2 Any non-exact answers to this question should be given correct to four significant figures.\\
A thin, straight beam AB is 2 m long. It has a weight of 600 N and its centre of mass G is 0.8 m from end A. The beam is freely pivoted about a horizontal axis through A.

The beam is held horizontally in equilibrium.\\
Initially this is done by means of a support at B, as shown in Fig.2.1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-3_222_805_644_669}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}

(i) Calculate the force on the beam due to the support at B .

The support at B is now removed and replaced by a wire attached to the beam 0.3 m from A and inclined at $50 ^ { \circ }$ to the beam, as shown in Fig. 2.2. The beam is still horizontal and in equilibrium.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-3_275_803_1226_671}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}

(ii) Calculate the tension in the wire.

The forces acting on the beam at A due to the hinge are a horizontal force $X \mathrm {~N}$ in the direction AB , and a downward vertical force $Y \mathrm {~N}$, which have a resultant of magnitude $R$ at $\alpha$ to the horizontal.\\
(iii) Calculate $X , Y , R$ and $\alpha$.

The dotted lines in Fig. 2.3 are the lines of action of the tension in the wire and the weight of the beam. These lines of action intersect at P .

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-3_460_791_2074_678}
\captionsetup{labelformat=empty}
\caption{Fig. 2.3}
\end{center}
\end{figure}

(iv) State with a reason the size of the angle GAP.

\hfill \mbox{\textit{OCR MEI M2 2011 Q2 [17]}}