Читать книгу Experimental Mechanics - Robert S. Ball - Страница 22

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66. Having now examined the subject experimentally, we proceed to investigate what may be learned from the results we have proved.

Fig. 22.

The weight of the bar being allowed for in the way we have explained, by subtracting one-half of it from each of the strains indicated by the spring balance (Fig. 20), we may omit it from consideration. As the balances are pulled downwards by the bar when it is loaded, so they will react to pull the bar upwards. This will be evident if we think of a weight—say 14 lbs.—suspended from one of these balances: it hangs at rest; therefore its weight, which is constantly urging it downwards, must be counteracted by an equal force pulling it upwards. The balance of course shows 14 lbs.; thus the spring exerts in an upward pull a force which is precisely equal to that by which it is itself pulled downwards.

67. Hence the springs are exerting forces at the ends of the bar in pulling them upwards, and the scales indicate their magnitudes. The bar is thus subject to three forces, viz.: the suspended weight of 20 lbs., which acts vertically downwards, and the two other forces which act vertically upwards, and the united action of the three make equilibrium.

68. Let lines be drawn, representing the forces in the manner already explained. We have then three parallel forces ap, bq, cr acting on a rod in equilibrium (Fig. 22). The two forces ap and bq may be considered as balanced by the force cr in the position shown in the figure, but the force cr would be balanced by the equal and opposite force cs, represented by the dotted line. Hence this last force is equivalent to ap and bq. In other words, it must be their resultant. Here then we learn that a pair of parallel forces, acting in the same direction, can be compounded into a single resultant.

69. We also see that the magnitude of the resultant is equal to the sum of the magnitudes of the forces, and further we find the position of the resultant by the following rule. Add the two forces together; divide the distance between them into as many equal parts as are contained in the sum, measure off from the greater of these two forces as many parts as there are pounds in the smaller force, and that is the point required. This rule is very easily inferred from that which we were taught by the experiments in Art. 51.