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(
T = Mg / 2 sin(
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.