Concurrency of three coplanar forces for a body in equilibrium 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.
 A: 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 $\frac{360^{\circ}}{n}$. 
A: 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
A: 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.
A: 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.
