Varignon’s theorem also called Law of Moment. The practical application of varignon’s theorem is to find out the position of the resultant from any point of the body.
It states “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”
Proof: Let us consider, for the sake of simplicity, two concurrent forces P and Q represented in magnitude and direction by AB and AC as shown in fig(a)
Let ‘O’ be the point, about which the moment are taken, through O draw a line OD parallel to the direction of force P, to meet the line of action of the force Q at C. Now with AB and AC as two adjacent sides, complete the Parallelogram ABDC as shown in fig(a). Joint the diagonal AD of the parallelogram and OA and OB. From the parallelogram law of forces, We know that the diagonal AD represents in magnitude and direction, the resultant of two forces P and Q. Now we see that the moment of the force P about O: = 2. Area of the triangle AOB ...(i)
Similarly, moment of the force Q about O: = 2. Area of the triangle AOC ...(ii)
And moment of the resultant force R about O: = 2.Area of the triangle AOD ...(iii)
But from the geometry of the fig.ure, we find that Area of triangle AOD = Area of triangle AOC + Area of triangle ACD But Area of triangle ACD = Area of triangle ABD = Area of triangle AOB (Because two “AOB and ADB are on the same base AB and between the same // lines) Now Area of triangle AOD = Area of triangle AOC + Area of triangle AOB Multiply both side by 2 we get; 2. Area of triangle AOD = 2.Area of triangle AOC + 2. Area of triangle ACD, i.e.
Moment of force R about O = Moment of force P about O + Moment of force Q about O or,
Where R ⋅ d = ∑M
∑M = Sum of the moment of all forces d = Distance between the resultant force and the point where moment of all forces are taken. This principle is extended for any number of forces.
