Varignon’s Theorem states that the moment of a force about any point is equal to the *algebraic sum *of the moments of its components about that point.

Principal of moments states that the moment of the resultant of a number of forces about any point is equal to the *algebraic sum *of the moments of all the forces of the system about the same point.

**Proof of Varignon’s Theorem**

**Fig. 2.34 (a)**

**Fig.2.34 (b)**

Fig2.34 *(a) *shows two forces Fj and *F2 *acting at point O. These forces are represented in magnitude and direction by *OA *and *OB. *Their resultant R is represented in magnitude and direction by OC which is the diagonal of parallelogram OACB. Let O’ is the point in the plane about which moments of F_{1}, F_{2}* *and R* *are to be determined. From point *O’, *draw perpendiculars

on OA, OC and OB.

Let r_{1}= Perpendicular distance between F_{1} and *O’. *

r_{2}= Perpendicular distance between R and *O’. *

r_{3}= Perpendicular distance between F_{2} and *O’.*

Then according to Varignon’s principle;

Moment *of R *about O’ must be equal to algebraic sum of moments of F_{1}* and *F_{2}* *about *O’. *

R × r = F_{1} × r_{1} + F_{2} × r_{2}

Now refer to Fig. 2.34 *(b). *Join *OO’ *and produce it to *D. *From points C, *A *and *B *draw perpendiculars on *OD *meeting at *D, E *and *F *respectively. From *A *and *B *also draw perpendiculars on *CD *meeting the line *CD *at G and *H *respectively.

Let θ_{1} = Angle made by *F; *with *OD, *

θ = Angle made by *R *with *OD, *and

θ_{2} = Angle made by F_{2}* *with *OD. *

In Fig. 2.34 *(b), **OA *= *BC *and also *OA *parallel *to BC, *hence the projection *of OA and BC *on the same vertical line *CD *will be equal *i.e., GD *= *CH *as *GD *is the projection of *OA *on *CD *and *CH *is the projection of *BC *on *CD. *

Then from Fig. 2.34 *(b), *we have

P_{1} sin θ_{1} = AE = GD = CH

F_{1} cos θ_{1} = OE

F_{2} sin θ_{1} = BF = HD

F2 cos θ_{2} = OF = ED

* ** (OB *= *AC *and also *OB *|| *AC. *Hence projections of *OB *and *AC *on the same horizontal line *OD *will be equal *i.e., OF *= *ED) *

*R *sin θ *=CD
R *cos θ

*=OD*

Let the length *OO’ *= *x. *

Then x sin θ_{1} = r, x sin θ = r and x sin θ_{2} = r_{2}

Now moment of *R *about O’

*= **R *× (distance between O’ and *R) *= *R *× *r *

*= **R *× *x *sin θ ( r = *x *sin θ)

=*(R *sin θ) × *x *

*= **CD *× *x * ( *R *sin θ = *CD) *

= (*CH +HD)**×** x *

= (F_{1} sin θ_{1} + F2 sin θ_{2}) × x ( CH = F_{1} sin θ_{1} and HD = F_{2} sin θ_{2})

= F_{1}* *× *x *sin θ_{1} + F_{2}* *× *x *sin θ_{2}

= F_{1} × r_{1 }+ F_{2 }× r_{2} ( *x* sin θ_{1 }= r_{1 }and *x* sin θ_{2 }= r_{2})

= Moment of F_{1}* *about O’ + Moment of F_{2}* *about *O’. *

Hence moment of *R *about any point in the algebraic sum of moments of its components *i.e., *F_{1}* *and F_{2})* *about the same point. Hence Varignon’s principle is proved.

The principle of moments (or Varignon’s principle) is not restricted to only two concurrent forces but is also applicable to any coplanar force system, *i.e., *concurrent or non-concurrent or parallel force system.

**Problem 2.17 ***A force of 100 N is acting at a point A as shown in Fig. *2.35. *Determine the moments of this force about *O.

**Sol. **Given:

Force at *A *= 100 N

Draw a perpendicular from O on the line of action of force 100 N. Hence *OB *is the perpendicular on the line of action of 100 N as shown in Fig. 2.35.

**1st Method **

Triangle *OBC *is a right-angled triangle. And angle

OCB= 60^{o}

Sin60^{o} = 0B/OC

OB = OC sin 60°

= 3 × 0.866 = 2.598 m

Moment of the force 100 N about O

= 100 × OB = 100 × 2.598

= 259.8 N m (clockwise). **Ans.**

**2nd Method **

The moment of force 100 N about O, can also be determined by using Varignon’s principle. The force 100 N is replaced by its two rectangular components at any convenient point. Here the convenient point is chosen as *C. *The horizontal and vertical components of force 100 N acting at C are shown in Fig. 2.36.

*(i) *The horizontal component

= 100 × cos 60° = 50 N

But this force is passing through O and hence has no moment about O.

The vertical component

= 100 × sin 60° =-100 × 0.866 = 86.6 N

This force is acting vertically downwards at *C. *Moment of this force about O

=86.6 × OC = 86.6 × 3

= 259.8 N (clockwise). **Ans.**

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