Edexcel M2 2003 January — Question 3 9 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2003
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeFriction inequality derivation
DifficultyStandard +0.3 This is a standard M2 statics problem requiring three equilibrium equations (horizontal forces, vertical forces, and moments) with straightforward geometry. While it involves multiple steps and careful bookkeeping of forces, it follows a well-practiced template that M2 students drill extensively. The 'least coefficient' aspect is routine (friction at limiting equilibrium), and the given tan α simplifies trigonometry. Slightly easier than average due to its predictable structure.
Spec3.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.04b Equilibrium: zero resultant moment and force

\includegraphics{figure_1} A uniform ladder \(AB\), of mass \(m\) and length \(2a\), has one end \(A\) on rough horizontal ground. The other end \(B\) rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall. The ladder makes an angle \(\alpha\) with the horizontal, where \(\tan \alpha = \frac{4}{3}\). A child of mass \(2m\) stands on the ladder at \(C\) where \(AC = \frac{1}{2}a\), as shown in Fig. 1. The ladder and the child are in equilibrium. By modelling the ladder as a rod and the child as a particle, calculate the least possible value of the coefficient of friction between the ladder and the ground. [9]

Main Solution:
AnswerMarks
\(R = 3mg\)B1
Moments about B:
AnswerMarks
\(mga \cos \alpha + 2mg \times \frac{1}{3}a \cos \alpha + Fr \times 2a \sin \alpha = R \times 2a \cos \alpha\)M1 A2 1.0
Solving to \(Fr = \frac{3}{4}mg\):M1 A1
Friction constraint:
AnswerMarks Guidance
\(Fr \leq \mu R \Rightarrow \frac{3}{4}mg \leq \mu \times 3mg\)M1
\(\mu \geq \frac{1}{4}\) (least value is \(\frac{1}{4}\))M1 A1 (9 marks)
**Main Solution:**
$R = 3mg$ | B1 |

**Moments about B:**
$mga \cos \alpha + 2mg \times \frac{1}{3}a \cos \alpha + Fr \times 2a \sin \alpha = R \times 2a \cos \alpha$ | M1 A2 1.0 |

**Solving to $Fr = \frac{3}{4}mg$:** | M1 A1 |

**Friction constraint:**
$Fr \leq \mu R \Rightarrow \frac{3}{4}mg \leq \mu \times 3mg$ | M1 |

$\mu \geq \frac{1}{4}$ (least value is $\frac{1}{4}$) | M1 A1 | **(9 marks)**

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\includegraphics{figure_1}

A uniform ladder $AB$, of mass $m$ and length $2a$, has one end $A$ on rough horizontal ground. The other end $B$ rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall. The ladder makes an angle $\alpha$ with the horizontal, where $\tan \alpha = \frac{4}{3}$. A child of mass $2m$ stands on the ladder at $C$ where $AC = \frac{1}{2}a$, as shown in Fig. 1. The ladder and the child are in equilibrium.

By modelling the ladder as a rod and the child as a particle, calculate the least possible value of the coefficient of friction between the ladder and the ground. [9]

\hfill \mbox{\textit{Edexcel M2 2003 Q3 [9]}}