Varignon’s Theorem 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:- 1.let us consider two concurrent forces F1 and F2 represented in magnitude and direction by AB and AC as shown in diagram. 2. let 'O' be the point about which the moments are taken . 3.though 'O' draw a line OD parallel to the direction of force F1 to meet the line of action of the force F2 at C. 4. with AB and AC as two adjacent slides ,complete the parallelogram ABDC as shown in diagram.
5. where AD is the diagonal of the parallelogram. 6. from the parallelogram law of force , the resultant of F1 and F2 is AD as shown in dia. 7. moment of force =2 x area of triangle, Whose base line represented force and whose vertex is the pt. About which the moment is taken. * moment of force F1 about O = 2 x area of AOB moment of force F2 about O= 2 x area of AOC Moment of resultant R about O = 2 x area of AOD => from same base and equal height , Area of AOB = area of ADB ---------------1 Because ABDC is a parallelogram , Area of ADB= area of ACD---------------2
From eq 1 & eq 2 Area of AOB = area of ACD--------3 From the diagram Area of AOD = area of AOC + area of ACD ----------- 4 From eqn no -----3 Area of AOD = Area of AOC + area of AOB Multiply both side by 2 2 x area of AOD = 2 x area of AOC + 2 x area of AOB Moment of force R about O = moment of force F1 about O + moment of force F2 about O
QUE: four coplanar forces 2KN,3KN,5KN& 7KN are acting on a square body of side 1m.determine magnitude ,direction & position of resultant force from point A, which will keep the body in equilibrium