OF PhffSICS iBi B w m MWii*w *i i ri i hif ii rM ifc W iiiii "iw i I™* i L*l«.lll I The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924076055783 PRINCIPLES OF PHYSICS BY FRANK M. GILLEY CHELSEA fflOH SCHOOL mXKoo ALLYN AND BACON Sogton antf Cfjicago COPYRIGHT, 19 1. BY FRANK M. CILLEY. NotiDOolj ^xim 3. 8. Gushing & Co. — Berwick Si Smith Norwood MasB. U.S.A. CORNELL UNIVERSITY LIBRARY 3 1924 076 055 783 PREFACE. In addition to acknowledgments made in the text, the author desires to express his thanks to many who have assisted him in the preparation of this work; more especially to Charles D. Jenkins, Massachusetts State Inspector of Gas Metres and Illuminating Gas, for suggestions in Photometry; to F. C. Morton, in Steam Engines and Friction ; to N. H. Black, Rox- bury Latin School, apparatus in Conjugate Foci ; to I. 0. Palmer, Newton High School, apparatus for Composition of Forces ; the Westinghouse Air Brake Company, in Coefficient of Friction. Mr. I. 0. Palmer and Mr. E. P. Churchill have read portions of the manuscript, and Mr. C. B. Hersey has read the manuscript and proofs of the entire book. While the great body of scientific literature has been drawn upon for suggestions, the author would mention the works of John Perry and S. P. Thompson as particularly helpful. FRANK M. GILLEY. Chelsea, Mass., August, 1901. HI REFERENCE BOOKS. Carhart's University Physics. Watson's Text-Book of Physics. Ame's Laboratory Manual. Trowbridge's What is Electricity ? How Two Boys made their own Electrical Apparatus. Taylor's Optics of Photography. Orford's Lens Grinding. Bottone's Wireless Telegraphy. Lodge's Signalling without Wires. C. V. Boy's Soap Bubbles. Perry's Spinning Tops. Avery's A. B. 0. of Dynamo Design. Ayrton's Practical Electricity. Thompson's Elementary Lessons in Electricity and Magnetism. Thompson's Light, Visible and Invisible. The following journals contain articles of Interest to students of Physics : — American Machinist.- Scientific American. Electrical Eeview. American Electrician. Model Engineer and Amateur Electrician. The foUovring are a few of the many firms that issue trade catalogues containing instructive illustrations and descriptions : — Carborundum Co., Pittsburg, Pa. Bausch & Lomh, Rochester, N.T. i _. , „ , ,„ . „ r.1 1 J c\ \ Binocular Telescopes. Warner & Swazey, Cleveland, O. J Nat. Tufts Meter Co., Boston, Mass. Photometer. Economy Hot-air Engine, 4 Oliver St., Boston, Mass. \ rr , • p ■ Rider Ericson Engine Co., New York. / " ?*'*««• Holophane Glass Co. , New York. Luxfer Prism Co., Boston, Mass. General Electric Co. Westinghouse Electric Co. The address of many others are in the various scientific papers. Iv CONTENTS. I. Density — Specific Gkatitt. (Exercises 1-7) II. Pressure III. Liquids and Gases. (Exercises 8-11) IV. Forces. (Exercise 12) . V. Friction. (Exercise 13) VI. Parallel Forces. (Exercises 14, 15) VII. Machines — Pulleys VIII. Work. (Exercises 16, 17) . IX. Centre of Gravity. (Exercise 18) X. Weight and Mass .... XI. Velocity. (Exercise 19) XII. Elasticity. (Exercises 20-2-3) Xill. Heat. (Exercise 24) . . . XIV. Thermometers. (Exercises 25-29) XV. Evaporation and Boiling. (Exercises 30-33) XVI. Expansion of Gases — Law of Charles. (Exercises 34, 35) XVII. Thermodynamics. (Exercise 36) . XVIII. Light — Reflection. (Exercises 37, 38) XIX. Light — Refraction. (Exercises 39-47) XX. Lens.es. (Exercises 48-54) XXI. Curved Mirrors. (Exercises 55, 56) . XXII. Dispersion of Light .... XXIII. Photometry. (Exercise 57) . XXIV. Optical Instruments .... XXV. Sound. (Exercises 58-60) XXVI. Electricity ^ — -Magnets. (Exercises 61-64) XXVII. Batteries. (Exercises 65, 66) XXVIII. Magnetic Action of Electric Current. 67, 68) XXIX. Measurement of Electric Current . XXX. Ohm's Law — Resistance. (Exercise 69) XXXI. Measurement of Resistance. (Exercises 70, 71) XXXII. Internal Resistance of Batteries — Grouping of Cells — Storage Cells. (Exercises 72, 73) v (Exercises PAGK 1 19 34 66 76 85 99 107 119 130 140 162 175 190 212 236 244 264 276 298 325 .331 340 348 362 .S82 404 424 433 443 459 468 VI CONTENTS. CHAPTER XXXIII. XXXIV. XXXV. PAGE Electromagnets — Induced Cureents — Dynamos AND Motors — The Induction Coil. (Exercises 74-80) 482 Signalling throdgh Ocean Cables .... 512 Practical Applications of Electricity . . . 516 Appendix "''* Index ^^ LIST OF EXERCISES COKRESPONDING TO THOSE REQUIRED FOR ADMISSION TO HARVARD COLLEGE. (Numerals at the left of the page, not preceded by §, refer to Exercises.) MECHANICS AND HYDROSTATICS. 1. Weight of Unit Volume of a Substance .... 2. Lifting Effect of Water upon a Body entirely immersed in it 3. Specific Gravity of a Solid Body that will sink in Water . . 9 i. Weight of Water displaced by a Floating Body .... 12 § 16. Specific Gravity by Flotation Method 13 5, 6. Specific Gravity of a Block of Wood by Use of a Sinker 15, 16 7. §45. Specific Gravity of a Liquid : Tv70 Methods . . 17,36 12. Parallelogram of Forces 66 13 a. Friction between Solid Bodies (on a level) .... 76 13 6. Coefficient of Friction (by sliding on incline) .... 80 14. The Straight Lever : First Class 85 14. Levers of the Second and Third Classes 85 14. Force exerted at the Fulcrum of a Lever 85 App. Errors of a Spring-balance 536 18. Centre of Gravity and Weight of a Lever 126 LIGHT. 37. Images in a Plane Mirror 39. Index of Refraction of Water .... 41. Index of Refraction of Glass .... 48. Focal Length of a Converging Lens . 50. Conjugate Foci of a Lens 54. Shape and Size of a Real Image formed by a Lens §351. Virtual Image formed by a Lens . 