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Two persons are holding a light rope tightly at its ends so that it is horizontal .A 15 kg weight is attached to the rope at the mid point which now no longer remains horizontal .The minimum tension required to completely straighten the rope is -
To keep it horizontal tensional force must be equal to the applied force 15kg weight= 15g=15×10 N=150 N.(Taking g=10m/s2 approx)
All right.So when it is not horizontal,let it make an angle A with the horizontal.At this point Horizontal component of T or TcosA will cancel out.Vertical component2TsinA=mg=15gBut for the wire to be horizontal A must be zero.SinA=0.T=15g/2sinA=15g/0=infinity.Hope this helps.
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