96 8. 



Let tlic straight lim\-< AH and .17;' represent (lie two forces, 

 AA' being a I sine at i 



angles to both. Suppose two paral- 

 lel lines Ax, A'x drawn, each at 

 right angles t< > .1.1 . ami .-!/, J // '. 

 respectively at right angles to Ax, 

 ami also at right angles to 

 . Let 



1 let T and T denote the int. 



i of the forces in All and 

 respectively. Then T may be resolved into jfcos $ 

 Tain < acting at A along Ax and Ay respectively, and T 

 into T'cos<t>, jTsin^' acting at A along A'x and J //' 

 respectively. Let a be the inclination of AB and J'JT, so 

 that <f> = < -f a. Now determine ^ by the conation 



rcos< + T'cos^O (1), 



that is jTcos <f> -f T cos (< + a) = 0. 



Then by (1) the forces Tcos <j> and T cos 6' will form a couple 

 in the plane xAA'x' ; and jTsin <^ and T sin d>' will have a 

 single resultant perpendicular to the plane of this couple, 

 for they cannot form a couple since then the whole syst< 

 forces would reduce to a single couple which is contrary to 

 supposition. Let P denote the intensity of this single force 

 so that 



P= Tain < +!T sin f (2). 



The moment of the couple is AA 1 x Tcos <f>. Hence, by 



the latter part of Art. 98, AA xPx Tcos<f> is constant 



ver be the position and magnitude of the forces T and 



T', so long as they are equivalent to a given system of forces. 



Now the volume of the tetrahedron of which AB and A'B' 

 are opposite edges is ^AB.A'B'.AA' sin a. For the base 

 may DC considered to be the triangle AA'Jt' ', the area of 



which is ^AA'.A'J;' ; and the height will then be Attain a. 



But from (1) and (2) we have T' sin a = P cos </>. Hence the 

 volume of the tetrahedron becomes %AA' . T. Pcos<f>, which 

 iiavc just seen to be constant. 



This result is due to Chasles; see Mobius, Lchrluch der 

 Statik. i. 



