THE RESOLUTION OF FORCES

Experimental Mechanics by Robert S. Ball is part of the HackerNoon Books Series. You can jump to any chapter in this book here. THE RESOLUTION OF FORCES

LECTURE II. THE RESOLUTION OF FORCES.

Introduction.—One Force resolved into Two Forces.—Experimental Illustrations.—Sailing.—One Force resolved into Three Forces not in the same Plane.—The Jib and Tie-rod.

INTRODUCTION.

Fig. 9.

24. As the last lecture was principally concerned with discussing how one force could replace two forces, so in the present we shall examine the converse question, How may two forces replace one force? Since the diagonal of a parallelogram represents a single force equivalent to those represented by the sides, it is obvious that one force may be resolved into two others, provided it be the diagonal of the parallelogram formed by them.

25. We shall frequently employ in the present lecture, and in some of those that follow, the spring balance, which is represented in Fig. 9: the weight is attached to the hook, and when the balance is suspended by the ring, a pointer indicates the number of pounds on a scale. This balance is very convenient for showing the strain along a cord; for this purpose the balance is held by the ring while the cord is attached to the hook. It will be noticed that the balance has two rings and two corresponding hooks. The hook and ring at the top and bottom will weigh up to 300 lbs., corresponding to the scale which is seen. The hook and ring at the side correspond to another scale on the other face of the plate: this second scale weighs up to about 50 lbs., consequently for a weight under 50 lbs. the side hook and ring are employed, as they give a more accurate result than would be obtained by the top and bottom hook and ring, which are intended for larger weights. These ingenious and useful balances are sufficiently accurate, and can easily be tested by raising known weights. Besides the instrument thus described, we shall sometimes use one of a smaller size, and we shall be able with this aid to trace the existence and magnitude of forces in a most convenient manner.

ONE FORCE RESOLVED INTO TWO FORCES.

26. We shall first illustrate how a single force may be resolved into a pair of forces; for this purpose we shall use the arrangement shown in Fig. 10 (see next page).

The ends of a cord are fastened to two small spring balances; to the centre e of this cord a weight of 4 lbs. is attached. At a and b are pegs from which the balances can be suspended. Let the distances ae, be be each 12", and the distance ab 16". When the cord is thus placed, and the weight allowed to hang freely, each of the cords ea, eb is strained by an amount of force that is shown to be very nearly 3 lbs. by the balances. But the weight of 4 lbs. is the only [Pg 18]weight acting; hence it must be equivalent to two forces of very nearly 3 lbs. each along the directions ae and be. Here the two forces to which 4 lbs. is equivalent are each of them less than 4 lbs., though taken together they exceed it.

Fig. 10.

27. But remove the cords from ab and hang them on cd, the length cd being 1' 10", then the forces shown along fc and d are each 5 lbs.; here, therefore, one force of 4 lbs. is equivalent to two forces each of 5 lbs. In the last lecture (Art. 19) we saw that one force could balance two greater forces; here we see the analogous case of one force being changed into two greater forces. Further, we learn that the number of pairs of forces into which one force may be decomposed is unlimited, for with every different distance between the pegs different forces will be indicated by the balances.

Whenever the weight is suspended from a point half-way between the [Pg 19]balances, the forces along the cords are equal; but by placing the weight nearer one balance than the other, a greater force will be indicated on that balance to which the weight is nearest.

EXPERIMENTAL ILLUSTRATIONS.

Fig. 11.

28. The resolution or decomposition of one force into two forces each greater than itself is capable of being illustrated in a variety of ways, two of which will be here explained. In Fig. 11 an arrangement for this purpose is shown. A piece of stout twine ab, able to support from 20 lbs. to 30 lbs., is fastened at one end a to a fixed support, and at the other end b to the eye of a [Pg 20]wire-strainer. A wire-strainer consists of an iron rod, with an eye at one end and a screw and a nut at the other; it is used for tightening wires in wire fencing; and is employed in this case for the purpose of stretching the cord. This being done, I take a piece of ordinary sewing-thread, which is of course weaker than the stout twine. I tie the thread to the middle of the cord at c, catch the other end in my fingers, and pull; something must break—something has broken: but what has broken? Not the slight thread, it is still whole; it is the cord which has snapped. Now this illustrates the point on which we have been dwelling. The force which I transmitted along the thread was [Pg 21]insufficient to break it; the thread transferred the force to the cord, but under such circumstances that the force was greatly magnified, and the consequence was that this magnified force was able to break the cord before the original force could break the thread. We can also see why it was necessary to stretch the cord. In Fig. 10 the strains along the cords are greater when the cords are attached at c and d than when they are attached at a and b; that is to say, the more the cord is stretched towards a straight line, the greater are the forces into which the applied force is resolved.

