ILLUSTRATIVE EXAMPLES 55 



. From this second triangle of forces we obtain 



1 =*!=*!, (b) 



oc OQ CQ 



where T', R' represent the tension and reaction acting on Q. 



Since the ring at O is supposed to be smooth, the tension in the string POQ 

 is the same at all points. Thus T = T'. We now obtain, from equations (a) 



and (6), 



w OP = w' OQ, (c) 



since each product is equal to T OC. If Ms the whole length of the string, 

 we have 



w w' w + w' , ,v 



- =i - = - ) (en 



OQ OP I 



showing that the string arranges itself so that it is divided by the ring at 

 in the inverse ratio of the two weights. We notice also from equations (a) 

 and (b) that 



B -^ 



w w' 



for each ratio is that of the radius of the sphere to OC. Thus the reactions are 

 in the direct ratio of the weights. 



If the string is inextensible, the length I is known, so that equations (d) 

 determine the lengths OP, OQ completely. Suppose, however, that the string is 

 an extensible string, say of natural length a and modulus X. Then, instead of I 

 being a known quantity, we have one additional equation between unknown 

 quantities, namely 



Remembering that equation (c) gives two values for the quantity T> OC, 

 we have 



so that 



This equation determines the value of I, and having found this we proceed 

 as before. 



5. A weight W is supported by strings of which the tensions are TI, T 2 , > - T n . 

 The strings do not hang vertically, but the angles between the different pairs of 

 strings are known, being e 12 , ei 3 , etc. Find the weight W in terms of the tensions 

 and of these angles. 



