92 



MECHANICS 



It is necessary, of course, that OB be at right angles to BA, 

 so that all the effect of OA in the required direction may be 

 represented by BA and none of it by OB. Thus, OB' and 

 B'A, although components of OA, and although one of them 

 is in the required direction, would not be a correct solution 

 of the problem because, besides B'A, OA exerts some more 

 effect in the line BA, namely, a part of OB' is in that direc- 

 tion with an amount BB' . 



EXAMPLE. If a body weighing 200 Ib. rests on an inclined 

 plane whose angle of inclination to the horizontal is 18, 

 what force does it exert perpendicular to the plane, and what 

 force does it exert parallel to the plane, tending to slide it 

 downwards ? 



SOLUTION. Let ABC, Fig. 7, be the plane, the angle A 

 being 18, and let W be the weight. Draw a vertical line 



FIG. 7 



FD = 200 Ib., to represent the magnitude of the weight. 

 Through F draw FE parallel to AB, and through D draw DE 

 perpendicular to EF, the two lines intersecting at E. FD is 

 now resolved into two components, one FE tending to pull 

 the weight down the incline, and the other ED acting as a 

 perpendicular pressure on the plane. On measuring FE 

 with the same scale by which the weight FD was laid off, 

 it is found to be about 61.8 Ib., and the perpendicular pres- 

 sure ED on the plane is found to measure 190.2 Ib. 



As it is often necessary to resolve a force into two com- 

 ponents at right angles to each other, a simple method of 

 solution is employed. In Fig. 8, let OA be the force it is 

 desired to resolve into two components at right angles to 





