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

by Robert S. Ball

INTRODUCTION.

48. The parallelogram of forces enables us to find the resultant of two forces which intersect: but since parallel forces do not intersect, the construction does not avail to determine the resultant of two parallel forces. We can, however, find this resultant very simply by other means.

Fig. 19.

49. [Fig. 19] represents a wooden rod 4' long, sustained by resting on two supports a and b, and having the length a b divided into 14 equal parts. Let a weight of 14 lbs. be hung on the rod at its middle point c; this weight must be borne by the supports, and it is evident that they will bear the load in equal shares, for since the weight is at the middle of the rod there is no reason why one end should be differently circumstanced from the other. Hence the total pressure on each of the supports will be 7 lbs., together with half the weight of the wooden bar.

50. If the weight of 14 lbs. be placed, not at the centre of the bar, but at some other point such as d, it is not then so easy to see in what proportion the weight is distributed between the supports. We can easily understand that the support near the weight must bear more than the remote one, but how much more? When we are able to answer this question, we shall see that it will lead us to a knowledge of the composition of parallel forces.

PRESSURE OF A LOADED BEAM
ON ITS SUPPORTS.

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.

EQUILIBRIUM OF A BAR SUPPORTED
ON A KNIFE-EDGE.

60. The weight of the bar has hitherto somewhat complicated our calculations; the results would appear more simply if we could avoid this weight; but since we want a strong bar, its weight is not so small that we could afford to overlook it altogether. By means of the arrangement of [Fig. 21], we can counterpoise the weight of the bar. To the centre of A B a cord is attached, which, passing over a fixed pulley D, carries a hook at the other end. The bar, being a pine rod, 4 feet long and 1 inch square, weighs about 12 ounces; consequently, if a weight of twelve ounces be suspended from the hook, the bar will be counterpoised, and will remain at whatever height it is placed.

Fig. 21.

61. a b is divided by lines drawn along it at distances of 1" apart; there are thus 48 of these divisions. The weights employed are furnished with rings large enough to enable them to be slipped on the bar and thus placed in any desired position.

62. Underneath the bar lies an important portion of the arrangement; namely, the knife-edge c. This is a blunt edge of steel firmly fastened to the support which carries it. This support can be moved along underneath the bar so that the knife-edge can be placed under any of the divisions required. The bar being counterpoised, though still unloaded with weights, may be brought down till it just touches the knife-edge; it will then remain horizontal, and will retain this position whether the knife-edge be at either end of the bar or in any intermediate position. I shall hang weights at the extremities of the rod, and we shall find that there is for each pair of weights just one position at which, if the knife-edge be placed, it will sustain the rod horizontally. We shall then examine the relations between these distances and the weights that have been attached, and we shall trace the connection between the results of this method and those of the arrangement that we last used.

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.