Edexcel M1 2004 January — Question 4 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2004
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeParticle suspended by strings
DifficultyStandard +0.3 This is a standard M1 equilibrium problem with friction and moments. Part (a) requires resolving forces on ring C (straightforward vertical equilibrium). Part (b) involves resolving horizontally and vertically on rings A or B, then applying friction law F=μR. The geometry is given (tan θ = 3/4), so students just need to find sin θ and cos θ, then systematically apply equilibrium conditions. While it requires multiple steps and careful bookkeeping across 10 marks total, it follows a predictable template for M1 statics problems with no novel insight required.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces

\includegraphics{figure_2} Two small rings, \(A\) and \(B\), each of mass \(2m\), are threaded on a rough horizontal pole. The coefficient of friction between each ring and the pole is \(\mu\). The rings are attached to the ends of a light inextensible string. A smooth ring \(C\), of mass \(3m\), is threaded on the string and hangs in equilibrium below the pole. The rings \(A\) and \(B\) are in limiting equilibrium on the pole, with \(\angle BAC = \angle ABC = \theta\), where \(\tan \theta = \frac{3}{4}\), as shown in Fig. 2.
  1. Show that the tension in the string is \(\frac{5}{2}mg\). [3]
  2. Find the value of \(\mu\). [7]

Part (a)
Answer/Working:
- For \(C\): \(27 \sin \theta = 3mg\)
- \(\sin \theta = \frac{3}{5} \Rightarrow T = \frac{5}{3}mg\) (*)
Marks: M1, A1, A1 (3 marks)
Part (b)
Answer/Working:
- For \(A\) or \(B\): \(R \uparrow = 2mg + T \sin \theta = 2mg + \frac{5}{3}mg \cdot \frac{3}{5} = \frac{7}{3}mg\)
- \(R \rightarrow\) for \(A\) or \(B\): \(T \cos \theta = \mu R\)
- Solve to get \(\mu\) as number: \(\frac{5}{3}mg \cdot \frac{4}{5} = \mu \cdot \frac{7}{3}mg \Rightarrow \mu = \frac{4}{7}\) (Accept 0.57 awrt)
Marks: M1, A1, M1, A1, M1, A1 (7 marks)
## Part (a)
**Answer/Working:**
- For $C$: $27 \sin \theta = 3mg$
- $\sin \theta = \frac{3}{5} \Rightarrow T = \frac{5}{3}mg$ (*)

**Marks:** M1, A1, A1 (3 marks)

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## Part (b)
**Answer/Working:**
- For $A$ or $B$: $R \uparrow = 2mg + T \sin \theta = 2mg + \frac{5}{3}mg \cdot \frac{3}{5} = \frac{7}{3}mg$
- $R \rightarrow$ for $A$ or $B$: $T \cos \theta = \mu R$
- Solve to get $\mu$ as number: $\frac{5}{3}mg \cdot \frac{4}{5} = \mu \cdot \frac{7}{3}mg \Rightarrow \mu = \frac{4}{7}$ (Accept 0.57 awrt)

**Marks:** M1, A1, M1, A1, M1, A1 (7 marks)

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\includegraphics{figure_2}

Two small rings, $A$ and $B$, each of mass $2m$, are threaded on a rough horizontal pole. The coefficient of friction between each ring and the pole is $\mu$. The rings are attached to the ends of a light inextensible string. A smooth ring $C$, of mass $3m$, is threaded on the string and hangs in equilibrium below the pole. The rings $A$ and $B$ are in limiting equilibrium on the pole, with $\angle BAC = \angle ABC = \theta$, where $\tan \theta = \frac{3}{4}$, as shown in Fig. 2.

\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string is $\frac{5}{2}mg$. [3]
\item Find the value of $\mu$. [7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2004 Q4 [10]}}