CAIE FP2 2015 November — Question 1 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod with end on ground or wall supported by string
DifficultyStandard +0.8 This is a multi-step statics problem requiring resolution of forces in two directions, taking moments about a strategic point, and using the geometric constraint tan θ = 2 tan α. While it involves standard mechanics techniques (equilibrium conditions, moments), the presence of the rope at an intermediate point C rather than at an end, combined with the geometric relationship between angles, requires careful setup and algebraic manipulation. This is more demanding than routine statics problems but doesn't require exceptional insight—typical of harder Further Maths mechanics questions.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.10a Vectors in 2D: i,j notation and column vectors3.03m Equilibrium: sum of resolved forces = 03.04b Equilibrium: zero resultant moment and force

1 \includegraphics[max width=\textwidth, alt={}, center]{a8e37fb1-14c7-4004-b186-d607878e200d-2_744_504_255_824} A uniform ladder \(A B\), of length \(3 a\) and weight \(W\), rests with the end \(A\) in contact with smooth horizontal ground and the end \(B\) against a smooth vertical wall. One end of a light inextensible rope is attached to the ladder at the point \(C\), where \(A C = a\). The other end of the rope is fixed to the point \(D\) at the base of the wall and the rope \(D C\) is in the same vertical plane as the ladder \(A B\). The ladder rests in equilibrium in a vertical plane perpendicular to the wall, with the ladder making an angle \(\theta\) with the horizontal and the rope making an angle \(\alpha\) with the horizontal (see diagram). It is given that \(\tan \theta = 2 \tan \alpha\). Find, in terms of \(W\) and \(\alpha\), the tension in the rope and the magnitudes of the forces acting on the ladder at \(A\) and at \(B\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R_B = T\cos\alpha\)M1 A1 Resolve horizontally
\(R_A = W + T\sin\alpha\)M1 A1 Resolve vertically
\(R_B \cdot 3a\sin\theta = W(3a/2)\cos\theta + Ta(\sin\alpha\cos\theta + \cos\alpha\sin\theta)\) or \(+Ta\sin(\alpha+\theta)\) or \(+T3a\cos\theta\sin\alpha\)M1 A1 Moments about A
\(R_A \cdot 3a\cos\theta = W(3a/2)\cos\theta + T2a(\sin\alpha\cos\theta + \cos\alpha\sin\theta)\) or \(+T2a\sin(\alpha+\theta)\) or \(+T3a\sin\theta\cos\alpha\)(M1 A1) Moments about B
\(R_A \cdot a\cos\theta + W(a/2)\cos\theta = R_B \cdot 2a\sin\theta\)(M1 A1) Moments about C
\(R_A \cdot 3a\cos\theta - W(3a/2)\cos\theta = R_B \cdot 3a\sin\theta\)(M1 A1) Moments about D
\(T = W/2\sin\alpha\) or \(\frac{1}{2}W\csc\alpha\)B1
\(R_A = 3W/2\)B1
\(R_B = W/2\tan\alpha\) or \(\frac{1}{2}W\cot\alpha\)B1 Total: 9
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_B = T\cos\alpha$ | M1 A1 | Resolve horizontally |
| $R_A = W + T\sin\alpha$ | M1 A1 | Resolve vertically |
| $R_B \cdot 3a\sin\theta = W(3a/2)\cos\theta + Ta(\sin\alpha\cos\theta + \cos\alpha\sin\theta)$ or $+Ta\sin(\alpha+\theta)$ or $+T3a\cos\theta\sin\alpha$ | M1 A1 | Moments about A |
| $R_A \cdot 3a\cos\theta = W(3a/2)\cos\theta + T2a(\sin\alpha\cos\theta + \cos\alpha\sin\theta)$ or $+T2a\sin(\alpha+\theta)$ or $+T3a\sin\theta\cos\alpha$ | (M1 A1) | Moments about B |
| $R_A \cdot a\cos\theta + W(a/2)\cos\theta = R_B \cdot 2a\sin\theta$ | (M1 A1) | Moments about C |
| $R_A \cdot 3a\cos\theta - W(3a/2)\cos\theta = R_B \cdot 3a\sin\theta$ | (M1 A1) | Moments about D |
| $T = W/2\sin\alpha$ or $\frac{1}{2}W\csc\alpha$ | B1 | |
| $R_A = 3W/2$ | B1 | |
| $R_B = W/2\tan\alpha$ or $\frac{1}{2}W\cot\alpha$ | B1 | Total: 9 |

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\includegraphics[max width=\textwidth, alt={}, center]{a8e37fb1-14c7-4004-b186-d607878e200d-2_744_504_255_824}

A uniform ladder $A B$, of length $3 a$ and weight $W$, rests with the end $A$ in contact with smooth horizontal ground and the end $B$ against a smooth vertical wall. One end of a light inextensible rope is attached to the ladder at the point $C$, where $A C = a$. The other end of the rope is fixed to the point $D$ at the base of the wall and the rope $D C$ is in the same vertical plane as the ladder $A B$. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, with the ladder making an angle $\theta$ with the horizontal and the rope making an angle $\alpha$ with the horizontal (see diagram). It is given that $\tan \theta = 2 \tan \alpha$. Find, in terms of $W$ and $\alpha$, the tension in the rope and the magnitudes of the forces acting on the ladder at $A$ and at $B$.

\hfill \mbox{\textit{CAIE FP2 2015 Q1 [9]}}