Year: 1991
Paper: 1
Question Number: 10
Course: LFM Pure and Mechanics
Section: Newton's laws and connected particles
No solution available for this problem.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
$\ $\vspace{-1cm}
\noindent \begin{center}
\psset{xunit=1.1cm,yunit=0.7cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-0.32,-0.43)(11.1,8.55) \pspolygon[linewidth=0pt,linecolor=white,hatchcolor=black,fillstyle=hlines,hatchangle=45.0,hatchsep=0.11](0,0)(0,-0.2)(10,-0.2)(10,0) \psline(0,0)(10,0) \psline(0,8)(1,5.5) \psline(1,5.5)(2,4) \psline(3,3)(4,2.5) \psline(5,2.3)(6,2.5) \psline(2,4)(3,3) \psline[linestyle=dashed,dash=1pt 1pt](4,2.5)(5,2.3) \parametricplot[linestyle=dashed,dash=1pt 1pt]{5.080052976030177}{5.934024592002003}{1*5.25*cos(t)+0*5.25*sin(t)+4.11|0*5.25*cos(t)+1*5.25*sin(t)+7.4} \psline(9,5.5)(10,8) \psline(1,5.5)(1,0) \psline(2,4)(2,0) \psline(3,3)(3,0) \psline(4,2.5)(4,0) \psline(5,2.3)(5,0) \psline(6,2.51)(6,0) \psline(9,5.5)(9,0) \rput[tl](0.02,8.53){$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\end{center}
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.