Timeline for Does the weight of a steel ball changes when submerged in water?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 23, 2019 at 4:12 | comment | added | Bob D | @David White Thanks David. I was in product safety engineering for nearly 50 years specializing in electrical safety. | |
| May 23, 2019 at 2:48 | comment | added | David White | @BobD, I'm glad that you saw my comment as an attempt to help. And note - I am also a retired engineer (chemical), so I guess we have something in common. Note 2 - I read your revised answer and upvoted it. Take care. | |
| May 22, 2019 at 21:33 | comment | added | Bob D | @DavidWhite See my revised answer thanks to your help. | |
| May 22, 2019 at 21:33 | comment | added | Bob D | @JMac See my revised answer thanks to your help. | |
| May 22, 2019 at 20:56 | history | edited | Bob D | CC BY-SA 4.0 |
clarification
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| May 22, 2019 at 20:27 | comment | added | Bob D | @DavidWhite See my comment to JMac | |
| May 22, 2019 at 20:26 | comment | added | Bob D | @JMac You have a good point. I didn't pick up on that since I was having a hard time following the OPs question. | |
| May 22, 2019 at 20:25 | comment | added | David White | @JMac, I agree ... the problem statement could have been somewhat clearer. | |
| May 22, 2019 at 20:20 | comment | added | JMac | @DavidWhite To be fair, the question does a poor job IMO making that clear. I only noticed it when re-reading after seeing your comment. It mentions the ball under water vs atmospheric conditions, then uses i.e. and seems to compare water vs atmosphere to water vs vacuum. The only reason it's clear to me that he meant all 3 was because he says "all 3 cases" at the end, but that's easy to miss. | |
| May 22, 2019 at 20:15 | comment | added | David White | Bob D, it's not "critical" for this discussion, but the OP asked about weighing the steel ball in water, in the atmosphere, and in vacuum. With a sufficiently accurate and precise scale, there would be a slight difference in weight between weight in air and weight in vacuum. | |
| May 22, 2019 at 20:11 | history | edited | Bob D | CC BY-SA 4.0 |
clarification
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| May 22, 2019 at 20:10 | comment | added | Bob D | @JMac Yes, I agree. I will edit to clarify that. Thanks for the input. | |
| May 22, 2019 at 20:09 | comment | added | Bob D | @DavidWhite Yes, I agree. It's my understanding that scales making critical measurements are calibrated to account for this. Would you agree, however, that this is not a critical factor in this discussion? | |
| May 22, 2019 at 20:07 | comment | added | JMac | "That displacement, in turn, will depend on the density of the ball." Only if the ball is floating. For the steel ball case in this question, the displacement only cares about the total volume of the ball. It just so happens that the water displaced by the ball's full volume still weighs less than the ball (i.e. steel is more dense than water). | |
| May 22, 2019 at 19:59 | comment | added | David White | There is in fact a very small buoyant force on the ball when it is weighed in air, equal to the weight of the air that was displaced by the ball. If this force didn't exist, helium balloons wouldn't float in air. | |
| May 22, 2019 at 19:56 | history | answered | Bob D | CC BY-SA 4.0 |