Читать книгу Experimental Mechanics. A Course of Lectures Delivered at the Royal College of Science for Ireland онлайн

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18. It is easy, by the application of a set square, to prove that in this case the cords attached to the 3 lb. and 4 lb. weights are at right angles to each other. We could have inferred, from the parallelogram of force, that this must be the case, for the sides of the triangle o p r are 3, 4, and 5 respectively, and since the square of 5 is 25, and the squares of 3 and of 4 are 9 and 16 respectively, it follows that the square of one side of this triangle is equal to the sum of the squares of the two opposite sides, and therefore this is a right-angled triangle (Euclid, i. 48). Hence, since p r is parallel to o q, the angle p o q must also be a right angle.

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19. Cases might be multiplied indefinitely by placing various amounts of weight on the hooks, constructing the parallelogram on cardboard, and comparing it with the cords as before. We shall, however, confine ourselves to one more illustration, which is capable of very remarkable applications. Attach 1 lb. to each of the hooks e and f; the cord joining them remains straight until drawn down by placing a weight on the centre hook. A very small weight will suffice to do this. Let us put on half-a-pound; the position the cords then assume is indicated in ssss1. As before, each force is equal and opposite to the resultant of the other two. Hence a force of half-a-pound is the resultant of two forces each of 1 lb. The apparent paradox is explained by noticing that the forces of 1 lb. are very nearly opposite, and therefore to a large extent counteract each other. Constructing the cardboard parallelogram we may easily verify that the principle of the parallelogram of forces holds in this case also.