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.

Answer 1

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.