328 Prof. Twining on the Parallelogram of Forces. 



BC 2 

 Then AC : BC : : IJ--=BC : BI= j^-. But BI=BG+GH+HI. 



BC 2 AB 2 

 Therefore -^r=^-+BF-fAC. Then BC 2 =AB 2 +AC . BF 



+AC 2 , 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 3 = 

 AB 2 +AC 2 . Or if, in the result we substitute the values suppo- 

 sed, in the mechanical problem just considered, we shall deduce 



y=(l+2acos.A-{-a 2 ) , 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 

 to be m. Its only residual effect, if but one, Fig. 4. 



is at right angles to AB, and may be called n. 

 Now the effect of m, in the direction AB is m 2 , 

 and that of n in AB is n 2 . Wherefore m 2 -\-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' ; BAE", &c. are the double, the triple, &c. Drop 

 E6 normal to A6, E'6' to A6', &c. ; also Ec normal to BE 7 , E'c' 

 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, E6, &c. 

 is spoken of, the term includes not intensity alone, but that spe- 

 cific direction which the force, thus symbolized, shall have been 

 previously defined to have. 



This understood, — the pairs of lines AE, E6, AE 7 , E'B, AE", 

 E"B, &c. may, by what was before shown, represent the intens- 

 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 



