# The Principles of Mechanics

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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. 1899 edition. Excerpt: ...R T-vh (1) we have x_R A cos M A B yR C sin MA B _ R A x sin MA B _x tan MA B R Cy cos MA B y But-= tan MAB y.-..R-4, = tan2MAB. When, as in the example, MA B = 30 degrees, tan SMA B =. Hence when A C=2 feet, RA = 6 inches, as obtained before. We now proceed to show how to find the resultant of two parallel forces. Once that can be done, the resultant of any number of them can be found by compounding them one by one, as has been pointed out in the case of convergent forces on page 66. We have already seen that two forces may have cooperating or antagonistic moments, and that we denote which is the case by algebraical signs given to the forces, or, what comes to the same thing, to their moments. If two parallel forces tend to produce rotation the same way round a point which is not between them, or would be concurrent if they were in the same straight line, they are said to be like, otherwise they are considered unlike. Let P, Q, be two parallel forces, and let their points of application, A and B, be considered to be rigidly connected. Take moments about A, letting fall A T perpendicular to Q B. Call this perpendicular a. Then we have--Moment of P about A = o, Moment of Q about A = Q a, .: the total moment about A=Q a, and this must evidently be the moment of the resultant of the system. Now, it is clear that the magnitude of the resultant is P+ Q if the forces are like, and P-Q if they are unlike. Here it will be observed that we take Q, and therefore its moment, as negative, whence P and its moment must be regarded as positive. Now, if the components are like, we have--Moment of the resultant about A = Q a, Magnitude of the resultant = P+ Q, .: the length of the perpendicular from A on to the resultant = pQ-Hence the line of action of the