Edexcel M2 2008 January — Question 5 10 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2008
SessionJanuary
Marks10
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Mark schemeDownload PDF ↗
TopicFriction
TypeFriction inequality derivation
DifficultyStandard +0.3 This is a standard M2 ladder equilibrium problem requiring resolution of forces, taking moments, and applying friction law. While it involves multiple steps (resolving horizontally/vertically, taking moments about a point, combining equations), the approach is entirely routine and well-practiced. The 30° angle gives clean trigonometry, and the setup follows the standard textbook template, making it slightly easier than average.
Spec3.03m Equilibrium: sum of resolved forces = 03.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force
\includegraphics{figure_2} A ladder \(AB\), of mass \(m\) and length \(4a\), has one end \(A\) resting on rough horizontal ground. The other end \(B\) rests against a smooth vertical wall. A load of mass \(3m\) is fixed on the ladder at the point \(C\), where \(AC = a\). The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load is modelled as a particle. The ladder rests in limiting equilibrium making an angle of 30° with the wall, as shown in Figure 2. Find the coefficient of friction between the ladder and the ground. [10]
Part (a)
M(A): \(N \times 4a\cos 30° = 3mg \times a\sin 30° + mg \times 2a\sin 30°\)
AnswerMarks Guidance
\(N = \frac{5}{4}mg\tan 30°\) (\(= \frac{5}{4\sqrt{3}}mg = 7.07\ldots \text{m}\))M1 A2(1,0) DM1 A1 B1 B1 B1 (10 marks)
\(\Rightarrow F_f = N\), \(\uparrow R = 4mg\)
Using \(F_f = \mu R\):
\(\frac{5}{4\sqrt{3}}mg = \mu R\) for their \(R\)
AnswerMarks Guidance
\(\mu = \frac{5}{16\sqrt{3}}\) awrt 0.18M1 A1 [10 marks]
Alternative method:
M(B): \(mg \times 2a\sin 30° + 3mg \times 3a\sin 30° + F \times 4a\cos 30° = R \times 4a\sin 30°\)
\(1 lmga\sin 30° + F\frac{4\sqrt{3}}{2} = 2R\)
\(\uparrow R = 4mg\)
Using \(F_f = \mu R\):
AnswerMarks
\(8\mu\sqrt{3} = \frac{5}{2}\), \(\mu = \frac{5}{16\sqrt{3}}\)M1 A3(2,1,0) DM1 A1 B1 B1 M1 A1 
**Part (a)**

M(A): $N \times 4a\cos 30° = 3mg \times a\sin 30° + mg \times 2a\sin 30°$

$N = \frac{5}{4}mg\tan 30°$ ($= \frac{5}{4\sqrt{3}}mg = 7.07\ldots \text{m}$) | M1 A2(1,0) DM1 A1 B1 B1 B1 | (10 marks)

$\Rightarrow F_f = N$, $\uparrow R = 4mg$

Using $F_f = \mu R$:

$\frac{5}{4\sqrt{3}}mg = \mu R$ for their $R$

$\mu = \frac{5}{16\sqrt{3}}$ awrt 0.18 | M1 A1 | [10 marks]

**Alternative method:**
M(B): $mg \times 2a\sin 30° + 3mg \times 3a\sin 30° + F \times 4a\cos 30° = R \times 4a\sin 30°$

$1 lmga\sin 30° + F\frac{4\sqrt{3}}{2} = 2R$

$\uparrow R = 4mg$

Using $F_f = \mu R$:

$8\mu\sqrt{3} = \frac{5}{2}$, $\mu = \frac{5}{16\sqrt{3}}$ | M1 A3(2,1,0) DM1 A1 B1 B1 M1 A1 |

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\includegraphics{figure_2}

A ladder $AB$, of mass $m$ and length $4a$, has one end $A$ resting on rough horizontal ground. The other end $B$ rests against a smooth vertical wall. A load of mass $3m$ is fixed on the ladder at the point $C$, where $AC = a$. The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load is modelled as a particle. The ladder rests in limiting equilibrium making an angle of 30° with the wall, as shown in Figure 2.

Find the coefficient of friction between the ladder and the ground. [10]

\hfill \mbox{\textit{Edexcel M2 2008 Q5 [10]}}
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