coucher avez-vous? 18. Nous en avons deux. 19. Avez-vous une bouteille de vin? 20. Non, Monsieur, mais j'ai une bouteille à vin (wine-bottle) [§ 81]. 21. Voyez-vous les chats-huants? 22. Non, mais je vois les chauves-souris.

EXERCISE 112.

1. Is your father in England? 2. No, Sir, he is in France with my brother. 3. Have they taken passports? 4. Yes, Sir, they have taken two. 5. Is it necessary to have a passport to travel in America ? 6. No, Sir, but it is necessary to have one to travel in Italy. 7. Is there a steamboat from Calais to Dover (Douvres)? 8. There are several. 9. Is there a railroad from Paris to Brussels (Bruxelles)? 10. There is one from Paris to Brussels, and one from Paris to Tours. 11. Has your brother bought a windmill ? 12. No, Sir, but he has built a steam-mill. 13. Are there many wind-mills in America ? 14. No, Sir, but there are many water and steam-mills. 15. Does your cousin learn drawing? 16. He does not learn it, he cannot find a drawing-master. 17. Is the fencing-master in the .dining-room? 18. No, Sir, he is in the drawing-room. 19. Is your cousin in his bedroom? 20. No, Sir, he is out (sorti). 21. How many rooms are there in your house? 22. Five: a kitchen, a dining-room, a drawing-room, and two bedrooms. 23. Are there owls here? 24. Yes, Sir, and bats too.

KEY TO EXERCISES IN LESSONS IN FRENCH.

EXERCISE 18 (Vol. I., page 78).

1. Avez-vous mes tables ou les vôtres ? 2. Je n'ai ni les vôtres ni les miennes, j'ai celles de l'aubergiste. 3. Les avez-vous? 4. Non, Monsieur, je ne les ai pas. 5. Votre sœur a-t-elle mes chevaux ? 6. Oui, Monsieur, elle a vos deux chevaux et ceux de votre frère. 7. Avez-vous raison ou tort? 8. J'ai raison, je n'ai pas tort. 9. Le ferblantier a-t-il mes chandeliers d'argent ou les vôtres ? 10. Il n'a ni vos chandeliers d'argent ni les miens. 11. Qu'a-t-il? 12. Il a les tables de bois de l'ébéniste. 13. A-t-il vos chaises d'acajou? 14. Non,

Monsieur, il a mes tables de marbre blanc. 15. Avez-vous ces tables

ci ou celles-là? 16. Je n'ai ni celles-ci ni celles-là, j'ai celles de l'ébéniste. 17. Avez-vous de bons porte-crayons? 18. Non, Monsieur, mais j'ai de bons crayons. 19. Le voyageur a-t-il des fusils de fer? 20. Oui, Monsieur, il a les miens, les vôtres et les siens. 21. N'a-t-il pas ceux de votre frère? 22. Il n'a pas ceux de mon frère. 23. L'ouvrier a-t-il mes marteaux de fer? 21. Oui, Monsieur, il les a. 25. Mon frère a-t-il vos plumes ou celles de mon cousin? 26. Il a les miennes et les vôtres. 27. Avez-vous les habits des enfants? 28. Oui, Madame, je les ai. 29. Avez-vous le chapeau de votre sœur? 30. J'ai celui de ma cousine. 31. Votre frère a-t-il quelque chose? 32. Il a froid et faim. 33. Avez-vous des chevaux ? 34. Oui, Monsieur, j'ai deux chevaux. 35. J'ai deux matelas de crin et un matelas de laine..

EXERCISE 19 (Vol. I., page 79).

1. Is that lady pleased? 2. No, Sir, that lady is not pleased. 3. Is your daughter quick ? 4. My son is very quick, and my daughter is idle. 5. Is she not wrong? 6. She is not right. 7. Is your cousin happy? 8. Yes, Madam, she is good, beautiful, and happy. 9. Has she friends? 10. Yes, Sir, she has relations and friends. 11. Has she

a new dress and old shoes ? 12. She has old shoes and an old dress. 13. Has not your brother a handsome coat? 14. He has a handsome coat and a good cravat. 15. Have you good meat, Sir? 16. I have excellent meat. 17. Is this meat better than that? 18. This is better than that. 19. Has your friend the beautiful china inkstand? 20. His inkstand is beautiful, but it is not china. 21. Is any one hungry? 22. No one is hungry. 23. Are the generals here? 21. The generals and the blacksmiths are here. 25. I have your parasols and your umbrellas, and your children's.

