36 APPLIED MECHANICS 
The lines of action of OP and OQ intersect at O, and the parallelogram 
POQT determines OT, their resultant. The lines of action of OT and 
O,R intersect at O,, and O,T, being made equal to OT, the parallelo- 
gram T,O,RU determines O,U, the resultant of OT and O,R. The lines 
of action of O,U and O,S intersect at O,, and O,U, being made equal 
to O,U, the parallelogram U,O,SV determines O,V, the resultant of 
O,U and O,S, and therefore also the resultant of OP, OQ, O,R, and 
O,S. By this method the resultant of the given forces is completely 
determined: 
The polygon of forces ABCDE may be drawn as before, and the 
closing line AE will represent in magnitude and direction the resultant 
of the given forces, but the line of action is undetermined. A point in 
the line of action of the resultant may however be found by drawing 
another polygon, called the funicular polygon, which is discussed in the 
next Article but one. 
When the given forces are all parallel, the polygon of forces becomes 
a straight line. 
55. Lettering of Forces—Bow’s Notation.—In Fig. 31 the diagram 
(m) shows the lines of action of a number of forces which act at a point 
and which are in equilibrium. The ; 
diagram (n) is the corresponding poly- b 
gon of forces. In one system of letter- 
ing, each force is denoted by a single 
letter, as P. In Bow’s notation, each 
force is denoted by two letters, which 
are placed on opposite sides of the 
line of action of the force in diagram 
(m), and at the angular points of the 
polygon in diagram (z). In Bow’s Fig. 31. 
notation the force P is referred to as 
the force AB. In like manner the force Q is referred to as the force BC. 
The diagram (m), which shows the lines of action of the forces, is called 
the spice diagram, and the diagram (7), which shows the polygon of forces, 
is called the farce diagram. 
In this work, when Bow’s notation is used, capital letters will be 
placed on the space diagram, and the corresponding small letters on the 
force diagram. 
56. The Funicular Polygon.—Let P, Q, R, and § (Fig. 32) be four 
forces which act on a rigid body, and which are balanced by a fifth force 
T, which is at present unknown. Draw the polygon of forces abcde, 
then from what has already been shown the,line ea which closes the 
polygon will represent the magnitude and direction of the fifth force T. 
Take any point o and join it to a, 6,¢c,d,ande. Take any point 2 in 
the line of action of P and draw the line 2B3 parallel to ob to meet the 
line of action of Q at 3. Draw 3C4 parallel to oc to meet the line of 
action of R at 4. Draw 4D5 parallel to od to meet the line of action 
of S at 5. Draw 5El1 parallel to oe and 2A1 parallel to oa. The latter 
two lines will meet at a point 1 on the line of action of T. 
Conceive that the lines A, ae C, D, and E represent bars jointed to 
one another at the points 1, 2, $, 4, and 5. Then these bars may be 
supposed to take the place of the rigid body upon which the five forces 
