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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. 2010 Excerpt: ... BT: TC: . C: B; join AT, and take TE: EA:: A: Jb+ C, or TE: TA:: A: A + B + C, then will E be the center of gravity of the system (Art. 161.) Suppose the momentum communicated to A would cause it to move from A to a? in T, and at x let the body be stopped; join Tx, and take TF: Tx:: A: A+B+C, then Fis the center of gravity of the bodies when they are at x, B, C; join EF, and since TE: TA:: A: A+B+C:: TF i Tx, EF is parallel to Ax (Euo. 2. vi.), and consequently the triangles TEF, TAx, are similar; therefore EF-. Ax.: A: A + B + C. Hence if one body A in the system be moved from A to x, the center of gravity is moved from E to F; which point may be thus determined; draw EF parallel to Ax, and take EF: Ax:: A: A + B + C. Next let a momentum be communicated to B, which would cause it to move from B to y in T ; ai y let the body be stopped; then, according to the rule above laid down, draw FG parallel to By, and take FG: By:: B: A+B+C, and G will be the center of gravity of the bodies when they are at x, y, C. In the same manner, let a momentum be communicated to C, which would cause it to move from C to z in T, and at z let' the body be stopped; draw GH parallel to Cz, and take GH: Cz:: C: A + B+C, then H is the center of gravity of the bodies when they are at x, y, z. If now the momenta, instead of being communicated separately, be communicated at the same instant to the bodies, at the end of T they will be found in x, y, z, respectively; therefore, at the end of T, their common center of gravity will be in H. Now let E be a body equal to A + B+C, and let the same momentum be communicated to it that was before communicated to A, and in the same direction; then since EF is parallel to Ax, EF is in the direction in which the body E ...

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