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by Robert S. BallApril 16th, 2023

*Experimental Mechanics by Robert S. Ball is part of the HackerNoon Books Series. You can jump to any chapter in this book **here**. PARALLEL FORCES
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*Introduction.—Pressure of a Loaded Beam on its Supports.—Equilibrium of a Bar supported on a Knife-edge.—The Composition of Parallel Forces.—Parallel Forces acting in opposite directions.—The Couple.—The Weighing Scales.*

**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 [Pg 35]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 [Pg 36]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, [Pg 38]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 [Pg 39]that we could afford to overlook it altogether. By means of the arrangement of Fig. 21, we can counterpoise the weight of the bar. [Pg 40]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 [Pg 41]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. [Pg 42]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 [Pg 43]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.[Pg 44]

PARALLEL FORCES ACTING IN

OPPOSITE DIRECTIONS.

70. Since the forces ap, bq, cr (Fig. 22) are in equilibrium, it follows that we may look on bq as balancing in the position which it occupies the two forces of ap and cr in their positions. This may remind us of the numerous instances we have already met with, where one force balanced two greater forces: in the present case ap and cr are acting in opposite directions, and the force bq which balances them is equal to their difference. A force bt equal and opposite to bq must then be the resultant of cr and ap, since it is able to produce the same effect. Notice that in this case the resultant of the two forces is not between them, but that it lies on the side of the larger. When the forces act in the same direction, the resultant is always between them.

71. The actual position which the resultant of two opposite parallel forces occupies is to be found by the following rule. Divide the distance between the forces into as many equal parts as there are pounds in their difference, then measure outwards from the point of application of the larger force as many of these parts as there are pounds in the smaller force; the point thus found determines the position of the resultant. Thus, if the forces be 14 and 20, the difference between them is 6, and therefore the distance between their directions is divided into six parts; from the point of application of the force of 20, 14 parts are measured outwards, and thus the position of the resultant is determined. Hence we have the means of compounding two parallel forces in general.

THE COUPLE.

72. In one case, however, two parallel forces have no resultant; this [Pg 45]occurs when the two forces are equal, and in opposite directions. A pair of forces of this kind is called a couple; there is no single force which could balance a couple,—it can only be counterbalanced by another couple acting in an opposite manner. This remarkable case, may be studied by the arrangement of Fig. 23.

Fig. 23.

A wooden rod, a b 48" × 0"·5 × 0"·5, has strings attached to it at points a and d, one foot distant. The string at d passes over a fixed pulley e, and at the end p a hook is attached for the purpose of receiving weights, while a similar hook descends from a; the weight of the rod itself, which only amounts to three ounces, may be neglected, as it is very small compared with the weights which will be used.

73. Supposing 2 lbs. to be placed at p, and 1 lb. at q, we have two parallel forces acting in opposite directions; and since their difference is 1 lb., it follows from our rule that the point f, where d f is equal to a d, is the point where the resultant is applied. You see this is easily verified, for by placing my finger over the rod at f it remains horizontal and in equilibrium; whereas, when I move my finger to one side or the other, equilibrium is impossible. If I move it nearer to b, the end a ascends. If I move it towards a, the end b ascends.[Pg 46]

74. To study the case when the two forces are equal, a load of 2 lbs. may be placed on each of the hooks p and q. It will then be found that the finger cannot be placed in any position along the rod so as to keep it in equilibrium; that is to say, no single force can counteract the two forces which form the couple. Let o be the point midway between a and d. The forces evidently tend to raise ob and turn the part o a downwards; but if I try to restrain o b by holding my finger above, as at the point x, instantly the rod begins to turn round x and the part from a to x descends. I find similarly that any attempt to prevent the motion by holding my finger underneath is equally unsuccessful. But if at the same time I press the rod downwards at one point, and upwards at another with suitable force, I can succeed in producing equilibrium; in this case the two pressures form a couple; and it is this couple which neutralizes the couple produced by the weights. We learn, then, the important result that a couple can be balanced by a couple, and by a couple only.

75. The moment of a couple is the product of one of the two equal forces into their perpendicular distance. Two couples tending to turn the body to which they are applied in the same direction will be equivalent if their moments are equal. Two couples which tend to turn the body in opposite directions will be in equilibrium if their moments are equal. We can also compound two couples in the same or in opposite directions into a single couple of which the moment is respectively either the sum or the difference of the original moments.

THE WEIGHING SCALES.

76. Another apparatus by which the nature of parallel forces may be [Pg 47]investigated is shown in Fig. 24; it consists of a slight frame of wood a b c, 4' long. At e, a pair of steel knife-edges is clamped to the frame. The knife-edges rest on two pieces of steel, one of which is shown at o f. When the knife-edges are suitably placed the frame is balanced, so that a small piece of paper laid at a will cause that side to descend.

Fig. 24.

77. We suspend two small hooks from the points a and b: these are made of fine wire, so that their weight may be left out of consideration. With this apparatus we can in the first place verify the principle of equality of moments: for example, if I place the hook a at a distance of 9" from the centre o and load it with 1 lb., I find that when b is laden with 0·5 lb. it must be at a distance of 18" from o in order to counterbalance a; the moment in the one case is 9 × 1, in the other 18 × 0·5, and these are obviously equal.

78. Let a weight of 1 lb. be placed on each of the hooks, the frame [Pg 48]will only be in equilibrium when the hooks are at precisely the same distance from the centre. A familiar application of this principle is found in the ordinary weighing scales; the frame, which in this case is called a beam, is sustained by two knife-edges, smaller, however, than those represented in the figure. The pans p, p are suspended from the extremities of the beam, and should be at equal distances from its centre. These scale-pans must be of equal weight, and then, when equal weights are placed in them, the beam will remain horizontal. If the weight in one slightly exceed that in the other, the pan containing the heavier weight will of course descend.

79. That a pair of scales should weigh accurately, it is necessary that the weights be correct; but even with correct weights, a balance of defective construction will give an inaccurate result. The error frequently arises from some inequality in the lengths of the arms of the beam. When this is the case, the two weights which really balance are not equal. Supposing, for instance, that with an imperfect balance I endeavour to weigh a pound of shot. If I put the weight on the short side, then the quantity of shot balanced is less than 1 lb.; while if the 1 lb. weight be placed at the long side, it will require more than 1 lb. of shot to produce equilibrium. The mode of testing a pair of scales is then evident. Let weights be placed in the pans which balance each other; if the weights be interchanged and the balance still remains horizontal, it is correct.

80. Suppose, for example, that the two arms be 10 inches and 11 inches long, then, if 1 lb. weight be placed in the pan of the 10-inch end, its moment is 10; and if ¹⁰/₁₁ of 1 lb. be placed in the pan belonging to the 11-inch end, its moment is also 10: hence 1 lb. at the short end balances ¹⁰/₁₁ of 1 lb. at the long end; and therefore, if the shopkeeper placed his weight in the short arm, his customers would lose [Pg 49]¹/₁₁ part of each pound for which they paid; on the other hand, if the shopkeeper placed his 1 lb. weight on the long arm, then not less than ¹¹/₁₀ lb. would be required in the pan belonging to the short arm. Hence in this case the customer would get ¹/₁₀ lb. too much. It follows, that if a shopman placed his weights and his goods alternately in the one scale and in the other he would be a loser on the whole; for, though every second customer gets ¹/₁₁ lb. less than he ought, yet the others get ¹/₁₀ lb. more than they have paid for.[Pg 50]

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