CAIE FP2 2012 June — Question 11 EITHER

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod on peg or cylinder
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple equilibrium conditions (forces and moments), careful geometry with the cylinder contact, and algebraic manipulation to find both a friction coefficient expression and a range of k values. It demands systematic application of statics principles with non-trivial algebra, placing it well above average difficulty but not at the extreme end for Further Maths content.
Spec6.04e Rigid body equilibrium: coplanar forces

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The diagram shows a uniform rod \(A B\), of length \(4 a\) and weight \(W\), resting in equilibrium with its end \(A\) on rough horizontal ground. The rod rests at \(C\) on the surface of a smooth cylinder whose axis is horizontal. The cylinder rests on the ground and is fixed to it. The rod is in a vertical plane perpendicular to the axis of the cylinder and is inclined at an angle \(\theta\) to the horizontal, where \(\cos \theta = \frac { 3 } { 5 }\). A particle of weight \(k W\) is attached to the rod at \(B\). Given that \(A C = 3 a\), show that the least possible value of the coefficient of friction \(\mu\) between the rod and the ground is \(\frac { 8 ( 2 k + 1 ) } { 13 k + 19 }\). Given that \(\mu = \frac { 9 } { 10 }\), find the set of values of \(k\) for which equilibrium is possible.

Question 11(a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R_C \cdot 3a = kW \cdot 4a\cos\theta + W \cdot 2a\cos\theta\)M1 A1 Take moments about \(A\) for rod to find \(R_C\)
\(R_C = (2k+1)(2W/5)\)A1
\(F_A = R_C\sin\theta = (2k+1)(8W/25)\)M1 A1 Resolve horizontal forces on rod to find \(F_A\)
\(R_A = W + kW - R_C\cos\theta = (13k+19)(W/25)\)M1 A1 Resolve vertical forces on rod to find \(R_A\)
*OR:* Take moments about \(C\) for rod:
\(F_A \cdot 3a\sin\theta - R_A \cdot 3a\cos\theta = kWa\cos\theta - Wa\cos\theta\)(M1 A1)
\(4F_A - 3R_A = (k-1)W\)(A1)
Resolve forces parallel to rod: \(F_A\cos\theta + R_A\sin\theta = kW\sin\theta + W\sin\theta\)(M1)
\(3F_A + 4R_A = 4(k+1)W\)(A1)
Combine to find \(F_A\), \(R_A\): \(F_A = (2k+1)(8W/25)\) and \(R_A = (13k+19)(W/25)\)(M1 A1)
Use \(F_A \le \mu R_A\) to find \(\mu_{\min}\): \(\mu_{\min} = 8(2k+1)/(13k+19)\) A.G.M1 A1 (No use of inequality loses A1)
\(160k + 80 \le (\text{or} =)\ 117k + 171\)M1 Find bound on \(k\) (allow use of equality)
Bound \(= 91/43\) or \(2.12\)A1
\(k \le 91/43\) or \(2.12\)A1 Correct use of inequality
Total: 12 marks
Question 11(b):
Find correlation coefficient \(r\):
AnswerMarks Guidance
\(r^2 = (-0.5)(-1.2) = 0.6 \text{ or } 0.775^2\)M1 A1
\(r = -0.775\)*A1 3 marks
State both hypotheses (B0 for \(r\)...):
AnswerMarks
\(H_0: \rho = 0,\ H_1: \rho \neq 0\)B1
State or use correct tabular one-tail \(r\) value:
AnswerMarks
\(r_{5,\ 2.5\%} = 0.878\)*B1
Valid method for reaching conclusion:
AnswerMarks Guidance
Accept \(H_0\) if \(r <\) tabular value
Correct conclusion (AEF, dep *A1, *B1):
AnswerMarks Guidance
Coeff. does not differ from zeroA1 4 marks
EITHER: Find two indep. eqns for \(p, q\):
AnswerMarks
\(14 + q = (-0.5)(13 + p) + 5 \times 5\)M1 A1
\(or\ (14 + q)/5 = (-0.5)(13 + p)/5 + 5\)
AnswerMarks
\(13 + p = (-1.2)(14 + q) + 5 \times 7.6\)
\(or\ (13 + p)/5 = (-1.2)(14 + q)/5 + 7.6\)A1
Combine to find \(p, q\):
AnswerMarks
\(p = 7,\ q = 1\)A1, A1
OR: Combine regression lines to find \(\bar{x},\ \bar{y}\):
AnswerMarks
\(\bar{x} = 4;\ \bar{y} = 3\)M1 A1; A1
Use sample data to find \(p, q\):
AnswerMarks Guidance
\(p = 7,\ q = 1\)A1, A1 5 marks
Total: [12]
## Question 11(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_C \cdot 3a = kW \cdot 4a\cos\theta + W \cdot 2a\cos\theta$ | M1 A1 | Take moments about $A$ for rod to find $R_C$ |
| $R_C = (2k+1)(2W/5)$ | A1 | |
| $F_A = R_C\sin\theta = (2k+1)(8W/25)$ | M1 A1 | Resolve horizontal forces on rod to find $F_A$ |
| $R_A = W + kW - R_C\cos\theta = (13k+19)(W/25)$ | M1 A1 | Resolve vertical forces on rod to find $R_A$ |
| *OR:* Take moments about $C$ for rod: | | |
| $F_A \cdot 3a\sin\theta - R_A \cdot 3a\cos\theta = kWa\cos\theta - Wa\cos\theta$ | (M1 A1) | |
| $4F_A - 3R_A = (k-1)W$ | (A1) | |
| Resolve forces parallel to rod: $F_A\cos\theta + R_A\sin\theta = kW\sin\theta + W\sin\theta$ | (M1) | |
| $3F_A + 4R_A = 4(k+1)W$ | (A1) | |
| Combine to find $F_A$, $R_A$: $F_A = (2k+1)(8W/25)$ and $R_A = (13k+19)(W/25)$ | (M1 A1) | |
| Use $F_A \le \mu R_A$ to find $\mu_{\min}$: $\mu_{\min} = 8(2k+1)/(13k+19)$ **A.G.** | M1 A1 | (No use of inequality loses A1) |
| $160k + 80 \le (\text{or} =)\ 117k + 171$ | M1 | Find bound on $k$ (allow use of equality) |
| Bound $= 91/43$ or $2.12$ | A1 | |
| $k \le 91/43$ or $2.12$ | A1 | Correct use of inequality |

