WJEC Further Unit 6 2022 June — Question 3 14 marks

Exam BoardWJEC
ModuleFurther Unit 6 (Further Unit 6)
Year2022
SessionJune
Marks14
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeParticle attached to lamina - find mass/position
DifficultyChallenging +1.2 This is a standard Further Maths centre of mass problem requiring systematic application of composite body techniques. While it involves multiple shapes (rectangle, triangle, semicircular hole) and two parts (finding CoM, then equilibrium with added mass), the methods are routine for FM students. The calculations are somewhat lengthy but straightforward, with part (b) answer given to verify working. The equilibrium analysis in part (c) follows standard procedures without requiring novel geometric insight.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

3. The diagram below shows a lamina \(A B C D E\) which is made of a uniform material. It consists of a rectangle \(A B D E\) with \(A B = 6 a\) and \(A E = 8 a\), together with an isosceles triangle \(B C D\) with \(B C = D C = 5 a\). A semicircle, with its centre at the midpoint of \(A E\) and radius \(3 a\), is removed from \(A B D E\). \includegraphics[max width=\textwidth, alt={}, center]{b9c63cb4-d446-4548-be42-e30b10cb4b99-3_606_703_603_680}
  1. Write down the distance of the centre of mass of the lamina \(A B C D E\) from \(A B\).
  2. Show that the distance of the centre of mass of the lamina \(A B C D E\) from \(A E\) is \(\frac { 140 } { 40 - 3 \pi } a\).
  3. The lamina \(A B C D E\) is freely suspended from the point \(D\) and hangs in equilibrium.
    1. Calculate the angle that \(B D\) makes with the vertical.
    2. The mass of the lamina is \(M\). When a particle of mass \(k M\) is attached at the point \(C\), the lamina hangs in equilibrium with \(A B\) horizontal. Determine the value of \(k\).

3. The diagram below shows a lamina $A B C D E$ which is made of a uniform material. It consists of a rectangle $A B D E$ with $A B = 6 a$ and $A E = 8 a$, together with an isosceles triangle $B C D$ with $B C = D C = 5 a$. A semicircle, with its centre at the midpoint of $A E$ and radius $3 a$, is removed from $A B D E$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b9c63cb4-d446-4548-be42-e30b10cb4b99-3_606_703_603_680}
\begin{enumerate}[label=(\alph*)]
\item Write down the distance of the centre of mass of the lamina $A B C D E$ from $A B$.
\item Show that the distance of the centre of mass of the lamina $A B C D E$ from $A E$ is $\frac { 140 } { 40 - 3 \pi } a$.
\item The lamina $A B C D E$ is freely suspended from the point $D$ and hangs in equilibrium.
\begin{enumerate}[label=(\roman*)]
\item Calculate the angle that $B D$ makes with the vertical.
\item The mass of the lamina is $M$. When a particle of mass $k M$ is attached at the point $C$, the lamina hangs in equilibrium with $A B$ horizontal. Determine the value of $k$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 6 2022 Q3 [14]}}