322 On the Equilibrium of a Comlinatlon of Beams g 



Let the beams AB, BC, and CD, fig, 10. be connectea 



Ingelher by cords BB and CC, and freely suspended from 

 A and D, and suppose them by their i^ravity to have taken 

 the ligiuc represented, and consequently that of equili- 

 brium. Continue the lines BB and CC both ways, and 

 their intersection at t will necessarily be in the vertical 

 that passes through the centre of gravity of the beam BC 

 at b. Let a and c be the centres of gravity of the beams 



AB and CD, from which let fall the verticals as and c?<, 



cutttmg the hnes BB and CC continued, in s and u. Dravi^ 

 *A and ?<D. The directions of ilie forces which sustain 

 the beams AB will be sA. and *B ; of those which 



sustain BC, / B and t C ; and of those which sustain 



111 



CD, uC and ?<D. But obviously and necessarily these are 

 not parallel to the beams, as generally shown in diagrams 

 «»f this kind : for takins:; any individual intermediate beam, 

 tlie directions of those which adjoin it not necessarily in- 

 tersecting each other in the vertical that passes through its 

 centre of gravity, any forces supp<->sed to act in their di- 

 rections to sustain it, are not qualified to produce the equi- 

 librium, and therefore do not subsist in fact. 



From s continue the line tl^s to w, and parallel to uCt, 



and Du; draw sj) and sr. The sides Atz, ns and ^A of 

 the triangle Ans are proportional to the weight of the 



beam AB, to the force acting at B (= the tension of the 



It 

 cord BB) and to that acting at A, and in the same direc- 

 tions. The sides np, p s, and sn of the triangle vps, are 

 proportional to the weight of the beam BC, to the force 

 acting at C (= the tension of the cord CC) and to that 

 acting at B, equal and conirary to that acting at B (= the 



tension of the cord BB) and in parallel directions. Also 



lit I . II 



the sides pr, r s, and s p oi the triangle p r a are propor- 



I 

 tional to the weight of the beam CD, to the force acting at 



D, and to that at C equal and contrary to that acting at 



C (= the tension of the cord CC) being in parallel di- 

 rections. 



Now, 



