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Exercise 5.5 · Q5

Q.Parabolic cable of a 60 m60\,\text m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical Cables are to be spaced every 6 m6\,\text m along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.

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Put the vertex (the cable's lowest, 3 m3\,\text m-above-roadbed point) at the origin; the towers, 30 m30\,\text m either side at 16 m16\,\text m, fix aa; then a vertical cable's total length is 3 m3\,\text m (base) plus the parabola's rise at that point.

Step 1. Set up. Vertex at the origin (cable's lowest point), x2=4ayx^2=4ay (yy measured up from the vertex). Half the 60 m60\,\text m span is 30 m30\,\text m, where the tower height is 16 m16\,\text m; since the vertex itself sits 3 m3\,\text m above the roadbed, the parabola's rise there is 16−3=13 m16-3=13\,\text m.

Step 2. Find aa.

302=4a(13)⇒900=52a⇒a=90052=22513≈17.3130^2=4a(13) \Rightarrow 900=52a \Rightarrow a=\dfrac{900}{52}=\dfrac{225}{13}\approx17.31.

Step 3. Rise at x=6x=6 (first cable). …

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