EXERCISE 20 (Vol. I., page 79).

1. Votre petite sœur est-elle contente ? 2. Oui, Madame, elle est contente. 3. Cette petite fille est-elle belle ? 4. Cette petite fille n'est pas belle, mais elle est bonne. 5. Avez-vous de bon drap et de

bonne soie ? 6. Mon drap et ma soie sont ici. 7. Votre sœur est

elle heureuse? 8. Ma soeur est bonne et heureuse. 9. La sœur de ce médecin a-t-elle des amis? 10. Non, Madame, elle n'a pas d'amis. 11. Votre viande est-elle bonne? 12. Ma viande est bonne, mais mon fromage est meilleur. 13. L libraire a-t-il un bel encrier de porcelaine? 14. Il a un bel encrier d'argent et une paire de souliers de cuir. 15. Avez-vous mes parasols de soie ? 16. J'ai vos parapluies de coton. 17. L'habit de votre frère est-il beau ? 18. Mon frère a un bel habit et une vieille cravate de soie. 19. Avez-vous des parents et des amis ? 20. Je n'ai pas de parents, mais j'ai des amis. 21. Cette belle dame a-t-elle tort ? 22. Cette belle dame n'a pas tort, 23. Avez-vous de belle porcelaine? 24. Notre porcelaine est belle et bonne. 25. Elle est meilleure que la vôtre. 26. Cette petite fille n'a-t-elle pas faim?

27. Cette belle petite fille n'a ni faim ni soif. 28. Qu'a-t-elle ? 29.
30. Cette montre d'or est-elle bonne ?
Elle n'a ni parents ni amis.
32. L'avez-vous ?
34. Je n'ai ni la vôtre

31. Celle-ci est bonne, mais celle-là est meilleure.
33. Je l'ai, mais je n'ai pas celle de votre sœur.
ni la mienne, j'ai celle de votre mère.

MECHANICS.-XIV.

ILLUSTRATIONS OF PRECEDING PRINCIPLES-KITE, BOAT, ETC.

-ELEMENTS OF MACHINERY.

WE have now to trace the practical application of the principles already laid down, and the best way of doing this is to take some common instances and carefully examine them, and we shall see that the same rules will apply to other and more complicated cases.

A heavy box is resting on a four-legged table. What are the forces that act on it, and what on the table? On the box there are only two-its own weight acting downwards, and the upward pressure of the table, exactly counterbalancing this weight. We turn, then, to the table, the forces acting on which are not quite so easily determined. There are its own weight and the weight of the box acting through their respective centres of gravity. These are parallel forces, and, as we have seen, have a resultant equal to their sum, and acting at a point in the line joining them, so taken that their distances from it are in the inverse proportion to their intensities. The other force which acts on the table is the resistance of the floor on which it rests, which resistance is transmitted upward through the four legs. If the weight act at a point equally distant from these, each bears an equal share; but if not, it is divided between them in the inverse proportion to their distances. To make this more clear, we will suppose these distances to be 2, 6, 6, and 8 feet respectively. Find the least common multiple of these numbers, that is, the least number each will divide without any remainder. In this case it is 24. the quotients 12, 4, 4, and 3; and these numbers represent the This we divide successively by the distances, and obtain proportion in which the weight is divided between the legs.

Now suppose the weight of the table and box to be 207 pounds. Since 12, 4, 4, and 3, added together make 23, the leg 2 feet off supports 12 parts out of every 23, i.e., of the weight, or 108 pounds. Those 6 feet off support, or 36 pounds each; and the other sustains only, or 27 pounds. A calculation of this sort is mine the relative strength the difvery frequently required to deter R ferent parts of a building should have.

Fig. 86.