55. Images formed by a Convex Cylindrical Mirror . § 356. Images formed by a Concave Cylindrical Mirror § 380. Use of Rumf ord Photometer .... 277 285 300 309 320 823 326 328 345 LIST OF EXERCISES. Vll MECHANICS. 8. Specific Gravity of a Liquid by balancing Columns 9 a, 6. § 49. Compressibility of Air : Boyle's Law 10. Density of Air 17. Four Forces at Right Angles in One Plane § 148. Comparison of Masses by Acceleration-test . 19. Action and Reaction : Elastic Collision § 180. Elastic Collision continued ; Inelastic Collision 20. Elasticity : Stretching 21. Brealiing-strength of a Wire .... § 188. Comparison of Wires in Breaking Tests 22 a. Elasticity : Bending ; Effect of Varying Load 22 6. Elasticity : Bending ; Effect of Varying Dimensions 23. Elasticity : Twisting . 34 37, 39, 40 . 43 . 116 . 130 . 156 . 159 . 162 . 164 . 165 . 168 . 170 . 173 HEAT. 24. Linear Expansion of a Solid ....... 184 25. Testing a Mercury Thermometer 191 29. Determination of the Dew-point 208 31 6. Specific Heat of a Solid 222 32. Latent Heat of Melting 227 33. Latent Heat of Vaporization . . .... 231 34. Increase of Volume of a Gas heated at Constant Pressure . . 237 35. Increase of Pressure of a Gas heated at Constant Volume . . 242 SOUND. 58. Velocity of Sound in Open Air 369 59. Number of Vibrations of a Tuning-fork 370 § 421. Wave-length of Sound 375 ELECTRICITY AND MAGNETISM. 61. Lines of Force near a Bar Magnet 386 §§ 460-463. Study of a Single-fiuid Galvanic Cell . . . .404 66. Study of a Two-fluid Galvanic Cell 418 68. Lines of Force about a Galvanoscope 426 69 (I). Resistance of Wires by Substitution : Various Lengths . 447 69 (II). Resistance of Wires by Substitution : Cross-section and Multiple Arc 448 70. Resistance by Wheatstone's Bridge: Specific Resistance of Copper 465 71. Temperature-coefficient of Resistance in Copper . . 467 72. §§ 540, 541. Battery Resistance 468 VUl LIST OF EXERCISES. PAGE §§658-56Q. Putting together the Parts of a Telegraph Key and Sounder . . 483 76. Putting together the Parts of a Small Motor .... 497 78. Putting together the Parts of a Small Dynamo .... 602 ADDITIONAL EXERCISES. EXBRCISE 11. Pressure in a Liquid due to its Weight ..... 52 15. Moments of a Force . . . .... 91 16. Inclined Plane 109 26 a, b. Temperatures corresponding to Pressure of Steam . 196, 198 27. Testing a Thermometer for Points between 0° and 100° C. . 204 28. Effect of Dissolved Substances on the Freezing-point of Water 206 30. Boiling-points of Liquids 216 31 a. Specific Heat of a Solid 221 36. Weight and Volume of a Gas 244 38. Mirrors at Right Angles 272 40. Critical Angle of Water 280 42. Critical Angle of Glass &88 43. Law of Internal Reflection 293 44. Index of Refraction of Glass by Parallax 294 45. Path of a Ray through Glass having Parallel Sides . . . 295 46. Path of a Ray through a Prism 295 47. Measurement of Angle of Minimum Deviation . . . 297 49. Measurement of Curvature of Lenses 305 51. Real Conjugate Foci — Parallax Method 314 62. Virtual Foci 315 53. Relative Size of Object and Image 317 56. Principal Focus of a Concave Mirror 329 57. Photometry . . . . . . 343 60. Overtones in Strings . 378 62. Effects of Heat on a Magnet .... .397 63. The Simple Pendulum 399 64. Distribution of Magnetism in a Magnet — Vibration Method . 402 65. Study of a Simple Cell . . .... 411 67. Magnetic Action of a Current . 425 73. Storage Batteries — Polarization . .... 477 74. Current induced by a Bar Magnet 486 75. Current induced by an Electromagnet .... 494 77. Lines of Force in the Armature of Dynamo or Motor . . 500 79. Principle of the Induction Coil 606 80. Thermo-electricity . . 509 PRINCIPLES OF PHYSICS. CHAPTER I. DENSITY. - SPEOIPIO GEAVITT. 1. Physics is the study of those laws of nature that govern the forms of substances and their movements. These laws, so far as they are known, were discovered originally by experi- ment. Every one knows that when water gets very cold it freezes, and when it gets very hot it boils. An experiment will show exactly how cold it must be to freeze or how hot it must be to boil under different pressures. The process of ex- perimenting is simply that of testing some principle to find out exactly what its limitations are. It is necessary, then, at the very outset, to distinguish be- tween the ordinary and the scientific ideas of such matters. For example, the terms hard and soft, which are accurate enough for everyday use, need definite qualification when used in science. Steel is commonly called hard ; yet an armor plate will splash like water when struck by a projectile from a mod- ern cannon. Similarly, in scientific language it is not enough to say that a body moves fast or slow, that it is heavy or light; we must measure its speed or its weight, and know exactly how fast it moves or how heavy it is. The student of Physics should learn, first of all, to use such words with precise scien- tific meaning. He must realize that the common meaning of 1 2 PRINCIPLES OF PHYSICS. these words includes only a few degrees along the middle of the scale, while science uses the entire scale. 2. An Experiment is an effort to determine the answer to a question. Since the whole field of Physics is to measure pre- cisely things that we estimate vaguely every day, its whole value lies in its accuracy. From every process of an experi- ment comes a certain unchangeable result. The results of an experiment have a definite meaning in every case, and if the experiment has been properly conducted, they tell the student what lie wishes to know ; otherwise they will tell him some- thing else. He must therefore have clearly in view the end he is seeking, and know the exact value of every step he takes towards it. Accuracy is equally necessary in conducting the experiment and in interpreting the results. 