Study of Steam And the Marine Engine for Young Sea Officers in H.M. Navy ...
Study of Steam And the Marine Engine for Young Sea Officers in H.M. Navy ...
S. M. Saxby
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C by the law of forces. — Given A cAb and jLdAb, and the weight twenty tons. Draw the parallelogram Abdc A, making A 6 = 20 from any scale of equal parts, make b d parallel to A c, and d c parallel to A 6. Then A d divides the parallelogram into two triangles, in which, in AAcd, the side c d = A 6 = 20, and we wish to find A c, which will represent the tension of the chain A G. Let L cAd = 38°3' Z. d A 6 = 22^ 24' Then, because the right line A d falls on the two parallel 112 THE MARINE ENGINE.... right lines Ac,dby the alternate angles are equal (Euclid I. 29) ; therefore Z. A d c = 22^ 24' (24> Then, by rule of sines (45), Aa mxi L cAdSff" y . arco 0-210173 Is to side cd 20"* ... 1-301030 So is sin Z. cd A 22° 24' . . . 9-581005 ^^^ . To side A c . . . = 1-092208 = 12-37 N.B. — If the angles had not otherwise been given, A B, A C, B C being known would have given them (62). 2nd. To find the tension of AG by the rules for the lever. — In this case it will be obvious, that whether 20 tons be suspended from A towards D, or from D itself, its effect would be the same, and because two forces are act- ing, one on each side of A B, so, to preserve equilibrium, the axis of moment must be the point B.
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