Edexcel M2 2010 June — Question 3

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2010
SessionJune
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 question requiring systematic application of learned techniques: finding the centre of mass of a uniform wire frame using symmetry and the formula for composite bodies (part a), then applying equilibrium conditions for a suspended body (part b). While it involves multiple steps and careful coordinate work, it follows predictable patterns taught in M2 with no novel problem-solving required.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_1} A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle \(ABC\), where \(AB = AC = 10\) cm and \(BC = 12\) cm, as shown in Figure 1.
  1. Find the distance of the centre of mass of the frame from \(BC\). (5)
The frame has total mass \(M\). A particle of mass \(M\) is attached to the frame at the mid-point of \(BC\). The frame is then freely suspended from \(B\) and hangs in equilibrium.
  1. Find the size of the angle between \(BC\) and the vertical. (4)

\includegraphics{figure_1}

A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle $ABC$, where $AB = AC = 10$ cm and $BC = 12$ cm, as shown in Figure 1.

\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the frame from $BC$.
(5)
\end{enumerate}

The frame has total mass $M$. A particle of mass $M$ is attached to the frame at the mid-point of $BC$. The frame is then freely suspended from $B$ and hangs in equilibrium.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the size of the angle between $BC$ and the vertical.
(4)
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2010 Q3}}