24 Prof. Barton and Miss Browning on 



On resolving parallel to C'C" and reducing, we find 



W (a-c + d) 



-T7- = ~, -/\z(a — c)-iva}. ... (8) 



Mg (a-c)cd ' v ; J 



Again, for the point C, true length of B'C is (c f — b)/sm i, 

 and that of CD' is (d — c')/sin t. The displacement of B' has 

 already been shown to be 2ud/(d+a — b), that of C to.be 

 2wdj(d + a — c), while that of D' is iv + {y -w){d—a-\-c)jc. 

 Hence the sines of deflexions of the various, portions from 

 their undisturbed positions can be written. 



Thus, at the point C the forces acting are W from C to 

 C n , Mg/sin l from C to B' and from C to D'. Resolving 

 these parallel to CC" and reducing, we find 



W (a-c + d) ■ 



wr (b-c)cd M-™-K*-«)}- • • (9) 



Equating right sides of (8) and (9), we eliminate W and d, 

 and find 



uc(a — c) = — (y + z)(a — c)(b~c)-\-iv(2ab — bc — ca). (10) 



We may note here that the disappearances of d from 

 (7) and (10) show that the couplings are independent of d. 



Then, from (7) and (10) we obtain u and w in terms of x, 

 y, and z. These values may be written : 



_ (a — b){x(2ab — bc—ca) + {y±z)b{a — c)} ( 



a(4a£-36c-ca) ~ 5 * ^ } 



_ (a — c){xc{a — b) -f (y + z) (2ab — bc—ca)} . 



a(4:ab — '3bc — ca) 



Of the seven variables, x, ?/, r, 2<, i#, Z7, and II 7 , we have 

 thus eliminated U and II 7 , leaving five equations, viz. : (11) 

 and (12) and those denoted by (2). Hence, putting (11) and 

 (12) in (2), we may obtain three equations involving only 

 the three variable displacements x, y, and z. These may be 

 written : 



d 2 x 7 gx __ g(a — b){x{2ab — be— ca) + (y + z)b{a-c)} 

 dt 2 b ab(4tab—'dbc — ca) 



d 2 y gy _g(a — c)\xc(a — h) + (y-\-z)(2ab — bc — ca)} 



dt 2 c ca{Aab — '6bc — ca) 



d 2 z gz _g(a — c){xc(a — b) 4- (y + z) (2ab — bc — ca)} 



dt 2 c ca(4:ab — 3bc — ca) 



K13) 



