Timeline for Why cars don't accelerate according to $F = ma$?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| S Apr 26 at 11:34 | vote | accept | tharindu nilanga | ||
| S Apr 26 at 11:33 | vote | accept | tharindu nilanga | ||
| S Apr 26 at 11:34 | |||||
| Apr 26 at 11:33 | vote | accept | tharindu nilanga | ||
| S Apr 26 at 11:33 | |||||
| Apr 19 at 19:49 | history | reopened |
KDP John Rennie gandalf61 |
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| Apr 19 at 18:33 | review | Reopen votes | |||
| Apr 19 at 19:49 | |||||
| Apr 19 at 18:25 | history | closed |
Dale Miyase hft |
Needs details or clarity | |
| Apr 19 at 17:46 | comment | added | Cort Ammon | Something I find important and is often confusing. If using $F=ma$, $F$ is always the combined force that comes from summing all forces together. Sometimes I write it $\Sigma F=ma$ to remind myself. If I feel like F=ma isn't working, its a strong indication that I'm failing to account for one of the forces that needs to be summed in. | |
| Apr 19 at 17:42 | answer | added | Stevan V. Saban | timeline score: 0 | |
| Apr 19 at 13:51 | answer | added | KDP | timeline score: 0 | |
| Apr 19 at 13:46 | history | edited | KDP | CC BY-SA 4.0 |
Clarified intent of the question
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| Apr 19 at 11:07 | review | Close votes | |||
| Apr 19 at 13:46 | |||||
| Apr 19 at 9:24 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited title
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| Apr 19 at 8:26 | answer | added | John Alexiou | timeline score: 3 | |
| Apr 19 at 8:22 | comment | added | Farcher | Related as there is a difference between the toque and power produced by a car engine. How does torque work with a Lamborghini vs a truck? | |
| Apr 19 at 8:21 | comment | added | Eric Lippert | A question which presupposes a falsehood is hard to answer. Cars, like all other objects, do accelerate proportionally to the net force applied. Why do you think they do not? | |
| Apr 19 at 8:18 | comment | added | John Alexiou | The torque produced by the engine is not the only force that applies to a car. Remember that in $F = m a$ the $F$ stands for the net force (total force). | |
| Apr 19 at 7:47 | comment | added | Dheeraj Gujrathi | Toning this down to highest approximations, When you stop putting your leg on Accelerator(Which you say stable engine), Friction on tires reduces the speed of car, to maintain constant speed despite being the friction working, One must use te accelerator,Also, I still have not considered air resistance | |
| S Apr 19 at 7:38 | history | suggested | Sancol. | CC BY-SA 4.0 |
Improving formulae
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| Apr 19 at 7:31 | answer | added | Thomas Fritsch | timeline score: 8 | |
| Apr 19 at 7:27 | comment | added | Sancol. | This may be your answer: physics.stackexchange.com/q/709480 | |
| Apr 19 at 7:05 | comment | added | nammerkage | What is your question? Your car output is not constant first of all, and the car does not exert a constant force on the floor. Unless we are talking about a car already accelerated up to a speed, and the engine maintain that speed on the road. Are you aware of frictional forces? | |
| Apr 19 at 7:04 | review | Suggested edits | |||
| S Apr 19 at 7:38 | |||||
| S Apr 19 at 7:02 | review | First questions | |||
| Apr 19 at 7:04 | |||||
| S Apr 19 at 7:02 | history | asked | tharindu nilanga | CC BY-SA 4.0 |