414 - Mr. Bowditch on the motion of a pendulum 
point D suspended by the line ABCD void of gravity. Suppose that — | 
at the end of the time ¢ the pendulous body is at L, and the centreC’ 
at ec; the co-ordinates of the point L being GM=x, MK=z, KL=y. 
Continue MK to meet the arch EDF in I, and on MK let fall the per- 
pendicular CS. Draw Lea, ICa, which will evidently meet in the 
point a of the line AB, making Ie=La and IM=LM. Put CD=r, 
CG=r', the force of gravity=g, and the tension of the thread Le 
at the point L equal to as ee compound quantity is taken 
Vrr—x2x0 ; : 
rather than a simple one, in order to render the equations found by 
this method identical with those investigated in a different manner in 
Ma, LK, KM 
in the next article. This tension being multiplied by —— = che 
will give the corresponding forces in directions of the axes x, Y, Te 
pectively. Now by the similar triangles ICS, IaM we have Ia(=La)= 
ee rJzztyy 
SSG / pre? 
because LM=,/MK? + KL?= JBTW & 
S ry 
IS=,/CP?—CS' cpocs a 1T—XLL5 also Ma= pans _iMcs_ =v, 
which substituted give the forces in directions of these axes 
ae =< bei prefixed 
Se pee alas yy? Vf azyy’ the negative sign beimg 
because these forces tend to decrease the co-ordinates: This a 
force is to be increased by g, because gravity acts in direction of 
AZ , 
axis z, and the force in direction of that axis becomes Jae 
Now (by the usual rules for variable ‘qnations) these forces beim 
ye- 
multiplied by the fluxion of the time d¢ will give sis eae of a 
locities in direction of those axes, namely, d. =, d i yOeGe 
herice by the transposition we have the following system peo 
