Part 3

36. An important point may be brought out by contrasting the arrangements of Figs. 15 and 16. In the one case three cords are used, and in the other three rods. Three rods would have answered for both, but three cords would not have done for the tripod. In one the cords are strained, and the tendency of the strain is to break the cords, but in the other the nature of the force down the rods is entirely different; it does not tend to pull the rod asunder, it is trying to crush the rod, and had the weight been large enough the rods would bend and break. I hold one end of a pencil in each hand and then try to pull the pencil asunder; the pencil is in the condition of the cords of Fig. 15; but if instead of pulling I push my hands together, the pencil is like the rods in Fig. 16.

37. This distinction is of great importance in mechanics. A rod or cord in a state of tension is called a “tie”; while a rod in a state of compression is called a “strut.” Since a rod can resist both tension and compression it can serve either as a tie or as a strut, but a cord or chain can only act as a tie. A pillar is always a strut, as the superincumbent load makes it to be in a state of compression. These distinctions will be very frequently used during this course of lectures, and it is necessary that they be thoroughly understood.

THE JIB AND TIE ROD.

38. As an illustration of the nature of the “tie” and “strut,” and also for the purpose of giving a useful example of the decomposition of forces, I use the apparatus of Fig. 17 (see next page).

It represents the principle of the framework in the common lifting crane, and has numerous applications in practical mechanics. A rod of wood B C 3' 6" long and 1" × 1" section is capable of turning round its support at the bottom B by means of a joint or hinge: this rod is called the “jib”; it is held at its upper end by a tie A C 3' long, which is attached to the support above the joint. A B is one foot long. From the point C a wire descends, having a hook at the end on which a weight can be hung. The tie is attached to the spring balance, the index of which shows the strain. The spring balance is secured by a wire-strainer, by turning the nut of which the length of the wire can be shortened or lengthened as occasion requires. This is necessary, because when different weights are suspended from the hook the spring is stretched more or less, and the screw is then employed to keep the entire length of the tie at 3'. The remainder of the tie consists of copper wire.

39. Suppose a weight of 20 lbs. be suspended from the hook W, it endeavours to pull the top of the jib downwards; but the tie holds it back, consequently the tie is put into a state of tension, as indeed its name signifies, and the magnitude of that tension is shown to be 60 lbs. by the spring balance. Here we find again what we have already so often referred to; namely, one force developing another force that is greater than itself, for the strain along the tie is three times as great as the strain in the vertical wire by which it was produced.

40. What is the condition of the jib? It is evidently being pushed downwards on its joint at B; it is therefore in a state of compression; it is a strut. This will be evident if we think for a moment how absurd it would be to endeavour to replace the jib by a string or chain: the whole arrangement would collapse. The weight of 20 lbs. is therefore decomposed by this contrivance into two other forces, one of which is resisted by a tie and the other by a strut.

41. We have no means of showing the magnitude of the strain along the strut, but we shall prove that it can be computed by means of the parallelogram of force; this will also explain how it is that the tie is strained by a force three times that of the weight which is used. Through C (Fig. 18) draw C P parallel to the tie A B, and P Q parallel to the strut C B then B P is the diagonal of the parallelogram whose sides are each equal to B C and B Q. If therefore we consider the force of 20 lbs. to be represented by B P, the two forces into which it is decomposed will be shown by B Q and B C; but A B is equal to B Q, since each of them is equal to C P; also B P is equal to A C. Hence the weight of 20 lbs. being represented by A C, the strain along the tie will be represented by the length A B, and that along the strut by the length B C. Remembering that A B is 3' long, C B 3' 6", and A C 1', it follows that the strain along the tie is 60 lbs., and along the strut 70 lbs., when the weight of 20 lbs. is suspended from the hook.

42. In every other case the strains along the tie and strut can be determined, when the suspended weight is known, by their proportionality to the sides of the triangle formed by the tie, the jib, and the upright post, respectively.

43. In this contrivance you will recognize, no doubt, the framework of the common lifting crane, but that very essential portion of the crane which provides for the raising and lowering is not shown here. To this we shall return again in a subsequent lecture (Art. 332). You will of course understand that the tie rod we have been considering is entirely different from the chain for raising the load.

44. It is easy to see of what importance to the engineer the information acquired by means of the decomposition of forces may become. Thus in the simple case with which we are at present engaged, suppose an engineer were required to erect a frame which was to sustain a weight of 10 tons, let us see how he would be enabled to determine the strength of the tie and jib. It is of importance in designing any structure not to make any part unnecessarily strong, as doing so involves a waste of valuable material, but it is of still more vital importance to make every part strong enough to avoid the risk of accident, not only under ordinary circumstances, but also under the exceptionally great shocks and strains to which every machine is liable.

45. According to the numerical proportions we have employed for illustration, the strain along the tie rod would be 30 tons when the load was 10 tons, and therefore the tie must at least be strong enough to bear a pull of 30 tons; but it is customary, in good engineering practice, to make the machine of about ten times the strength that would just be sufficient to sustain the ordinary load. Hence the crank must be so strong that the tie would not break with a tension less than 300 tons, which would be produced when the crane was lifting 100 tons. So great a margin of safety is necessary on account of the jerks and other occasional great strains that arise in the raising and the lowering of heavy weights. For a crane intended to raise 10 tons, the engineer must therefore design a tie rod which not less than 300 tons would tear asunder. It has been proved by actual trial that a rod of wrought iron of average quality, one square inch in section, can just withstand a pull of twenty tons. Hence fifteen such rods, or one rod the section of which was equal to fifteen square inches, would be just able to resist 300 tons; and this is therefore the proper area of section for the tie rod of the crane we have been considering.

46. In the same way we ascertain the actual thrust down the jib; it amounts to 35 tons, and the jib should be ten times as strong as a strut which would collapse under a strain of 35 tons.

47. It is easy to see from the figure that the tie rod is pulling the upright, and tending, in fact, to make it snap off near B. It is therefore necessary that the upright support A B (Fig. 17) be secured very firmly.

LECTURE III. _PARALLEL FORCES._

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.

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.

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.

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.

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.

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.

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 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.