MOMENTS 61 



Resolving the force into two components, one parallel to L and one 

 perpendicular to Z, it is clear that the former will not give any tendency 

 to turn about L, so that the whole tendency to turn comes from the second 

 component. 



The two definitions which have now been given suffice to deter- 

 mine the moment of any force F about any line L. It may be 

 noticed that the moment vanishes 



(a) if the line of action of F is parallel to L ; 



(b) if the line of action of F intersects L. 



Obviously, in either of these cases, the tendency to turn about L is zero. 



47. Let the line L be at right angles to the plane of the paper, 

 and intersect it in the point M. Let PA be the line of action of 

 a force F in the plane of the paper, acting on a particle at A, and 

 let MN be the perpendicular from M on 

 to PA. Then, by definition, the moment 

 of the force F about L is F x MN. 



Let the angle PAS be drawn equal 

 to the angle NMA, say equal to 0, so 

 that AS is perpendicular to MA. Then 

 the moment of the force F about L 



= Fx MN 



= Fx AM cos 



= AMx Fcos 6 



= AM x resolved part of F along SA. 



Instead of F being the actual force acting at A, suppose that F 

 is the resolved part, in the plane perpendicular to the line L, of 

 some other force R. Then the moment of R about L is, by defini- 

 tion, the same as the moment of F 9 and the resolved part of R 

 along AS = Fcos 6. Hence what has just been proved may be 

 put in the form 



moment about L of any force R acting at A 



= AM x resolved part of R along SA, 



and SA is now determined as the direction which is perpendicular 

 to L, and also to AM, the perpendicular. from A on to L. 



