272.] ATTRACTION. !6 5 



Resolving this force along and at right angles to PO, and considering 

 that owing to symmetry the resultant R must pass through O, we have 



If the radius vector PQ meet the surface of the cylinder at M and N, 

 the integration with respect to r gives 



With PO = p, we have MN 2vV / 2 sin 2 0, and the limits of the 

 integration are sin" 1 (#//). Hence 



271. Homogeneous Circular Plate. To determine the attraction of 

 a homogeneous circular area on a point P situated on the axis through 

 the centre O at right angles to its plane at the distance PO=p, we 

 may resolve the plate into ring-shaped elements of radii r and r -f- dr. 

 The mass of such a ring is 2 irprdr ; all points of the ring are at the 

 same distance Vr 2 +/ 2 from P, and their attractions make the same 

 angle < = \axr l (r/p) with the axis PO. Hence the attraction of the 

 ring is 



cos <f> = 2 TTKp sin <pd<pj 



since dr =/^>/cos 2 < = (/ 2 + 



Let 2 be the vertical angle of the cone that subtends the plate at 

 P; then 



/<x a 



^? = 2 7TK/3 I Sin <</< = 4 TTKp SUi*-, (l l) 



or, in terms of/, 



R=2TTKp(l--^^\ (ll') 



V V^ +// 



272. Homogeneous Spherical Shell : Geometrical Method, (a) Attrac- 

 tion at an internal point P. Let C be the centre, a the radius of the 

 sphere (Fig. 81). A thin double cone having its vertex at Pcuts the 

 sphere in two elements, AB = dS, A'' = dS 1 , which can be shown to 

 exert equal and opposite attractions at P. 



