| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or object resting on curved surface |
| Difficulty | Challenging +1.8 This is a challenging statics problem requiring multiple equilibrium conditions (forces and moments) for two connected bodies, geometric reasoning to find contact angles, and careful resolution of forces in 3D geometry. However, it's a standard Further Maths mechanics question with clear structure and guided parts, making it harder than typical A-level but not exceptionally difficult for FM students. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(R_P\,r\cos 60° = W\,r\sin 60°\) | M1 A1 | Find \(R_P\) by moments about \(Q\) for disc |
| \(R_P = W\tan 60° = \sqrt{3}\,W\) A.G. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(R_Q\cos 60° = W,\ R_Q = 2W\) | B1 | Find \(R_Q\) by resolving vertically for disc |
| \(R_B\,3a\sin 60° = 2W(3a/2)\cos 60° + R_Q\,a\) | M1 A1 | Find \(R_B\) by moments about \(A\) for \(AB\) |
| \(R_B = W(3/2 + 2)/(3\sqrt{3}/2)\) | — | |
| \(= (7\sqrt{3}/9)\,W\) A.G. | A1 | |
| \(X_A = R_B - R_Q\sin 60°\) | M1 | Resolve horizontally for rod |
| \(= -(2\sqrt{3}/9)\,W\) | A1 | |
| \(Y_A = 2W + R_Q\cos 60° = 3W\) | A1 | Resolve vertically (or for rod and disc) |
| \(R = \sqrt{(4 \times 3/81 + 9)}\,W\) | M1 | Find magnitude \(R\) of reaction at \(A\) |
| \(= \sqrt{(247/27)}\,W = 3.02\,W\) | A1 | |
| Part total: 4+5=9 (with part i: 3) | Total: 12 |
## Question 5:
### Part (i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $R_P\,r\cos 60° = W\,r\sin 60°$ | M1 A1 | Find $R_P$ by moments about $Q$ for disc |
| $R_P = W\tan 60° = \sqrt{3}\,W$ **A.G.** | A1 | |
**Part total: 3**
### Part (ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $R_Q\cos 60° = W,\ R_Q = 2W$ | B1 | Find $R_Q$ by resolving vertically for disc |
| $R_B\,3a\sin 60° = 2W(3a/2)\cos 60° + R_Q\,a$ | M1 A1 | Find $R_B$ by moments about $A$ for $AB$ |
| $R_B = W(3/2 + 2)/(3\sqrt{3}/2)$ | — | |
| $= (7\sqrt{3}/9)\,W$ **A.G.** | A1 | |
| $X_A = R_B - R_Q\sin 60°$ | M1 | Resolve horizontally for rod |
| $= -(2\sqrt{3}/9)\,W$ | A1 | |
| $Y_A = 2W + R_Q\cos 60° = 3W$ | A1 | Resolve vertically (or for rod and disc) |
| $R = \sqrt{(4 \times 3/81 + 9)}\,W$ | M1 | Find magnitude $R$ of reaction at $A$ |
| $= \sqrt{(247/27)}\,W = 3.02\,W$ | A1 | |
**Part total: 4+5=9 (with part i: 3) | Total: 12**5\\
\includegraphics[max width=\textwidth, alt={}, center]{38694ab3-44cd-48d1-922a-d5eb09b62826-3_650_698_248_721}
Two parallel vertical smooth walls $E F$ and $C D$ meet a horizontal plane at $E$ and $C$ respectively. A uniform smooth rod $A B$, of weight $2 W$ and length $3 a$, is freely hinged to the horizontal plane at the point $A$, between $E$ and $C$. The end $B$ rests against $C D$. A uniform smooth circular disc of weight $W$ is in contact with the wall $E F$ at the point $P$ and with the rod at the point $Q$. It is given that angle $B A C$ is $60 ^ { \circ }$ and that $A Q = a$ (see diagram). The rod and the disc are in equilibrium in the same vertical plane, which is perpendicular to both walls. Show that\\
(i) the magnitude of the reaction at $P$ is $\sqrt { } 3 W$,\\
(ii) the magnitude of the reaction at $B$ is $\frac { 7 \sqrt { } 3 } { 9 } W$.
Find, in the form $k W$, the magnitude of the reaction on $A B$ at $A$, giving the value of $k$ correct to 3 significant figures.
\hfill \mbox{\textit{CAIE FP2 2013 Q5 [12]}}