Timeline for Concurrency of three coplanar forces for a body in equilibrium
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 7, 2017 at 19:32 | history | edited | Kira | CC BY-SA 3.0 |
Corrected answer
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| May 7, 2017 at 19:19 | comment | added | Kira | @sammy gerbil thanks for pointing out.I apologize for my ignorance and forgetting the lami's theorem which clearly explains the ways any three forces can be in equilibrium.Hence I support your 2nd and 3rd argument.Now as far as your 1st argument goes you seem to have misinterpreted what I stated in my answer.When the user asked why the forces had to pass through a common point I had to explain it to him the meaning of vector translation but I have clearly stated how this is not related to his question and in this case the forces have to pass through a common point, I'd suggest you reread it. | |
| May 7, 2017 at 18:23 | comment | added | sammy gerbil | -1. Wrong. (1) You cannot arbitrarily change the line of action of one of the 3 forces. If you do, so that the forces no longer pass through a common point, you will create a torque and the object will rotate. (2) No, the forces in equilibrium do not have to be equal in magnitude and do not have to make equal angles with each other. (3) Your comment to alephzero contradicts what you said in your penultimate sentence. Yes 3 forces of equal magnitude can be in equilibrium. The resultant of any 2 is equal in magnitude and opposite in direction to the 3rd force. | |
| May 7, 2017 at 3:36 | comment | added | alephzero | @Kira You seem to be assuming the forces all have equal magnitude. In general, that is not true. | |
| May 6, 2017 at 15:40 | comment | added | Elliott Macneil | This might sound like a silly question, but why do the forces have to be at angles of 120 degrees to one another in the case of n = 3? (I will drop the generalisation part, because it adds little to the question) | |
| May 6, 2017 at 15:32 | history | answered | Kira | CC BY-SA 3.0 |