Question 65 of 70intermediate🔧 ApplyLong Answer4 marks

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100100 m long is supported by vertical wires attached to the cable, the longest wire being 3030 m and the shortest being 66 m. Find the length of a supporting wire attached to the roadway 1818 m from the middle.

Correct Answer

9.11 m

Exercise: Miscellaneous Exercise on Chapter 10 | Q: 3 | (Page 29)
For More Understanding

Explanation

This is a direct application of parabola geometry. The cable forms a parabola with its vertex at the lowest point (middle of the bridge). The shortest supporting wire (66 m) is at the vertex, and the longest wire (3030 m) is at the ends. By finding the equation of the parabola and substituting x=18x = 18 m, we can find the required wire length.

Solution Steps

  1. Step 1: Let the vertex of the parabola be at the middle of the bridge. The shortest wire is 66 m at the vertex, so the vertex is 66 m above the roadway.

  2. Step 2: The roadway is 100100 m long, so the distance from middle to each end is 5050 m. The longest wire (3030 m) is at the ends.

  3. Step 3: At x=50x = 50 m from the middle, the cable height above the vertex level =30−6=24= 30 - 6 = 24 m. So the parabola passes through point (50,24)(50, 24).

  4. Step 4: Using the equation x2=4ayx^2 = 4ay for a parabola with vertex at origin and axis vertical:

    502=4a(24)50^2 = 4a(24) 2500=96a2500 = 96a a=250096 ma = \frac{2500}{96} \text{ m}

  5. Step 5: At x=18x = 18 m from the middle:

    y=x24a=1824×250096=324×9610000=3.11 m (height above vertex)y = \frac{x^2}{4a} = \frac{18^2}{4 \times \frac{2500}{96}} = \frac{324 \times 96}{10000} = 3.11 \text{ m (height above vertex)}

  6. Step 6: Wire length at 1818 m from middle:

    6+3.11=9.11 m6 + 3.11 = \mathbf{9.11 \text{ m}}

    ∴ The length of the supporting wire is 9.11 m.

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