°\) OR \(10Cor S\alpha = F\) \(10\sin\alpha = 6\) \(\alpha = 36.9°\) OR \(\tan\theta = F/8\) \(\tan\alpha = 6/8\) \(\alpha = 36.9°\) | M1 A1 A1 [3] | Component of 10 = 8 Accept 37 36.8 and 37 from 36.7 Using value of F(ii) Using F(=6) from (ii) OR \(\tan\theta = 8/F\), using value of F from (ii) |
| \(F = 10\sin36.9\) \(F = 6 \text{ N}\) OR \(F^2 + 8^2 = 10^2\) \(F = 6 \text{ N}\) | M1 A1ft A1 [3] | \(F = 10Cor S\alpha\) Allow \(10Cos53.1\) Accept 6.01 (or from \(10Cos53.1\)) or 6.0 Pythagoras, 3 squared terms |
| anything rounding to 6.0 from correct working. Accept \(F^2 = 8^2 + 10^2\) |
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $10Cor S\alpha = 8$ $10\cos\alpha = 8$ $\alpha = 36.9°$ OR $10Cor S\alpha = F$ $10\sin\alpha = 6$ $\alpha = 36.9°$ OR $\tan\theta = F/8$ $\tan\alpha = 6/8$ $\alpha = 36.9°$ | M1 A1 A1 [3] | Component of 10 = 8 Accept 37 36.8 and 37 from 36.7 Using value of F(ii) Using F(=6) from (ii) OR $\tan\theta = 8/F$, using value of F from (ii) |
| $F = 10\sin36.9$ $F = 6 \text{ N}$ OR $F^2 + 8^2 = 10^2$ $F = 6 \text{ N}$ | M1 A1ft A1 [3] | $F = 10Cor S\alpha$ Allow $10Cos53.1$ Accept 6.01 (or from $10Cos53.1$) or 6.0 Pythagoras, 3 squared terms | **Cor S is Cos or Sin (passim)** Do not accept 36.7 |
| | | **anything rounding to 6.0 from correct working.** Accept $F^2 = 8^2 + 10^2$ |2\\
\includegraphics[max width=\textwidth, alt={}, center]{4c6c9323-8238-4ec2-94a1-6e8188a34521-02_597_885_676_630}
Three horizontal forces of magnitudes $F \mathrm {~N} , 8 \mathrm {~N}$ and 10 N act at a point and are in equilibrium. The $F \mathrm {~N}$ and 8 N forces are perpendicular to each other, and the 10 N force acts at an obtuse angle $( 90 + \alpha ) ^ { \circ }$ to the $F \mathrm {~N}$ force (see diagram). Calculate\\
(i) $\alpha$,\\
(ii) $F$.
\hfill \mbox{\textit{OCR M1 2011 Q2 [6]}}