OCR Further Mechanics 2018 December — Question 7

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2018
SessionDecember
TopicCentre of Mass 1

7 Particles \(A , B\) and \(C\) of masses \(2 \mathrm {~kg} , 3 \mathrm {~kg}\) and 5 kg respectively are joined by light rigid rods to form a triangular frame. The frame is placed at rest on a horizontal plane with \(A\) at the point ( 0,0 ), \(B\) at the point ( \(0.6,0\) ) and \(C\) at the point ( \(0.4,0.2\) ), where distances in the coordinate system are measured in metres (see Fig. 1). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-5_304_666_434_251} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} \(G\), which is the centre of mass of the frame, is at the point \(( \bar { x } , \bar { y } )\).
  1. - Show that \(\bar { x } = 0.38\).
    • Find \(\bar { y }\).
    • Explain why it would be impossible for the frame to be in equilibrium in a horizontal plane supported at only one point.
    A rough plane, \(\Pi\), is inclined at an angle \(\theta\) to the horizontal where \(\sin \theta = \frac { 3 } { 5 }\). The frame is placed on \(\Pi\) with \(A B\) vertical and \(B\) in contact with \(\Pi\). \(C\) is in the same vertical plane as \(A B\) and a line of greatest slope of \(\Pi . C\) is on the down-slope side of \(A B\). The frame is kept in equilibrium by a horizontal light elastic string whose natural length is \(l \mathrm {~m}\) and whose modulus of elasticity is \(g \mathrm {~N}\). The string is attached to \(A\) at one end and to a fixed point on \(\Pi\) at the other end (see Fig. 2). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-5_611_842_1649_248} \captionsetup{labelformat=empty} \caption{Fig. 2}
    \end{figure} The coefficient of friction between \(B\) and \(\Pi\) is \(\mu\).
  2. Show that \(l = 0.3\).
  3. Show that \(\mu \geqslant \frac { 14 } { 27 }\). \section*{OCR} Oxford Cambridge and RSA