About the Book
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1887 Excerpt: ...the way just explained by the construction of the special polar polygon, 8D lc 2Q c0. Secondly, by the graphic construction of the product of the resultant force acting along, FF, and the arm or perpendicular distance. aa d0, of any point, aa. The constructive process necessary for this purpose is given in Fig. 115. Set off along the horizontal line, 00 f, a distance, 00 t, equal to the unit-length of scale, OP. Along a line, tt', at right angles to, 00 T, lay off a distance, 11', equal in length to the arm of the resultant force; or to a0 d0. Make the length, 002, equal to, If, the magnitude of the resultant of the system of forces, (8), (1), (2); and from the point, 2, so determined erect a perpendicular, 22', meeting the line, 00t', in a point, 2'; then by similar triangles, 22'_ (V2 tt' 00t' that is, 22' x 001 = 00 2 x tt'. But, 00t = 0 P = unit-length of scale, therefore, 22' = 002 x tt' = Xc x a0d0 = magnitude of resultant x arm = moment about a0 Since the lines, 22' and, x, yv both represent one quantity, they must be equal in length, which fact can be used as a check upon the two methods employed. But the value of the same moment can be found in a third way. Decompose the system, (1), (2), (3), (4), (5), (6), (7), (8) into two series of forces, horizontal and vertical. The vertical series is represented on the polygon of forces, by the divisions (1), (2), (3), (4), (5), (6), and the two equal reactions at the abutments, (7') and (8'). The horizontal components are two equal and opposite thrusts applied at the abutments and represented in magnitude by the line, X P, on the polygon of forces. Name these thrusts respectively, (72) and (82). The moment about, a0, due to the forces, (8), (1), (2), on the left of the plane of section, can be then found by...
