| 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 multiple equilibrium conditions (forces and moments) applied to a composite rigid body system with both rough and smooth contacts. Part (i) involves geometric reasoning with the constraint that spheres touch surfaces at specific points. Parts (ii) and (iii) require setting up and solving simultaneous equations from resolving forces horizontally/vertically and taking moments, with limiting equilibrium adding complexity. While the techniques are standard Further Maths mechanics, the multi-body system, careful geometry, and algebraic manipulation across three interconnected parts make this substantially harder than typical A-level questions. |
| 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 |
|---|---|---|
| \(\sin\theta = 6a/10a = 3/5\) | B1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_A = (k+2)W\) | M1 A1 | |
| \(F_A\,7a + Wa + kWa(1 + 5\cos\theta) + (W - R_A)\,a(1 + 10\cos\theta) = 0\) | M1 A1 | moments about \(B\) |
| \(7F_A = 9R_A - 5kW - 10W = 4kW + 8W\) | A1 | |
| \(7F_A = 4kW + 8W\) | ||
| \(\mu = F_A / R_A = 4/7\) or \(0.\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((k+2)^2 + (4/7)^2(k+2)^2 = 65\) | M1 A1 | |
| \(k^2 + 4k - 45 = (k-5)(k+9) = 0\) | ||
| or \((k+2)^2 = 49\) so \(k = 5\) | A1 | A.G. |
# Question 3:
## Part (i)
| $\sin\theta = 6a/10a = 3/5$ | B1 | A.G. |
|---|---|---|
## Part (ii)
| $R_A = (k+2)W$ | M1 A1 | |
|---|---|---|
| $F_A\,7a + Wa + kWa(1 + 5\cos\theta) + (W - R_A)\,a(1 + 10\cos\theta) = 0$ | M1 A1 | moments about $B$ |
| $7F_A = 9R_A - 5kW - 10W = 4kW + 8W$ | A1 | |
| $7F_A = 4kW + 8W$ | | |
| $\mu = F_A / R_A = 4/7$ or $0.$ | M1 A1 | |
## Part (iii)
| $(k+2)^2 + (4/7)^2(k+2)^2 = 65$ | M1 A1 | |
|---|---|---|
| $k^2 + 4k - 45 = (k-5)(k+9) = 0$ | | |
| or $(k+2)^2 = 49$ so $k = 5$ | A1 | A.G. |
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\includegraphics[max width=\textwidth, alt={}, center]{184020e1-7ff2-4172-8d33-baff963afa76-3_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]}}