3. Metric System of Measurement. — A decimal system of weights and measures, called the Metric System, is commonly used in experiments in Physics, as it is easier to attain a high degree of accuracy with it than with the English system. Examine a meter stick. Count the number of spaces that are a little larger than the diameter of a lead pencil. Notice that the meter is several inches — about a finger's length — longer than a yard. To be exact, a meter = 39.37 inches. This number (which need not be committed to memory) is to be used in changing from one scale to the other, i.e. from inches to meters and from meters to inches. On the meter stick there are one hundred divisions, somewhat larger than the diameter of a lead pencil. One of these divisions is there- fore one one-hundredth of a meter. Just as we call the hundredth of a dollar a cent, the hundredth of a meter is called a centimeter. The abbreviation for, or short way of writing, 'centimeter' is to use the letters cm. In like man- ner, m. stands for meter. One meter contains one hundred centimeters. DENSITY. — SPECIFIC GRAVITY. 3 Measure in centimeters the diameter of a nickel ; the length and breadth of your book ; the length of a pencil. 4. Length. — Try to measure exactly the diameter of a cent piece ; of a dime. Express the amount in tenths of a centi- meter, as near as can be guessed. In scientific work this is called estimating. As the tenth of a cent is called a mill, so the tenth of a centimeter is called a millimeter. The abbre- viation for millimeter is mm. One centimeter equals how many millimeters ? What is the number of millimeters in a meter ? Compare with the number of mills in a dollar. Make up a table as follows : — So many millimeters make a centimeter So many centimeters make a meter This is the table of length. A piece of paper 20 cm. by 12 cm. has an area of 20 x 12, or 240, square centimeters ; written, for brevity, 240 sq. cm. Find the area of a leaf of the note-book. 5. Volume. — What is the volume of a cube 10 cm. on an edge ? How many little cubes 1 cm. on each edge could be laid in a column 10 cm. long ? How many of these columns must be placed side by side to make the width 10 cm. ? How many little cubes in all have been so far used ? How many of the large squares formed must be piled one on top of the other to make the pile 10 cm. high ? How many of the little cubes in the whole pile ? The volume of the pile is called a liter. Of course it may be put into any shape, and will still have the same volume, — one liter, — and contain the same number of cubic centimeters. In working problems, be sure to change all the dimensions in a question to the same unit ; that is, to have all the numbers millimeters, centimeters, or meters before multi- plying. PRINCIPLES OF PHYSICS. Problems. 1. What is the height in meters of a man 5 ft. 11 in. tall ? 2. How many meters long is a 600-inch fish line ? 3. In 2.5 m. how many centimeters ? 4. 500 cm. = how many meters? 5. 0.01 m. equals how many centimeters? 6. 15 cm. equals how many meters ? 7. In 30 cm. how many millimeters ? 8. A pole measuring 750 mm. is how many meters long? 9. A boy 5 ft. tall is how high in millimeters? 10. If a postal card is 14 cm. long and 9 cm. broad, what is its area? 11. A board, area 300 sq. cm., is 15 cm. long ; what is its breadth ? 12. Find the area of a strip of paper 1.5 cm. wide and 40 m. long. 13. Find the area of a strip of board 2 mm. wide and 6.5 m. long. 14. How many liters in a tank 50 cm. wide, 30 cm. deep, and 100 cm. long? 15. Give the volume, in cubic centimeters, of a box 60 cm. long, 50 cm. wide, 10 cm. deep. How many liters of milk would the box hold? 16. If a cubical reservoir is 6 m. on each edge, what number of liters will it hold ? 17. A cubic centimeter of water weighs approximately one gram ; how many grams of water in a liter ? 18. What is the volume of 3500 grams of water? of 9750 grams of water? 19. If a box 8 cm. by 4 cm. by 10.5 cm. were filled with water, what would the contents weigh ? 6. Density is the weight of a unit volume. Where grams and centimeters are used, density is a number giving the weight, in grams, of one cubic centimeter. When pounds and cubic feet are used, density gives the weight, in pounds, of one cubic foot. B L I /\ . / / 1 / C E / B F DENSITY. ^SPECIFIC GRAVITY. 5 Bxercise 1. DENSITY OF A SOLID. Apparatus: A 250-gram spring balance; centimeter rule; block of pine, spruce, oak, ash, whitewood, ebony, lignum vitae, or other wood. No two dimensions of the block should be alike, and in any one dimension there should be measurable variation in different parts of the block. Find the dimensions of a rectangular block, making several meas- urements of the height, width, and length. Record the measurements on a skeleton diagram (Fig. 1). Find the average height by adding all i t n the heights, AB, CD, etc., and divid- ing by the number of measurements taken. In the same way find the average length and width. Multiply these three average dimensions to- gether to find the volume. In measuring, the graduated edge ^'^■'"' of the rule should be placed directly upon the block, and the reading made in centimeters and tenths. Try to estimate to tenths of a millimeter. -As the entry in the note-book is to be in centimeters, tenths of a millimeter would be written in the hundredths' place. Do not measure from the very end of the rule, for the corner is apt