On the Parallelogram of Forces. 309 
makes the angles 5 6, with y’ and y’, we have by (1) and (2) = 
i P y’ ys a ak 3 i 
—— Oats W Byes g = SSS _ =——— SS SS e)) AMA gt 
a1) : =el5 Treas ie = (8) 
es is satisfied, and (2’) becomes —ti=2 or 2° y?=27, (4’)$ 
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+dz, y+dy, and let 2’ be their resultant which makes the 
indefinitely small angle dd with z. Let x, y: denote the values of 
x+dx;, ytdy, resolved at right angles to z, and suppose that a: 
is \y", then it is evident that 2’ is between z and « ; by (1) and (2) 
. P x y, 
h Wh Sey Gi =f = Cn at d. 
we have mers is ) tate pees state ia 
=<(yt+4y), and x*— y: ot the whole force which is 
2 
eile to z. 
It is evident that x —y'=z' resolved at right angles to z, .°. by 
SY 5 (7 —a1) or by (1’), and substituting the value of 
z*—y", we have xdy ae = kd); but by neglecting quantities of 
the orders dz?, dz*, &c. we may use z for 2’, .". by substituting the 
a. 
value of z? from (4’), we shall have —“— ” ~ — __“ =d), whose 
integral gives 2 = tan. (k)-+-c), or by (4/) e=z cos. (kJ-+c); where 
r 
=| (he COnrecton. |) Viren) 41.) 2)).". Cos. ¢ == 1, and ¢—O; 
hence r=z cos. ki; put ae then cos. k3=0, .*. e=0; but when 
c=0, we evidently have - i ae which gives k=1; hence 
“=z cos.4; 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. 