**Total: 12 marks**

## Question 11(b):

**Find correlation coefficient $r$:**

$r^2 = (-0.5)(-1.2) = 0.6 \text{ or } 0.775^2$ | M1 A1 |

$r = -0.775$ | *A1 | 3 marks

---

**State both hypotheses (B0 for $r$...):**

$H_0: \rho = 0,\ H_1: \rho \neq 0$ | B1 |

---

**State or use correct tabular one-tail $r$ value:**

$r_{5,\ 2.5\%} = 0.878$ | *B1 |

---

**Valid method for reaching conclusion:**

Accept $H_0$ if $|r| <$ tabular value | M1 |

---

**Correct conclusion (AEF, dep *A1, *B1):**

Coeff. does not differ from zero | A1 | 4 marks

---

**EITHER: Find two indep. eqns for $p, q$:**

$14 + q = (-0.5)(13 + p) + 5 \times 5$ | M1 A1 |

$or\ (14 + q)/5 = (-0.5)(13 + p)/5 + 5$

$13 + p = (-1.2)(14 + q) + 5 \times 7.6$ | |

$or\ (13 + p)/5 = (-1.2)(14 + q)/5 + 7.6$ | A1 |

---

**Combine to find $p, q$:**

$p = 7,\ q = 1$ | A1, A1 |

---

**OR: Combine regression lines to find $\bar{x},\ \bar{y}$:**

$\bar{x} = 4;\ \bar{y} = 3$ | M1 A1; A1 |

**Use sample data to find $p, q$:**

$p = 7,\ q = 1$ | A1, A1 | 5 marks

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**Total: [12]**
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\includegraphics[max width=\textwidth, alt={}]{71a3b842-9d31-4c25-b894-ca6d1f47d84b-5_474_796_479_676}
\end{center}

The diagram shows a uniform rod $A B$, of length $4 a$ and weight $W$, resting in equilibrium with its end $A$ on rough horizontal ground. The rod rests at $C$ on the surface of a smooth cylinder whose axis is horizontal. The cylinder rests on the ground and is fixed to it. The rod is in a vertical plane perpendicular to the axis of the cylinder and is inclined at an angle $\theta$ to the horizontal, where $\cos \theta = \frac { 3 } { 5 }$. A particle of weight $k W$ is attached to the rod at $B$. Given that $A C = 3 a$, show that the least possible value of the coefficient of friction $\mu$ between the rod and the ground is $\frac { 8 ( 2 k + 1 ) } { 13 k + 19 }$.

Given that $\mu = \frac { 9 } { 10 }$, find the set of values of $k$ for which equilibrium is possible.

\hfill \mbox{\textit{CAIE FP2 2012 Q11 EITHER}}