| 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 requiring resolution of forces in two perpendicular directions and application of F=μR. The question guides students by asking them to show a specific value, making it easier than average. The calculations involve basic trigonometry (sin/cos 45°) and simple algebra with no conceptual challenges beyond standard equilibrium conditions. |
| 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 |
|---|---|---|
| (i) \([R + 7\sin45° = 0.8g]\) | M1 | For resolving forces vertically (needs 3 terms) |
| Normal component is \(3.05 \text{ N}\) | A1 | AG |
| [2] | ||
| (ii) \(F = 7\cos45°\) | B1 | |
| M1 | For using \(\mu = \frac{F}{3.05}\) | |
| Coefficient is \(1.62\) | A1 | |
| [3] |
**(i)** $[R + 7\sin45° = 0.8g]$ | M1 | For resolving forces vertically (needs 3 terms)
Normal component is $3.05 \text{ N}$ | A1 | AG
| | | **[2]**
**(ii)** $F = 7\cos45°$ | B1 |
| | M1 | For using $\mu = \frac{F}{3.05}$
Coefficient is $1.62$ | A1 |
| | | **[3]**
---\includegraphics{figure_3}
A small ring of mass 0.8 kg is threaded on a rough rod which is fixed horizontally. The ring is in equilibrium, acted on by a force of magnitude 7 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 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]}}