to be worn. Figure 2 shows the correct position for the meas- uring rule. The reading is one' centi- meter, six-tenths, and perhaps six- ^'e- ^* tenths of a tenth, or six hundredths. This would be written 1.66 cm. Find the weight of the block in grams, and compute the weight of one cubic centimeter of it. If the dimensions of the block are 4 cm., 5.1 cm., and 6i2 cm., the volume is 126 cc. (cubic centimeters), if we omit the figures in the product (which is 126.48) that are beyond the limit of accuracy of the measurement. If the block weighs 200 grams, one cubic centimeter would weigh -^ of 200 = 1.58 g. 7. Ezperiments on Density. — Find the density of a cylinder, a half cylinder, or a hexagonal prism, such as may be obtained IINlllll iliillllrllllll UijiuiJ 6 PRINCIPLES OF PHYSICS. from a set of drawing models. Find the density of a hard brick, and also of a soft burned brick. (Have them weighed at the nearest grocer's, if they are too heavy for the scales in the laboratory.) Find the size and weight of different kinds of bricks. Compute their density. Soak them in water, weigh again, and compute the density. The best bricks absorb the least water. Change pounds to grams by multiplying by 453.6, which is the number of grams in a pound. Find the weight of one cubic centimeter of each brick. Find the density in the English system, i.e. the weight of one cubic foot. The dimen- sions, if measured in centimeters, are to be changed to feet by dividing by 30.6, — the number of centimeters in a foot. 8. Formula for Density. — To find the density of a body, or, what is the same thing, to find the weight of one cubic centimeter of it, we divide the weight by the number of cubic centimeters in the body. The density of a lump that has a volume of 6 cubic centimeters and weighs 30 grams is ^ of 30, or 5. The rule, then, is : — Density equals weight divided by volume, or, Density = S — • Volume Abbreviating by using the first letter of each word for the whole-word, we have the formula, 9. Fonnula for Weight. — Knowing the density and the vol- ume of a body, the weight is found by multiplication. The weight of a body the density of which is 3 grams and the volume 6 centimeters, is 3 x 5 = 15 grams. Density times volume equals weight. DV= W. DENSITY. — SPECIFIC GRAVITY. 1 10. Formula for Volume. — Suppose we wish to know the size (volume) of a lump, the density of which is 3, and the weight 15 grams. As density is the weight of one cubic centimeter, the number of cubic centimeters in the lump will equal the number of times that 3 is contained in 15, or 6. To find the volume, divide the weight by the density. D _^ Problems. 1. If a rectangular block, 5 cm. by 6 cm. by 20 cm., weighs 500 g., what is the weight of 1 cc. ? What is its density ? 2. What is the weight of the contents of a box 8 cm. by 12 cm. by 30 cm., when filled with water? when filled with mercury, den- sity = 13.6? when filled with kerosene, density = .8? 3. What does 1 cc. of mercury weigh ? How much does a liter of mercury weigh ? a liter of kerosene ? 4. What is the density of a block of marble 1 m. by .3 m. by .2 m., weighing 162,000 g.? What is the weight of 50 cc. of the same marble ? 5. If a cube of zinc 4 cm. on an edge weighs 448 g., what does .1 cc. weigh ? What is its density ? How much would a block of zinc 5 cm. by 3 cm. by 2 cm. weigh? How many cubic centimeters in a block of zinc weighing 28 g. ? 6. How large is a piece of glass, weight = 100 g., density = 2.5? 7. What is the volume of 14 g. of zinc, density = 7 ? "^ 8. How much space is taken up by 900 g. of coal, density = 1.5? ■' 9. What is the size of a piece of pure gold, density = 19.4, weight = 500 g. ? What is the size of a rock, density = 5, weight = 30 g. ? 10. What is the weight of 6.5 cc. of lead, density = 11.5? 11. What is the weight of a glass paper-weight 1 cm. thick, 8 cm. long, and 12 cm. wide, density = 2.5? 12. Give the size of a 400 g. lead sinker, density = 11.4. 13. How many gi'ams of sulphuric acid, density = 1.8, can be put in a box 8 cm. by 4 cm. by 3 cm. ? 8 PRINCIPLES OF PHTSICS. 14. What is the density of a lump of wood containing 600 cc, ■weight = 400 g.? How much does 1 cc. of it weigh? 15. How big is a lump of cork, weight = 100 g., density = .25 ? What is the weight of 1 cc. of this cork ? 16. Find the weight of 1 liter of tin, density = 7.3. 17. What is the weight of 1 cc. of a body, 5 cm. by 4 cm. by 3 cm., weight = 80 g. ? Compute its density. 18. Compute the size of a piece of wood, weight = 80 g., den- sity = I. Exercise 2. EFFECT OF WATER ON A SOLID THAT SINKS. Apparatus: Overflow can or steam boiler; catch bucket; weighted block; spring balance or platform scales, reading to 250 g. Fill the overflow can. When dropping stops, slowly lower the block into the water. Catch the overflow in the bucket, and weigh. Also weigh the empty bucket, and find how many grams of water overflowed. Measure the block, and compute its volume. ' Weigh the block in air, then in water, and find the loss of weight. As one cubic centimeter of water weighs one gram, a sinking body displaces as many grams of water as there are cubic centimeters of volume in the body. When finding, by a spring balance, the weight in water of a sub- stance that sinks, do not allow any part of the balance to touch the water. 