1 62 



STATICS. 



[266. 



Denoting the chord AB of the given arc by c, the result can be put into 

 the form 



R = *?-<, (6 f ) 



which might have been found directly from Fig. 78, without integration, 

 since ds cos = qq\ 



Formula (6 f ) shows that, if the chord AB were covered with mass of 

 the same density p as the arc, and if this mass were concentrated at the 

 middle point C of the arc, it would produce at P the same attraction as 

 the axcACB. 



266. Homogeneous Straight Rod. To determine the attraction at 

 any point P produced by a mass distributed uniformly along a segment of 

 a straight line, AB, let QQ' = ds (Fig. 79) be an 

 element, OQ = s its distance from the foot of the 

 perpendicular PO=p dropped from P on AB, 

 and let r, be the polar co-ordinates of Q with 

 respect to P as pole and PO as polar axis. Re- 

 solving the attraction Kpds/r* of the element QQ 1 

 along PO and at right angles to it, we find the 

 components 





, v 



a A =. 



cos Oils 



f 



sin Oils 



Fig. 79. 



The figure gives r=//cos0, j=/tan0; hence 

 ; substituting these values, we have 



/ / 



and integrating between the limits ft = OPB and a = OPA : 



X= 5 (sin/? + sin a), Y= & (cos/8 - cosa). 

 / P 



The resultant attraction 



(7) 



makes with PO an angle $, for which we have 



