CAIE FP2 2013 June — Question 1 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod with end on ground or wall supported by string
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem requiring resolution of forces, taking moments about a point, and applying friction conditions. While it involves multiple steps (finding geometry, resolving forces in two directions, taking moments, and finding the limiting friction case), the approach is methodical and follows standard techniques taught in FM mechanics. The geometry is straightforward (3-4-5 triangle), and the algebra, though somewhat involved, is routine manipulation to reach the given answer.
Spec6.04e Rigid body equilibrium: coplanar forces
1 \includegraphics[max width=\textwidth, alt={}, center]{137d2806-f45c-4121-8ee9-bf89580e1cca-2_684_714_246_717} A uniform \(\operatorname { rod } A B\), of mass \(m\) and length \(4 a\), rests with the end \(A\) on rough horizontal ground. The point \(C\) on \(A B\) is such that \(A C = 3 a\). A light inextensible string has one end attached to the point \(P\) which is at a distance \(5 a\) vertically above \(A\), and the other end attached to \(C\). The rod and the string are in the same vertical plane and the system is in equilibrium with angle \(A C P\) equal to \(90 ^ { \circ }\) (see diagram). The coefficient of friction between the rod and the ground is \(\mu\). Show that the least possible value of \(\mu\) is \(\frac { 24 } { 43 }\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(CP = 4a\) or \(\sin\theta = 3/5\) or \(\cos\theta = 4/5\)B1 State or imply length of \(CP\) or equivalent, e.g. angle \(CPA = \theta\)
2 moment equations for \(R\) and \(F\), e.g. about \(P\): \(5aF = 2amg\cos\theta\); \(F = 8mg/25\)M1;A1
about \(C\): \(3aR\cos\theta - 3aF\sin\theta = amg\cos\theta\)M1
\(R = (4mg + 9F)/12 = 43mg/75\)M1 A1 Solve for \(R\)
OR: \(3aT = 2amg\cos\theta\), \([T = 8mg/15]\)(M1) Take moments about \(A\) to give tension \(T\)
\(F = T\sin\theta = 8mg/25\)(M1;A1) Resolve horizontally to find friction at \(A\)
\(R = mg - T\cos\theta = 43mg/75\)(M1;A1) Resolve horizontally to find reaction at \(A\)
\(\mu_{\min} = 24/43\) A.G.M1 A1 Use \(F = \mu_{\min}R\) to find \(\mu_{\min}\)
Total: 8 marks
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $CP = 4a$ or $\sin\theta = 3/5$ or $\cos\theta = 4/5$ | B1 | State or imply length of $CP$ or equivalent, e.g. angle $CPA = \theta$ |
| 2 moment equations for $R$ and $F$, e.g. about $P$: $5aF = 2amg\cos\theta$; $F = 8mg/25$ | M1;A1 | |
| about $C$: $3aR\cos\theta - 3aF\sin\theta = amg\cos\theta$ | M1 | |
| $R = (4mg + 9F)/12 = 43mg/75$ | M1 A1 | Solve for $R$ |
| OR: $3aT = 2amg\cos\theta$, $[T = 8mg/15]$ | (M1) | Take moments about $A$ to give tension $T$ |
| $F = T\sin\theta = 8mg/25$ | (M1;A1) | Resolve horizontally to find friction at $A$ |
| $R = mg - T\cos\theta = 43mg/75$ | (M1;A1) | Resolve horizontally to find reaction at $A$ |
| $\mu_{\min} = 24/43$ **A.G.** | M1 A1 | Use $F = \mu_{\min}R$ to find $\mu_{\min}$ |

**Total: 8 marks**

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\includegraphics[max width=\textwidth, alt={}, center]{137d2806-f45c-4121-8ee9-bf89580e1cca-2_684_714_246_717}

A uniform $\operatorname { rod } A B$, of mass $m$ and length $4 a$, rests with the end $A$ on rough horizontal ground. The point $C$ on $A B$ is such that $A C = 3 a$. A light inextensible string has one end attached to the point $P$ which is at a distance $5 a$ vertically above $A$, and the other end attached to $C$. The rod and the string are in the same vertical plane and the system is in equilibrium with angle $A C P$ equal to $90 ^ { \circ }$ (see diagram). The coefficient of friction between the rod and the ground is $\mu$. Show that the least possible value of $\mu$ is $\frac { 24 } { 43 }$.

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