| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Two jointed rods in equilibrium |
| Difficulty | Challenging +1.8 This is a challenging statics problem requiring analysis of a two-rod system with multiple equilibrium conditions. Students must apply moments about strategic points for both rods, resolve forces in two directions, and handle the smooth hinge constraint. The problem demands systematic application of statics principles across connected bodies, geometric reasoning with trigonometry, and finally friction analysis. While the techniques are standard Further Maths mechanics, the multi-body setup and the need to eliminate variables strategically across several equations makes this substantially harder than typical single-body equilibrium problems. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(R_C\) by moments for \(BC\) about \(B\): \(R_C \cdot 2a\sin\beta = mg\cdot a\cos\beta\) | ||
| \(R_C = \frac{1}{2}mg\cot\beta\) | M1 A1 | A.G. Total part: [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: Moments for system about \(A\): \(R_C(2a\sin\alpha + 2a\sin\beta)\) | ||
| \(= mg(3a\cos\alpha + a\cos\beta)\) | M1 A1 | |
| Substitute for \(R_C\) from (i): \(\frac{1}{2}\cos\beta(2\sin\alpha + 2\sin\beta)\) | ||
| \(= \sin\beta(3\cos\alpha + \cos\beta)\) | M1 A1 | |
| \(\tan\alpha = 3\tan\beta\) | A1 | A.G. |
| OR: Moments for \(AB\) about \(B\): \(R_A \cdot 2a\cos\alpha = F_A \cdot 2a\sin\alpha + mg\cdot a\cos\alpha\) | M1 A1 | |
| Substitute \(R_A = 2mg\), \(F_A = R_C\): \(4\cos\alpha = (\frac{1}{2}\cot\beta)\sin\alpha + \cos\alpha\) | M1 A1 | |
| \(\tan\alpha = 3\tan\beta\) | A1 | A.G. Total part: [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(\mu_{min}\) using \(F_A \leq \mu R_A\): \(\mu_{min} = \frac{1}{4}\cot\beta = \frac{3}{4}\cot\alpha = \frac{1}{4}\sqrt{3}\) | M1 A1 | Total part: [2], Question total: [9] |
## Question 3(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $R_C$ by moments for $BC$ about $B$: $R_C \cdot 2a\sin\beta = mg\cdot a\cos\beta$ | | |
| $R_C = \frac{1}{2}mg\cot\beta$ | M1 A1 | **A.G.** Total part: [2] |
## Question 3(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER:** Moments for system about $A$: $R_C(2a\sin\alpha + 2a\sin\beta)$ | | |
| $= mg(3a\cos\alpha + a\cos\beta)$ | M1 A1 | |
| Substitute for $R_C$ from (i): $\frac{1}{2}\cos\beta(2\sin\alpha + 2\sin\beta)$ | | |
| $= \sin\beta(3\cos\alpha + \cos\beta)$ | M1 A1 | |
| $\tan\alpha = 3\tan\beta$ | A1 | **A.G.** |
| **OR:** Moments for $AB$ about $B$: $R_A \cdot 2a\cos\alpha = F_A \cdot 2a\sin\alpha + mg\cdot a\cos\alpha$ | M1 A1 | |
| Substitute $R_A = 2mg$, $F_A = R_C$: $4\cos\alpha = (\frac{1}{2}\cot\beta)\sin\alpha + \cos\alpha$ | M1 A1 | |
| $\tan\alpha = 3\tan\beta$ | A1 | **A.G.** Total part: [5] |
## Question 3(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $\mu_{min}$ using $F_A \leq \mu R_A$: $\mu_{min} = \frac{1}{4}\cot\beta = \frac{3}{4}\cot\alpha = \frac{1}{4}\sqrt{3}$ | M1 A1 | Total part: [2], Question total: [9] |
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\includegraphics[max width=\textwidth, alt={}, center]{bcd7ee99-e382-4cb6-aa39-d8b385b01319-2_506_623_977_760}
Two uniform rods $A B$ and $B C$, each of length $2 a$ and mass $m$, are smoothly hinged at $B$. They rest in equilibrium with $C$ in contact with a smooth vertical wall and $A$ in contact with a rough horizontal floor. The rods are in a vertical plane perpendicular to the wall. The rods $A B$ and $B C$ make angles $\alpha$ and $\beta$ respectively with the horizontal (see diagram). Show that\\
(i) the reaction at $C$ has magnitude $\frac { 1 } { 2 } m g \cot \beta$,\\
(ii) $\tan \alpha = 3 \tan \beta$.
The coefficient of friction at $A$ is $\mu$. Given that $\alpha = 60 ^ { \circ }$, find the least possible value of $\mu$.
\hfill \mbox{\textit{CAIE FP2 2012 Q3 [9]}}