Equilibrium of an Inextensible String. 433 



s the length of a portion of the string intercepted between a 

 fixed point in the string and the point (xyz). 



Y [the resolved parts of the forces at the point {xyz) J 

 2 | parallel to Trj 



R the normal reaction, making angles « /3 y with the axes. 



Then T being the tension of the string at one extremity of 

 the element 85, T d g x will be the resolved part in x, and Td g x 

 + d s (T d s x)$s will be the resolved part in x of the tension at 

 the other extremity. Hence d (Td x)$s will be the differ- 

 ence of the resolved parts in x. 



The other forces acting on Is parallel to x are X8s and 

 R Is cos a. : hence by the equations of equilibrium, supposing 

 85 to be a rigid substance, we have (dividing by Is) 



d s (Td s x) + X + Rcosa = 0. . . . (1.) 



So d a {Td s y) + Y + Rcos/3 = 0, . . . (2.) 



d s {Td s z) +Z + Rcosy = 0; . . . (3.) 



and we have to eliminate T and R between these equations. 



Now (1.) d s x+ (2.) d s y + (3.) d g z = 0, gives 



dT{(d s x)* + (d,y)* + (d s zy} + T{d s xd?x + d s yd s *y+d s zd s *z} 



+ Xd s x + Yd s y+Zd s z 



+ R {cos u d s x + cos fid s y + cosy d 8 z} = 0. 



Now {d s xf + {d s yf + {d s zf = 1 ; 



.:d s xd*x+d s yd*y + d s zd*z = 0. 



Also, since the tangent to the curve is perpendicular to the 

 normal to the surface, we have 



cos a d s x + cos /3 d s y + cos y d s z = 0. 

 Hence the above equation becomes 



d s T + Xd s x + Y d s y + Zd s z = 0; 

 or if X d s x + Y d s y + Zd s z = d s v, 



T + w = 0, . (4.) 



supposing the arbitrary constant included in v. 

 Again, (1.) d s y — (2.) d s x = gives 



T {d s y d s * x-d s x d*y) + Xd s y-Yd t x 

 + R (cos u d s y — cos fid s x) — 0. 

 Now if V = {(d,uf + (d y uf + (d x u?}-\ 



cos « = V d» n cos (& = V d y w cos 7 = V & w, 

 the differential coefficients being partial. 

 Phil. Mag. S. 3. Vol. 24. No. 161. June 1844. 2 F 



