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I found this image on one of the answers I was going through. In this picture since the man balances himself in the middle of the rope, there's equal tension throughout it. However if he would have been standing at the end there must have been unequal magnitude of tension at both the ends of the rope, if the angles would have been of different magnitude. But since it is an ideal string it must have equal tension throughout so why there's different tension in its different parts?enter image description here

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    $\begingroup$ Tension is only equal throughout if no force is exerted along the direction of the rope. If the man's foot exerts a frictional force on the rope, the tension changes there. $\endgroup$ – Ben51 May 1 '18 at 16:58
  • $\begingroup$ The tension experienced by the rope doesn't have to be the same throughout the length of the rope. It can be different on either side of the man. What's required is that the three red force vectors add up to zero. If the man is at the exact middle of the rope, then symmetry requires that the rope tensions on either side of the man are the same. But if the man is closer to one pole than the other, the requirement that the three forces sum to zero will mean that the magnitudes of the tensions $T_L$ and $T_R$ are different. $\endgroup$ – Samuel Weir May 1 '18 at 17:18
  • $\begingroup$ But we are taught that an ideal string has constant tension throughout its length? $\endgroup$ – Physics freak May 1 '18 at 17:34
  • $\begingroup$ Hang a rope or string AB freely from one end A. Tie a weight W to its midpoint M. The tension in section AM is W. the tension in section MB is zero. $\endgroup$ – sammy gerbil May 1 '18 at 23:44
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since it is an ideal string it must have equal tension throughout

Why is that the case? If we imagine a small segment of string (massless and inextensible), then for it to be stationary, the acceleration must be zero and the net forces must be zero. In the middle of a string, the only forces are the tension from either side, and gravity. Since we've declared the string to be massless, then we ignore gravity and declare the tension on both sides equal.

So for all parts of the string that are not impacted by some other force, this holds. We can even add an ideal pulley, a device that can provide forces normal to the string, but will add no forces along the string. In such a case, tension remains equal throughout.

But the walker here adds forces to a portion of the string, and some of those forces are parallel to the line of force from either end. This modifies the forces felt by that segment, and the tension to keep the net force zero is no longer necessarily equal.

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  • $\begingroup$ That means because of the external force( weight of the man) the tension no longer is constant. I get it thanks! $\endgroup$ – Physics freak May 1 '18 at 17:38
  • $\begingroup$ @RituKhanna - Well, even even if there are no external forces and you have a string with mass then in general the tension will not be constant throughout the string. For example, if you hold a string by one end then the portion of the string nearest to your hand will be under the most tension while the other end of the string which is free and dangling downward will have very little tension in it near its end. I think that "constant tension" must refer to a situation where there are no external forces and gravitational forces on the string are negligible. $\endgroup$ – Samuel Weir May 1 '18 at 18:24

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