| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Ring on vertical rod equilibrium |
| Difficulty | Moderate -0.3 This is a straightforward statics problem requiring resolution of forces in two perpendicular directions and application of the friction law F=μR. The 'show that' part guides students to the answer, and the coefficient of friction calculation is a standard textbook exercise. Slightly easier than average due to the scaffolding and routine nature of the mechanics. |
| Spec | 3.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([F + 5\cos 30° = 0.6g]\) | M1 | For resolving forces vertically (3 terms needed) |
| Frictional force is 1.67 N | A1 | 2 |
| (ii) \(R = 5\sin 30°\) \([= 2.5]\) | B1 | Can be scored in (i) |
| Coefficient is 0.668 | M1, A1 | 3 |
(i) $[F + 5\cos 30° = 0.6g]$ | M1 | For resolving forces vertically (3 terms needed)
Frictional force is 1.67 N | A1 | 2
(ii) $R = 5\sin 30°$ $[= 2.5]$ | B1 | Can be scored in (i)
Coefficient is 0.668 | M1, A1 | 3 | For using $1.67 = \mu R$2\\
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A small ring of mass 0.6 kg is threaded on a rough rod which is fixed vertically. The ring is in equilibrium, acted on by a force of magnitude 5 N pulling upwards at $30 ^ { \circ }$ to the vertical (see diagram).\\
(i) Show that the frictional force acting on the ring has magnitude 1.67 N , correct to 3 significant figures.\\
(ii) The ring is on the point of sliding down the rod. Find the coefficient of friction between the ring and the rod.
\hfill \mbox{\textit{CAIE M1 2006 Q2 [5]}}