90 



MECHANICS 



EXAMPLE. Find the resultant of all the forces acting on 

 the point O, Fig. 4, the length of the lines being proportional 

 to the magnitude of the forces. 



SOLUTION. Draw OE parallel and equal to AO, and EF 

 parallel and equal to BO; then OF is the resultant of these 

 two forces, and its direction is from O to F. Consider OF as 

 replacing OE and EF, and draw FG parallel and equal to 

 CO; OG will be the resultant of OF and FG; but OF is the 

 resultant of OE and EF; hence, OG is the resultant of OE, EF, 

 and FG, and likewise of AO, BO, and CO. The line FG, 

 parallel to CO, could not be drawn from the point O to the 

 right of OE, for in that case it would be opposed in direction 

 to OF; but FG must have the same direction as OF. 



(b) 



F 



FIG. 4 



For the same reason, draw GL parallel and equal to DO. 

 Join O and L, and OL will be the resultant of all the forces 

 AO, BO, CO, and DO (both in magnitude and direction) 

 acting at the point O. If L'O is drawn parallel and equal to 

 OL, and having the same direction, it will represent the effect 

 produced on the body by the combined action of the forces 

 AO, BO, CO, and DO. For brevity, the terms forces AO, 

 BO, etc. and resultants OF, OG, and OL have been used in 

 this solution. It should be remembered, however, that these 

 are merely lines that represent the forces in magnitude and 

 direction. 



