| 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, part (ii) requires resolving forces and using limiting equilibrium (F=μR), and part (iii) combines force resolution with a resultant condition to find k. The multi-component system and three interconnected parts requiring extended reasoning place this well above average difficulty, though the techniques themselves are standard for Further Maths mechanics. |
| 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. Verify \(\sin\theta\) from triangle \(CDE\) where \(E\) is level with \(C\) and vertically below \(D\) |
| 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 \(B\) |
| *or D:* \(F_A\,7a + kW\,5a\cos\theta + (W - R_A)\,10a\cos\theta = 0\) | ||
| *or A:* \(R_B\,7a = kW\,5a\cos\theta + W\,10a\cos\theta\) | ||
| *or C:* \(F_A\,a + R_B\,6a = W\,10a\cos\theta + kW\,5a\cos\theta\) | ||
| Using \(R_B = F_A\), \(\sin\theta = 3/5\), \(\cos\theta = 4/5\): | ||
| \(B\): \(7F_A = 9R_A - 5kW - 10W = 4kW + 8W\) | A1 | |
| \(7F_A = 4kW + 8W\) | ||
| \(7F_A = 4R_A\) [so \(R_A\) is not required here] | A1 | |
| \(\mu = F_A / R_A = 4/7\) or \(0.\) | M1 A1 | Find \(\mu\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((k+2)^2 + (4/7)^2(k+2)^2 = 65\) | M1 A1 | Equate resultant force at \(A\) to \(W\sqrt{65}\) |
| \(k^2 + 4k - 45 = (k-5)(k+9) = 0\) | A1 | A.G. |
| *or* \((k+2)^2 = 49\) so \(k = 5\) |
# Question 3:
## Part (i)
| $\sin\theta = 6a / 10a = 3/5$ | B1 | A.G. Verify $\sin\theta$ from triangle $CDE$ where $E$ is level with $C$ and vertically below $D$ |
|---|---|---|
## Part (ii)
| $R_A = (k+2)W$ | 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 $B$ |
| *or D:* $F_A\,7a + kW\,5a\cos\theta + (W - R_A)\,10a\cos\theta = 0$ | | |
| *or A:* $R_B\,7a = kW\,5a\cos\theta + W\,10a\cos\theta$ | | |
| *or C:* $F_A\,a + R_B\,6a = W\,10a\cos\theta + kW\,5a\cos\theta$ | | |
| Using $R_B = F_A$, $\sin\theta = 3/5$, $\cos\theta = 4/5$: | | |
| $B$: $7F_A = 9R_A - 5kW - 10W = 4kW + 8W$ | A1 | |
| $7F_A = 4kW + 8W$ | | |
| $7F_A = 4R_A$ [so $R_A$ is not required here] | A1 | |
| $\mu = F_A / R_A = 4/7$ or $0.$ | M1 A1 | Find $\mu$ |
## Part (iii)
| $(k+2)^2 + (4/7)^2(k+2)^2 = 65$ | M1 A1 | Equate resultant force at $A$ to $W\sqrt{65}$ |
|---|---|---|
| $k^2 + 4k - 45 = (k-5)(k+9) = 0$ | A1 | A.G. |
| *or* $(k+2)^2 = 49$ so $k = 5$ | | |
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\includegraphics[max width=\textwidth, alt={}, center]{62d0d8cb-8f8c-4298-9705-71a735a9a4e7-3_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]}}