| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| 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 moments about a point, resolution of forces, and Hooke's law for elastic strings. While it involves multiple steps (finding tension using extension/Hooke's law, taking moments about B, resolving forces at the hinge), these are routine techniques for FM students with no novel insight required. The geometry is straightforward with given angles, making it slightly above average difficulty but well within standard FM mechanics territory. |
| Spec | 1.05g Exact trigonometric values: for standard angles3.04b Equilibrium: zero resultant moment and force6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W \times a\cos 30° + 3W \times 2a\cos 30°\) | Take moments for rod about \(B\) | |
| \(= T \times 2a\cos 30°\) | M1 A1 | (or with \(\cos 30° = \sqrt{3}/2\)) |
| \(T = 7W/2\) | A1 | Can earn M1 A0 A1 if e.g. \(\sin 30°\) wrongly used |
| \(T = \lambda(2a - 3a/5)/(3a/5)\) | Find modulus \(\lambda\) using Hooke's Law | |
| \(\lambda = (3/7)(7W/2) = 3W/2\) | M1 A1 | |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(X = T\cos 30° = (7\sqrt{3}/4)W\) *or* \(3.03W\) | B1\(\checkmark\) | Find horizontal component of force \(F\) at \(B\) |
| \(Y = 4W - T\sin 30° = 9W/4\) | B1\(\checkmark\) | Find vertical component |
| \(F = \sqrt{(X^2 + Y^2)} = (\sqrt{57}/2)W\) *or* \(3.77[5]W\) | B1 | Find magnitude of \(F\) |
| \(\tan^{-1}Y/X = \tan^{-1}3\sqrt{3}/7 = 36.6°\) *or* \(0.639\) radians | M1 A1 | Find direction of \(F\) (AEF); A0 if direction unclear |
| OR: \(F_1 = (4W+T)\sin 30° = 15W/4\) | (B1\(\checkmark\)) | Find component along \(CB\) |
| \(F_2 = (4W-T)\cos 30° = (\sqrt{3}/4)W\) | (B1\(\checkmark\)) | Find normal component |
| \(F = (\sqrt{57}/2)W\) *or* \(3.77[5]W\) | (B1) | |
| \(\tan^{-1}F_2/F_1 = \tan^{-1}\sqrt{3}/15 = 6.6°\) *or* \(0.115\) radians | (M1 A1) | Upward force at angle to \(CB\); A0 if direction unclear |
| OR: \(\pm F_1 = T - 4W\sin 30° = 3W/2\) | (B1\(\checkmark\)) | Find component parallel to string \(CA\) |
| \(\pm F_2 = 4W\cos 30° = 2\sqrt{3}W\) | (B1\(\checkmark\)) | Find normal component |
| \(F = (\sqrt{57}/2)W\) *or* \(3.77[5]W\) | (B1) | |
| \(\tan^{-1}F_2/F_1 = \tan^{-1}4/\sqrt{3} = 66.6°\) *or* \(1.16\) radians | (M1 A1) | Upward force at angle to \(AC\); A0 if direction unclear |
| Total: 5 marks |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W \times a\cos 30° + 3W \times 2a\cos 30°$ | | Take moments for rod about $B$ |
| $= T \times 2a\cos 30°$ | M1 A1 | (or with $\cos 30° = \sqrt{3}/2$) |
| $T = 7W/2$ | A1 | Can earn M1 A0 A1 if e.g. $\sin 30°$ wrongly used |
| $T = \lambda(2a - 3a/5)/(3a/5)$ | | Find modulus $\lambda$ using Hooke's Law |
| $\lambda = (3/7)(7W/2) = 3W/2$ | M1 A1 | |
| **Total: 5 marks** | | |
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER:** $X = T\cos 30° = (7\sqrt{3}/4)W$ *or* $3.03W$ | B1$\checkmark$ | Find horizontal component of force $F$ at $B$ |
| $Y = 4W - T\sin 30° = 9W/4$ | B1$\checkmark$ | Find vertical component |
| $F = \sqrt{(X^2 + Y^2)} = (\sqrt{57}/2)W$ *or* $3.77[5]W$ | B1 | Find magnitude of $F$ |
| $\tan^{-1}Y/X = \tan^{-1}3\sqrt{3}/7 = 36.6°$ *or* $0.639$ radians | M1 A1 | Find direction of $F$ (AEF); A0 if direction unclear |
| **OR:** $F_1 = (4W+T)\sin 30° = 15W/4$ | (B1$\checkmark$) | Find component along $CB$ |
| $F_2 = (4W-T)\cos 30° = (\sqrt{3}/4)W$ | (B1$\checkmark$) | Find normal component |
| $F = (\sqrt{57}/2)W$ *or* $3.77[5]W$ | (B1) | |
| $\tan^{-1}F_2/F_1 = \tan^{-1}\sqrt{3}/15 = 6.6°$ *or* $0.115$ radians | (M1 A1) | Upward force at angle to $CB$; A0 if direction unclear |
| **OR:** $\pm F_1 = T - 4W\sin 30° = 3W/2$ | (B1$\checkmark$) | Find component parallel to string $CA$ |
| $\pm F_2 = 4W\cos 30° = 2\sqrt{3}W$ | (B1$\checkmark$) | Find normal component |
| $F = (\sqrt{57}/2)W$ *or* $3.77[5]W$ | (B1) | |
| $\tan^{-1}F_2/F_1 = \tan^{-1}4/\sqrt{3} = 66.6°$ *or* $1.16$ radians | (M1 A1) | Upward force at angle to $AC$; A0 if direction unclear |
| **Total: 5 marks** | | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{baea9836-ea05-442f-9e87-a2a1480dc74c-2_338_957_1482_593}
A uniform rod $B C$ of length $2 a$ and weight $W$ is hinged to a fixed point at $B$. A particle of weight $3 W$ is attached to the rod at $C$. The system is held in equilibrium by a light elastic string of natural length $\frac { 3 } { 5 } a$ in the same vertical plane as the rod. One end of the elastic string is attached to the rod at $C$ and the other end is attached to a fixed point $A$ which is at the same horizontal level as $B$. The rod and the string each make an angle of $30 ^ { \circ }$ with the horizontal (see diagram). Find\\
(i) the modulus of elasticity of the string,\\
(ii) the magnitude and direction of the force acting on the rod at $B$.
\hfill \mbox{\textit{CAIE FP2 2015 Q4 [10]}}