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Question

A cable is supported between two points 30 m horizontally apart. The left support is 3 m above the right support . The cable carries a load of 2 kN/m on the horizontal span . The lowest point of the cable is 6 m below the left support. The maximum tension in cable (in kN)

A
86.6 kN
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B
51.4 kN
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C
35.2 kN
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D
62.3 kN
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Solution

The correct option is D 62.3 kN

Let X = horizontal distance between A and C.
Consider equilibrium of part AC of the cable

Resolving vertically, Va=2x

Taking moment about A, H×6=2x22

H=x26

Now consider equilibrium of whole cable :

Taking moment about B, Va×30=H×3+2×3022

30Va=3H+900

10Va=H+300...(1)

Substituting Va and H,

10(2x)=x26+300

x2120x+1800=0

x=17.5736 m

Va=2x=2×17.5736=35.1472 kN

Also, H=x26=17.573626=51.4719 kN

Hence, maximum tension,

Tmax=35.14722+51.47192

Tmzx=62.327 kN

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