Nothing was said as to how F, 0, II, and K were drawn in a and b. These forces simply represent one of an infinite number of combinations of forces which would produce the same result. The point o is chosen at random, and lines (called rays) are drawn to the extremities of all the forces. The lines of force (A, B, and C) in diagram b (which is called the force diagram), are together called the load line. The line of forces (F, G, II, and K) in diagram a, together with the closing line yz, is called an equilibrium polygon.
39S. Statics of a Linear Arch. We shall assume that the lines in Fig. 215 by which we have represented forces F, G, II, and K represent struts which are hinged at their intersections with the forces A, B, and C, which represent loads; and that the two end struts F and K are hinged at two abutments located at y and z. Then all of the struts will be in compression, and the rays of the force dia gram will represent, at the same scale as that employed to repre sent forces or loads A, B, and C, the compression in each of the struts. In the force diagram, draw a line from o, parallel with the line yz. It intersects the load line in the point n. Considering the triangle opn as a force diagram, op represents the force F, while pn and on may represent the direction and amount of two forces which will hold P in equilibrium. Therefore pn would represent the amount and direction of the vertical component of the abutment reaction at y, and on would represent the component in the direction of yz. Similarly we may consider the triangle onq as a force diagram; that nq repre sents the vertical component R", and that on represents the component in the direction zy. Since on is common to both of these force tri angles, they neutralize each other, and the net resultant of the two forces F and K is the two vertical forces R and R"; but since the resultant R is the resultant of F and K, we may say that R' and R" are two vertical forces whose combined effect is the equal and opposite of the force R. Although an indefinite number of com binations of forces could begin and end at the points y and z, and could produce equilibrium with the forces A, B, and C, the forces R' and R" are independent of that particular combination of struts, F, G, IT, and K.
399., Graphical Demonstration of Laws of Statics by Student. The student should test all this work in Statics by drawing figures very carefully and on a large scale, in accordance with the general instructions as described in the sections, and should purposely make some variation in the relative positions and amounts of the forces from those indicated by the figures. By this means the student will
he able to obtain a virtual demonstration of the accuracy of the laws of Statics as formulated. The student should also remember that the laws are theoretically perfect; and when it is stated, for example, that certain lines should be parallel, or that a certain line drawn in a certain way should intersect some certain point, the mathematical laws involved are perfect; and if the drawing does not result in the expected way, it either proves that a blunder has been made, or it may mean that the general method is correct, but that the drawing is more or less inaccurate.
400. Equilibrium Polygon with Horizontal Closing Lines. In Fig. 216, have been drawn the same forces A, B, and C, having the same relative positions as in Fig. 215. The lines of action of the two vertical forces R' and R" have also been drawn in the same relative position as in Fig. 215. The point n has also been located on the load line in the same position as in Fig. 215. Thus far the lines are a repetition of those already drawn in Fig. 215, the remainder of the figure being omitted for simplicity. Since the point n in Fig. 215 is the end of the line from the trial pole o, which is parallel to the closing line yz, and since the point n is a definitely fixed point and determines the abutment reactions regardless of the position of the trial pole o, we may draw from n an indefinite horizontal line, such as no', and we know that the pole of any force diagram must be on this line if the closing line of the corresponding equilibrium polygon is to be a horizontal line. For example, we shall select a point o' on this line at random. From o' we shall draw rays to the points p, s, r, and q. From the point y, we shall draw a line parallel to o'p. Where this line intersects the force A, draw a line parallel to the ray o's. Where this intersects the force B, draw a line parallel to the ray o'r. Where this intersects the force C, draw a line parallel to the ray o'q. This line must intersect the point z', which is on a horizontal line from y. The student should make some such drawing as here described, and should demonstrate for himself the accuracy of this law. This equilibrium polygon is merely one of an infinite number which, if acting as struts, would hold these forces in equilibrium, but it com bines the special condition that it shall pass through the points y and z'. There are also an infinite number of equilibrium polygons which will hold these forces in equilibrium and which will pass through the points y and z'.