252 Dr. Stevelly on the Composition of Forces. 



V. The general theorem is now easily proved in the ordinary 

 way. Let A B and A C or B D represent any two forces P and 



Q acting together on the material point A; then shall AD 

 represent both in direction and magnitude their resultant. For 

 from B let fall the perpendicular B x on AD, then may P be 

 replaced by the two forces represented by A x with x B, and Q 

 (represented by B D) by the two forces represented by B x with 

 x D, and P with Q together by the four represented by A x, 

 xB, Bx, and xD ; of these four, xB equal and opposite to Bx 

 produce no effect and may be removed, and therefore A x with 

 x D or A D must represent the entire resultant of P with Q. 



This demonstration seems to me the simplest extant, with the 

 exception of the dynamic proof given in Dr. Robinson's too much 

 overlooked treatise on Mechanics, the fallacy in which I have 

 never been able to detect. The foregoing static proof, however, 

 is shorter and more direct than Poisson's or D'Alembert's, and 

 does not need the introduction of the differential or integral cal- 

 culus as used by Laplace, nor the principle of transference of 

 forces as used in the very beautiful proof of Duchayla. 



same as both A D»_ 1 with D n - \X, and of A D» with D n x. But since A x 



AD»-i D»— \x 

 bisects the angle D»_i AD», ~TJ\ — = p x (Euclid VI. 3); hence 



A T)n— 1 A Dn 



Yi ~= r> Z > therefore (bv II.) the resultant of AD»-i with D n -\ x 



±Jn — 1 •*• UftX " 



must make the same angle with AD»_i as that of AD» with D n % does 

 with AD». Hence A x must be the direction of both these (as 

 ~D n -iAx=D n Ax); and therefore Ax must be the direction of the result- 

 ant of A B with B x ; and Laplace's principle proves the length of A a; to 

 represent its magnitude. Exactly similarly we prove the same for A x' and 

 A a?", and so on to the limit A z, as in the text above. 



