56 



REACTIONS FOR A SINGLE FORCE. 



Art. 46. 



This value of E^ substituted in the equation above will give the 



value of R 2 = P P~J~ P~J~' As a check, with center at A, 

 from 3 moments = 0, 



R 2 L PZ X or R 2 = P~Y~ 



This illustrates the law of the lever, to which the student 

 should become accustomed so that it ivill be unnecessary to think 

 of the moment equations. Thus, if (Fig. 31) L = 20 ft., and J T = 

 12 ft, andZ 2 ^=8ft, R t =^P=% P, andR*=l s $P= |P. Asa 

 check, Ri+Rf=(l + l)P = P. 



Graphically, the problem is to find two vertical forces R^ and 

 which will be in equilibrium with P. 



It is only necessary to draw a closed string polygon and a 



corresponding force polygon. 

 The three forces P, R 19 and R 2 

 must act through the apexes of 

 the string polygon. In Fig. 32, 

 draw the strings 1 and 2 from 

 C parallel to the equilibrants 

 (bO and Oa) of P; where these 

 strings intersect R^ and R 2 in D 

 ^ ^ _ and E, there must be apexes of 



K 



fb/yyofr the string polygon ; hence DCED 

 is the string polygon. Draw- 

 ing nO parallel to D E, it is the 

 ray corresponding to the string 

 D E and therefore cuts off the 

 Fi 8- 32 - forces R! and R 2 in the force 



polygon which is a b n a, a straight line, since the forces are par- 

 allel. The student should prove that there is equilibrium as was 

 done in Art. 37. 



It is evident that P = R -{-R 2 ; the line nO, drawn parallel to 

 the closing line, divides P into two parts equal to R^ and R 2 . 



Exactly the same pro- e t o \ p 



cedures apply to any kind of 

 a truss supported at its ends, 

 because reactions are inde-a 

 pendent of the form of the^T 



structure. 

 In Fig. 



33, the panels 



