24 FORCES IN THE SAME PLANE. [CHAP. IL 



polygon from a to 0, to C and C to a, and referring back to 

 the frame, we find strain in in n acting towards apex m, there- 

 fore compressive / strain in m a acting away from m, therefore 

 t< nxile. In like manner ., i.% i , are in tension, while S 4 or d e 

 is in compression and S 5 or e n in tension. 



Hence we may,^ any two points of the equilibrium polygon 

 by joining them by a line. The forces acting at these points 

 are at once found by drawing from C in the force polygon a 

 parallel to this line to intersection with resultant. Thus a C 

 (since we have taken m n parallel to S t ) is the force in m n and 

 a 0, 5 a, are the forces opposed to the resultant at m and n. 



23. Influence of a Couple. Among the forces in Fig. 12 

 there are two, P 2 and P 3 which are equal, parallel and opposite, 

 the direction of rotation being as indicated by the arrow. Ex- 

 amining the equilibrium polygon, we see that the influence of 

 the couple is to shift S t through a certain distance parallel to 

 itself, to S 3 . Now suppose the forces composing the couple 

 were not given, but the value of the couple known, from the 

 direction of rotation and the area of the triangle A 2 P 2 P 8 , 

 which has its base equal to one of the forces and a height equal 

 to their perpendicular distance. In this case the lines 1 2, and 

 83 in the force polygon, would disappear, but we can none the 

 less find the point d, and from this point continue the polygon 

 by drawing S 4 and S s , and thus find the same points e and o' as 

 before. To do this we have simply to apply the principle 

 deduced in Art. 21, that one couple can be replaced by another 

 provided the area of the triangle is constant. 



In the present case we must replace the given couple by 

 another whose forces are St and 83, having the same direction 

 of rotation. 



Lay off then from a, a i equal by scale to S t as given in the 

 force polygon. Describe upon Sj the triangle a g h equal to 

 the given area A 2 P 2 PS. Draw g i, and then through h, h k 

 parallel to g i. The point k is upon the line of direction of 

 S 3 , or in other words the area of the triangle i k a is equal to 

 a g h. The proof is easy. The two triangles i g h and i g k 

 are equal, since they have the same base i g, and height. But 

 if from the triangle a i g we subtract i g h, we obtain a g h. 

 If from the same triangle a i g we subtract i g k, which is equal 

 to i g h, we obtain i k a. Equals subtracted from equals leave 

 equals. Hence i k a is equal to a g h. 