B

We will take another case. A body, a (Fig. 86), rests on an inclined plane, the angle of which, A CA B, is 30°, and the co-efficient of friction is. What forces act on G, what are their amount, and what other force must be applied to keep it in its place? We will try and solve these questions. As already seen, three forces act on a-its weight acting along a w, the resistance of the plane acting along G R, and the force of friction, which is of the weight and acts along FG. We found, when considering the inclined plane, that the power necessary to sustain G must bear the same proportion to its weight that B C does to a C. This, then, is the first thing we must find out, and we must have a slight acquaintance with mathematics for this; there is, however, no difficulty in the matter. Produce C B to D, so as to make B D the same length as B C, and join A D. The triangles ABC and ABD are exactly equal. For as B C is equal to B D, and each is at right angles with A B, A B C would, if we were to turn it over, exactly lie on A B D. AD is, then, equal to A c. Now in any and every triangle the three interior angles are together equal to 180°, or two right angles, and in A B C we know that the angle CAB is 30°, and A B C, being a right angle, is 90°; therefore A C B must be 60°, and A D B is equal to it, and therefore is also 60°. The angle B A D is likewise equal to в A C, and as each is 30°, the angle c AD is 60°. We see thus that each of the angles of the triangle CAD is 60°, and therefore they are equal to one another; and, since the angles are equal, the sides are also equal, for there is no reason why one should be greater than another. The triangle is thus equilateral and equiangular. We have now found out what we wanted; for, if BC be repre

sented in length by 1, CD will be 2; and AC is equal to CD, therefore it is also 2, and the proportion BC bears to A c is 1 to 2, or. On an incline of 30°, then, the power must be half the weight; but, in this case, friction sustains one-fourth, and therefore a power must be applied, acting in the direction GP, and equal to one-fourth of the weight, in order to maintain equilibrium.

We have now to solve the remainder of the question. We know how much G W, G F, and GP are; but we want to know what portion of the weight is borne by the plane, that is, what proportion AB, which represents the resistance of the plane, bears to AC. To find this, we need another very important geometrical proposition, which you will find fully proved in Lessons in Geometry, Problem XXX., Vol. I., page 337.

In every right-angled triangle the square described on the side opposite the right angle is equal to the sum of the squares on the sides containing it. If we represent the sides by numbers expressing their lengths, the rule holds equally true. Suppose, for example, we measure off from one of the sides containing the right angle a length of 4 inches, and from the other a length of 3 inches, we shall find the line joining these two points is equal to 5 inches.

The square of 4 (which means 4 multiplied by itself) is 16, and the square of 3 is 9. These added together make 25, which is the square of 5. The usual way of writing this is 42+32= 52. In this way, if we know the length of any two sides of a rightangled triangle, we can always calculate the third. Now in the case we are examining, we know that the side A c is equal to 2 and the side B C to 1; but the square of AC is equal to the sum of the squares A B and B C; the square of A B must, therefore, be equal to the difference between those of AC and B C. Now these are 4 and 1; the square of A B is, then, equal to 3, and the length of A B must be represented by the quantity which, multiplied by itself, will make 3. This is called the square root of 3, and is written 3. By arithmetic we can easily find exactly what this number is, but you can see that it is very nearly 13. The proportion, then, of AB to A c is 12 to 2, or 7 to 8, and the plane sustains a pressure equal to about of the weight. We have thus discovered the magnitude of all the forces as required.

When, as in our last lesson, we have resolved all the forces acting on a body along two lines at right angles, we can in this way find the magnitude of the resultant without the trouble and possible inaccuracy of actual measurement. Suppose we have a remainder of 12 pounds acting along one of the lines, and one of 5 pounds along the other, the resultant will be equal to 122+52; that is, to the square root of 144 + 25, or 169, which is 13. In the same way we can solve many questions frequently met with. Here is an example. Two forces act on a body; the resultant is 34 pounds, and one of the forces is 16 pounds; what is the other? We first find the square of 34, which is 1,156; from this we take the square of 16, or 256, and we have left 900. The square root of this is 30, and this accordingly is the intensity of the other force.

