A weight Mg is suspended from the middle of a rope whose ends are at the same level. The rope is no longer horizontal. The minimum tension required to completely straighten the rope is ????(1) Mg/2 ......(2) Mgcos (theta).......(3) 2Mgcos ( theta) ....(4) Infinitely large......

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Draw horizontal and vertical components of Tension.

You will see that Vertical components T sin(Q) + T sin(Q) will balance the weight Mg

Mg = 2T sin(Q)

T = Mg / 2 sin(Q)


In order to straighten the rope, theta must be 0

T = Mg / 2 sin(0) = Mg/0 = infinite


Comon! hang a weight in the middle of a thread and try to straighten it, you won't be able to straighten it. It will always bend a little at the mid point.

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OMG !!! nicely explained !!! tOtAlly impressed !!! sir- greAT :)
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byeee choco..

and i want to say that.. your friends are really missing u..


  • -18

wow! that was a really nice explanation sir! thank u soo much!

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Thank u so much!!!
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Sir plz give a diagram of it
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T=mg/2cosΦ T=mg/2cos90° T=mg/0 T=infinity
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Thanku sir
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Avabe hat be gandu
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