| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Challenging +1.8 This is a challenging statics problem requiring careful geometric analysis of a composite rigid body (rod + two spheres), resolution of forces in two directions, taking moments about a strategic point, and working with limiting equilibrium friction. Part (i) requires geometric reasoning with the constraint that B is at height 7a. Parts (ii) and (iii) involve systematic application of equilibrium conditions with multiple unknowns. While the techniques are standard Further Maths mechanics (moments, friction, resolution), the composite structure and multi-part coordination make this significantly harder than typical A-level questions but not exceptionally difficult for Further Maths students. |
| Spec | 1.05g Exact trigonometric values: for standard angles3.03m Equilibrium: sum of resolved forces = 03.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin\theta = 6a/10a = 3/5\) | B1 | A.G. Verify \(\sin\theta\) from triangle \(CDE\) where \(E\) is level with \(C\) and vertically below \(D\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R_A = (k+2)W\) | M1 A1 | Resolve forces on object vertically |
| \(F_A\ 7a + Wa + kWa(1 + 5\cos\theta) + (W - R_A)a(1 + 10\cos\theta) = 0\) | M1 A1 | Take moments about \(B\) |
| \(7F_A = 9R_A - 5kW - 10W = 4kW + 8W\) | A1 | Using \(R_B = F_A\), \(\sin\theta = 3/5\), \(\cos\theta = 4/5\) |
| \(7F_A = 4kW + 8W\) | ||
| \(\mu = F_A/R_A = 4/7\) or \(0.\) | M1 A1 | Find \(\mu\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((k+2)^2 + (4/7)^2(k+2)^2 = 65\) | M1 A1 | Equate resultant force at \(A\) to \(W\sqrt{65}\) |
| \(k^2 + 4k - 45 = (k-5)(k+9) = 0\) | Solve to verify value of \(k\) | |
| \(or\ (k+2)^2 = 49\) so \(k = 5\) | A1 | A.G. |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin\theta = 6a/10a = 3/5$ | B1 | A.G. Verify $\sin\theta$ from triangle $CDE$ where $E$ is level with $C$ and vertically below $D$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R_A = (k+2)W$ | M1 A1 | Resolve forces on object vertically |
| $F_A\ 7a + Wa + kWa(1 + 5\cos\theta) + (W - R_A)a(1 + 10\cos\theta) = 0$ | M1 A1 | Take moments about $B$ |
| $7F_A = 9R_A - 5kW - 10W = 4kW + 8W$ | A1 | Using $R_B = F_A$, $\sin\theta = 3/5$, $\cos\theta = 4/5$ |
| $7F_A = 4kW + 8W$ | | |
| $\mu = F_A/R_A = 4/7$ or $0.$ | M1 A1 | Find $\mu$ |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(k+2)^2 + (4/7)^2(k+2)^2 = 65$ | M1 A1 | Equate resultant force at $A$ to $W\sqrt{65}$ |
| $k^2 + 4k - 45 = (k-5)(k+9) = 0$ | | Solve to verify value of $k$ |
| $or\ (k+2)^2 = 49$ so $k = 5$ | A1 | A.G. |
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\includegraphics[max width=\textwidth, alt={}, center]{58728f93-bfdb-4f76-a9b9-3a1d1592bfc9-3_898_1116_258_518}
The end $P$ of a uniform rod $P Q$, of weight $k W$ and length $8 a$, is rigidly attached to a point on the surface of a uniform sphere with centre $C$, weight $W$ and radius $a$. The end $Q$ is rigidly attached to a point on the surface of an identical sphere with centre $D$. The points $C , P , Q$ and $D$ are in a straight line. The object consisting of the rod and two spheres rests with one sphere in contact with a rough horizontal surface, at the point $A$, and the other sphere in contact with a smooth vertical wall, at the point $B$. The angle between $C D$ and the horizontal is $\theta$. The point $B$ is at a height of $7 a$ above the base of the wall (see diagram). The points $A , B , C , D , P$ and $Q$ are all in the same vertical plane.\\
(i) Show that $\sin \theta = \frac { 3 } { 5 }$.
The object is in limiting equilibrium and the coefficient of friction at $A$ is $\mu$.\\
(ii) Find the numerical value of $\mu$.\\
(iii) Given that the resultant force on the object at $A$ is $W \sqrt { } ( 65 )$, show that $k = 5$.
\hfill \mbox{\textit{CAIE FP2 2016 Q3 [11]}}