11. Experiments on Volume. — Describe a method of finding the volume of an irregular lump. Describe another method of finding the size or volume of an irregular body, by using a spring balance and a dish of water. Find the volume of pieces of marble, sulphur, coal, and glass, by one or both of these methods. How does the v^eight of a rock in water compare with the weight in air ? Bodies in water appear to lose weight. They are buoyed up by a force equal to what ? Can a man lift a greater load under watei than he could lift in the air ? DENSITY. — SPECIFIC GRAVITY. Problems. 1. A piece of rock weighs 25 g. in air and 15 g. in water-; what is its volume ? Calculate its density. How does this compare with the volume of the blocli in cubic centimeters ? 2. Find the weight of one cubic centimeter of platinum, density = 22. 3. Find the weight of a block of platinum that has the volume of a liter. 4. A lump weighs 40 g. in air and 32 g. in water ; find its volume. Find its loss of weight in water. How much water would run over if the lump were immersed in a dish full of water? 5. Compute the density of a lump of ore 2 cm. by 4 cm. by 1 cm., weighing 25 g. 6. Find the weight of a liter of milk, density = 1.03. 7. If a body 10 cm. long, 3 em. wide, and 6 cm. thick weighs 500 g. in water, how much will it weigh in air ? Exercise 3. SPECIFIC GRAVITY OF A BODY THAT SINKS IN WATER. Apparatus: A 250-gram spring -balance ; a piece of sulphur, coal, glass (a bottle, for instance), marble, iron, brass, or lead. Find the density of one of these substances. Weigh in air ; then in water. Compute the volume, and divide the weight by the volume. Suppose a lump weighs 15 g. in air and 10 g. in water; the loss of weight, or the buoyant effect, is 5 g. It must, therefore, displace 5 g. of water ; and as 5 g. of water have a volume of 5 cc, the lump, also, has a volume of 5 cc. The density, or the weight of one cubic centi- meter, is one-fifth of 15 g., or 3 g. The lump is three times as heavy as water. The specific gravity of the lump, then, is 3. 12. Specific Gravity tells how many times heavier a body is than an equal bulk of water. 10 PRINCIPLES OF PHYSICS. 13. Specific Gravity by the English System. — Using the metric system, in which one cubic centimeter of water weighs one gram, the density and specific gravity are represented by the same number ; but, in the English system of measures, a cubic foot of water weighs 62.6 pounds. A cubic foot of a substance the specific gravity of which is 3, weighs three times as much as a cubic foot of water, or 187.5 pounds per cubic foot, — the density expressed in pounds and cubic feet. If, in Exercise 3, page 9, balances reading in ounces had been used, the lump would have weighed about one-thirtieth as many ounces as it did grams. The weight in air, then, would have been .5 oz. ; in water, .33 oz. ; the .„ ., weight in air speemc gravity = , j . , , . r — loss of weight in water .5 -.33 '.17 = 3. Thus, since the lump is always the same number of times heavier than water, the figure representing its specific gravity remains the same, no matter what system of weights and measures is used. 14. Formula for Specific Gravity. — -To shorten the formula, write sp. gr. for specific gravity; Wior weight in air; w for weight in water. W but as loss of weight in water = W—w, W sp. gr. =•= W—w DENSITY. — SPECIFIC GRAVITY. 11 Problems. 1. If a substance has a density of 6, what is the weight of 1 cc. of it? What is its specific gravity? What is its density in the English system ? 2. Calculate the specific gravity of a piece of ore weighing 40 g. in air and 25 g. in water. Of a stone weighing 14 pounds in air and 8 J pounds in water. What is the density in the metric system? 3. A body weighs 12 g. in air and 8 g. in water ; find its size, density, and the weight of 1 cc. of it. How many times heavier than water is it? What is its specific gravity ? 4. A rock weighs 80 g. in air and 60 g. in water; what is its specific gravity, size, and density? If put in a full pail of water, what would happen ? 5. If it takes twelve men to lift a rock on land and nine men to hoist it when under water, what is its specific gravity ? What is its density? Can you tell its size? Why not? 6. A body of rectangular form is 8 cm. by 5 cm. by 10 cm., and weighs 500 g. ; find its density. If it were immersed in a dish full of water, how much water would run over? 7. What is the density of a substance of which .25 liter weighs 200 g.? 8. A lump has a specific gravity of S.5 ; what is its weight per cubic centimeter ? its density ? If it weighs 70 g. in air, what is its volume? How much, then, is it buoyed up in water? What would it weigh in water ? 9. A piece of stone weighing 127 g. in water and 234 g. in air is put in a full dish of water; how much water runs over? 10. Find the volume of a body, weight = 80 g., density = 5. 11. Find the weight of a body, volume = 60, density = 10.5. 12. Find the size of a lump, weight = 200 g., density = 1.5. 13. What is the size of a lump of cork, density = .25, weight = 300 g.? 14. If 32 cc. of a substance weigh 128 g., what is its specific gravity? 12 PRINCIPLES OF PHYSICS. 