VIRTUAL WOEK 



159 



1 inch, we shall shorten each of these s ropes by 1 inch, and so lengthen 



W 



the power end by s inches. Thus n = s, so that, in this case, P = 



S 



For instance, with two pulleys in the lower block and three in the 

 upper block the value of n will be 5, so that each pound of power will 

 support 5 pounds of weight, a man pulling with a vertical pull of 

 100 pounds could support a weight of 500 pounds, and as soon as his pull 

 exceeds 100 pounds, he will raise the weight of 500 pounds. 



ILLUSTRATIVE EXAMPLES 



1. As a first example of the principle of virtual work, let us suppose that we 

 have an endless elastic string of natural length a, modulus X, placed over a 

 sphere of radius 6, and allowed to stretch under gravity. We might, of course, 

 find the amount of stretching in the equilibrium 

 position by resolving forces, but we can get it 

 more readily by the method of virtual work. Let 

 us suppose that, when in equilibrium, the string 

 lies on a small circle of angular radius 6. Let a 

 small displacement be given, this consisting of 

 each element of the string being displaced down 

 the surface of the sphere, so that the string forms 

 a new circle of angular radius 6 + d&. The length 

 of the string when forming a circle of angle was 

 2 Trb sin ; the increase in this when 6 is changed 



to e + dd is d6 (2 Trb sin 0) or 2 Trb cos d0. The 



d0 



work done in stretching the string by this amount 

 is T 2 -n-b cos dd, where T is the tension. Work 



is also done against (or, in this particular case, with) the force of gravity. The 

 height of the center of gravity of the string when forming a circle of angle is 

 b cos ; on increasing to + dd, this increases by b sin dd, so that the work 

 done against gravity is - wb sin d0. We have now calculated all the work 

 performed in the small displacement; by the principle of virtual work, the 

 total amount of this work must be nil, so that 



FIG. 92 



Thus, 



- wb sin de + T 2 Trb cos d0 = 0. 



T= tan*, 



2ir 



and the length of the string corresponding to tension T is, as we have seen, 



Hence 



an equation giving 0. 



all + -^-tan0) = 



\ 27T\ / 



2 -rrb sin 0, 



