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)
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 F1, F2 and R are to be determined. From point O’, draw perpendiculars
on OA, OC and OB.
Let r1= Perpendicular distance between F1 and O’.
r2= Perpendicular distance between R and O’.
r3= Perpendicular distance between F2 and O’.
Then according to Varignon’s principle;
Moment of R about O’ must be equal to algebraic sum of moments of F1 and F2 about O’.
R × r = F1 × r1 + F2 × r2
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 θ1 = Angle made by F; with OD,
θ = Angle made by R with OD, and
θ2 = Angle made by F2 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
P1 sin θ1 = AE = GD = CH
F1 cos θ1 = OE
F2 sin θ1 = BF = HD
F2 cos θ2 = 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 θ1 = r, x sin θ = r and x sin θ2 = r2
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
= (F1 sin θ1 + F2 sin θ2) × x ( CH = F1 sin θ1 and HD = F2 sin θ2)
= F1 × x sin θ1 + F2 × x sin θ2
= F1 × r1 + F2 × r2 ( x sin θ1 = r1 and x sin θ2 = r2)
= Moment of F1 about O’ + Moment of F2 about O’.
Hence moment of R about any point in the algebraic sum of moments of its components i.e., F1 and F2) 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.
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.
Triangle OBC is a right-angled triangle. And angle
Sin60o = 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.
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.