Three concurrent co-planar forces 1 ,N,2 ,N,a | vedclass

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(C) For a body to be in equilibrium under the action of three concurrent forces,the vector sum of the forces must be zero. This implies that the magni

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cannot keep the body in equilibrium

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(C) For a body to be in equilibrium under the action of three concurrent forces,the vector sum of the forces must be zero. This implies that the magnitude of any one force must be less than or equal to the sum of the other two forces and greater than or equal to the difference of the other two forces.Let the forces be $F_1 = 1 \,N$,$F_2 = 2 \,N$,and $F_3 = 3 \,N$.The condition for equilibrium is that the resultant of any two forces must be equal and opposite to the third force.Here,$F_1 + F_2 = 1 + 2 = 3 \,N$,which is equal to $F_3$.However,for the resultant of $F_1$ and $F_2$ to be $3 \,N$,they must act in the same direction (angle $	heta = 0^{\circ}$).If they act in the same direction,they are not acting along 'different directions' as specified in the problem statement.If the forces act along different directions,the resultant of $1 \,N$ and $2 \,N$ will always be less than $3 \,N$.Therefore,the three forces cannot form a closed triangle,and the body cannot be in equilibrium.

Three concurrent co-planar forces $1 \,N$,$2 \,N$,and $3 \,N$ acting along different directions on a body:

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