ILLUSTRATIVE EXAMPLES 51 



The resultant of all these forces vanishes, so that the sum of their com- 

 ponents in any direction vanishes. Resolving normal to the plane, we obtain 



R + .Fsin0 Wcosa = 0, 

 and resolving along the plane, 



F cos + Wsma- fj.R = 0. 



Eliminating the unknown reaction B, we obtain 



F(n sin + cos 6) W(fj. cos a sin a) = 0, 



sothat F= H^coso--sina)_ x 



H sin 6 + cos 8 



Replacing /* by tan e, we obtain 



_ W(cos a tan e sin a:) _ Wsin(e a) 

 cos 6 + tan e sin cos (0 e) 



The value of F is a minimum when cos (0 e) is a maximum, and this occurs 

 when cos (0 e) = 1 ; i.e. when = e. In this case the value of F is 



F= "FT sin (e- a). 



Thus this is the smallest force by which motion can be produced, and it must 

 act so as to make an angle with the plane equal to the angle of friction e. 

 Since, by hypothesis, the weight rests without slipping when no force is applied, 

 the angle e must be greater than a. Thus the direction of the force F must 

 always be inclined in an upward direction. The function performed by the 

 force F is twofold : it supports part of the weight of the body (through its 

 component normal to the plane), and so lessens the amount of friction to be 

 overcome ; and it also supplies (through its component in the inclined plane) 

 the motive power for overcoming the frictional resistance. When 

 these two parts of the force are balanced in the most advantageous 

 way the value of F is a minimum, and this, as we have proved, 

 occurs when = e. 



An interesting and instructive solution of this problem can 

 also be obtained geometrically. For equilibrium, the three forces 

 already enumerated must satisfy the condition of forming a tri- 

 angle of forces. 



Let AB represent the weight and BC the reaction between the 

 mass and the plane, then CA must represent the applied force F. 

 If the body is on the point of motion, the reaction must make an 

 angle e with the normal to the plane, so that the angle ABC must 

 be e a. Thus the line BC is fixed in direction, and the problem 

 is that of finding the direction and magnitude of AC, when the length AC is a 

 minimum. Obviously the minimum occurs when A C is perpendicular to BC, so 

 that A C must be in a direction making an angle e a with the horizontal, as 

 already found, and since A C = AB sin A BC = AB sin (e a), the magnitude of 

 the force required will be TF"sin(e a). 



