Читать книгу Experimental Mechanics - Robert S. Ball - Страница 20

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51. To study this question we shall employ the apparatus shown in Fig. 20. An iron bar 5' 6" long, weighing 10 lbs., rests in the hooks of the spring balances a, c, in the manner shown in the figure. These hooks are exactly five feet apart, so that the bar projects 3" beyond each end. The space between the hooks is divided into twenty equal portions, each of course 3" long. The bar is sufficiently strong to bear the weight b of 20 lbs. suspended from it by an S hook, without appreciable deflection. Before the weight of 20 lbs. is suspended, the spring balances each show a strain of 5 lbs. We would expect this, for it is evident that the whole weight of the bar amounting to 10 lbs. should be borne equally by the two supports.

52. When I place the weight in the middle, 10 divisions from each end, I find the balances each indicate 15 lbs. But 5 lbs. is due to the weight of the bar. Hence the 20 lbs. is divided equally, as we have already stated that it should be. But let the 20 lbs. be moved to any other position, suppose 4 divisions from the right, and 16 from the left; then the right-hand scale reads 21 lbs., and the left-hand reads 9 lbs. To get rid of the weight of the bar itself, we must subtract 5 lbs. from each. We learn therefore that the 20 lb. weight pulls the right-hand spring balance with a strain of 16 lbs., and the left with a strain of 4 lbs. Observe this closely; you see I have made the number of divisions in the bar equal to the number of pounds weight suspended from it, and here we find that when the weight is 16 divisions from the left, the strain of 16 lbs. is shown on the right. At the same time the weight is 4 divisions from the right, and 4 lbs. is the strain shown to the left.

Fig. 20.

53. I will state the law of the distribution of the load a little more generally, and we shall find that the bar will prove the law to be true in all cases. Divide the bar into as many equal parts as there are pounds in the load, then the pressure in pounds on one end is the number of divisions that the load is distant from the other.

54. For example, suppose I place the load 2 divisions from one end: I read by the scale at that end 23 lbs.; subtracting 5 lbs. for the weight of the bar, the pressure due to the load is shown to be 18 lbs., but the weight is then exactly 18 divisions distant from the other end. We can easily verify this rule whatever be the position which the load occupies.

55. If the load be placed between two marks, instead of being, as we have hitherto supposed, exactly at one, the partition of the load is also determined by the law. Were it, for example, 3·5 divisions from one end, the strain on the other would be 3·5 lbs.; and in like manner for other cases.

56. We have thus proved by actual experiment this useful and instructive law of nature; the same result could have been inferred by reasoning from the parallelogram of force, but the purely experimental proof is more in accordance with our scheme. The doctrine of the composition of parallel forces is one of the most fundamental parts of mechanics, and we shall have many occasions to employ it in this as well as in subsequent lectures.

57. Returning now to Fig. 19, with which we commenced, the law we have discovered will enable us to find how the weight is distributed. We divide the length of the bar between the supports into 14 equal parts because the weight is 14 lbs.; if, then, the weight be at d, 10 divisions from one end a, and 4 from the other b, the pressure at the corresponding ends will be 4 and 10. If the weight were 2·5 divisions from one end, and therefore 11·5 from the other, the shares in which this load would be supported at the ends are 11·5 lbs. and 2·5 lbs. The actual pressure sustained by each end is, however, about 6 ounces greater if the weight of the wooden bar itself be taken into account.

58. Let us suspend a second weight from another point of the bar. We must then calculate the pressures at the ends which each weight separately would produce, and those at the same end are to be added together, and to half the weight of the bar, to find the total pressure. Thus, if one weight of 20 lbs. were in the middle, and another of 14 lbs. at a distance of 11 divisions from one end, the middle weight would produce 10 lbs. at each end and the 14 lbs. would produce 3 lbs. and 11 lbs., and remembering the weight of the bar, the total pressures produced would be 13 lbs. 6 oz. and 21 lbs. 6 oz. The same principles will evidently apply to the case of several weights: and the application of the rule becomes especially easy when all the weights are equal, for then the same divisions will serve for calculating the effect of each weight.

59. The principles involved in these calculations are of so much importance that we shall further examine them by a different method, which has many useful applications.