Triple Pendulums with Mutual Interaction. 23 



Let the bridle cords be inclined i to the horizontal, 

 see fig. 2. 



Then true length of AB = 6/sint and 



true lengths BD = *=± = , , (< *~ a t *¥ ■ .' (4) 



° suit [d-\-a—0)smt 



The tensions on the bridle cords BD and AB are each 



4-^- and their deflexions from their undisturbed position 



Sim 



can be found from (3) and (4). 



Thus, the three forces acting at B are Mgjsin i along BA, 

 Mg/sim along BD, and U along B'B. 



Then, by resolving parallel to B'B and reducing, we find 



U a—b+d { , 



w 9 = {^w {tlia ~ } ~ ua} ' ' ' ' (5) 



In like manner for the point B', we have the forces U from 

 B to B', Mff/sin v from B 7 to A', and Mg/ sin i from B' to C. 



Now, displacement of C is 2ivd/(d + a — c), the true length 

 of A'B' is b'/sini. and that of B'C is (V — b')jsin i, whence 

 the deflexions of each from their undisturbed positions as 

 before. 



We thus find for the point W , by resolving parallel to BB' 

 and reducing, that 



U __(a — b + d)(u.a — uc — wa + wb) ,„. 



Mg~~ (a-b){b-c)d ~' ' ' ( } 



Equating right sides of (5) and (6), we eliminate ZJand d, 

 obtaining 



u(2ab — ac — bc)—iob{a — b)rrx{a — b){b — c). . (7) 

 Let us next consider the point C" on the bridge C'C". 



J he displacement ok D" = iv+ v — . 



The displacement of C" = wc'jc = 2wd/(d -\-a—c). 



The true length of A^'OW/sin i, that of CD" 



= (d— c')/sin t. 



Thus, we can write the sines of the deflexions from the 

 undisturbed positions of the cords A"C" and C"D". 



At the point C" we have the three forces : W from C" to 

 C, Mg/sint, from C" to A", and Mg [sin i from 0" to D". 