29. We give a second example, in illustration of the same principle.

In Fig. 12 is shown a chain 8' long, one end of which b is attached to a wire-strainer, while the other end is fastened to a small piece of pine a, which is 0"·5 square in section, and 5" long between the two upright irons by which it is supported. By means of the nut of the wire-strainer I straighten the chain as I did the string of Fig. 11, and for the same reason. I then put a piece of twine round the chain and pull it gently. The strain brought to bear on the wood is so great that it breaks across. Here, the small force of a few pounds, transmitted to the chain by pulling the siring, is magnified to upwards of a hundredweight, for less than this would not break the wood. The explanation is precisely the same as when the string was broken by the thread.

Fig. 12.

SAILING.

30. The action of the wind upon the sails of a vessel affords a very instructive and useful example of the decomposition of forces. By the parallelogram of forces we are able to explain how it is that a vessel [Pg 22]is able even to sail against the wind. A force is that which tends to produce motion, and motion generally takes place in the line of the force. In the case of the action of wind on a vessel through the medium of the sails, we have motion produced which is not necessarily in the direction of the wind, and which may be to a certain extent opposed to it. This apparent paradox requires some elucidation.

Fig. 13.

31. Let us first suppose the wind to be blowing in a direction shown by the arrows of Fig. 13, perpendicular to the line ab in which the ship’s course lies.

In what direction must the sail be set? It is clear that the sail must not be placed along the line ab, for then the only effect of the wind would be to blow the vessel sideways; nor could the sail be [Pg 23]placed with its edge to the wind, that is, along the line o w, for then the wind would merely glide along the sail without producing a propelling force. Let, then, the sail be placed between the two positions, as in the direction p q. The line o w represents the magnitude of the force of the wind pressing on the sail.

We shall suppose for simplicity that the sail extends on both sides of o. Through o draw o r perpendicular to p q, and from w let fall the perpendicular w x on p q, and w r on o r. By the principle of the parallelogram of forces, the force o w may be decomposed into the two forces o x and o r, since these are the sides of the parallelogram of which o w, the force of the wind, is the diagonal. We may then leave o w out of consideration, and imagine the force of the wind to be replaced by the pair of forces o x and o r; but the force o x cannot produce an effect, it merely represents a force which glides along the surface of the sail, not one which pushes against it; so far as this component goes, the sail has its edge towards it, and therefore the force produces no effect. On the other hand, the sail is perpendicular to the force o r, and this is therefore the efficient component.

The force of the wind is thus measured by o r, both in magnitude and direction: this force represents the actual pressure on the mast produced by the sail, and from the mast communicated to the ship. Still o r is not in the direction in which the ship is sailing: we must again decompose the force in order to find its useful effect. This is done by drawing through r the lines r l and r m parallel to o a and o w, thus forming the parallelogram o m r l. Hence, by the parallelogram of forces, the force o r is equivalent to the two forces o l and o m.

The effect of o l upon the vessel is to propel it in a direction perpendicular to that in which it is sailing. We must, therefore, endeavour to counteract this force as far as possible. This [Pg 24]is accomplished by the keel, and the form of the ship is so designed as to present the greatest possible resistance to being pushed sideways through the water: the deeper the keel the more completely is the effect of o l annulled. Still o l would in all cases produce some leeway were it not for the rudder, which, by turning the head of the vessel a little towards the wind, makes her sail in a direction sufficiently to windward to counteract the small effect of o l in driving her to leeward.

Thus o l is disposed of, and the only force remaining is o m, which acts directly to push the vessel in the required direction. Here, then, we see how the wind, aided by the resistance of the water, is able to make the vessel move in a direction perpendicular to that in which the wind blows. We have seen that the sail must be set somewhere between the direction of the wind and that of the ship’s motion. It can be proved that when the direction of the sail supposed to be flat and vertical, is such as to bisect the angle w o b, the magnitude of the force o m is greater than when the sail has any other position.

