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Suppose a string of uniformly distributed mass $M$ is hanging from a ceiling. Now I was asked to calculate tension at middle of string and I answered it correctly as $0.5Mg$. Now suppose if I were asked to calculate tension at lowermost point on string. I have learnt that tension is common magnitude of forces with which two parts of same string on opposite sides of a cross-section of string pull each other at that cross section, At lowermost part i.e. the lowermost cross section has string on upper part only , so how do we tension at that point.

I think it should be undefined but my teacher told me it will be zero. Please explain

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To hold the very last atom at the bottom of the string requires an upward force (tension) equal to the weight of that atom which is effectively zero (when compared with the tensions in other much higher parts of the string).

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  • $\begingroup$ you want to convey that to hold the last cross-section the string needs to pull it upwards with a force equal to it's weight which is negligibly zero $\endgroup$ Jun 30, 2021 at 7:12
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I am going to assume that the string is attached to the ceiling at only one end, and that it hangs vertically.

The tension in the string at any point must be sufficient to support the weight of the string below that point. So the tension varies linearly along the length of the string, from $Mg$ at the top end of the string, through $0.5Mg$ in the middle, to $0$ at the bottom end.

A tension of zero seems odd at first, but you can think of the tension at the bottom end as being the limit of the tension at a point $P$ as $P$ approaches the bottom end of the string. This limit is zero.

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