CE 201 - Statics

1. CE 201 - Statics Lecture 11

2. MOMENT OF A FORCE-VECTOR FORMULATION In scalar formulation, we have seen that the moment of a force F about point O or a bout a moment axis passing through point O is equal to: + MO = F  d where d is the perpendicular distance between the force and point O. In vector form, the moment can be expressed as: + MO = rF where r is the position vector drawn from O to any point on the line of action of F. The moment has both magnitude and direction.

3. Moment Axis Moment Axis MO MO F F d  A A O O r r Magnitude + MO = rF = r  F  sin  = F  (r  sin ) = F  d

4. z C B rC MO rB rA A y O F x Transmissibility of a Force F has a moment about point O equals to: MO = rF We have seen that r can be extended from point O to any point on the line of action of force F. Then: MO = rAF = rBF = rCF Then F has the property of a sliding vector and, therefore can act at any point along its line of action. F is therefore a transmissible force.

5. Cartesian Vector Formulation If F and r are expressed as Cartesian vectors, then i j k + MO = rF = rx ry rz Fx Fy Fz If the determinant was determined, then, +MO = (ryFz - rzFy) i – (rxFz – rzFx) j + (rxFy – ryFx) k

6. z Fz F Moment Axis F z MO Fy r rz Fx r y O y ry rx x x The vector formulation method has the benefit of solving three-dimensional problem easier. Position vectors can be easily formed rather than perpendicular distance ( d ).

7. F1 z F2 r1 r2 y r3 x F3 Resultant Moment of a System of Forces + MRO =  ( rF )

8. F1 F F2 r O Principle of Moments If F = F1 + F2+ MO = ( rF1 ) + ( rF2 ) = r (F1 + F2 ) = rF The principle of moments states that “the moment of a force about a point is equal to the sum of the moments of the force’s components about the point”.