MECHANICAL PHILOSOPHY. STATIC* [MBOLTAXT or PARALLM. IMBCB. 



went*\ in magnitude and direction by OR', C K' being 

 - \ i: 



vrir T may be traiwferred from B to C, and re- 

 presontod in magnitude and direction by C T, I'T being 

 equal to B T. 



Through C draw the lino 8 C S parallel to A B. 

 Abo through T draw T Q/ parallel to B A, meeting 

 S parallel to P 01 meeting S C in S. 



aw R/F parallel to AB, 

 I' in !', jin.1 U'S parallel to DO, meeting 



The parallelogram 8CQT is similar and eqaal in all 

 napeota to U parallelogram sltyT, and the porallel- 

 ogram C P R'Sto the parallelogram A 1' 1 



111 , the force R acting at C, in the 

 direction C R', may be replaced by the forces S aud P, 

 npratented iu magnitude and direction by C S and C P'; 

 also the force T acting at C in direction C T, may be 

 replaced by the force* S and Q' acting iu the directions 

 CSandCQ'. 



The equal and opposite forces 8 and S, acting in the 

 line S C S, will destroy each other. 



And the parallel forces V mid Q acting at A ami B 

 are replaced by the forces P and Q acting at C in tliu 

 direction C D, and these forces again may be transferred 

 fn.m (J to D. 



> the resultant of the forces P and Q acting at A 

 and B, will be a force P + Q acting at D in a direction 

 parallel to the directions of P and Q. 



To determine tht position of the point D. 



Since P'R' is parallel to AD, the triangles CP'R', 

 C D A are equiangular, and therefore, Euc. , B. VI., P. 4 



Similarly because O/ T is parallel to D B, the triangles 

 C I' Q', C B D are equiangular, and 



BD : CD:: TQ- Ctt - 



Hence C . D . BD -.. C -. 



\ l. (D l: !' CQ' 



BntR'P'-r^thervfore^;^ 



And 



11 ' 



therefore 



BD + AD P + Q 



AD 



ButBD + AD=AE. 



A B, which determines the point D. 



Or, in other words, the distance of the fulcrum from 



nt of application of one of the forces P, is equal 



to the other force U multiplied by the distance between 



the two forces, aud divided by the sum of the two forces. 



When A B is perpendicular to A P or B O., wo have, 



BD P 



\ I>~Q 

 QXBD-1'X AD; 



but in this case Q X B D U the moment of Q with re- 

 spect to I*, and P x A D the moment of P with respect 

 T 1 1 II. in v in this due, as in the last proposition, tho 

 moments of P and Q with respect to D, are equal to each 



In the cue we have iust considered, we supposed the 

 two forces P and Q to be acting in the same direction ; 

 when they act in the opposite directions, our construc- 

 tion will be nodifod as represented by the accompany- 

 iogdiagram. (Fig. 43). 



To A and B, as before, apply two equal and opposite 

 (ores* rpi<Btd in magnitude and direction by A 8 

 and B& 



CompUte the parallelograms A P R 8 and B Q T 8. 



Join U T mid A U, aud produce them till they meet 



Through C draw C D parallel to A P or B Q, meeting 

 B A produced in D. 



The resultant of the forces S and Q acting at B will 

 be represented in n. . 



-Q 



. 



magnitude and direc- 

 tion by B T, and this 

 force may be trans- 

 ferred from B to C, 

 and represented by 



C-F-BT. 



The resultant of 

 the forces P an.l S 

 acting at A will also 

 be represented by 

 A R, and this may be 

 transferred from A to 

 C, and represented by 

 CR'-AR. 



Through C draw 

 80S parallel to BA, 

 through r, T'O/ pa- 

 rallel to AB, TS 

 parallel to DC, aud 

 through R 1 , R' V parallel to AB, meeting DO pro- 

 duced in P 1 , and R' S parallel to D C. 



The force C T may be replaced by the forces repre- 

 sented by C Q' and S, and the force C R' by the f 

 represented by C S and C P*. 



But C8 = BS, and CS=AS, and A8=BS by con- 

 struction. 



Hence the two equal forces acting on C in opposite 

 directions in the lino S C S destroy each other, and the 

 forces P acting at A, and Q acting at B, are repi 

 by the two forces C Q' and C P' acting on C iu op[. 

 directions. 



Bdrl i <r=BQ=Q,andCP'-AP=P. 



Hence these two forces may be represented by a single 

 force P-Q acting at in the direction O P', an.l 

 force may be transferred from O to D in the line C D. 



The resultant of the two parallel forcrs, P and Q, 

 acting in opposite directions at the points A and B, will 

 be a force P Q acting at D in B A produced. 



To determine the position of the fo'nt D. 



Since A P is parallel to D C, the angle A R 8 angle 

 A C D, aud the angles at C and D are common to the 

 two triangles A R S, A C D. 



Therefore the triangles A R S, A C D are equiangular 

 and similar. 



And AS :SR:: AD :DO, or A S 



SR DO 



Also because BQ is parallel to CD, the angle TBQ 

 = angle BCD. 



Again, TQ is parallel to BD, therefore the angle 

 1,1 T It = angle DEC. 



And the two triangles T Q B, B D C are equiangular 

 and similar. 



And BQ :QT ::DC : BD, or B ^= DC . 



Multiplying the equal fractions together, wu have 

 \S BQ_AD DC 

 SR' QT~DC' BD' 

 ButQT-BS- AS, andSR=AP 



Hence 11 '. 



A. X J 



AD 



Now BD-AB + AD, therefore = __ . . 



P A B + A 1) 



And Q. A B + Q. A D - P. A D, or Q. A B - (P-Q) 

 AD. 



O 



Therefore AD - 



p_ , 



This result might have been obtained from the pre- 

 vious case by substituting Q for Q, which w.mM give 

 P Q for the resultant, but the expression for AD 



O A Tt 

 would be 10 negative sign signifying that the 



l.int D in this case would lie on a different xidu of A 



