- \hspace{0pt} [The centre of mass of a semicircular arc of radius \(r\) is \(\frac { 2 r } { \pi }\) from the centre.]
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0762451f-b951-4d66-9e01-61ecb7b30d95-20_668_371_358_790}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Uniform wire is used to form the framework shown in Figure 2.
In the framework,
- \(A B C\) is straight and has length \(25 a\)
- \(A D E\) is straight and has length \(24 a\)
- \(A B D\) is a semicircular arc of radius \(7 a\)
- \(E C = 7 a\)
- angle \(A E C = 90 ^ { \circ }\)
- the points \(A , B , C , D\) and \(E\) all lie in the same plane
The distance of the centre of mass of the framework from \(A E\) is \(d\).
- Show that \(d = \frac { 53 } { 2 ( 7 + \pi ) } a\)
The framework is freely suspended from \(A\) and hangs in equilibrium with \(A C\) at angle \(\alpha ^ { \circ }\) to the downward vertical.
- Find the value of \(\alpha\).