232 NOTES ON STATICS. 



3. Let a set of forces be reducible to a resultant R acting along OA 

 and a couple G, axis parallel to OB. 



Then these can be transformed into R acting at a point 0' and a 

 couple G', axis parallel to 00', only when the plane BOO' is perpen- 

 dicular to the plane A 00', that is, when 



cos A OB = cos AOO'. cos BOO', 

 as is evident by describing a sphere round 0. 



4. To find the positions of the momental-plaaes as the moment-centre 

 (the origin) moves along a given line OA. 



Here the forces are supposed reduced to R at and a couple G, 

 ■whose plane, the momental plane, passes through 0. 



Let OA and R be in the plane of the paper, OB tbe intersection of 

 the momental-plane with the plane of the paper, and C the projection 

 of the axis of G on the plane of the paper. 



Let 0' be the new position of the origin on OA, 00' = r, d = 

 distance of O'B', parallel to OB, from OB; d, (f, 4' the angles which 

 the direction of ^, the axis of G, and OC make, respectively, with OA. 



Then on transferring to 0' we have R and the couples G, Rr sin 0, 

 the resultant of which will be a couplewhose axis lies in a plane parallel 

 to 00 and perpendicular to the plane of the paper. 



Now let s = distance of the line of intersection of the momental- 

 planes at and 0' from 0; a the angle between OC and axis of G, 

 fi the angle between the momental-planes at and 0', y the angle 

 between the momental plane at 0' and the plane of the paper. 



Then s cos (a -{-/') = tZ cos (a -j- /5), or s sin /3 ^= d sin y 



, sin y , ^ , ^ G COS di 



.-. s = a r= d. — = /■ cos </' = ■ , 



sm (i Rr sin Q Rr sin Q R am 9 



which is independent of the distance of 0' from 0. 



Therefore, as the moment-centre moves along OA, the momental- 

 planes all pass through the same line, the distance of which from 

 ^ G cos Tp 



R sine' 



If a plane be drawn through this line parallel to the plane of the 



paper, and p = distance between these planes, then 



G cos 1^ cos a G cos ^ 



p = scOBa= ^^.^^ = ^-^^, 



since cos <f = cos <p cos a. 



