$25 
the Density of the Earth . 
To compute from hence the attraction of the case on the 
ball, let the box DCBA, (fig. 1.) in which the ball plays, be 
divided into two parts, by a vertical section, perpendicular to 
the length of the case, and passing through the centre of the 
ball ; and, in fig. 9, let the parallelepiped ABDE abde be one 
of these parts, ABDE being the abovementioned vertical sec- 
tion ; let x be the centre of the ball, and draw the parallelogram 
fi np mix parallel to B b d D, and xgrp parallel to fiB bn, and 
bisect fi $ in c. Now, the dimensions of the box, on the inside, 
are B6= 1,75; BD = 3,6" ; B /2 = 1,75 ; and fiA — 5 ; whence 
I find, that if x c and fix are taken as in the two upper lines of 
the following table, the attractions of the different parts are as 
set down below. 
xc 
’75 
>5 
,25 
fix " — 
BO 5 
E 3 
B 55 
Excess of attract, of Ddrg above B brg 
.2374 
,1614 
,0813 
m d rp above n b rp 
.2374 
,1614 
,0813 
- — mesp above nasp 
,37°5 
,251 6 
,1271 
Sum of these 
>84,53 
>5744. 
>2897 
Excess of attract, of B b n fi above D dml 
,5007 
. 327 1 
,160b 
" 1 - — — A an fi above E eml 
> 4,677 
>3079 
,1325 
Whole attraction of the inside surface’'! 
of the half box j 
,1231 
,0606 
,0234 
It appears, therefore, that the attraction of the box on x in- 
creases faster than in proportion to the distance xc. 
The specific gravity of the wood used in this case is ,61, and 
its thickness is -J of an inch ; and therefore, if the attraction of 
the outside surface of the box was the same as that of the in- 
side, the whole attraction of the box on the ball, when cx = ,75, 
