Experimental Mechanics / A Course of Lectures Delivered at the Royal College of Science for Ireland

by Robert S. Ball

63. Supposing that 6 lbs. be hung at each end of the rod, we might easily foresee that the knife-edge should be placed in the middle, and we find our anticipations verified. When the edge is exactly at the middle, the rod remains horizontal; but if it be moved, even through a very small distance, to either side, the rod instantly descends on the other. The knife-edge is 24 inches distant from each end; and if I multiply this number by the number of pounds in the weight, in this case 6, I find 144 for the product, and this product is the same for both ends of the bar. The importance of this remark will be seen directly.

64. If I remove one of the 6 lb. weights and replace it by 2 lbs., leaving the other weight and the knife-edge unaltered, the bar instantly descends on the side of the heavy weight; but, by slipping the knife-edge along the bar, I find that when I have moved it to within a distance of 12 inches from the 6 lbs., and therefore 36 inches from the 2 lbs., the bar will remain horizontal. The edge must be put carefully at the right place; a quarter of an inch to one side or the other would upset the bar. The whole load borne by the knife-edge is of course 8 lbs., being the sum of the weights. If we multiply 2, the number of pounds at one end, by 36, the distance of that end from the knife-edge, we obtain the product 72; and we find precisely the same product by multiplying 6, the number of pounds in the other weight, by 12, its distance from the knife-edge. To express this result concisely we shall introduce the word moment, a term of frequent use in mechanics. The 2 lb. weight produces a force tending to pull its end of the bar downwards by making the bar turn round the knife-edge. The magnitude of this force, multiplied into its distance from the knife-edge, is called the moment of the force. We can express the result at which we have arrived by saying that, when the knife-edge has been so placed that the bar remains horizontal, the moments of the forces about the knife-edge are equal.

65. We may further illustrate this law by suspending weights of 7 lbs. and 5 lbs. respectively from the ends of the bar; it is found that the knife-edge must then be placed 20 inches from the larger weight, and, therefore, 28 inches from the smaller, but 5 × 28 = 140, and 7 × 20 = 140, thus again verifying the law of equality of the moments.

From the equality of the moments we can also deduce the law for the distribution of the load given in [Art. 53]. Thus, taking the figures in the last experiment, we have loads of 7 lbs. and 5 lbs. respectively. These produce a pressure of 7 + 5 = 12 lbs. on the knife-edge. This edge presses on the bar with an equal and opposite reaction. To ascertain the distribution of this pressure on the ends of the beam, we divide the whole beam into 12 equal parts of 4 inches each, and the 7 lb. weight is 5 of these parts, i.e., 20 inches distant from the support. Hence the edge should be 20 inches from the greater weight, which is the condition also implied by the equality of the moments.

THE COMPOSITION OF PARALLEL FORCES.

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.