(2) For equilibrium of rotation the funicular polygon must close with a vertex on every force.
The following, solutions show how to apply these conditions.
I. To find two unknown magnitudes. Sup pose that a weightless beam carries a load W and rests on two supports as in Fig. 3. It is required to find the reactions R. and R.
which are evidently vertical. The answers are obtained as follows; it is left to the reader to supply the underlying statical reasons. In ii draw W and any rays a, b, and in i parallels to a, b, intersecting on W. RI, as yet unknown, will be determined by some ray c also un known. Now from the two triangles which any assumed ray c forms with a and b in ii we can find the two points of concurrence through which c must pass in i ; this gives the true direction of c.
II. To find two unknown magnitudes and one unknown direction. The rod in Fig. 4 weights 200 pounds, rests on a knife edge, is hinged at the left, and is subjected to 200 pounds at the right. Find the reactions. In this problem we shall use a slightly modified form, due to H. T. Eddy, of the ingenious nota tion devised by R. H. Bow in The Eco nomics of Construction in Relation to Framed Structures,' (1873) ; the full power of the nota tion will not be evident until we take up frames. Letter the forces ab, bc, cd, da, nam ing first all those completely known, then cd, known only in direction, and finally the least known force — the hinge reaction da — of which only a point of application is given. The forces are to be used in this order. In ii draw ab, bc and rays from any pole 0; this is as far as we can go. However, assume a ray Od and imagine ii completed. In i draw oa, ob, oc. Now od must go through the in tersection of oc and cd and of oa and da since it forms triangles with them. But the inter section of oa and da cannot be found unless oa passes through the hinge. Hence to solve a problem of this type, start the first ray through the only given point — usually a hinge —of the force whose direction is unknown; the solution is then easily finished. Observe
that, for example, the sides ob and oc of the funicular polygon meet on the force bc.
III. To find three unknown magnitudes. Two of the unknown magnitudes can be re placed by a single force through their inter section, unknown in direction as well as in magnitude. The problem therefore is reduced to type II.
Properties of the funicular polygon. If a string, fastened at A and B is in equilibrium under forces ab and bc as in Fig. 5 it is evi dent that if 0 is properly chosen the sides of the funicular polygon will coincide with the lines of the string. (Latin, funiculus). The string polygon gives also the lines of thrust in an arch. The determination of the forces in a loaded cable or an arch requires the poly gon to pass through two and sometimes three given points. Consider first the effect of shift ing the pole. Assume, for brevity, two forces bc and construct two funicular polygons with any poles 0 and 0', Fig. 6.
The pairs of forces O'a and O'b and Oa and Ob in ii have the same resultant, and if their senses are properly chosen they will be in equilibrium and form the quadrilateral OaO'bO. By the theorem of four forces the line through P and R must be parallel to 00'; by similar reasoning Q is also on this line. Hence cor responding pairs of sides of two polygons in tersect on a line, called the polar axis, parallel to that connecting the poles. Consequently to pass a funicular polygon through two given points, say P and P', proceed thus: draw any polygon through P and pass one side o'b, of another through P'. This determines a point R on the polar axis through P. Then 0' must lie on a parallel to PR and on a ray O'b parallel to side o'b. Furthermore if the polygon is to go through another point P" (not shown) besides P and P', the new polar axis must pass through P, P'. Hence from 0' draw a parallel to PP'; the new pole 0" will lie on it and its position will be determined by the direction of the side which is arbitrarily chosen to pass through P".