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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 being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

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The problem can be represented diagrammatically as follows:
Here AB is the longest wire (30m) and (OC) is the shortest at 6m. LN is the supporting wire, 18m from the middle.
We can see that BC = half of AB = 50m.
The shape of the parabola opens upwards, therefore the equation is $x^2 = 4ay$.
We can see that coordinates of A is $(50, 24)$ (i.e, $y$ coordinate is $AB - OC = 30 - 6 = 24$)
Since $A$ is a point on the parabola, $50^2 = 4\times a \times 24^2 \rightarrow a = \large\frac{625}{24}$
Therefore, the equation of the parabola is $x^2 = 4\times\large\frac{625}{24}$$y \rightarrow 6x^2 = 625y$
$x$ coordinate at point L $=18 \rightarrow 6 \times 18^2 = 625 y \rightarrow y $ coordinate at point L $ = \large\frac{6\times 18^2}{625}$$ \approx 3.11$
We need to calculate LN, which is $3.11 + 6 = 9.11m$.
answered Apr 8, 2014 by balaji.thirumalai

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