Part 2

Let O P, O Q (Fig. 5) be the directions of the cords; O P and O Q being each of the length which corresponds to 1 lb., while O S corresponds to 1·5 lb. Here, as before, O P and O Q together may be considered to counteract O S. But O S could have been counteracted by an equal and opposite force O R. Hence O R may be regarded as the single force equivalent to O P and O Q, that is, as their resultant; and thus it is proved experimentally that these forces have a resultant. We can further verify that the resultant is the diagonal of the parallelogram of which the equal forces are the sides. Construct a parallelogram on a piece of cardboard having its four sides equal, and one of the diagonals half as long again as one of the sides. This may be done very easily by first drawing one of the two triangles into which the diagonal divides the parallelogram. The diagonal is to be produced beyond the parallelogram in the direction O S. When the cardboard is placed close against the cords, the two cords will lie in the directions O P, O Q, while the produced diagonal will be in the vertical O S. Thus the application of the parallelogram of force is verified.

16. The same experiment shows that two unequal forces may be compounded into one resultant. For in Fig. 5 the two forces O P and O S may be considered to be counterbalanced by the force O Q; in other words, O Q must be equal and opposite to a force which is the resultant of O P and O S.

17. Let us place on the central hook G a weight of 5 lbs., and weights of 3 lbs. on the hook E and 4 lbs. on F. This is actually the case shown in Fig. 3. The weights being unequal, we cannot immediately infer anything with reference to the position of the cords, but still we find, as before, that the cords assume a definite position, to which they return when temporarily displaced. Let Fig. 6 represent the positions of the cords. No two of the angles are in this case equal. Still each of the forces is counterbalanced by the other two. Each is therefore equal and opposite to the resultant of the other two. Construct the parallelogram on cardboard, as can be easily done by forming the triangle O P R, whose sides are 3, 4, and 5, and then drawing O Q and R Q parallel to R P and O P. Produce the diagonal O R to S. This parallelogram being placed behind the cords, you see that the directions of the cords coincide with its sides and diagonal, thus verifying the parallelogram of forces in a case where all the forces are of different magnitudes.

18. It is easy, by the application of a set square, to prove that in this case the cords attached to the 3 lb. and 4 lb. weights are at right angles to each other. We could have inferred, from the parallelogram of force, that this must be the case, for the sides of the triangle O P R are 3, 4, and 5 respectively, and since the square of 5 is 25, and the squares of 3 and of 4 are 9 and 16 respectively, it follows that the square of one side of this triangle is equal to the sum of the squares of the two opposite sides, and therefore this is a right-angled triangle (Euclid, i. 48). Hence, since P R is parallel to O Q, the angle P O Q must also be a right angle.

A SMALL FORCE SOMETIMES BALANCES TWO LARGER FORCES.

19. Cases might be multiplied indefinitely by placing various amounts of weight on the hooks, constructing the parallelogram on cardboard, and comparing it with the cords as before. We shall, however, confine ourselves to one more illustration, which is capable of very remarkable applications. Attach 1 lb. to each of the hooks E and F; the cord joining them remains straight until drawn down by placing a weight on the centre hook. A very small weight will suffice to do this. Let us put on half-a-pound; the position the cords then assume is indicated in Fig. 7. As before, each force is equal and opposite to the resultant of the other two. Hence a force of half-a-pound is the resultant of two forces each of 1 lb. The apparent paradox is explained by noticing that the forces of 1 lb. are very nearly opposite, and therefore to a large extent counteract each other. Constructing the cardboard parallelogram we may easily verify that the principle of the parallelogram of forces holds in this case also.

20. No matter how small be the weight we suspend from the middle of a horizontal cord, you see that the cord is deflected: and no matter how great a tension were applied, it would be impossible to straighten the cord. The cord could break, but it could not again become horizontal. Look at a telegraph wire; it is never in a straight line between two consecutive poles, and its curved form is more evident the greater be the distance between the poles. But in putting up a telegraph wire great straining force is used, by means of special machines for the purpose; yet the wires cannot be straightened: because the weight of the heavy wire itself acts as a force pulling it downwards. Just as the cord in our experiments cannot be straight when any force, however small, is pulling it downwards at the centre, so it is impossible by any exertion of force to straighten the long wire. Some further illustrations of this principle will be given in our next lecture, and with one application of it the present will be concluded.

21. One of the most important practical problems in mechanics is to make a small force overcome a greater. There are a number of ways in which this may be accomplished for different purposes, and to the consideration of them several lectures of this course will be devoted. Perhaps, however, there is no arrangement more simple than that which is furnished by the principles we have been considering. We shall employ it to raise a 28 lb. weight by means of a 2 lb. weight. I do not say that this particular application is of much practical use. I show it to you rather as a remarkable deduction from the parallelogram of forces than as a useful machine.

A rope is attached at one end of an upright, A (Fig. 8), and passes over a pulley B at the same vertical height about 16' distant. A weight of 28 lbs. is fastened to the free end of the rope, and the supports must be heavily weighted or otherwise secured from moving. The rope AB is apparently straight and horizontal, in consequence of its weight being inappreciable in comparison with the strain (28 lbs.) to which it is subjected; this position is indicated in the figure by the dotted line AB. We now suspend from C at the middle of the rope a weight of 2 lbs. Instantly the rope moves to the position represented in the figure. But this it cannot do without at the same moment raising slightly the 28 lbs., for, since two sides of a triangle, CB, CA, are greater than the third side, AB, more of the rope must lie between the supports when it is bent down by the 2 lb. weight than when it was straight. But this can only have taken place by shortening the rope between the pulley B and the 28 lb. weight, for the rope is firmly secured at the other end. The effect on the heavy weight is so small that it is hardly visible to you from a distance. We can, however, easily show by an electrical arrangement that the big weight has been raised by the little one.

22. When an electric current passes through this alarum you hear the bell ring, and the moment I stop the current the bell stops. I have fastened one piece of brass to the 28 lb. weight, and another to the support close above it, but unless the weight be raised a little the two will not be in contact; the electricity is intended to pass from one of these pieces of brass to the other, but it cannot pass unless they are touching. When the rope is straight the two pieces of brass are separated, the current does not pass, and our alarum is dumb; but the moment I hang on the 2 lb. weight to the middle of the rope it raises the weight a little, brings the pieces of brass in contact, and now you all hear the alarum. On removing the 2 lbs. the current is interrupted and the noise ceases.

23. I am sure you must all have noticed that the 2 lb. weight descended through a distance of many inches, easily visible to all the room; that is to say, the small weight moved through a very considerable distance, while in so doing it only raised the larger one a very small distance. This is a point of the very greatest importance; I therefore take the first opportunity of calling your attention to it.

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.

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 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.

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 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.

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 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 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.

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 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.

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 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 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.

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 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.

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 (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 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.