18 BEST AND MOTION 



To find the direction of the resultant, we divide the correspond- 

 ing sides of (1) and (2) and obtain 



tane = ^ = ^ 8iDe ' + ^ sin *' + . 

 X RI cos e 1 + R z cos e 2 + 



If we nave only two vectors R lf R z , making an angle with 

 one another, we may put e 1 e 2 = 0, and obtain 



Since R is obviously the diagonal of *a parallelogram having two 

 edges of lengths B lt J2 2) meeting at the angle 0, this result can 

 be obtained' directly from the geometry of the triangle ADC, in 

 D which the angle at C is evidently TT 0. 

 _ , Thus 



^2> 



"^^ ' 7?2 T>2 i 7?2 O T> T> I u\ 



Jt = Ji 1 T -ti 2 .Z JK^Jt^ cos (TT i/j, 



which is clearly identical with the above 

 expression. 



We may take two examples to illustrate 

 the method of resolving vectors into rectangular components in a plane. 



1. In Ex. 2, p. 10, suppose that the direction of the ship (AB in fig. 5) 

 is taken for axis Ox, and that that in which the shot is to travel is taken 

 for axis Oy. Let the shot be fired with velocity V, making an angle e with 

 Ox, the velocity of the ship being v. The resultant velocity is to be along 

 Oy, so that the velocity along Ox, say X, is to be nil. We have, however, 



X = v + Fcos0, 



V 



so that we must have cos 6 = > giving 



the result already obtained. 



2. To find the acceleration of a point 

 moving with uniform velocity F in a circle 

 of radius a. Let A be the position of the 

 particle at time t = 0, and take OA for 

 axis of x. After time t the particle has 

 described a length Vt of arc, so that if 

 B is its position after time t, the angle 



BOA is in circular measure. The 

 a 



direction of velocity at B, namely the tangent at B, will accordingly 



