Question

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $100m$ long is supported by vertical wages attached to the cable, the longest wire being $30m$ and the shortest being $6m$. Find the length of a supporting wire attached to the roadway $18m$ from the middles.

Answer 1

Hint: Here since the roadway is horizontal then cables are supporting the roadway must be below the roadway. So the parabola formed from the cable of a uniformly loaded suspension bridge is symmetrical and facing upwards which means that the x-axis and y-axis both are positive. So the equation of parabola will be-${x^2} = 4ay$. Make a proper diagram by using the given conditions and then we can easily solve for the unknown values.

Complete step by step answer:

Let ACB be the roadway whose length=$100m$ and C be the middle point of the roadway so CB=$50m$
It is supported by vertical wages attached to the cable.
The longest wire=$30m$ and the shortest wire=$6m$
The cable hangs in the form of a parabola which is uniform. This means the parabola is facing upwards and is symmetrical. Its vertex is O which has co-ordinates$\left( {0,0} \right)$.

Then the equation of the parabola will be ${x^2} = 4ay$
Now from the diagram we can see that the co-ordinates of B are $\left( {50,24} \right)$ .
Since point B lies on the parabola hence putting values of coordinates x and y in the equation of parabola we get,
$ \Rightarrow {\left( {50} \right)^2} = 4a\left( {24} \right)$
On solving we get,
$ \Rightarrow 2500 = 4a \times 24$
$ \Rightarrow 4a = \dfrac{{2500}}{{24}}$
$ \Rightarrow 4a = \dfrac{{625}}{6}$
We can find the values of a,
$ \Rightarrow a = \dfrac{{625}}{{24}}$
Now we need to find the length of supporting cable attached to roadway $18m$ from the middle (DF).
From the diagram OE=$18m$.Let DF=d
Also, DE=DF-EF=$d - 6$ m
Then Coordinates of D are $\left( {18,d - 6} \right)$
Now point D also lies in the parabola then it will also satisfy the equation of the parabola.
On putting x=$18$ and y=$d - 6$ , a=$\dfrac{{625}}{{24}}$ we get
$ \Rightarrow {\left( {18} \right)^2} = 4 \times \dfrac{{624}}{{24}}\left( {d - 6} \right)$
On simplifying we get,
$ \Rightarrow 324 = \dfrac{{625}}{6}\left( {d - 6} \right)$
On adjusting we get,
$ \Rightarrow \left( {d - 6} \right) = \dfrac{{324 \times 6}}{{625}}$
On solving we get,
$ \Rightarrow d - 6 = 3.11$
$ \Rightarrow d = 3.11 + 6 = 9.11$
So DF=$9.11m$
Answer-The length of the supporting cable $18m$ from the middle of the roadway is $9.11m$.

Note: The parabola whose co-ordinates of focus are positive $\left( {0,a} \right)$ forms equation-${x^2} = 4ay$. The parabola whose co-ordinates of focus are $\left( {a,0} \right)$ forms equation- ${y^2} = 4ax$. So the student may go wrong if they use the${y^2} = 4ax$to solve the question because then the axis of the parabola will be along the negative x-axis which is not according to the statement given in the question about the structure of the cable. The values obtained using this equation will also be wrong.