Challenging +1.8 This is a challenging Further Maths mechanics problem requiring geometric analysis of a two-rod system, moment equations, and force resolution. The multi-part structure (finding angle, tension, and joint force) with 12 marks total, combined with the need to handle trigonometric relationships in the constraint equation and resolve forces at a joint, places it well above average difficulty but not at the extreme end for Further Maths content.
\includegraphics{figure_5}
Two uniform rods, \(AB\) and \(BC\), each have length \(2a\) and weight \(W\). They are smoothly jointed at \(B\), and \(A\) is attached to a smooth fixed pivot. A light inextensible string of length \((2\sqrt{2})a\) joins \(A\) to \(C\) so that angle \(ABC = 90°\). The system hangs in equilibrium, with \(AB\) making an angle \(\alpha\) with the vertical (see diagram). By taking moments about \(A\) for the system, or otherwise, show that \(\alpha = 18.4°\), correct to the nearest \(0.1°\). [3]
Find the tension in the string in the form \(kW\), giving the value of \(k\) correct to 3 significant figures. [3]
Find, in terms of \(W\), the magnitude of the force acting on the rod \(BC\) at \(B\). [6]
Equate moments of AB and BC about A: Wa sin α = Wa (cos α – 2 sin α) M1 A1
Evaluate α: tan α = ⅓, α = 18⋅4° A.G. A1
Take moments about B for rod BC: T 2a cos 45° = Wa cos α M1 A1
Find tension T in form kW: T = (cos 18⋅4° / √2) W = 0⋅671 W A1
EITHER: Resolve vertically for rod BC: RV = W – T sin (45° + α)
(ignore signs of all components RV etc) = W – T sin 63⋅4° [= 0⋅400 W] M1 A1
Resolve horizontally for rod BC: RH = T cos (45° + α) [= 0⋅300 W] M1 A1
OR: Resolve along AB for rod BC: RAB = W cos α – T sin 45°
(or take moments about C) [= 3W/2√10 = 0⋅474 W] (M1 A1)
Resolve along BC for rod BC: RBC = W sin α – T cos 45°
[= –W/2√10 = –0⋅158 W] (M1 A1)
Answer
Marks
Combine components, e.g √(RV 2 + RH 2 ): R = 0⋅5 W M1 A1
(3)
(3)
Answer
Marks
Guidance
(6)
[12]
Page 6
Mark Scheme: Teachers’ version
Syllabus
GCE A LEVEL – October/November 2009
9231
02
Question 5:
5 | Equate moments of AB and BC about A: Wa sin α = Wa (cos α – 2 sin α) M1 A1
Evaluate α: tan α = ⅓, α = 18⋅4° A.G. A1
Take moments about B for rod BC: T 2a cos 45° = Wa cos α M1 A1
Find tension T in form kW: T = (cos 18⋅4° / √2) W = 0⋅671 W A1
EITHER: Resolve vertically for rod BC: RV = W – T sin (45° + α)
(ignore signs of all components RV etc) = W – T sin 63⋅4° [= 0⋅400 W] M1 A1
Resolve horizontally for rod BC: RH = T cos (45° + α) [= 0⋅300 W] M1 A1
OR: Resolve along AB for rod BC: RAB = W cos α – T sin 45°
(or take moments about C) [= 3W/2√10 = 0⋅474 W] (M1 A1)
Resolve along BC for rod BC: RBC = W sin α – T cos 45°
[= –W/2√10 = –0⋅158 W] (M1 A1)
Combine components, e.g √(RV 2 + RH 2 ): R = 0⋅5 W M1 A1 | (3)
(3)
(6) | [12]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2009 | 9231 | 02
\includegraphics{figure_5}
Two uniform rods, $AB$ and $BC$, each have length $2a$ and weight $W$. They are smoothly jointed at $B$, and $A$ is attached to a smooth fixed pivot. A light inextensible string of length $(2\sqrt{2})a$ joins $A$ to $C$ so that angle $ABC = 90°$. The system hangs in equilibrium, with $AB$ making an angle $\alpha$ with the vertical (see diagram). By taking moments about $A$ for the system, or otherwise, show that $\alpha = 18.4°$, correct to the nearest $0.1°$. [3]
Find the tension in the string in the form $kW$, giving the value of $k$ correct to 3 significant figures. [3]
Find, in terms of $W$, the magnitude of the force acting on the rod $BC$ at $B$. [6]
\hfill \mbox{\textit{CAIE FP2 2009 Q5 [12]}}