15. Specific Gravity by Flotation. — To find the density of a substance we must know its weight and volume ; to find its specific gravity we must find the weight of the substance and the weight of an equal volume of water. How was this done with a solid that sank in water ? Exercise 4. SPECIFIC GEAVITY OF A BODY THAT FLOATS. Apparatus: A stick of one square centimeter cross-section, marked length- wise in centimeters, and loaded at the lower end so as to float upright ; a jar of water ; an overflow can. Weigh the stick. Place the stick in the jar of water; measure and record the number of divisions above the water line and the number below. As each section of the stick is a cubic centimeter, the stick displaces as many grams of water as there are sections of the stick under water. Compare this with the weight of the stick, and record the relation between the weight of a floating body and the volume im- mersed. Put a gram weight on top of the stick ; try a two-gram, then a flve-gram weight. Place an object of unknown weight on the stick, and determine its weight. The volume of the stick is in this case the number of divisions. These are counted, and the number re- corded. How can the weight of one cubic centimeter be found ? Weigh some larger object that will float (not neces- sarily of any regular shape); weigh also the water displaced by lowering it into an overflow can, and compare the two readings. Record as follows: Weight of object =: Weight of bucket to catch water overflowing = Fig. 3. Weight of bucket and overflow = Amount of water displaced — How much water does a floating body displace ? DENSITY. — SPECIFIC GEAVITT. 13 Refill the overflow can, and find the volume of the object by weigh- ing the water that flows over when the object is pushed under the surface. Find the density, i.e. the weight of one cubic centimeter. What does a body that sinks, or is made to sinlt , displace ? 16. Density of a Body that Floats. — Find the density or spe- cific gravity of a cylindrical rod. It can best be kept upright by a frame holding two rings, as in Fig. 4. Mark the water level. Remove the stick, measure, and record : — Length under water = Whole length = While the stick probably does not have a cross-section of one square centimeter, yet it floats exactly as far down in the water as would a stick of the same wood, of the same length, with a cross-section of one square centimeter, or any other diameter. The length under water, then, represents the weight, and the whole length, the volume. Fig. 4. Density : Weight Length under water Volume Whole length By this method the density of any volume that floats is quickly determined. A piece of cork, a stick of paraffin (a candle), a block of ice, and a piece of pure gum rubber, if obtainable, may be tried. How many times larger is the part of an iceberg under water than the part above water? What solid has the least density ? Is charcoal lighter or heavier than water ? Does it float or sink ? 17. Measurement of Displacement. — ^ A tall glass jar, or a stu- dent lamp-chimney closed at the neck by a cork and supported 14 PRINCIPLES OF PHYSIOS. on a block of wood (Fig. 6)/is graduated by pouring in, from a burette, 6 cc. of water at a time, and marking the level on a gummed strip of paper fastened to tbe jar or chimney. A little water is poured in and a zero mark made be- fore water from the burette is added. Fill the jar part full of water, as in Fig. 6, A; record the level. Put in the jar a substance that floats (Fig. 6, B). Eead the level and record. Push the substance under water with a wire, and record the level of the water (Fig. 6, C). The weight in grams equals Fie. ^^ the number of centimeters of water dis- placed, or the reading of B less the reading of A. "When the substance is pushed under water, its volume is dis- placed, and the increase of level in C over that in A is the volume of the body. Calculate its density. Problems. 1. A plank 8 cm. thick floats 3 cm. out of water ; find its density. 2. What is the density of a pole that floats 12 feet under water and 4 feet out of water ? 3. How does a piece of wood float, density = J ? Make a diagram showing how it floats. 4. A piece of cork floats with one-fourth of its length under water ; what is its density and specific gravity? 5. A stick 6 cm. long has a density of f ; make a diagram showing how it floats in water. 6. How much water runs over from a tub full of water, when a five-pound toy sail-boat is placed in it? 7. A beam, density = .4, is 8 cm. thick ; how does it float? DENSITY. — SPECIFIC GBA VITT. 15 8. A lump, volume = 60 cc, weight = 40 g., is put in a full dish of water; how much runs over? Does the body sink or float? What is its density ? 9. If, in jar A, Fig. 6, the reading of the water level is 15 cc, in B the level is 45 cc, and in C the level is 55 cc, what is the weight, volume, and weight per cubic centimeter of the body. 10. If a cake of soap, 8 cm. by 5 cm. by 3 cm., density = .96, is put in a full basin of water, how much water overflows ? 18. Displacement and Loss of Weight. — The volume of a body that sinks has as many cubic centimeters as the number of grams' loss in water. A body that floats loses all its weight without being wholly under water. But, unless the body sinks, or, as in the case of a floating body, is made to sink, it cannot displace its volume of water. Tor every cubic centi- meter of water displaced there is a buoyant effect of one gram. Exercise 5. SPECIFIC GEAVITY BY IMMERSION. Apparatus: Block of wood; spring balance; jar of water; small pulley attached to a vertical rod, which is fastened to a heavy cross-piece of wood resting on the edges of the jar. Weigh the block of wood, find its volume by measurement, and record. By means of the pulley, draw the block down into the water (Fig. 7). Record the force required to make the block sink, and compare with the difference between the volume and the weight. What is the force tend- ing to bnoy the block up, when im- mersed? What force beside the pull of the balance tends to hold the block down? If the block weighed 40 g. in air and required a pull of 60 g. to sink it, hovr large was the block? p. ^ 16 PRINCIPLES OF PHYSICS. Exercise 6. IMMERSION BY USE OF A SINKER. Apparatus: Spring balance; block, of wood; sinker; jar of water. Weigh the block in air. Weigh the sinker in water. Attach the sinker to the block (Fig. 8) and weigh the two in water. Record as follows ; — Weight of block = Weight of sinker in water = Weight of sinker and block in water = Find how much lighter the sinker appears to be with the block attached, and add this number to the weight of the block; compare it with the volume of the block. 