| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod against wall and ground |
| Difficulty | Challenging +1.2 This is a multi-part statics problem requiring resolution of forces, moments, and geometric reasoning with the rope perpendicular to the rod. While it involves several steps and coordinate geometry to find AC, the techniques are standard for Further Maths mechanics: taking moments about a point, resolving forces, and using given friction conditions. The geometric setup is more complex than basic mechanics questions but follows predictable solution patterns once the diagram is understood. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC = AD\cos\theta\) and \(AD = 2a\cos\theta\) [= \(6a/5\)] so \(AC = 2a\cos^2\theta = 2a(3/5)^2 = 18a/25\) | M1 A1 AG | Find \(AC\) (\(D\) denotes other end of rope from \(C\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\): \(R_B \times 2a\sin\theta - W \times a\cos\theta - T \times AC = 0\) \([R_B \times 8a/5 - W \times 3a/5 - T \times 18a/25 = 0]\) so \(40R_B - 15W - 18T = 0\) | M1 A1 | Take moments for rod about one chosen point (Note that a vertical resolution will then give \(T\), earning 6/6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(G\): \(F_A \times a\sin\theta - R_A \times a\cos\theta + R_B \times a\sin\theta + T \times (a - AC) = 0\) \([F_A \times 4a/5 - R_A \times 3a/5 + R_B \times 4a/5 + T \times 7a/25 = 0]\) so \(20F_A - 15R_A + 20R_B + 7T = 0\)] | (\(G\) is mid-point of \(AB\)) | |
| Horizontally: \(R_B - F_A = T\sin\theta\) [= \(4T/5\)] | B1 | Find two more indep. eqns, e.g. resolution of forces on rod |
| Vertically: \(R_A - W = T\cos\theta\) [= \(3T/5\)] | B1 | (a second moment eqn. may be used) |
| \(T = W/4\) | M1 A1 AG | Find or verify \(T\) using \(F_A = \frac{1}{4}R_A\), \(\sin\theta = 4/5\), \(\cos\theta = 3/5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R_A = 23W/20\) or \(1.15W\) [\(F_A = 23W/80 = 0.2875W\)] | B1 | Find \(R_A\), \(R_B\) (can assume \(T = W/4\)) |
| \(R_B = 39W/80\) or \(0.487_{[5]}W\) | B1 |
## Question 4:
### Part 4(i):
$AC = AD\cos\theta$ and $AD = 2a\cos\theta$ [= $6a/5$] so $AC = 2a\cos^2\theta = 2a(3/5)^2 = 18a/25$ | M1 A1 AG | Find $AC$ ($D$ denotes other end of rope from $C$)
### Part 4(ii):
$A$: $R_B \times 2a\sin\theta - W \times a\cos\theta - T \times AC = 0$ $[R_B \times 8a/5 - W \times 3a/5 - T \times 18a/25 = 0]$ so $40R_B - 15W - 18T = 0$ | M1 A1 | Take moments for rod about one chosen point (Note that a vertical resolution will then give $T$, earning 6/6)
$B$: $F_A \times 2a\sin\theta - R_A \times 2a\cos\theta + W \times a\cos\theta + T \times (2a - AC) = 0$ $[F_A \times 8a/5 - R_A \times 6a/5 + W \times 3a/5 + T \times 32a/25 = 0]$ so $15W + 32T = 3R_A - 40F_A = 20R_A$]
$C$: $F_A \times AC\sin\theta - R_A \times AC\cos\theta + R_B \times (2a - AC)\sin\theta - W \times (a - AC)\cos\theta = 0$ $[F_A \times 72a/125 - R_A \times 54a/125 + R_B \times 128a/125 - W \times 21a/125 = 0]$ so $128R_B - 21W = 54R_A - 72F_A = 36R_A$]
$D$: $R_A \times 2a\cos\theta - R_B \times 2a\sin\theta - W \times a\cos\theta = 0$ $[R_A \times 6a/5 - R_B \times 8a/5 - W \times 3a/5 = 0]$ so $6R_A - 8R_B - 3W = 0$]
$G$: $F_A \times a\sin\theta - R_A \times a\cos\theta + R_B \times a\sin\theta + T \times (a - AC) = 0$ $[F_A \times 4a/5 - R_A \times 3a/5 + R_B \times 4a/5 + T \times 7a/25 = 0]$ so $20F_A - 15R_A + 20R_B + 7T = 0$] | | ($G$ is mid-point of $AB$)
Horizontally: $R_B - F_A = T\sin\theta$ [= $4T/5$] | B1 | Find two more indep. eqns, e.g. resolution of forces on rod
Vertically: $R_A - W = T\cos\theta$ [= $3T/5$] | B1 | (a second moment eqn. may be used)
$T = W/4$ | M1 A1 AG | Find or verify $T$ using $F_A = \frac{1}{4}R_A$, $\sin\theta = 4/5$, $\cos\theta = 3/5$
### Part 4(iii):
$R_A = 23W/20$ or $1.15W$ [$F_A = 23W/80 = 0.2875W$] | B1 | Find $R_A$, $R_B$ (can assume $T = W/4$)
$R_B = 39W/80$ or $0.487_{[5]}W$ | B1 |
---4 A uniform $\operatorname { rod } A B$ has length $2 a$ and weight $W$. The end $A$ rests on rough horizontal ground and the end $B$ rests against a smooth vertical wall. The angle between the rod and the horizontal is $\theta$, where $\tan \theta = \frac { 4 } { 3 }$. One end of a light inextensible rope is attached to a point $C$ on the rod. The other end is attached to a point where the vertical wall and the horizontal ground meet. The rope is taut and perpendicular to the rod. The rope and rod are in a vertical plane perpendicular to the wall.\\
(i) Show that $A C = \frac { 18 } { 25 } a$.\\
The magnitude of the frictional force at $A$ is equal to one quarter of the magnitude of the normal reaction force at $A$.\\
(ii) Show that the tension in the rope is $\frac { 1 } { 4 } W$.\\
(iii) Find expressions, in terms of $W$, for the magnitudes of the normal reaction forces at $A$ and $B$.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q4 [10]}}