| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with elastic string or spring support |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem requiring resolution of forces, moments about a point, and Hooke's law. While it involves multiple steps (finding reactions, using limiting equilibrium, applying spring formula) and careful geometry with the given angle, it follows a well-established method for rod equilibrium problems. The 'show that' part provides scaffolding, and the techniques are all routine for FM students, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(F_A = \frac{1}{3}R_A\) | B1 | Relate \(F_A\) and \(R_A\) using \(F = \mu R\) |
| \(R_B = F_A\left[= \frac{1}{3}R_A\right]\) | B1 | Resolve horizontally |
| \(S = mg - R_A\) | B1 | Resolve vertically |
| \(R_B\frac{1}{4}l\sin\alpha + F_A\frac{3}{4}l\sin\alpha + mg\frac{1}{4}l\cos\alpha = R_A\frac{3}{4}l\cos\alpha\) | M1 A1 | Take moments about \(C\) |
| \(R_A + 3R_A + 4mg = 12R_A \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}\) | M1 A1 | Combine using \(\tan\alpha = \frac{3}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_Bl\sin\alpha + S\frac{3}{4}l\cos\alpha = mg\frac{1}{2}l\cos\alpha\) | (M1 A1) | |
| \(R_A + 3(mg - R_A) = 2mg \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}\) | (M1 A1) | Combine using \(\tan\alpha = \frac{3}{4}\) |
| Answer | Marks |
|---|---|
| \(F_Al\sin\alpha + mg\frac{1}{2}l\cos\alpha = R_Al\cos\alpha + S\frac{1}{4}l\cos\alpha\) | (M1 A1) |
| \(R_A + 2mg = 4R_A + mg - R_A \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}\) | (M1 A1) |
| Answer | Marks |
|---|---|
| \(R_A\frac{3}{4}l\cos\alpha = R_Bl\sin\alpha + mg\frac{1}{4}l\cos\alpha\) | (M1 A1) |
| \(3R_A = R_A + mg \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}\) | (M1 A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(S = \frac{1}{2}mg = \frac{2mge}{L},\ e = \frac{1}{4}L\) | B1 | Use Hooke's Law to relate extension \(e\) and natural length \(L\) |
| \(CD = \frac{3}{4}l\sin\alpha = \frac{9l}{20}\) | B1 | Find length of \(CD\) |
| \(L - \frac{1}{4}L = \frac{9l}{20},\ L = \frac{3l}{5}\) | M1 A1 | Combine to find \(L\) |
| Subtotal: 4 | Total: [11] |
## Question 4(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F_A = \frac{1}{3}R_A$ | B1 | Relate $F_A$ and $R_A$ using $F = \mu R$ |
| $R_B = F_A\left[= \frac{1}{3}R_A\right]$ | B1 | Resolve horizontally |
| $S = mg - R_A$ | B1 | Resolve vertically |
| $R_B\frac{1}{4}l\sin\alpha + F_A\frac{3}{4}l\sin\alpha + mg\frac{1}{4}l\cos\alpha = R_A\frac{3}{4}l\cos\alpha$ | M1 A1 | Take moments about $C$ |
| $R_A + 3R_A + 4mg = 12R_A \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}$ | M1 A1 | Combine using $\tan\alpha = \frac{3}{4}$ |
*OR: Take moments about $A$:*
| $R_Bl\sin\alpha + S\frac{3}{4}l\cos\alpha = mg\frac{1}{2}l\cos\alpha$ | (M1 A1) | |
|---|---|---|
| $R_A + 3(mg - R_A) = 2mg \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}$ | (M1 A1) | Combine using $\tan\alpha = \frac{3}{4}$ |
*OR: Take moments about $B$:*
| $F_Al\sin\alpha + mg\frac{1}{2}l\cos\alpha = R_Al\cos\alpha + S\frac{1}{4}l\cos\alpha$ | (M1 A1) | |
|---|---|---|
| $R_A + 2mg = 4R_A + mg - R_A \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}$ | (M1 A1) | |
*OR: Take moments about $D$:*
| $R_A\frac{3}{4}l\cos\alpha = R_Bl\sin\alpha + mg\frac{1}{4}l\cos\alpha$ | (M1 A1) | |
|---|---|---|
| $3R_A = R_A + mg \Rightarrow R_A = \frac{1}{2}mg\ \mathbf{A.G.}$ | (M1 A1) | |
**Subtotal: 7**
---
## Question 4(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $S = \frac{1}{2}mg = \frac{2mge}{L},\ e = \frac{1}{4}L$ | B1 | Use Hooke's Law to relate extension $e$ and natural length $L$ |
| $CD = \frac{3}{4}l\sin\alpha = \frac{9l}{20}$ | B1 | Find length of $CD$ |
| $L - \frac{1}{4}L = \frac{9l}{20},\ L = \frac{3l}{5}$ | M1 A1 | Combine to find $L$ |
**Subtotal: 4 | Total: [11]**4\\
\includegraphics[max width=\textwidth, alt={}, center]{2c6b6722-ebba-4ade-9a9d-dd70e61cf52b-3_513_643_260_749}
A uniform rod $A B$, of length $l$ and mass $m$, rests in equilibrium with its lower end $A$ on a rough horizontal floor and the end $B$ against a smooth vertical wall. The rod is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$, and is in a vertical plane perpendicular to the wall. The rod is supported by a light spring $C D$ which is in compression in a vertical line with its lower end $D$ fixed on the floor. The upper end $C$ is attached to the rod at a distance $\frac { 1 } { 4 } l$ from $B$ (see diagram). The coefficient of friction at $A$ between the rod and the floor is $\frac { 1 } { 3 }$ and the system is in limiting equilibrium.\\
(i) Show that the normal reaction of the floor at $A$ has magnitude $\frac { 1 } { 2 } m g$ and find the force in the spring.\\
(ii) Given that the modulus of elasticity of the spring is $2 m g$, find the natural length of the spring.
\hfill \mbox{\textit{CAIE FP2 2014 Q4 [11]}}