Question

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

Answer 1

Let X'OX be the bridge and PAQ be the suspension cable.
The suspension cable forms a parabola with the vertex at (0, 6).

Let the equation of the parabola formed by the suspension cable be ${\left(x-0\right)}^2=4a\left(y-6\right)$. (1)

It passes through P (−50, 30) and Q (50, 30).

∴ $2500=4a\left(30-6\right)$
⇒ $4a=\frac{2500}{24}$

Putting the value of 4a in equation (1):

$x^2=\frac{2500}{24}\left(y-6\right)$ (2)

Let LM be the supporting wire attached at M, which is 18 m from the mid-point (O) of the bridge.

Let the coordinates of L be (18, l).
It lies on the parabola (2).

∴ ${18}^2=\frac{2500}{24}\left(l-6\right)$
$\Rightarrow l=9.11\mathrm{m}$

Hence, the length of the supporting wire attached to the roadway 18 m from the middle is 9.11 m.