Now turn to another common thing. A ladder, A B (Fig. 87), leans against a wall. What are the forces acting on it? Its own weight acts vertically downwards through G, and the other forces which keep it at rest are the reaction of the wall and the ground at A and B respectively. Now there is but little friction at A, and we may therefore consider the reaction to be in the

P

direction AP perpendicular to the wall. GW and A P, then, represent two of the forces acting on the ladder; but, as it is at rest, all three forces must act through the same point. Now the only point in which wG and AP meet is that found by producing WG till it cuts AP. Let R be this point; the force at B must act in the direction B R. This force is the resultant of two other-the resistance of the ground acting vertically upwards, and the force of friction which acts along the ground and towards the base of the wall; and Fig. 87. we easily see that the more nearly vertical the ladder is, the greater is the former as compared with the latter, and therefore the less the amount of friction which is required to keep it in its place. We have thus the scientific reason for the well-known fact that the more a ladder is inclined, the greater need there is for the foot to be held, so as to keep it from slipping.

B

W

Here is another case, involving the same principle. A bracket A B (Fig. 88), projects from a wall, to which it is fastened by a screw. A strut, A C, supports the outer end, and a weight, w, rests on it. In what direction is the strain on the screw? The three forces here are, gravity acting along w G, the thrust of the beam, A C, and the strain on the screw. The two former act through the point o, and the direction of the third is found by drawing a line from this point to the screw. We may take a line, o c, of such a length as to represent the weight, and resolve it into o b and o a, one acting along the strut, the other perpendicular to the wall. These will represent two forces, which are together equivalent to o c, and of these o b will be overcome by the pressure of the strut, and the other force, o a, tends to draw the screw from the wall. A portion, however, of the pressure of the strut will be borne by the screw, and these two forces combining produce the resultant, which acts on the screw towards the point o.

Fig. 88.

It is frequently very important to be able thus to tell in what direction a strain will act, as the strength of our materials must be proportionate to it. In this way the direction of the tie-beams and king and queen posts of a roof are determined. We know, too, where to apply struts and braces to the framework of a building, so as to gain the greatest benefit from them.

Now there are one or two cases of the composition and resolution of forces that are frequently given as illustrations, and if we clearly understand them we shall be able to master most others. The first is that of a kite, for there is science to be learned even from this common plaything. Indeed, we shall frequently find, among the very commonest things, good illustra tions of any subject we may be studying.

Let K (Fig. 89) represent a kite. The forces which act on it are, the force of the wind, acting, we will suppose, in the direction of the arrow, the tension of the string, acting along it in the direc tion K s, and the weight of the kite; and by the action of these three forces it is kept at rest. We will consider them singly; and first, we will take the force of the wind. Take K w, of such a length as to represent this force. The kite is always so made as to present a large surface to the wind in proportion to its weight, and the string is fastened to the loop in such a way that it does not hang vertical, but inclined at an angle; the tai, however, prevents its being so acted on by the wind as to come

W

Fig. 89.

in the sam straight line with the cord. Let us, then, resolve the force of the wind into two, one acting edgewise on the kite, the other perpendicular to its surface. We draw the parallelogram KO W A, and thus have the two forces, K O and K A, instead of K W. Ko has no effect, as it acts on the edge, and we need, therefore, only consider the part K A.

Now we will introduce a second force, that of the string. Produce 8 K backward, draw A B perpendicular to в K, and complete the parallelogram DK BA. We can again resolve KA into K D and K B. The latter will be expended in stretching the string, and have no tendency to move the kite, and thus we have K D left as the effective resultant of these two forces. We now consider the third, which is the weight of the kite. Draw KG

S

of such a length as to represent this, and complete the parallelogram D C GK. We have then K C the resultant of K D and KG, and therefore of all the forces which act on the kite, and this is the direction in which the kite will move, but as it does so, the angle at which it is inclined varies till K D and K G become opposite and equal, and then the kite will remain at rest as long as the force of the wind remains unaltered.

Fig. 90.

The other case we will consider is that of a ship, which will sail within a few points of the wind. Let co (Fig. 90) represent the direction and intensity of the wind, and s v the direction in which it is desired that the vessel should advance. The sail is placed in the direction A B, which is midB way between that of the wind and that of the vessel. We must, as before, resolve c o into two forces, EO and F O. The part E o, which acts along the direction of the sail, has no effect in moving the vessel; FO, which acts perpendicularly to the sail, is the effective portion. We must now again resolve this force along two directions, one being that in which the boat moves, the other at right angles to it. We make a o equal to F o, and about it describe the parallelogram H GIO, and thus have two forces, represented by o H and O I, in the place of the original force, c o. Now, of these, o I has no tendency to cause the vessel to advance; it acts sideways on the vessel, usually inclining it, and causing a slight motion, but it is resisted by the pressure of the water against the side; the other portion, o H, represents the portion of the force of the wind which is effective, and produces motion. In the same way you can calculate what portion of the force of the wind is effective in turning a windmill. The vanes are always set at an inclination with the plane in which they turn, and you must resolve the force along two directions, one perpendicular to the surface, and the other along it. The former you again resolve, and thus find what part of it produces rotation, and what part presses against the face of the mill.

