| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem requiring moment equilibrium about two points and resolution of forces. While it involves multiple steps (moments about E and B, resolving horizontally and vertically, finding μ), the techniques are routine for FM students and the geometry is straightforward with tan θ = 1/4 given explicitly. The problem is harder than typical A-level mechanics due to the lamina context and FM syllabus placement, but follows a predictable solution pattern without requiring novel insight. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| 2(i) | R × 3a = W cos θ × 2a – W sin θ × 2a |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | M1 A1 | Take moments about B |
| Answer | Marks | Guidance |
|---|---|---|
| E | A1 | Find normal reaction at E. AEF |
| Answer | Marks | Guidance |
|---|---|---|
| B E | M1 A1 | Find normal reaction at B by resolving forces vertically |
| Answer | Marks | Guidance |
|---|---|---|
| 2(ii) | F = R sin θ = 2W/15 | |
| B E | M1 A1 | Find friction at B by resolving forces horizontally |
| µ = (2/15) / (3/5) = 2/9 | A1 | Find µ from F = µR |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(i) ---
2(i) | R × 3a = W cos θ × 2a – W sin θ × 2a
E
or R × 3a = W × 2a (1 – tan θ) cos θ
E
π
or R E × 3a = 2 2Wsin −θ
4 | M1 A1 | Take moments about B
R = 4W / 3√10
E | A1 | Find normal reaction at E. AEF
R = W – R cos θ = 3W/5
B E | M1 A1 | Find normal reaction at B by resolving forces vertically
5
--- 2(ii) ---
2(ii) | F = R sin θ = 2W/15
B E | M1 A1 | Find friction at B by resolving forces horizontally
µ = (2/15) / (3/5) = 2/9 | A1 | Find µ from F = µR
B B
3
Question | Answer | Marks | Guidance\includegraphics{figure_2}
A uniform square lamina $ABCD$ of side $4a$ and weight $W$ rests in a vertical plane with the edge $AB$ inclined at angle $\theta$ to the horizontal, where $\tan \theta = \frac{1}{4}$. The vertex $B$ is in contact with a rough horizontal surface for which the coefficient of friction is $\mu$. The lamina is supported by a smooth peg at the point $E$ on $AB$, where $BE = 3a$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find expressions in terms of $W$ for the normal reaction forces at $E$ and $B$.
[5]
\item Given that the lamina is about to slip, find the value of $\mu$.
[3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2019 Q2 [8]}}