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Question

A cable PQ of length 25 m is supported at two ends at same level as shown in the figure. The horizontal distance between the supports is 20 m. A point load of 150 kN is applied at point R, which divides it into two equal parts.





Neglecting the self-weight of the cable, the tension (in kN, in integer value) in the cable due to applied load will be____
  1. 125

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Solution

The correct option is A 125

OR=(12.5)2(10)2=7.5 m

V1=V2=75 kN

MR=0 (of left hand side)

H1×OR=V1×10

H1=75×107.5=100 kN

Tension in the cable

T=H21+V21

=1002+752

=125 kN

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