26 S .] 



ATTRACTION. 



161 



attraction at P produced by each element must be determined, 

 and then all these forces must be compounded into a single 

 resultant. This is always possible because the forces all pass 

 through the point P. 



Let dm be the element of mass situated at the point Q, P the 

 attracted point of mass i, PQ = rihe distance between them, 

 and a, ft, 7 the direction cosines of r\ then the attraction at P 

 due to dm is dm/r^\ its components are adm/r 1 , ftdm/fl, 

 and the components of the resultant attraction R at P are 



<>.. Y =S> z =$' 



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where the integrations must be extended over the whole given 

 mass. The resultant R .itself, and its direction cosines a, b, c, 

 are finally found from the formulae 



V V "7 



** 7. * *" 



a =R' b =R' C =ll 



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The following examples will illustrate the process. 



265. Homogeneous Circular Arc. To determine the attraction 

 exerted by a mass distributed uniformly along a circular arc ACB 

 (Fig. 78) of angle 2 a and radius a on a mass i situated at the centre 

 Pof the circle, let QQ = ds be an element of the 

 arc, dm = pds its mass ; then 



is its attraction at P, and this force has the direc- 

 tion PQ. 



Resolving this force parallel to the bisecting 

 radius PC of the arc and at right angles to it, it 

 will be seen that the latter component need not be 

 considered, since it is balanced by an equal and 

 opposite component arising from the element sit- 

 uated symmetrically to QQ with respect to PC. 

 If K CPQ = <9, the component along PC is 



ds cos Q/a 2 , or since ds adO, Kp cos Odd /a. Hence the resultant 

 attraction at P is 



J? K P C +a (Mti sin ft 

 a J a. Q 



PART II II 



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