ELEMENTS OF MACHINERY.

As it is our object to make these lessons as practical as possible, it will be well to look at a few of the simpler modes of altering and transmitting power. Sometimes this is advanced to the rank of a separate science, and called kinematics, or the science of motion, but it should be referred to here as a part of practical mechanics.

We seldom have our power available for use in the exact way we desire. Sometimes we have an alternate motion, like that of the piston-rod of an engine, and we want to derive from it a rotatory or progressive motion; or we want to transmit it along a direction making some angle with its course, or to make many other alterations in its mode of action.

In large factories there is frequently a long shaft running along the building, and driven by an engine. From this it is required to drive all the machines in the place. This is accomplished by fixing wheels on the shaft, and letting endless straps pass over these and then over the driving pulleys of the machines. The motion may often be greatly altered in this way. The strap itself merely transmits the power, and whether there is a gain or loss in speed or power depends on the comparative size of the sheaves. Frequently there are several of these wheels of different sizes fixed on the axle and on the machine, and thus the speed may be altered at pleasure. If the strap passes over a large one on the shaft and a small one on the machine, there will be an increase of speed, and if we reverse the condition there will be a loss. A common illustration of a similar arrangement is seen in a watch. The spring when fully wound up exerts a much greater power than it does when the watch has run nearly down. Now this would make it go irregularly, and therefore the fusee is introduced. When the force of the spring is greatest the chain acts Fig. 91. on the smallest part of the fusee, and therefore has only a short leverage, but as it unwinds and loses its force the chain acts at a greater leverage, and a uniform rate of motion is thus maintained.

Sometimes toothed wheels are used instead of straps, espe cially when the distance through which the power has to be transmitted is small. The advantage is that they do not slip, as straps are liable to do; the friction with them is, however, greater. If we want to transmit motion from a shaft to another placed at an angle with it (Fig. 91), we employ what are known as bevelled wheels. The action of these will be clear from the figure, without any explanation.

Often it is required to change a rotating motion into a progressive one, and we can accomplish this by means of a rack and pinion. A number of notches are cut in a bar of metal (Fig. 92a), of such a size and at such distances that the teeth of the wheel exactly fit into them, and as the wheel is turned the rack is moved onwards. This is very frequently employed when a slow and regular motion is required, as in the adjustment of the tube of a microscope. Instead of a rack a chain is sometimes used (Fig. 92 b), the links being made of a peculiar shape, so that the teeth of the wheel may catch in them.

Fig. 92.

The crank (Fig. 93) is, perhaps, one of the most common of these elements of machinery. The piston-rod of an engine is usually jointed, and the end of the jointed part fixed to an arm projecting from the axle to be turned, and called a crank. Sometimes it is fixed to a pin in one of the spokes of the wheel, but the action is just the same. The force, however, with which it drives the wheel is continually varying. When the piston is at the bottom of the cylinder the crank and piston-rod are in one straight line, and therefore all the power presses on the bearing of the axle, and is lost. As the wheel turns the power acts with a leverage which increases till the wheel has made nearly onefourth of a revolution; it is then at its maximum, and diminishes till the piston-rod reaches its highest point, when it is all again lost. Now it is clear that unless we have some means of regulating the speed the machine will work very unevenly, and at times stop altogether. To obviate this, a large and heavy wheel, called the fly-wheel, is fixed on the axle of the crank. This, when once started, acquires an amount of momentum or moving force which carries it over the dead points when the power is lost. On account of the weight of the wheel, its motion is but slightly accelerated when the piston acts at its greatest leverage, but the additional force is stored up in it, and thus ensures a steady motion. The heavy wheel of a foot-lathe serves precisely the same purpose. The power is here applied during rather less than one-half of the revolution, but the momentum then acquired carries it through the remaining part.

