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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 being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Solution
Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the roadway.
Now AL = BM = 30 m, OC = 6 m and
LM = 100 m
∴ LC = CM = `1/2` LM = 50 m
Let O be the vertex and axis of the parabola be y-axis. So, the equation of parabola in standard form is x2 = 4ay
Coordinates of point B are (50, 24)
Since point B lies on the parabola, x2 = 4ay
∴ (50)2 = 4a × 24
⇒ a = `2500/(4 xx 24)`
= `625/6`
So, equation of parabola is
x2 = `(4 xx 625)/24 y` = x2 = `625/6 y`
Let length of the supporting wire PQ at a distance of 18 m be h.
∴ OR = 18 m and PR = PQ - QR = h - 6.
Coordinates of point P are (18, h - 6)
Since the point P lies on parabola x2 = `625/6 y`
∴ (18)2 = `625/6 (h - 6)`
= 324 × 6 = 625h - 3750
= 625h = 1944 + 3750
= h = `5694/625`
= 9.11 m approx.
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