Question
The horizontal distance from A at one end of the river to frame C at the other end is 20 m. The cable carries a load W=50 kN.
a.) At what distance from A is the load W such that the tension in segment AD of the cable is equal to that in segment CD?
b.) When the load W is at distance x1= 5m from A, the sag in the cable is 1m. Calculate the tension in segment DC of the cable
c.) If the sag in the cable is 1m at a distance x1= 5m, what is the total length of the cable?

FIGURE SIT. CC: 20. m D W-50 KN SIT. CC: The horizontal distance from A at one end of the river to frame C at the other end i
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Answer #1

Solution :- Criven that i- Vc 2om Rc re * HA А. He ac AD XCO Н. W WEBORN 6 Criven, Tension in AD = Tension in CD TAD Teo RAand we know that VA and No will be canal When load. W is at Cender of Span. that is at ild =lom=H= And (6) Whem, x=5m and SMoment is too we know that everywhere in the Cable Hence EMD=0 V* Hq - Henyo 12-5*15 = 187.5km He = H = 12:5*15 He=H= 187.5KN

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