38 APPLIED MECHANICS 
P and Q is represented in magnitude and direction by ca, and that the 
point of intersection of the sides A and C of the funicular polygon is a 
point in the line of action of this equilibrant. Also the equilibrant of P, 
Q, and R is represented in magnitude and direction by da, and the point 
of intersection of the sides A and D of the funicular polygon is a point 
in the line of action of this equilibrant. 
Having shown that the funicular polygon together with the polygon 
of forces may be used to determine the equilibrant of a system of non- 
concurrent forces, it is obvious that the same construction will also deter- 
mine the resultant of that system of forces, since the resultant acts along 
the same line and has the same magnitude as the equilibrant, but acts in 
the opposite direction. 
57. Examples of the use of the Funicular Polygon.—T wo examples 
will now be worked out to further illustrate the use of the funicular polygon. 
(1) Three vertical forces, AB, BC, and CD, act on a horizontal beam, 
as shown in Fig. 34. The beam rests on supports at its ends where there 
are vertical reactions DE and EA. It is required to determine the 
magnitudes of these reactions. 
Since the forces are all parallel the polygon of forces will be a straight 
ke I-64 - 2-6"- oe - 2! - eg" 
A CONN B ot C ant D 
\ 
C E 
AJE 0 D 
B (6) 
0 ed 
/ oe a 
Ao 5 eee 
er te a 
Fig. 34. 
line abcdea, and the reactions will be represented by de and ea, the 
position of the point e being as yet unknown. 
Choose a pole 0. Join oa, ob, oc, and od. Draw OA, OB, OC, and 
OD parallel to oa, ob, oc, and od respectively as shown. These four 
lines, OA, OB, OC, and OD, will form four sides of the funicular polygon, 
of which OE will be the closing side. Draw oe parallel to OE to meet 
ad ate. This completes the solution. It will be found that DE=1-48 
tons, and EA=1-27 tons. 
(2) A horizontal beam AB (Fig. 35) is acted on by an inclined 
force P= 200 lbs., a vertical foree Q@=150 Ibs., and an inclined force 
R=500 lbs., as shown. There is also a vertical foree T,; whose magnitude 
is unknown, acting at A, and a force S acting at B, whose magnitude 
and direction are both unknown. These forces being in equilibrium, it is 
required to determine T and 8. 
By the polygon of forces (a) and the funicular polygon 1234 the line 
of action 4N of the resultant U of the forces P, Q, and R is found, 
Replacing P, Q, and R by U, there are now only three forces acting on 
