Edexcel M2 2001 June — Question 3 9 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2001
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeLadder against wall
DifficultyStandard +0.3 This is a standard M2 ladder equilibrium problem requiring resolution of forces, friction inequality (F ≤ μR), and taking moments about a suitable point. While it involves multiple steps and careful angle work, it follows a well-established method taught explicitly in M2 with no novel insight required. The 9 marks reflect routine application of mechanics principles rather than conceptual difficulty.
Spec3.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force

A uniform ladder \(AB\), of mass \(m\) and length \(2a\), has one end \(A\) on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.5. The other end \(B\) of the ladder rests against a smooth vertical wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, and makes an angle of 30° with the wall. A man of mass \(5m\) stands on the ladder which remains in equilibrium. The ladder is modelled as a uniform rod and the man as a particle. The greatest possible distance of the man from \(A\) is \(ka\). Find the value of \(k\). [9]

AnswerMarks Guidance
Resolve \(\rightarrow\) S = F; Resolve \(\uparrow\) R = 6mgB1
\(M(A)\): \(S \cdot 2a \cos 30° = mg \sin 30° (a + 5x)\)M1A1
"F \(\leq\) 0.5 R" \(\Rightarrow\) \(S \leq 3\)mgM1
\((a + 5x) \tan 30° \leq 6a\), \(x \leq \frac{(6\sqrt{3} - 1)a}{5}\) \(\Rightarrow\) \(k = \frac{(6\sqrt{3} - 1)}{5}\) or 1.88M1A1 (9)
Alternatives:
AnswerMarks
\(M(F)\): \(R \cdot 2a \sin 30° = F \cdot 2a \cos 30° + mga \sin 30° + 5mgd\sin 30°\)M1A1A1
\(d = 2a - x\)B1
"F \(\leq\) 0.5 R" \(\Rightarrow\) F \(\leq\) 3mgM1
rest as scheme
\(M(\text{centre})\): \(Ra \sin 30° + 5mg(x - a)\sin 30° = (F + S) a \cos 30°\); \(S \leq 3\)mg etc.Mark as scheme
Note (i): MR - 30° to the ground - gives \(k = \frac{(6 - \sqrt{3})}{5}\) or 0.493
(ii) The same answer is obtained if only error is sin/cos confusion; both score 7/9.
(iii) m used for mg throughout, no penalty; inconsistent, as scheme but max -2
Resolve $\rightarrow$ S = F; Resolve $\uparrow$ R = 6mg | B1 |

$M(A)$: $S \cdot 2a \cos 30° = mg \sin 30° (a + 5x)$ | M1A1 |

"F $\leq$ 0.5 R" $\Rightarrow$ $S \leq 3$mg | M1 |

$(a + 5x) \tan 30° \leq 6a$, $x \leq \frac{(6\sqrt{3} - 1)a}{5}$ $\Rightarrow$ $k = \frac{(6\sqrt{3} - 1)}{5}$ or 1.88 | M1A1 | (9)

**Alternatives:**

$M(F)$: $R \cdot 2a \sin 30° = F \cdot 2a \cos 30° + mga \sin 30° + 5mgd\sin 30°$ | M1A1A1 |

$d = 2a - x$ | B1 |

"F $\leq$ 0.5 R" $\Rightarrow$ F $\leq$ 3mg | M1 |

rest as scheme | |

$M(\text{centre})$: $Ra \sin 30° + 5mg(x - a)\sin 30° = (F + S) a \cos 30°$; $S \leq 3$mg etc. | Mark as scheme |

**Note (i):** MR - 30° to the ground - gives $k = \frac{(6 - \sqrt{3})}{5}$ or 0.493 | |

**(ii)** The same answer is obtained if only error is sin/cos confusion; both score 7/9. | |

**(iii)** m used for mg throughout, no penalty; inconsistent, as scheme but max -2 | |

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A uniform ladder $AB$, of mass $m$ and length $2a$, has one end $A$ on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.5. The other end $B$ of the ladder rests against a smooth vertical wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, and makes an angle of 30° with the wall. A man of mass $5m$ stands on the ladder which remains in equilibrium. The ladder is modelled as a uniform rod and the man as a particle. The greatest possible distance of the man from $A$ is $ka$.

Find the value of $k$.
[9]

\hfill \mbox{\textit{Edexcel M2 2001 Q3 [9]}}