Timeline for Tension of a massless string
Current License: CC BY-SA 4.0
7 events
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| Mar 21, 2021 at 16:20 | comment | added | MSKB | So basically we can summarize this fact as that the force exerted by each points of the string on the ring is proportional to the force exerted on the point of the string by the top most region of the string.since two adjacent points aren't moving apart we can conclude that the force exerted on those two points are the same and that is why we consider that the tension is equal all over a massless string. Is this assumption correct? | |
| Mar 21, 2021 at 15:37 | comment | added | Claudio Saspinski | @MohammadSakibShahriar I edited the answer about this issue. | |
| Mar 21, 2021 at 15:36 | history | edited | Claudio Saspinski | CC BY-SA 4.0 |
answering a comment
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| Mar 21, 2021 at 14:29 | comment | added | MSKB | I agree this for the ring being suspended exactly at the midpoint but if it isn't at the middle why would the tensions be same at the both parts? | |
| Mar 21, 2021 at 12:48 | comment | added | Claudio Saspinski | @MohammadSakibShahriar Think of a small portion of the string. The force from the left must be equal to the force to the right. Otherwise the string would be accelerating and it is at rest. As it happens to all portions, the tension must be the same. | |
| Mar 21, 2021 at 5:31 | comment | added | MSKB | Why are the tensions of both the part of the string same? Shouldn't those be different since the ring is bent at the point where the disk/ring is hanging/resting? | |
| Mar 20, 2021 at 20:06 | history | answered | Claudio Saspinski | CC BY-SA 4.0 |