EXAMPLES.

Fig. 93.

1. Forces of 9 and 12 act at right angles; what is their resultant? 2. The resultant of two forces which act at right angles is 10 pounds. One of the forces is 6; find the other.

3. Two men, one on each side of a stream, tow a barge. The angle the two ropes make is 60°, and each pulls with a force of 100 pounds. What is the total force exerted on the barge?

4. The tension of a wire in a piano is 100 pounds, its length is 5 feet. What force is required to draw its middle point 2 inches out of its position? 5. A weight of 90 pounds rests on a plane inclined at an angle of 30°. The co-efficient of friction is }. What force is required to keep it at rest?

weighs 2 pounds, these amounts are 1 pound, pound, pound, and pound, together 1. Take this from 12, and we have an effective power of 10 remaining; and as the gain is 16, the weight raised is 16 × 101, or 162 pounds.

3. Friction requires a strain of 9× 20, or 180 pounds, to overcome it, and of the weight has to be borne. The strain, therefore, is 180448, or 628 pounds.

4. Friction is here 3 of 27 cwt., which equals

cwt. The amount of the weight sustained by the horse is of 27 cwt., or #cwt. total strain is thus 13 cwt., or 144 pounds.

5. The weight of the carriage is 25 × 80, or 2,000 pounds.

6. The co-efficient of friction is, or nearly.

The

25

60 × 24

Therefore, 3 days 4 hours 25 minutes are 3 + 1 + or 3 of a day, or 1 of a day. Therefore, of 3 days 4 hours 25 minutes 917 1% of a day. = 19% 24 hours of an hour = 1 x 60 minutes Hence the required answer is 1 day 6 hours 34 minutes.

19% of a day

EXERCISE 47.

hours 6 hours. = 34 minutes.

of £1 of 40s. s.

3 x 8. = 4 × 108. 1038.

15s.

-3.

4 x 8 s.;

2.

3

3.

8d.

4.

of 1 ton;

5.

of 1 mile;

of 1 gallon; of 1 peck.

of 1 minute; of 1 degree.

7. 3 of of a mile; of of a week.

Reducing 1834 minutes to higher denominations,

Therefore, of 3 days 4 hrs. 25 min. is 1 day 6 hrs. 34 min.

3. To reduce one Compound Quantity to a Fraction of any other.

Finding what fractional part of one compound quantity another given compound quantity is, is called reducing the latter quantity to the fraction of the first.

Thus, finding what fraction of one pound 6s. is, is reducing 6s. to the fraction of a pound.

This is, in fact, only another name for performing the operation of dividing one compound quantity by another, or of finding the ratio of two compound quantities (see Art. 11, Lesson XXVII., page 101).

4. EXAMPLE.-What fraction of £1 7s. 6d. is 3s. 6d. ?

Reduce to the fraction of a pound

16. 438.; 48. 7d.; 9s. 2 d.

17. 13s. 0d.; 31d.; £3 15s. 9 d.

18. What part of £1 is of a penny? 19. What part of 1 lb Troy is 7 oz.? 20. Reduce

21. Reduce

of a quart to the fraction of a gallon. of 1 secon to the fraction of a week. 22. Reduce £3 17s. 11 d. to the fraction of £7 3s. 23. Reduce 3 pecks 2 gallons to the fraction of 2 bushels. 24. Reduce 15 cwt. 65 lbs. to the fraction of 2 tons 3 cwt. 25. Reduce 1° 15′ 10′′ to the fraction of a right angle. 26. Reduce 1 acre to the fraction of 5 acres 2 r. 40 p. 27. Reduce of £1 to the fraction of a penny.

=

when

136 gals. 2 qts. 178 gals. 3 qts.

33. Find the value of

1 lb. 7 oz. 4 dwts. 2 lb. 7 oz. 10 dwts. 77 dys. 4 hrs. 30 min. 6 dys. 12 hrs. of 517 square feet 72 inches. £3 18s. 8d. £6 12s. 94.

of 15 guineas.

of

1° 17 13"

of 104 yards 9 inches.

This might have been more shortly performed as follows: