On the Parallelogram of Forces. 309 



P • x / 



makes the angles - — 6, 6 9 with y' and y", we have by (1) and (2)_ 



2 ) x 



z a? * \2 J y z \2 j y z 



(3') is satisfied, and (2') becomes — 4~— '«* or a: 2 +y*=2 a , (4') ; 



2: z 



which shows that the resultant is represented in quantity by the dia- 

 gonal of the rectangle whose adjacent sides denote the components. 

 We will now find the direction of the resultant. Let x and y be- 

 come x-\-dx, y-\-dy, and let z' be their resultant which makes the 

 indefinitely small angle dh with z. Let or, y* denote the values of 

 a?-f-dj?, y-\-dy, resolved at right angles to 2, and suppose that x* 

 is Ay? then it is evident that z' is between z and x ; by (1) and (2) 



we have Jf. =*=J? - »), JL-=*= 9 (i), .-.'n ^(.-J-A?), 



y* —-(y+dy)) an d ar — y=- ^= the whole force which is 



perpendicular to z. 



It is evident that ar — y=z' resolved at right angles to z, • ". by 



(2) ~~y =s<p{_ — d$ j or by (I 7 ), and substituting the value of 



£•— y, we have *: -y — V x _ ^j . ^ ut ^ neglecting quantities of 



the orders dz 2 , dz z , he. we may use z for z f , .*. by substituting the 



™ y 



value of z a from (40, we shall have x ^ ~ V .'- = — —=k<tt 3 whose 



integral gives ? = tan. (Jfc0+c)> or by (4') a?=z cos. (ki+c) ; where 



c m the correction. When 6 = 0, x = z, .\ cos. c = 1, and c=0, 



P 



hence a?=z cos. B\ put&)== , then cos. &3=0, .'. a?=0; but when 



2 



P P P 



x=0, we evidently have d=- , AK =- , which gives k=l ; hence 



«=zcos. 0; which shows that the direction of the resultant is the 

 same as that of the diagonal of the rectangle whose adjacent sides 



denote the components. 

 We will now find the direction of the resultant in another manner. 





