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.. .

Make a skethch of the forces acting on the mass, which consist of its weight pulling down and the tension in the rope to the left and right at a slight upward angle. Now consider that for the mass to be held steady the vertical component of those three forces = 0. What do you conclude about the magnitude of the tension force if its direction is purely horizontal?

I got the value of 'T' that is "mg/2cos@ or mg/2sin@ ". But am not able to get the next step.please do that for me. Thank you.

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Originally Posted by Sagarika Rout

I got the value of 'T' that is "mg/2cos@ or mg/2sin@ ".

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Well, which is it? If you're confused try this: imagine that the angle is 90 degrees, meaning the weight hangs straight down and the tension in the ropes is mg/2 - which of these formulas would give that answer? Once you have figured out what the correct formula is, calculate T for an angle = 0 degrees.

You can probably guess what the answer is without doing any math at all. Ever tried to get a rope perfectly taut? How hard do you have to pull? Is it a lot or a little?