19. Volume and Loss of Weight. — The block in water loses its own weight and makes the sinker appear to lose some. This total loss in grams is numerically equal to the volume of the block in cubic centimeters. A block weighs, in air, 200 g. ; a sinker, in water, 80 g. ; block and sinker, in water, 30 g. What is the total loss of weight ? The block loses its own weight of 200 g., and makes the sinker lose 80 — 30, or 50 g. The total loss is 260 g. Therefore, the volume of the block is 250 cc. Since the density of the block = — = .8. •^ 250 Weight of block Weight of block + loss of weight of sinker Fig. 8. Density = Problems. 1. A piece of wood weighs, in air, 160 g. ; a sinker weighs, in water, 200 g. ; the wood and sinker, in water, weigh 100 g. Find the volume and density of the wood. DENSITY. — SPECIFIC GHAVITT. 17 2. What is the density of a substance that weighs 60 g. in air, which, put in water, attached to 40 g. of metal, weighs 15 g. ? 3. An anchor weighing ten pounds in water sinks a fifteen-pound block of wood underneath the water ; a force of four pounds brings them to the surface together. What is the density of the wood? 4. A lump of ore weighs 35 g. in air and 25 g. in water ; find its loss in water, and its volume. Exercise 7. DENSITY AND SPECIFIC GRAVITY OF A LIQUID. CAPACITY OF A BOTTLE. Apparatus : Glass tottle with stopper ; 250-gram spring balance ; kerosene oil or sulphate of copper solution. Weigh the empty glass bottle and stopper. Fill the bottle with water, and weigh. The latter weight should be within the capacity of a 250-gram spring balance. From the fact that a cubic centimeter of water weighs one gram, calculate the volume of liquid held by the bottle. Fill the bottle with kerosene oil or sulphate of copper solu- tion, and find the number of grams of liquid held by the bottle. Cal- culate the weight of one cubic centimeter of the liquid, by dividing the weight by the volume. Problems. 1. An empty can weighs 200 g.; when full of water, it weighs 600 g.; how large is the can? How many cubic centimeters of milk will it hold ? If it weighs 612.8 g. when filled with milk, how many grams of milk does it hold ? What does 1 cc. of milk weigh ? What is the density of milk ? 2. How much water does a cubic centimeter of lead displace? How much, then, is it buoyed up in water? What is its loss of weight in water? 3. How much does 1 cc. of lead weigh in water, its density being 11.4? How much does 1 cc. of aluminum, density 2.6, weigh in water? Which is the better for the keel of a boat ? Which is the better for the body and deck? 18 PRINCIPLES OF PHYSICS. 4. A bottle weighs, when empty, 70 g. ; filled with water, 210 g. ; how many cubic centimeters of a liquid does it hold? Filled with nitric acid it weighs 280 g. ; find the weight of one cubic centimeter of the acid. What is the density of the acid? 5. A 'bottle weighs, empty, 100 g. ; filled with water, 900 g. ; what is the capacity of the bottle ? How many cubic centimeters of mer- cury, density 13.6, will it hold? How many grams of mercury will it hold ? How rnany grams of oil, density .82, will it hold? 6. A perfume bottle, empty, weighs 80 g. ; filled with water, 280 g. ; filled with perfume (alcohol), 250 g. How large is the bottle, i.e. what is its capacity ? What is the specific gravity of the perfume ? 7. The same bottle, filled with chloroform, weighs 380 g. ; find the specific gravity of chloroform. 8. A bottle full of water weighs 180 g. ; empty, weighs 30 g. ; what is its capacity? Some dry pieces of jock are put in; the bottle and rock weigh 300 g. The bottle is then filled with water, and found to weigh 350 g. Find the volume of water added, the volume or size of the fragments of rock, and the density of the rock. 9. Tell how to find the specific gravity of sand. 10. A piece of oak 9 cm. long, 5.5 cm. wide, and 4.5 cm. thick, weighs 189.4 g. ; what is its density? How much water will it dis- place? Will it sink or float? If it were placed in water, how much of its volume would be under water? How much would be out of water? 11. How large a weight must be placed on top of the piece of oak of Problem 10 to make the top of the block just level with the surface of the water? If a block of cherry, density = .5, of the same size as the block of oak, were floated in water, would it take a larger or smaller weight to sink the block of cherry to the level of the water, than to sink the block of oak ? ' How large a weight must be used ? 12. If a lead sinker weighs 60 g. in air and 54.7 g. in water, what is its specific gravity? What is its loss of weight? its displacement? its volume ? 13. What is the buoyant effect of the water on the sinker of Prob- lem 12? What would be the buoyant effect of alcohol, density = .8? Would this sinker float or sink in mercury, density = 13.6 ? CHAPTER II. PEESSTIEE. 