■^^\ \o KfUA^ 'v' A acting at Q is A-- ,|?M5ting' at Q.j wi^ho sfs 



#^ 



^1-/g^4^^!^gUyi' 



»:t;.o 



.■1 JX)liv7 III .^ MV 



"Whetifc^" >:'fi'iitr.-'rr!irM> n'T; vd vnno brgn 9({t noq/j boDJuhmq si: 



and in like manner we may show tliat pS=Q?'' '' 



Again, the resultant of the equal forces A and B acts along 

 the line bisecting the angle POQ, that is, along the line joining 

 and R ; therefore we may suppose A and B to act at R instead 

 of 0, or, what is the same thing, y(e may .suppose, B to act at Q, 

 and A at B : hence 



A acting at P+B acting at Q[=sA aiid; B acting'at 0,' "^ 



or- '■• -■!: ■ -h. . :" •■.■:-:': - ■: - :■: • A, ■;■;::'. li. 



(l+i8)A + (l+«)B=A + B, 

 whence 

 hj J-jf ot U(J gOOTi/SA + aBasO, or ^Asaj—'U^. "kuw^ ^lUi',^A 



*"■'■' (V.f ^p;nMcflZ Proof of the- I^thMlo'^anfWf^f^-^''^^ '^ 

 ^^iet XA, YB, and ZC be "' ' ' "* ' ' ' " 



■ ao ban a 



'B 9dt Tfd aid. 



b in iJ bur, <x 



three forces acting at O and 

 balancing each other, ABC 

 being units of force acting in 

 these three directions, and 

 XYZ mere numbers: let a, 

 /3, 7 be three units of length 

 respectively parallel to A, B, 

 and C respectively, and xu, y/3, zy the sides of a triangle 

 formed by lines drawn parallel to these three directions, scyz 

 being mere numbers : then, since 7C = (by IV.), we have 

 {r7){ZC) = ^Z(7C)=0j-;'^ 



xu + y/S + zy—O, and /, zy= — {xu + y^) 



■ iifioq 

 H (.VI) 

 »rffl ,d)saal 



but rm-^K 



and 



XA + YB + ZC = 0,and.-. ZC=-(XA-tj; 



Hence the equation (2'7)(ZC) = becomes 



(.r« + ?//S)(XA + YB)=0. 

 Whence, since aA=0, /3B=0, and «B= -/9A (by IV.), we &d 



(.TY-yX)«B = 0, 

 and 



.-. xY-»/X = 0, orX:Y::a?:y. 



In this way it is clear that 'It,— ss At^ 



. X:Y:Z::.r:y:5, PO J^l , • 



^bich J8 virtually the parallelogram of forces. . n .[i • ' 



