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. 1914 Excerpt: ... III. APPLICATIONS OF LARGE ANGLES 57. Composition and Resolution of Forces. As in 48, the components on the axes of any force of magnitude F which makes an angle a with the posi-tive end of the -axis, are (1) Fx = Fcosa, Fy = F sin a. Given two forces, F1, F" which make angles a', a," respectively with the positive ...
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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. 1914 Excerpt: ... III. APPLICATIONS OF LARGE ANGLES 57. Composition and Resolution of Forces. As in 48, the components on the axes of any force of magnitude F which makes an angle a with the posi-tive end of the -axis, are (1) Fx = Fcosa, Fy = F sin a. Given two forces, F1, F" which make angles a', a," respectively with the positive end of the -axis, we may find the components of each of them on each of the axes. The sum of the two -components is F'x + F"x = F' cos a' + F" cosa," and is equal to the avcomponent of the resultant of F' and F," as is evident from a figure, since Proj, O =ON= OM + MN = Projx F' + Proj, F -. Hence the as-component Bx of is: (2) R, = F cos -' + F' cos -," and in like manner the /-component Ry of is (3) Ry = F' sin -' + F" sin a." These results hold, by (1), when F' and F" lie in any positions. From (2) and (3) the magnitude of the resultant and the angle which it makes with the positive -axis are given by (4) = tan e=B + Bx; where, in case of ambiguity, the quadrant in which lies is determined by the signs of Rx and Ry in an obvious manner. 58. The Projection Theorem. We can now generalize the preceding results and prove the following important theorem: The sum of the projections on any straight line I, of a broken line whose segments are taken in order so that the terminal point of each segment is the initial point of the next, is equal to the projection on I of a line segment joining the. initial point of the first segment of the broken line to the terminal point of the last segment. Proof. (1) The theorem is true when the broken line consists of two segments, for usi...
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