110 STATICS OF EIGID BODIES 



obviously gives a state of equilibrium the ring is standing at rest on the table, 

 acted on solely by its weight but this state of equilibrium is not the one with 

 which we are concerned in the present problem. 



Let us now consider .the meaning of alternative (). If the three lines of 

 action meet in A, the reaction at B must act along BA, and this must be true 

 no matter how great the pressure at B. Hence the reaction at B will always 

 make an angle a with the normal. 



If a is less than e', the angle of friction at B, this will be a possible line of 

 action for the reaction, and no slipping can take place at B, no matter how 

 great the pressure applied at B may be. 



On the other hand, if a is greater than e', equilibrium is impossible, no matter 

 how small the pressure applied at B may be. Thus, when there is equilibrium, 

 the pressure at B must vanish, and we are led to the same 

 state of equilibrium as was reached from case (j3). As soon 

 as the pressure at B becomes appreciable, equilibrium is 

 obviously broken by slipping taking place at .B, since for 

 equilibrium to be maintained at B, the reaction would 

 have to act at an angle greater than the actual angle of 

 friction. 



Thus the solution resolves itself into two different cases : 

 CASE I. If a > e', as soon as pressure is applied at Z?, 

 motion takes place. The ring slips at B, and consequently 

 rolls at A. 



CASE II. If a < e', we have seen that no matter how 

 great a pressure is applied at JB, there can never be 

 FIG 63 slipping at B. It remains to examine whether there can 



be slipping at A. 



To settle this question, we have to determine whether the reaction at A can 

 ever be made to act at an angle with the vertical which is as great as the angle 

 of friction at A, namely e. Now the ring is acted on by three forces, the 

 reactions at A and .B, say E A and R B , and its weight W. The lines of action 

 of these forces meet in the point A , and from Lami's theorem we can connect 

 the magnitudes of the forces with the angles between them. 



The lines of action of the three forces are represented in fig. 63. The angle 

 between W and BB is, as we have seen, always equal to a. Let the angle 

 between R A and the vertical be supposed to be 0. Then, by Lami's theorem, 



E A = ^B_ = w 

 sin a sin 6 sin (a 6) 



The value of EA is not given, but on equating the last two fractions we have 



W sin (a - 6) 



= 5 '- = sin a cot 6 cos a, 



E B sin 6 



(W\ 

 cos a H ) cosec a. 

 EB/ 



