This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1901 edition. Excerpt: ...It can easily be seen that this must lie at the centre of the rod. For the rod may be imagined to be composed of a number of small elements of mass, situated symmetrically with regard to the centre of the rod. Thus the centre of gravity of each pair of equal elements, equidistant from the centre, and ...
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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1901 edition. Excerpt: ...It can easily be seen that this must lie at the centre of the rod. For the rod may be imagined to be composed of a number of small elements of mass, situated symmetrically with regard to the centre of the rod. Thus the centre of gravity of each pair of equal elements, equidistant from the centre, and on opposite sides of it, will coincide with the centre of the rod. III. Centre of gravity of a thin, uniform circular ring. Imagine the ring to be divided into a number of small equal elements by means of numerous diameters of the circle. The centre of gravity of each pair of equal elements at opposite ends of a diameter will be at the centre of the circle; hence the centre of gravity of the whole will coincide with that point. IV. Since a circular lamina may be divided into a number of uniform concentric rings, each of which has its centre of gravity at the geometrical centre of the lamina, the centre of gravity of the whole lamina will coincide with its geometrical centre. The simple results obtained above will now be used to determine the centre of gravity of a number of solids of more complicated shapes. Problem.--To find the centre of gravity of a triangular lamina. Let m = the mass of unit area of the triangular lamina ABC (Fig. 39). Imagine the triangle to be divided into a number of narrow strips, similar to DEHG, by lines parallel to BC. Then, since the centre of gravity of each of these strips will lie on the straight line A M, which passes through the centres of all of them, it is obvious that the centre of gravity of the triangular lamina will also lie on that line. Let x be the distance from A along AM at which the centre of gravity lies, and let AM = b. From F, the point of intersection of DE and AM, draw FL perpendicular to DE...
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