Timeline for Tension force in kilograms
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2022 at 18:14 | vote | accept | techno tux | ||
| Mar 6, 2022 at 22:36 | comment | added | techno tux | it does seem to me that it's correct to Divide force tension by 9.8 to find an equivalent mass in kilograms, suppose the acceleration is 0, from: FT=ma+mg, Ft=0+mg, so FT=mg , so to get 'm' you divide by g. | |
| Mar 3, 2022 at 2:51 | comment | added | Eli Yablon | That's not the reason you should use $9.81$, I'm just pointing out that if you have a future predicament always stick with what is more general if you're measuring quantities or comparing units. The reason you should use $9.81$ is because if a rope is rated for 5kg of "force," we are supposed to interpret that as saying it is rated for $5\cdot 9.81$ Newtons of force. If I say I weigh 50kg, all I'm saying is that I experience a normal force of $50 \cdot 9.81$ newtons when I step on a scale. However, people get lazy and just say kilograms because the $9.81 m/s^2$ is implied. | |
| Mar 3, 2022 at 2:48 | history | edited | Eli Yablon | CC BY-SA 4.0 |
deleted 12 characters in body
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| S Mar 3, 2022 at 2:47 | history | suggested | user320397 | CC BY-SA 4.0 |
Modified formatting
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| Mar 3, 2022 at 2:31 | review | Suggested edits | |||
| S Mar 3, 2022 at 2:47 | |||||
| Mar 3, 2022 at 2:15 | comment | added | techno tux | thanks for your answer sir , but are suggesting that i should use 9.81 just because "it is more general & well known"?? the apparent weight will be much different if i chose wrong?? | |
| Mar 3, 2022 at 2:04 | history | answered | Eli Yablon | CC BY-SA 4.0 |