CE 201 - Statics Lecture 11. MOMENT OF A FORCE-VECTOR FORMULATION In scalar formulation, we have seen that the moment of a force F about point O or a.

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CE Statics Lecture 11

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: + M O = F  d where d is the perpendicular distance between the force and point O. In vector form, the moment can be expressed as: + M O = 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.

Magnitude + M O = r  F = r  F  sin  = F  (r  sin  ) = F  d A F r O MOMO Moment Axis A F r O MOMO d 

Transmissibility of a Force F has a moment about point O equals to: M O = 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: M O = r A  F = r B  F = r C  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. A F rArA O MOMO rBrB B C rCrC x y z

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

The vector formulation method has the benefit of solving three-dimensional problem easier. Position vectors can be easily formed rather than perpendicular distance ( d ). F r O MOMO Moment Axis x y z z y x r rxrx ryry rzrz F FxFx FyFy FzFz

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

Principle of Moments If F = F1 + F2 + M O = ( 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”. F F1 F2 r O