Edexcel M2 — Question 6 12 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeFrame with straight rod/wire components only
DifficultyStandard +0.3 This is a standard M2 centre of mass problem with straightforward geometry and equilibrium. Part (a) uses basic trigonometry/cosine rule, part (b) applies standard centre of mass formulas for composite bodies, and part (c) uses the principle that the centre of mass must be vertically below the pivot. While multi-step, all techniques are routine for M2 students with no novel problem-solving required.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

A uniform wire \(ABCD\) is bent into the shape shown, where the sections \(AB\), \(BC\) and \(CD\) are straight and of length \(3a\), \(10a\) and \(5a\) respectively and \(AD\) is parallel to \(BC\). \includegraphics{figure_6}
  1. Show that the cosine of angle \(BCD\) is \(\frac{3}{5}\). [2 marks]
  2. Find the distances of the centre of mass of the bent wire from (i) \(AB\), (ii) \(BC\). [6 marks]
The wire is hung over a smooth peg at \(B\) and rests in equilibrium.
  1. Find, to the nearest 0.1°, the angle between \(BC\) and the vertical in this position. [4 marks]

AnswerMarks Guidance
(a) \(\sin C = \frac{3}{5}\) so \(\cos C = \frac{4}{5}\)M1 A1 (3, 4, 5 triangle)
(b)(i) \(3a(0) + 10a(5a) + 5a(8g) = 18a\bar{x}\)M1 A1 A1 \(\bar{x} = 5a\)
(b)(ii) \(3a(1-5a) + 10a(0) + 5a(1-5a) = 18a\bar{y}\)M1 A1 A1 \(\bar{y} = \frac{2a}{3}\)
(c) \(\tan \alpha = \frac{2}{3} + 5a = \frac{15}{2}\)M1 A1 M1 A1 \(\alpha = 7.6°\)
Total: 12 marks
**(a)** $\sin C = \frac{3}{5}$ so $\cos C = \frac{4}{5}$ | M1 A1 | (3, 4, 5 triangle)

**(b)(i)** $3a(0) + 10a(5a) + 5a(8g) = 18a\bar{x}$ | M1 A1 A1 | $\bar{x} = 5a$

**(b)(ii)** $3a(1-5a) + 10a(0) + 5a(1-5a) = 18a\bar{y}$ | M1 A1 A1 | $\bar{y} = \frac{2a}{3}$

**(c)** $\tan \alpha = \frac{2}{3} + 5a = \frac{15}{2}$ | M1 A1 M1 A1 | $\alpha = 7.6°$
**Total: 12 marks**

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A uniform wire $ABCD$ is bent into the shape shown, where the sections $AB$, $BC$ and $CD$ are straight and of length $3a$, $10a$ and $5a$ respectively and $AD$ is parallel to $BC$.

\includegraphics{figure_6}

\begin{enumerate}[label=(\alph*)]
\item Show that the cosine of angle $BCD$ is $\frac{3}{5}$. [2 marks]
\item Find the distances of the centre of mass of the bent wire from (i) $AB$, (ii) $BC$. [6 marks]
\end{enumerate}

The wire is hung over a smooth peg at $B$ and rests in equilibrium.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, to the nearest 0.1°, the angle between $BC$ and the vertical in this position. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q6 [12]}}