Timeline for Why isn't work $Fd \sec \theta$? [closed]
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2023 at 10:45 | history | edited | zaper newlson | CC BY-SA 4.0 |
added 88 characters in body
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| Nov 27, 2023 at 10:43 | vote | accept | zaper newlson | ||
| Nov 27, 2023 at 10:43 | |||||
| Nov 27, 2023 at 5:19 | comment | added | naturallyInconsistent | When resolving one force into two other forces, the convenient way to do that is to choose for the two other forces to be perpendicular to each other. If you make the right angle be at P, then the two forces will not be perpendicular, then the leg PN will also have some parallel component w.r.t. the displacement, and thus would also have to be included in the computation. The result will be the same, with much more difficult computation. | |
| Nov 27, 2023 at 1:07 | comment | added | Ghoster | I understand why it is cosθ but why cant it be secθ also? How could the amount of work be two different things? | |
| Nov 26, 2023 at 19:09 | history | closed |
Voulkos Miyase Jon Custer |
Not suitable for this site | |
| Nov 26, 2023 at 18:41 | review | Close votes | |||
| Nov 26, 2023 at 19:09 | |||||
| Nov 26, 2023 at 18:31 | history | edited | gandalf61 | CC BY-SA 4.0 |
edited title
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| S Nov 26, 2023 at 18:30 | history | suggested | Aurelius | CC BY-SA 4.0 |
Corrected what the asker asked. He meant work not force
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| Nov 26, 2023 at 15:34 | review | Suggested edits | |||
| S Nov 26, 2023 at 18:30 | |||||
| Nov 26, 2023 at 15:34 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited tags; edited title
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| Nov 26, 2023 at 15:32 | answer | added | Aurelius | timeline score: 1 | |
| S Nov 26, 2023 at 15:16 | review | First questions | |||
| Nov 26, 2023 at 15:29 | |||||
| S Nov 26, 2023 at 15:16 | history | asked | zaper newlson | CC BY-SA 4.0 |