ILLUSTKATIVE EXAMPLES 



109 



Also on resolving horizontally and vertically, it is evident that the action 

 at A must consist of a horizontal force of amount T and of direction opposite 

 to that of T. 



2. A ring (e.g. a dinner napkin ring) stands on a table, and a gradually 

 increasing pressure is applied by a finger to one point on the ring. Having given 

 the coefficients of friction at the two contacts, examine how equilibrium will first 

 be broken. 



Let A be the point of contact of the ring and table, and let B be the point 

 of contact of the ring and ringer. Let e, e' be the angles of friction at A and B 

 respectively. Let the line BA make an angle a with the vertical. 



The forces applied to the ring from outside are 



(a) the reaction at A ; 



(b) the reaction at B ; 



(c) the weight of the ring. 



' Regarding the latter as a single force W acting along the vertical diameter 

 CA of the ring, we see that so long as the ring remains at rest it is in equilibrium 

 under the action of three forces. 



Hence, by the theorem of 52, the lines of action of the three forces must 

 meet in a point. 



The line of action of the weight is already known to be the vertical CA, 

 and the line of action of the reaction at A must pass through A. Hence either 



(a) the point in which the three lines of action meet must be A ; or 



(/3) the reaction at A must act along CA, so that the point in which the 

 three lines of action meet will be some point in CA, other than A. 



The second alternative may be dismissed at once. For if the reaction at A 

 acts along CA, this and the weight may be combined into a single force, and 

 there must now be equilibrium under this force and the reaction at B. This 

 requires that each force should vanish, i.e. there must be no pressure at B, 

 and the reaction at A must be just equal to the weight of the ring. This 



