542 
MR. G H. DARWIN ON PROBLEMS CONNECTED 
( 2 giver) (o-aSO) (cos 08 </>) or \gwa cr 3 cos 6808(f), 
and it acts towards the equator. 
Hence the whole force due to pressure on the element resolved along the meridian 
towards the equator is 
^gu'a$68(f)((T : cos 0 —Jq(ct~ sm 6)), or — gwa868(f> sin 6a 
cl6‘ 
But the mass of the elementary prism 8m=wd 2 sin 6868(f>.o-. 
Hence the meridional force due to pressure is — 
We will next resolve the pressures perpendicular to the meridian. 
The excess of pressure on the face <f)-{-8<f) over that on the face (f> (whose area is 
cra80), measured in the direction of <f> increasing, is 
Hence the force due to pressure perpendicular to the meridian is — -Sm-rr tt. 
1 x 1 a snip d(f> 
We have now to consider the impressed forces on the element. 
Since u is a surface harmonic of the second degree, the potential of the layer of 
matter cr at an external point is f 9 a (~^j • Therefore the forces along and perpen¬ 
dicular to the meridian on a particle of mass 8m, just outside the layer cr but infinitely 
near the prismatic element, are and -f-Sm —— and these are also the forces 
1 da 
’ a dd b cT"'' sin# d(f>’ 
acting on the element 8m due to the attraction of the rest of the layer cr. 
1 dV 
Lastly, the forces due to the external potential V are clearly 8m-— and 
* 1 dV 
dm—— — . 
cc sm 6 d(p 
Then collecting results we get for the forces due both to pressure and attraction, 
along the meridian towards the equator 
8 m 
—4_3.fi— _i_ — 
a dd^ 5 a d0^~ad0 
d 
and perpendicular to the meridian, in the direction of <ft increasing 
g da og da dV 
8m 
a sin 6 d<f> 5 a sin 6 d<f> a sin 0d<f)_ 
asiK0dcj) y ’ 
2 
Henceforward — will be written g, as in the previous papers. 
occ 
