Lord Rayleigh on a Statical Theorem. 185 



in which for the present application Y denotes the impressed force, 

 not including the weight of the bar itself, and y is the vertical 

 displacement due to Y. 



Let y, y 1 denote two sets of displacements corresponding to 

 the forces Y, Y\ Then 



where the integration extends over the whole length of the bar. 

 Now, integrating by parts, 



f / d 2 (vdh,\ , . d (^ d*y\ dy' ^ dh, C^ d*y d*y' , 



in which the integrated terms always vanish in virtue of the ter- 

 minal conditions. In the present case, for example, y, y 1 , -—^ -7-^- 



vanish at each extremity. Thus the left-hand member of (B) 

 vanishes, and we derive 



j{2/'Y-2/Y'}& = (C) 



Let us now suppose that Y vanishes at all points of the bar ex- 

 cept in the neighbourhood of A, and also that Y' vanishes except 

 in the neighbourhood of B. Then from (C), 



or if JY(fa=JY'<fe, 



^a=2/b, (D) 



as was to be proved. 



A similar method is applicable to all such cases. 



I may here add that, corresponding to each of the statical pro- 

 positions in my former paper, there are others relating to initial 

 motions in which impulses and velocities take the place of forces 

 and displacements. Thus, to take an example from electricity, 

 if A and B represent two circuits, the sudden generation of a 

 given current in one of them gives rise to an electromotive im- 

 pulse in the other, which is the same, whether it be A or B in 

 which the current is generated. Or, to express what is really 

 the same thing in another way, the ratio of the currents in A 

 and B due to an electromotive impulse in B is the negative of the 

 ratio of the impulses in B and A necessary in order to prevent 

 the development of a current in B. These statements are not 

 affected by the presence of other circuits, C, D, &c, in which 

 induced currents are at the same time excited. 



Terling Place, Witham, 

 January 16, 1875. 



