Timeline for Components of normal reaction in banking of roads
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2018 at 15:14 | comment | added | Steeven | @TheMathemagician I am definitely not irritated 🙂 Bring on the questions 💪 Ah, I might have been unclear. The equation itself will naturally not be the exact same on - because that force, which holds the object stationary must be included as well. What I simply meant was that this method of reaching the equation, namely via Newton's 1st law, is correct when at rest. | |
| Jun 6, 2018 at 10:28 | comment | added | The Mathemagician | Sorry if i am irritating you by asking so many questions but when the body is at rest then according to you, both equations 3 and 2 would have been correct that is $R=mgcos\theta$ and $R = mg / cos\theta$ how is this possible??? | |
| Jun 6, 2018 at 10:23 | comment | added | Steeven | @TheMathemagician Yes. Resting means no acceleration, I assume, so Newton's 1st law would be valid. (That second equation is derived via Newton's 1st law; that's the whole point.) | |
| Jun 6, 2018 at 10:03 | comment | added | The Mathemagician | Would the second equation be true if the body is resting on an inclined plane | |
| Jun 6, 2018 at 8:56 | history | edited | Steeven | CC BY-SA 4.0 |
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| Jun 6, 2018 at 8:51 | history | edited | Steeven | CC BY-SA 4.0 |
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| Jun 6, 2018 at 8:47 | comment | added | Steeven | @TheMathemagician I just updated the answer to include this question from the comments. I hope it helps and clears out the confusion. | |
| Jun 6, 2018 at 8:45 | history | edited | Steeven | CC BY-SA 4.0 |
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| Jun 6, 2018 at 8:42 | vote | accept | The Mathemagician | ||
| Jun 6, 2018 at 8:39 | comment | added | Steeven | @TheMathemagician The acceleration is horizontally sideways, right? Therefore, a horizontal direction is parallel to and must include the acceleration, while a vertical direction is perpendicular to and does not include any acceleration. If this is not clear, let me know. Now, any "tilted" non-vertical direction contains some acceleration component. Not the full acceleration as if parallel but also not zero as if perpendicular. But a "bit" of it, so to say. A component of it. The direction perpendicular to the slope is exactly such a tilted non-vertical direction. | |
| Jun 6, 2018 at 7:44 | comment | added | The Mathemagician | So should it be $N=mgcos\theta + ma$ where a is the component of acceleration in that direction. | |
| Jun 6, 2018 at 7:20 | comment | added | The Mathemagician | Can you please elaborate what you are trying to say | |
| Jun 6, 2018 at 7:04 | history | answered | Steeven | CC BY-SA 4.0 |