| 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), the techniques are routine for FM students and the geometry is straightforward with tan α = 3/4 given. The 'show that' part guides students to the answer, reducing problem-solving demand. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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 \(C\) |
| \(R_A + 3R_A + 4mg = 12R_A\) | Combine using \(\tan\alpha = \frac{3}{4}\) | |
| \(R_A = \frac{1}{2}mg\) A.G. | M1 A1 | |
| *OR:* \(R_Bl\sin\alpha + S\frac{3}{4}l\cos\alpha = mg\frac{1}{2}l\cos\alpha\) | (M1 A1) | Take moments about \(A\) |
| \(R_A + 3(mg - R_A) = 2mg \Rightarrow R_A = \frac{1}{2}mg\) A.G. | (M1 A1) | |
| *OR:* \(F_Al\sin\alpha + mg\frac{1}{2}l\cos\alpha = R_Al\cos\alpha + S\frac{1}{4}l\cos\alpha\) | (M1 A1) | Take moments about \(B\) |
| \(R_A + 2mg = 4R_A + mg - R_A \Rightarrow R_A = \frac{1}{2}mg\) A.G. | (M1 A1) | |
| *OR:* \(R_A\frac{3}{4}l\cos\alpha = R_Bl\sin\alpha + mg\frac{1}{4}l\cos\alpha\) | (M1 A1) | Take moments about \(D\) |
| \(3R_A = R_A + mg \Rightarrow R_A = \frac{1}{2}mg\) A.G. | (M1 A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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):
| Answer/Working | 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 $C$ |
| $R_A + 3R_A + 4mg = 12R_A$ | | Combine using $\tan\alpha = \frac{3}{4}$ |
| $R_A = \frac{1}{2}mg$ **A.G.** | M1 A1 | |
| *OR:* $R_Bl\sin\alpha + S\frac{3}{4}l\cos\alpha = mg\frac{1}{2}l\cos\alpha$ | (M1 A1) | Take moments about $A$ |
| $R_A + 3(mg - R_A) = 2mg \Rightarrow R_A = \frac{1}{2}mg$ **A.G.** | (M1 A1) | |
| *OR:* $F_Al\sin\alpha + mg\frac{1}{2}l\cos\alpha = R_Al\cos\alpha + S\frac{1}{4}l\cos\alpha$ | (M1 A1) | Take moments about $B$ |
| $R_A + 2mg = 4R_A + mg - R_A \Rightarrow R_A = \frac{1}{2}mg$ **A.G.** | (M1 A1) | |
| *OR:* $R_A\frac{3}{4}l\cos\alpha = R_Bl\sin\alpha + mg\frac{1}{4}l\cos\alpha$ | (M1 A1) | Take moments about $D$ |
| $3R_A = R_A + mg \Rightarrow R_A = \frac{1}{2}mg$ **A.G.** | (M1 A1) | |
**Subtotal: 7**
## Question 4(ii):
| Answer/Working | 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]{5d40f5b4-e3d4-482c-8d8d-05a01bd3b43f-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]}}