46 
Mr. Kine on the Parallelogram of Forces. 
Now (1). & cannot involve p, for the angle of inclination remain- 
(2). 
(3). 
(4). 
ing the same, the resultant must necessarily be pro- 
portional to the component. 
k cannot involve 6: for if it does, it must evidently 
be such a function of @ as has no possible root: it 
must moreover involve no negative powers of 0, for 
then would r+ become infinite when @ vanished: nor 
yet any odd powers of 6, for then would r be altered 
by changing the sign of @: it must therefore involve 
the even powers of @ only; and as all the roots are 
impossible, the terms must all have the same sign, 
and consequently such a value of @ may easily be 
assumed as will make &.p. cos. @ greater than 2p, 
which is impossible. 
cos. @ cannot be raised to any power as cos. 6\”, for 
then the equation could not be made identical both 
when @=0 and when 6 = = 
the factors composing cos. @ cannot be raised to dif- 
ferent powers, for the equation could not then be 
made identical when @ equals any term of the series 
To determine k, let 6=0, in which case r ought to be equal 
to 2p; 
Eos? pus leperame. .¢ sit == 2: 
Hence r = 2p.cos. 0. 
The resultant of two unequal rectangular forces, and con- 
sequently of any two forces whatever inclined to each other at 
any angle, may easily be deduced from the expression for the 
resultant of two equal forces by processes already known, which 
require neither elucidation nor abbreviation. 