32. The same principles show how a vessel is able to sail against the wind: she cannot, of course, sail straight against it, but she can sail within half a right angle of it, or perhaps even less. This can be seen from Fig. 14.

The small arrows represent the wind, as before. Let o w be the line parallel to them, which measures the force of the wind, and let the sail be placed along the line p q; o w is decomposed into o x and o y, o x merely glides along the sail, and o y is the effective force. This is decomposed into o l and o m; o l is counteracted, as already explained, and o m is the force that propels the vessel onwards. Hence we see that there is a force acting to push the vessel onwards, even though the movement be partly against the wind.

Fig. 14.

It will be noticed in this case that the force o l acting to leewards exceeds o m pushing onwards. Hence it is that vessels with a very deep keel, and therefore opposing very great resistance to moving leewards, can sail more closely to the wind than others not so constructed; a vessel should be formed so that she shall move as freely as possible in the direction of her length, for which reason she is sharpened at the bow, and otherwise shaped for gliding through the water easily; this is in order that o m may have to overcome as little resistance as possible. If the sail were flat and vertical it should bisect the angle a ow for the wind to act in the most efficient manner. Since, then, a vessel can sail towards the wind, it follows that, by taking a zigzag course, she can proceed from one port to another, even though the wind be blowing from the place to which she would go towards the place from which she comes. This well known manœuvre is called “tacking.” You will understand that in [Pg 26]a sailing-vessel the rudder has a more important part to play than in a steamer: in the latter it is only useful for changing the direction of the vessel’s motion, while in the former it is not only necessary for changing the direction, but must also be used to keep the vessel to her course by counteracting the effect of leeway.

ONE FORCE RESOLVED INTO THREE FORCES NOT IN THE SAME PLANE.

Fig. 15.

33. Up to the present we have only been considering forces which lie in the same plane, but in nature we meet with forces acting in all directions, and therefore we must not be satisfied with confining our inquiries to the simpler case. We proceed to show, in two different ways, how a force can be decomposed into three forces not in the same plane, though passing through the same point. The first mode of doing so is as follows. To three points a, b, c[Pg 27] (Fig. 15) three spring balances are attached; a, b, c are not in the same straight line, though they are at the same vertical height: to the spring balances cords are attached, which unite in a point o, from which a weight w is suspended. This weight is supported by the three cords, and the strains along these cords are indicated by the spring balances. The greatest strain is on the shortest cord and the least strain on the longest. Here the force w lbs. produces three forces which, taken together, exceed its own amount. If I add an equal weight w, I find, as we might have anticipated, that the strains indicated by the scales are precisely double what they were before. Thus we see that the proportion of the force to each of the components into which it is decomposed does not depend on the actual magnitude of the force, but on the relative direction of the force and its components.

34. Another mode of showing the decomposition of one force into three forces not in the same plane is represented in Fig. 16. The tripod is formed of three strips of pine, 4' × 0"·5 × 0"·5, secured by a piece of wire running through each at the top; one end of this wire hangs down, and carries a hook to which is attached a weight of 28 lbs. This weight is supported by the wire, but the strain on the wire must be borne by the three wooden rods: hence there is a force acting downwards through the wooden rods. We cannot render this manifest by a contrivance like [Pg 28]the spring scales, because it is a push instead of a pull. However, by raising one of the legs I at once become aware that there is a force acting downwards through it. The weight is, then, decomposed into three forces, which act downwards through the legs; these three forces are not in a plane, and the three forces taken together are larger than the weight.

35. The tripod is often used for supporting weights; it is convenient on account of its portability, and it is very steady. You may judge of its strength by the model represented in the figure, for though the legs are very slight, yet they support very securely a considerable weight. The pulleys by means of which gigantic weights are raised are often supported by colossal tripods. They possess stability and steadiness in addition to great strength.

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 [Pg 29]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 [Pg 30]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.

Fig. 17.

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 [Pg 31]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.

Fig. 18.

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', [Pg 32]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 [Pg 33]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.

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