2

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{14403602-94a6-4441-a673-65f9b98180e5-2_501_752_1133_356}
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\caption{Fig. 1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{14403602-94a6-4441-a673-65f9b98180e5-2_519_558_1183_1231}
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\caption{Fig. 2}
\end{center}
\end{figure}

Two uniform rods $A B$ and $B C$, of weights 70 N and 110 N respectively, are freely jointed at $B$. The rods are in equilibrium in a vertical plane with $A$ and $C$ at the same horizontal level and $A C = 2 \mathrm {~m}$. The $\operatorname { rod } A B$ is freely jointed to a fixed point at $A$ and the rod $B C$ is freely jointed to a fixed point at $C$. The horizontal distance between $B$ and $A$ is 4 m and $B$ is 4 m below $A C$; angle $B A C$ is obtuse (see Fig. 1). The force exerted on the $\operatorname { rod } A B$ at $B$, by the $\operatorname { rod } B C$, has horizontal and vertical components as shown in Fig. 2.\\
(i) By taking moments about $A$ for the $\operatorname { rod } A B$ find the value of $X - Y$.\\
(ii) By taking moments about $C$ for the rod $B C$ show that $2 X - 3 Y + 165 = 0$.\\
(iii) Find the magnitude of the force acting between $A B$ and $B C$ at $B$.

\hfill \mbox{\textit{OCR M3 2009 Q2 [8]}}