Question:

A cable of span \(10\,\mathrm{m}\) carries a uniformly distributed load of \(6\,\mathrm{kN/m}\) over the entire span. If the central dip is \(3\,\mathrm{m}\), the horizontal tension is

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For a cable subjected to uniformly distributed load, \[ \boxed{H=\frac{wL^2}{8d}.} \]
Updated On: Jul 24, 2026
  • \(20\,\mathrm{kN}\)
  • \(25\,\mathrm{kN}\)
  • \(30\,\mathrm{kN}\)
  • \(35\,\mathrm{kN}\)
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The Correct Option is B

Solution and Explanation

Step 1: Use the horizontal tension formula for a cable carrying UDL. \[ H=\frac{wL^2}{8d}, \] where \[ w=6\,\mathrm{kN/m},\qquad L=10\,\mathrm{m},\qquad d=3\,\mathrm{m}. \]

Step 2:
Calculate the horizontal tension. \[ H=\frac{6\times10^2}{8\times3} =\frac{600}{24} =25\,\mathrm{kN}. \] Hence, \[ \boxed{25\,\mathrm{kN}} \] is the correct answer.
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