Edexcel M2 2012 January — Question 5 11 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2012
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod with string perpendicular
DifficultyStandard +0.3 This is a standard M2 moments equilibrium problem requiring resolution of forces and taking moments about a point. Part (a) involves straightforward moment calculation about point A to find tension. Part (b) requires resolving horizontally and vertically, then using F=μR at limiting friction. While it has multiple steps (11 marks total), the techniques are routine for M2 students with no novel problem-solving required—slightly easier than average overall.
Spec3.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_2} A uniform rod \(AB\) has mass \(4\) kg and length \(1.4\) m. The end \(A\) is resting on rough horizontal ground. A light string \(BC\) has one end attached to \(B\) and the other end attached to a fixed point \(C\). The string is perpendicular to the rod and lies in the same vertical plane as the rod. The rod is in equilibrium, inclined at \(20°\) to the ground, as shown in Figure 2.
  1. Find the tension in the string. [4]
Given that the rod is about to slip,
  1. find the coefficient of friction between the rod and the ground. [7]

(a)
AnswerMarks
Taking moments about A:M1
\(4g \times 0.7 \times \cos 20° = 1.4T\)A1 A1
\(T = 18.4\) NA1
Total: 4 marks
(b)
AnswerMarks
\(\uparrow\) \(R + T\cos 20 = 4g\)M1
\(R = 4g - T\cos 20°\)M1 A1
\(\rightarrow\) \(F = T\sin 20\)M1 A1
\(F = \mu R \Rightarrow T\sin 20° = \mu(4g - T\cos 20°)\)DM1 A1
\(\mu = \frac{T\sin 20°}{4g - T\cos 20°} = 0.29\)A1
Total: 7 marks
Question 5 Total: 11 marks
## (a)

| Taking moments about A: | M1 |
| $4g \times 0.7 \times \cos 20° = 1.4T$ | A1 A1 |
| $T = 18.4$ N | A1 |

**Total: 4 marks**

## (b)

| $\uparrow$ $R + T\cos 20 = 4g$ | M1 |
| $R = 4g - T\cos 20°$ | M1 A1 |
| $\rightarrow$ $F = T\sin 20$ | M1 A1 |
| $F = \mu R \Rightarrow T\sin 20° = \mu(4g - T\cos 20°)$ | DM1 A1 |
| $\mu = \frac{T\sin 20°}{4g - T\cos 20°} = 0.29$ | A1 |

**Total: 7 marks**

**Question 5 Total: 11 marks**

---
\includegraphics{figure_2}

A uniform rod $AB$ has mass $4$ kg and length $1.4$ m. The end $A$ is resting on rough horizontal ground. A light string $BC$ has one end attached to $B$ and the other end attached to a fixed point $C$. The string is perpendicular to the rod and lies in the same vertical plane as the rod. The rod is in equilibrium, inclined at $20°$ to the ground, as shown in Figure 2.

\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string.
[4]
\end{enumerate}

Given that the rod is about to slip,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the coefficient of friction between the rod and the ground.
[7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2012 Q5 [11]}}