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**Principle of Moments or Varignonâ€™s Theorem**

Varignonâ€™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 Varignonâ€™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 Oâ€™ is the point in the plane about which moments of F_{1}, F_{2}and Rare to be determined. From point *Oâ€™, *draw perpendiculars on OA, OC and OB.

Let r_{1}= Perpendicular distance between F_{1} and *Oâ€™. *

r_{2}= Perpendicular distance between R and *Oâ€™. *

r_{3}= Perpendicular distance between F_{2} and *Oâ€™.*

Then according to Varignonâ€™s principle;

Moment *of R *about Oâ€™ must be equal to algebraic sum of moments of F_{1}* and *F_{2}about *Oâ€™. *

R * r = F_{1} * r_{1} + F_{2} * r_{2}

Now refer to Fig. 2.34 (b). Join OOâ€™ 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 Î¸_{1} = Angle made by F; with OD,

Î¸ = Angle made by R with OD, and

Î¸_{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 Î¸_{1} = AE = GD = CH

F_{1} cos Î¸_{1} = OE

F_{2} sin Î¸_{1} = BF = HD

F2 cos Î¸_{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 Î¸ *=CD** R *cos Î¸

Let the length *OOâ€™ *= *x. *

Then x sin Î¸_{1} = r, x sin Î¸ = r and x sin Î¸_{2} = r_{2}

Now moment of *R *about Oâ€™

*= R ** (distance between Oâ€™ and *R) *= *R ** *r *

*= R ** *x *sin Î¸ (r = *x *sin Î¸)

=*(R *sin Î¸) * *x *

*= CD ** *x * (*R*sin Î¸ = *CD) *

= (*CH +HD)* x *

= (F_{1} sin Î¸_{1} + F2 sin Î¸_{2}) * x ( CH = F_{1} sin Î¸_{1} and HD = F_{2} sin Î¸_{2})

= F_{1}* *x *sin Î¸_{1} + F_{2}* *x *sin Î¸_{2}

= F_{1} *r_{1 }+ F_{2 }*r_{2} ( *x* sin Î¸_{1 }= r_{1 }and *x* sin Î¸_{2 }= r_{2})

= Moment of F_{1}about Oâ€™ + Moment of F_{2}about *Oâ€™. *

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 Varignonâ€™s principle is proved.

The principle of moments (or Varignonâ€™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.

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