| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| 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 statics problem with friction requiring resolution of forces in two perpendicular directions and application of F=μR. The question guides students by giving the normal force answer in part (i), making part (ii) a simple calculation. It involves standard mechanics techniques with no conceptual subtlety or problem-solving insight required. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| \([R + 7\sin45° = 0.8g]\) | M1 | For resolving forces vertically (needs 3 terms) |
| Normal component is 3.05 N | A1 | AG |
| [2] |
| Answer | Marks |
|---|---|
| F = 7cos45° | B1 |
| M1 | For using \(\mu = F/3.05\) |
| Coefficient is 1.62 | A1 |
| [3] |
## (i)
$[R + 7\sin45° = 0.8g]$ | M1 | For resolving forces vertically (needs 3 terms)
Normal component is 3.05 N | A1 | AG
| | [2]
## (ii)
F = 7cos45° | B1 |
| |
| M1 | For using $\mu = F/3.05$
Coefficient is 1.62 | A1 |
| | [3]\includegraphics{figure_3}
A small ring of mass $0.8 \text{ kg}$ is threaded on a rough rod which is fixed horizontally. The ring is in equilibrium, acted on by a force of magnitude $7 \text{ N}$ pulling upwards at $45°$ to the horizontal (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the normal component of the contact force acting on the ring has magnitude $3.05 \text{ N}$, correct to 3 significant figures. [2]
\item The ring is in limiting equilibrium. Find the coefficient of friction between the ring and the rod. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q3 [5]}}