328 Prof. Twining on the Parallelogram of Forces, 
Then AC: BC::IJ=BC: BI But BI=BG+GH-+HL 
-s 
~ AO: 
. BC? AB? 
Therefore AG =acg tBF+AC. Then BC? =AB?+AC.BF 
+AC?, which gives the relation sought. This result includes 
the relation of the sides of a right angled triangle; for, if BAC 
be supposed a right angle, BF disappears, and we have BC?= 
AB?+AC?. Or if, in the result we substitute the values suppo- 
sed, in the mechanical problem just considered, we shall deduce 
y=(14+2a cos. A+La?), as was proposed.—I now return to the 
last of the two methods mentioned, in the outset, for investiga 
ting the resultant of any two forces. 
- Method second. 
Let AB=1, (Fig. 4,) represent a force, in direction and intens- 
ity. Suppose the entire effect of the force, in the direction AC 
tobe m. Its only residual effect, if but one, Fig. 4. 
is at right angles to AB, and may be called no 8 
Now the effect of m, in the direction AB is m?, ‘ A 
- andthat of nin ABisn?. Wherefore m?-+-n?2 
=1, which determines the intensity of the resultant of two forces, 
m and n, acting at right angles to each other. 
It remains to determine the direction of a resultant in relation 
to that of its resolved or component forces. For this purpose let 
ABE, ABE’, ABE”, &c. (Fig. 5,) be right angled triangles hav- 
ing a common hypothenuse AB; and let BAE be a unit angle, 
whereof BAE’, BAK”, &c. are the double, the triple, &c. Drop 
Ké normal to Ab, E’b’ to Ab’, &c.; also Ee normal to BE’, BE’? 
to BE”, &c. As these lines are to be made the representatives 
of forces, let it be observed that they are so only in respect of 
intensity, and not of direction. Yet, when a force AE, Eb, &¢ 
is spoken of, the term includes not intensity alone, but that spe 
cific direction which the force, thus symbolized, shall have beet 
previously defined to have. 
This understood,—the pairs of lines AE, Eb, AE’, E/B, AE’ 
EvB, &c. may, by what was before shown, represent the indens 
ity of forces normal to each other, whose resultant is AB; and, 
if those forces would not have the direction, also, of the lines AE, 
EB, &c., let the direction in which the effect of AB shall be AE 
_ lie in the line AO. Then the residual effect will be EB, normal 
