Varignon’s Principle of Moments (Law of Moments)

Varignon’s theorem states that “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

Case (i) When the forces are concurrent

Let ‘P’ and ‘Q’ be any two forces acting at a point O along lines OX and OY respectively and let D be any point in their plane as shown in Fig. 1. Line DC is drawn parallel to OX to meet OY at B. The line OB represent the force Q in magnitude and direction and OA represent the force P in magnitude and direction.

Fig. 1

With OA and OB as the adjacent sides, parallelogram OACB is completed and OC is joined. Let ‘R’ be the resultant of forces P and Q. 

Then, according to the “Theorem of parallelogram of forces”, R is represented in magnitude and direction by the diagonal OC of the parallelogram OACB.

The point D is joined with points O and A. The moments of P, Q and R about D are given by 2 x area of ΔAOD, 2 x area of ΔOBD and 2 x area of ΔOCD respectively.

With reference to Fig1. (a), the point D is outside the <AOB and the moments of P, Q and R about D are all anti-clockwise and hence these moments are treated as positive.

Now, the algebraic sum of the moments of P and Q about

      D = 2ΔAOD + 2ΔOBD

          = 2 (ΔAOD + ΔOBD)

          = 2 (ΔAOC + ΔOBD)

          = 2 (ΔOBC + ΔOBD)

          = 2ΔOCD

          = Moment of R about D

[As AOC and AOD are on the same base and have the same altitude. ΔAOD = ΔOBC. As AOC and OBC have equal bases and equal altitudes. ΔAOC = ΔOBC]

With reference to Fig 1. (b), the point D is within the <AOB and the moments of P, Q and R about D are respectively anti-clockwise, clockwise and anti-clockwise.

Now, the algebraic sum of the forces P and Q about

      D = 2ΔAOD - 2 ΔOBD

          = 2 (ΔAOD-ΔOBD)

          = 2 (ΔAOC- ΔOBD)

          = 2(ΔOBC - ΔOBD)

          = 2ΔOCD

          = Moment of R about D

Case (ii) When the forces are parallel

Let P and Q be any two like parallel forces (i.e. the parallel forces whose lines of action are parallel and which act in the same sense) and O be any point in their plane.

Let R be the resultant of P and Q.

Then,

       R = P + Q

From O, line OACB is drawn perpendicular to the lines of action of forces P, Q and R intersecting them at A, B and C respectively as shown in Fig 2.

Fig 2

Now, algebraic sum of the moments of P and Q about O

     = P x OA + Q x OB

     = P x (OC - AC) + Q x (OC + BC)

     = P x OC – P x AC + Q x OC + Q x BC

  But P x AC = Q x BC

Algebraic sum of the moments of P and Q about O

     = P x OC + Q x OC

     = (P + Q) x OC

     = R x OC

     = Moment of R about O

In case of unlike parallel forces also it can be proved that the algebraic sum of the moments of two unlike parallel forces (i.e. the forces whose lines of action are parallel but which act in reverse senses) about any point in their plane is equal to the moment of their resultant about the same point.