If a body is in equilibrium under the action of three coplanar forces, those three forces pass through a common point. Why is this true? Could anyone provide a proof, and perhaps whether a generalisation for n forces is also true?
Thanks in advance.
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4 Answers
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The given statement is true for all bodies, it has no relation with the object to be a point mass, As we know for any body to be in equilibrium,
- Net External Force = 0
- Net External Torque about any point= 0
For the moment of given forces be zero all have to pass from same point.
To prove this, Let us consider that the 2 forces pass from point A and the line of action of 3rd Force does not pass through Point A, as the body is in equilibrium the net Torque about point A should be zero but as the third force is not passing through point A, the moment of the third force is not zero thus the body should not be in equilibrium which leads to a contradiction, thus proving our assumption wrong.
This statement can also be generalised for more than 3 forces as well
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Usually in newtonian mechanics, point masses are used. For point masses the three forces have to act on the same point in order to attain equilibrium, as there is no other point where they can act.
However, for a rigid body (i.e, a collection of many point masses) to attain equilibrium two different criteria must be met.
The net force must be $0$.
The net torque must be $0$.
For a point mass :-
For equilibrium, any two forces among the three equal forces must be inclined to each other at $120^{\circ}$. It can be proved by adding them vectorially. First you can find resultant of any 2 force using the formula, $R^2=F_1^2 + F_2^2 + 2F_1F_2 \cos{\theta}$. Then the resultant will cancel out the $3^{rd}$ force.
A generalization for $n$ equal forces can be made too. If there are $n$ forces, then two adjacent forces must be inclined at angle equal to $\frac{360^{\circ}}{n}$.
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$\begingroup$ Thanks for the clarity of the answer there, but I don't see the link between the angle of inclination between each of the forces and the concurrency of those three forces at a common point. $\endgroup$ Commented May 6, 2017 at 15:44
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$\begingroup$ It's an observation. You can prove it. $\endgroup$– MitchellCommented May 6, 2017 at 15:48
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$\begingroup$ That all makes sense now, thanks a lot for your answer! $\endgroup$ Commented May 6, 2017 at 15:49
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Your statement that forces pass through a common point is actually ambiguous as forces are vector quantities and can be moved anywhere in space indefinitely as far as we don't change its magnitude or direction so as long as the forces(vectors) are non parallel any number of forces can be made to pass through a common point,but if we are confined to your case of coplanar forces and that too on one particular body which is in equilibrium then the forces indeed pass through a common point so that the body is in equilibrium and indeed it can be generalized for any number of forces but if they are in equilibrium then we can always use lami's theorem to calculate the unknowns and check the equilibrium.In your case of 3 coplanar forces,the body can be in equilibrium in as many ways as lami's theorem allows. Edit:A big thanks to sammy gerbil for pointing out the error in the solution.I have made changes as per the requirement.
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$\begingroup$ 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) $\endgroup$ Commented May 6, 2017 at 15:40
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$\begingroup$ @Kira You seem to be assuming the forces all have equal magnitude. In general, that is not true. $\endgroup$ Commented May 7, 2017 at 3:36
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1$\begingroup$ -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. $\endgroup$ Commented May 7, 2017 at 18:23
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$\begingroup$ @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. $\endgroup$– KiraCommented May 7, 2017 at 19:19
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1.If on a rigid body three forces act at three points, say A, B, and C then we can't just displace them like regular vectors, especially not for this proof (as we could then simply make them concurrent forcefully; which is a moot point). 2. For a rigid body to be in equilibrium the net external force and the net external torque about any arbitrary point must be zero. Now assuming none of the three forces are concurrent- helps to visualise a triangle with the three forces as extension to each of its sides (which is a correct visualisation since any three forces in equilibrium are always coplanar- from F(a)cross(-F(a))=0 =>F(a)cross(F(b)+F(c))=0). 3. The proof :- for any point within the triangle the perpendicular distance to each line of action is non-zero therefore sum of moments/net torque about any point within this triangle is non- zero. As for point on any of its side if the torque becomes zero net force becomes non-zero and on vertex the there is always a torque about the opposite force. Hence the only case in which the sum of moments is zero when the area of the triangle is zero or it contracts to a point,i.e. Concurrent.
