| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Ring on horizontal rod equilibrium |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions and application of F=μR. The angle is given in a convenient form (tan α = 5/12 gives a 5-12-13 triangle), making calculations simple. All three parts follow standard mechanics procedures with no problem-solving insight required—purely routine application of equilibrium conditions and friction laws. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = 13\cos\alpha\) | M1 | For resolving forces horizontally |
| Frictional component is 12 N | A1 (×2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 1.1 \times 10 + 13\sin\alpha\) | M1 | For resolving forces vertically (3 terms needed) |
| Normal component is 16 N | A1 (×2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Coefficient of friction is 0.75 | B1 ft (×1) |
# Question 1:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = 13\cos\alpha$ | M1 | For resolving forces horizontally |
| Frictional component is 12 N | A1 (×2) | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 1.1 \times 10 + 13\sin\alpha$ | M1 | For resolving forces vertically (3 terms needed) |
| Normal component is 16 N | A1 (×2) | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Coefficient of friction is 0.75 | B1 ft (×1) | |
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\includegraphics[max width=\textwidth, alt={}, center]{e060fc3b-ae93-46b5-b476-dcecb14d6d06-2_200_588_267_781}
A ring of mass 1.1 kg is threaded on a fixed rough horizontal rod. A light string is attached to the ring and the string is pulled with a force of magnitude 13 N at an angle $\alpha$ below the horizontal, where $\tan \alpha = \frac { 5 } { 12 }$ (see diagram). The ring is in equilibrium.\\
(i) Find the frictional component of the contact force on the ring.\\
(ii) Find the normal component of the contact force on the ring.\\
(iii) Given that the equilibrium of the ring is limiting, find the coefficient of friction between the ring and the rod.
\hfill \mbox{\textit{CAIE M1 2004 Q1 [5]}}