[154] 

 NOTES ON MECHANICS- 



BY JAMES LOUDON, M.A., 



Mathematical Tutor and Dean, University College, Toronto. 



1. To find tlie resultant of two parallel forces. 



Let AB, CD he tlie given forces acting in the same direction- 

 Produce £A to JS, making AU equal to AB. Join AC, ED meeting 

 in F ; on CA take CG equal to AF ; and complete the figure by 

 drawing GHK parallel to AB or CD, and AH, BK parallel to ED. 

 Then on introducing the two equal and opposite forces AF, CG, the 

 forces AB, CD will be equivalent to AB, AF, CD, CG. Of these 

 the former two are evidently equivalent to HK, HD ; and the latter 

 to GH, DH. Hence the resultant is GK or AB + CD ; and its 

 line of action a line through G parallel to AB or CD. 



A point in the line of action of the resultant can also be found by 

 drawing the lines ALND, BLMC ; then if CM = BL, DN = AL,. 

 either M or N is the point in question. 



By reversing BAE we have the case where the forces act in oppo- 

 site directions. In this case (AB <C CD), CA, DE raeet in F; CG 

 is taken, in FAC produced, equal to AF; GKH is drawn parallel to 

 CD, and BK, AH parallel to FED. Then the resultant is GK = 

 CD — AB. 



2. In the figure of (1) since the parallelogram HC — DA = HE, it 

 follows that AB. p = CD.q, where p and q are the respective dis- 

 tances between the -resultant and AB, CD. 



From this relation it easily follows that the moment of the resultant 

 about any point in the plane of the forces is equal to the sum of 

 the moments of AB, CD. 



3. To. prove that the moment of the resultant of forces acting at 

 a point is equal to the sum of the moments of its components around 

 any line. ■ " . 



Let 00^ be any line, AD^ one of the component forces, AOO^ being 

 the plane of the paper and A perpendicular to 0^. Through A 

 draw AB;^ parallel to 00^ and let B^ and 6>i be the points where a 

 plane through D^ meets ABj^, 00^, respectively. Drop a perpen- 

 dicular O^F-^ ( =: jOj) on D^B^, and let d = O^B^ = O^B^ = . . . . 



Resolve AD^^ into AB^, B^C^ in the plane of the paper, and 

 CiZ?! perpendicular to the plane of the paper. Resolve the other 



