| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Ring on horizontal rod equilibrium |
| Difficulty | Moderate -0.3 This is a standard M1 friction problem requiring resolution of forces and application of F=μR at limiting equilibrium. The two-part structure guides students through the method (find components, then use friction law), making it slightly easier than average but still requiring proper force diagram understanding and algebraic manipulation. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces horizontally | |
| Horizontal component is \(T\cos25°\) (0.906T) | A1 | |
| M1 | For resolving forces vertically | |
| Vertical component is \(4g + T\sin25°\) \((40 + 0.423T)\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(F = 0.4R\) | |
| \(0.906T = 16 + 0.169T\) | A1ft | May be implied by correct answer for \(T\) |
| \(T = 21.7\) N | A1 | [3] |
# Question 4:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces horizontally |
| Horizontal component is $T\cos25°$ (0.906T) | A1 | |
| | M1 | For resolving forces vertically |
| Vertical component is $4g + T\sin25°$ $(40 + 0.423T)$ | A1 | [4] |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $F = 0.4R$ |
| $0.906T = 16 + 0.169T$ | A1ft | May be implied by correct answer for $T$ |
| $T = 21.7$ N | A1 | [3] |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{fa0e0e0d-b0a6-44e0-8b4f-4923e235c6c6-2_168_711_1612_717}
A ring of mass 4 kg is attached to one end of a light string. The ring is threaded on a fixed horizontal rod and the string is pulled at an angle of $25 ^ { \circ }$ below the horizontal (see diagram). With a tension in the string of $T \mathrm {~N}$ the ring is in equilibrium.\\
(i) Find, in terms of $T$, the horizontal and vertical components of the force exerted on the ring by the rod.
The coefficient of friction between the ring and the rod is 0.4 .\\
(ii) Given that the equilibrium is limiting, find the value of $T$.
\hfill \mbox{\textit{CAIE M1 2012 Q4 [7]}}