ILLUSTRATIVE EXAMPLES 



53 



The equation is most easily interpreted by noticing that if r I is replaced 

 by r, the equation becomes 



\ 2 \ 



TF 2 (sin2ar - ^ 2 cos 2 or) + r 2 + 2 - Wsin a - r cos = 0, (6) 



which is the polar equation of a circle. Thus the original locus represented by 

 equation (a) can be drawn by first drawing the circle represented by equation 

 (6), and then producing each radius vector through the origin to a distance I 

 beyond the circumference of this circle. 



FIG. 28 



The same result can be obtained by a geometrical treatment of the problem. 

 The particle is acted on by only three forces in the plane on which it rests, so 

 that lines parallel and proportional to these forces must form a triangle of forces. 



In fig. 29 let OP be the string, and let AP be a 

 length I measured off from P, so that AO is the 

 extension r I of the string. The tension is. always 

 proportional to AO and acts along AO. Let us 

 then agree that in the triangle of forces the tension 

 shall be represented by the actual line AO. On 

 the same scale let the component of the weight, 

 W sin a, be represented by the line OG, the direc- 

 tion of this being, of course, down the line of greatest 

 slope through 0. Then AOG must be the triangle 

 of forces, so that GA must represent the f rictional 

 reaction between the particle and the plane. The 

 maximum value possible for this is /j.W cos or, so that 

 if slipping is just about to occur, GA will represent 

 a force ^W cos a. Thus corresponding to a position 

 of P in which slipping is just about to occur, the positions of A are such that GA 

 represents the constant force /j,W cos or, in other words, the locus of A is a circle 

 of center G. This leads at once to the construction previously obtained. 



FIG. 29 



