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1991 Paper 1 Q10
D: 1500.0 B: 1484.0

\(\ \)\vspace{-1cm} \noindent

\psset{xunit=1.1cm,yunit=0.7cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-0.32,-0.43)(11.1,8.55) \pspolygon[linewidth=0pt,linecolor=white,hatchcolor=black,fillstyle=hlines,hatchangle=45.0,hatchsep=0.11](0,0)(0,-0.2)(10,-0.2)(10,0) \psline(0,0)(10,0) \psline(0,8)(1,5.5) \psline(1,5.5)(2,4) \psline(3,3)(4,2.5) \psline(5,2.3)(6,2.5) \psline(2,4)(3,3) \psline[linestyle=dashed,dash=1pt 1pt](4,2.5)(5,2.3) \parametricplot[linestyle=dashed,dash=1pt 1pt]{5.080052976030177}{5.934024592002003}{1*5.25*cos(t)+0*5.25*sin(t)+4.11|0*5.25*cos(t)+1*5.25*sin(t)+7.4} \psline(9,5.5)(10,8) \psline(1,5.5)(1,0) \psline(2,4)(2,0) \psline(3,3)(3,0) \psline(4,2.5)(4,0) \psline(5,2.3)(5,0) \psline(6,2.51)(6,0) \psline(9,5.5)(9,0) \rput[tl](0.02,8.53){\(A_0\)} \rput[tl](1.03,6.06){\(A_1\)} \rput[tl](2.01,4.52){\(A_2\)} \rput[tl](2.98,3.56){\(A_3\)} \rput[tl](3.98,3.12){\(A_n\)} \rput[tl](4.8,2.9){\(A_{n+1}\)} \rput[tl](5.75,3.2){\(A_{n+2}\)} \rput[tl](8.1,6.1){\(A_{2n+1}\)} \rput[tl](10.13,8.19){\(A_{2n+2}\)} \end{pspicture*} \par
The above diagram represents a suspension bridge. A heavy uniform horizontal roadway is attached by vertical struts to a light flexible chain at points \(A_{1}=(x_{1},y_{1}),\) \(A_{2}=(x_{2},y_{2}),\ldots,\) \(A_{2n+1}=(x_{2n+1},y_{2n+1}),\) where the coordinates are referred to horizontal and vertically upward axes \(Ox,Oy\). The chain is fixed to external supports at points \[ A_{0}=(x_{0},y_{0})\quad\mbox{ and }\quad A_{2n+2}=(x_{2n+2},y_{2n+2}) \] at the same height. The weight of the chain and struts may be neglected. Each strut carries the same weight \(w\). The horizontal spacing \(h\) between \(A_{i}\) and \(A_{i+1}\) (for \(0\leqslant i\leqslant2n+1\)) is constant. Write down equations satisfied by the tensions \(T_{i}\) in the portion \(A_{i-1}A_{i}\) of the chain for \(1\leqslant i\leqslant n+1\). Hence or otherwise show that \[ \frac{h}{y_{n}-y_{n+1}}=\frac{3h}{y_{n-1}-y_{n}}=\cdots=\frac{(2n+1)y}{y_{0}-y_{1}}. \] Verify that the points \(A_{0},A_{1},\ldots,A_{2n+1},A_{2n+2}\) lie on a parabola.