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
DifficultyStandard +0.8 This is a multi-step statics problem requiring resolution of forces in two directions, taking moments about a strategic point, and finding the limiting friction condition. While it involves several standard techniques (resolving forces, moments, friction), the geometry with the perpendicular string and the need to coordinate multiple equations to find μ makes it moderately challenging but still within typical Further Maths scope.
Spec3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces

1 \includegraphics[max width=\textwidth, alt={}, center]{a473cbb8-877f-48df-8751-c76d96396734-2_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 \(CPA = \theta\)
\(5aF = 2amg\cos\theta\); \(F = 8mg/25\)M1;A1 2 moment equations for \(R\) and \(F\), e.g. about \(P\)
\(3aR\cos\theta - 3aF\sin\theta = amg\cos\theta\)M1 About \(C\)
\(R = (4mg + 9F)/12 = 43mg/75\)M1 A1 Solve for \(R\)
\(\mu_{\min} = 24/43\) A.G.M1 A1 Use \(F = \mu_{\min}R\) to find \(\mu_{\min}\)
OR alternative:
AnswerMarks Guidance
\(3aT = 2amg\cos\theta\) \([T = 8mg/15]\)(M1) Take moments about \(A\)
\(F = T\sin\theta\) \(= 8mg/25\)(M1;A1) Resolve horizontally
\(R = mg - T\cos\theta = 43mg/75\)(M1;A1) Resolve horizontally for reaction at \(A\)
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 $CPA = \theta$ |
| $5aF = 2amg\cos\theta$; $F = 8mg/25$ | M1;A1 | 2 moment equations for $R$ and $F$, e.g. about $P$ |
| $3aR\cos\theta - 3aF\sin\theta = amg\cos\theta$ | M1 | About $C$ |
| $R = (4mg + 9F)/12 = 43mg/75$ | M1 A1 | Solve for $R$ |
| $\mu_{\min} = 24/43$ **A.G.** | M1 A1 | Use $F = \mu_{\min}R$ to find $\mu_{\min}$ |

**OR alternative:**
| $3aT = 2amg\cos\theta$ $[T = 8mg/15]$ | (M1) | Take moments about $A$ |
| $F = T\sin\theta$ $= 8mg/25$ | (M1;A1) | Resolve horizontally |
| $R = mg - T\cos\theta = 43mg/75$ | (M1;A1) | Resolve horizontally for reaction at $A$ |

**Total: 8 marks**

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\includegraphics[max width=\textwidth, alt={}, center]{a473cbb8-877f-48df-8751-c76d96396734-2_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 $\frac { 24 } { 43 }$.

\hfill \mbox{\textit{CAIE FP2 2013 Q1 [8]}}
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