20. Force, or pressure, is a push or a pull. If a body is stationary, a force applied to it tends to start it in motion ; or, if the body is already moving, the force tends to stop it, to make it go faster, or change the direction of its motion. In moving a train, when does the locomotive push, and when does it pull ? Does a horse push or pull on the collar ? on the wagon to which he is attached ? A Japanese carpenter uses his saw and plane in a way opposite to ours ; does he push or pull them ? In Holland, the dogs that move the milk carts are often hitched directly under the cart ; do they push or pull? 21. To set a Body in Motion, a force must be applied outside the body. It is the resistance of the rails that enables a loco- motive to move ahead. When the track is slippery, the driv- ing wheels slip round on the rails, and the locomotive stands still. That the force is applied outside a moving body is clearly indicated by the fact that, in rowing, limber oars are bent in the direction in which the boat moves, showing that the water presses against the blades in that direction. If a row of boys were standing on a frozen pond, and another boy tried to move himself along by pushing on the line, the boys in the line would feel that they were pushing him. If a boy standing in a basket pulls on the handles, does he lift himself from the ground ? Can other persons standing on the ground raise him by pulling upward on the handles ? Suppose that during a calm the helmsman should blow on the 19 20 PBINCIPLES OF PBT3ICS. sails of his boat with a pair of bellows, as he might on the sails of a toy boat in a tub, would the boat move as a toy one would ? 22. Weight is the downward force a body exerts because of its apparent attraction by the earth. If you hold a piece of iron or a book in your hand, you must press upward, — that is, exert an upward force, — to balance the downward force exerted by the iron or the book. This down- ward force, which is not exactly understood, appears to be an attraction exerted by the earth. 23. Weight of Air. — Air is a gas, and is free to move in any direction. The fact that it remains enveloping the earth, in- stead of flying off into space, shows that it is subject to the earth's attraction, — that is, it has weight. Since it has weight, it must exert a force downward. Imagine grapes piled many feet deep; where in the pile will the grapes be most com- pressed ? This downward pressure of the atmosphere is easily shown. 24. Transmission of Pressure. — Put a piece of paper over the mouth of a bottle filled with water, and invert the bottle. If the paper itself held the water in the bottle, could the paper be flat ? Put water in a large rubber cloth, and notice that the cloth bulges downward under the weight of the water. In place of paper, tie cloth over the mouth of the bottle, and repeat the experiment. Fill with water, and invert, a small bore tube, closed at one end, thus making a long, narrow bottle. When a bottle, or any other object, is pressed into a dish of water, the water transmits the pressure in every direction. If the dish is weak, the sides may give way, or the bottom be forced out ; otherwise, the water escapes upward. In driving many sticks of wood or piles into soft ground, the downward pressure exerted by the last ones often causes an upward press- ure in the ground sufficient to lift those previously driven. PRESSURE. 21 Fill a bottle with water ; cover the mouth with paper ; in- vert the bottle, place it mouth down in a dish of water, and remove the paper (Fig. 9). The downward pressure of air on the surface of the water in the dish causes an upward pressure in the bottle. This upward pressure keeps the water in the bottle. In place of a bottle, fill a tube, one end of which is Fig. 9. Fig. 10. covered with thin rubber (R, Fig. 10). The air pressure on the glass cannot be observed, as the strength of the glass prevents any bulging. 25. Suction. — A common attempt to explain the results of various experiments with water, is to call the force that holds the water in place suction. To determine how powerful that force is, put the empty tube, rubber down, on the surface of the water in the dish. Notice if any more force is needed to lift the tube from the water than is needed to lift it from the table. 26. Unbalanced Force. — To make a body move, a force must be applied from outside the body; but if two equal forces in opposite directions be applied, the body will not be moved by these forces. Therefore, if one force be observed to act on a body, and the body does not move, then there must be another force equal in amount acting in an opposite direction. In order, then, to set an object in motion there must be an unbalanced outside force, or pressure. 22 PRINCIPLES OF PHYSICS. B ! Fig. I 1, Fill a glass tube (Fig. 11) with water to the level BC. The air press- ure on B, being balanced by an equal air pressure on C, does not tend to move the column of water. Cover B with the thumb, and partially remove the air from A, by the mouth. The air pressure on C has been lessened, but the air pressure is kept from acting on B. With B uncovered, lessen the pressure on C. The press- ure on B is now unbalanced and forces the water down B and up the arm AC. 27. Magdeburgh Hemispheres are two brass cups, fitting together, with a joint made air-tight by tallow (Fig. 12). A valve permits air to be exhausted or admitted at will. Exhaust the air, and /p\\ KO Fig. 12. try to pull the hemispheres apart. Admit the air, and try it again. Exhaust the air again, and admit it while pulling. "When the hemispheres are full of air, the inside and outside pressures are balanced. When the air is exhausted, there is no inside pressure, and the outside forces, all pressing inward, hold the cups together. An " eight- in-one " apparatus even better serves the purpose of this experiment. Push in the piston, close the valve, and pull on the handles. Open the valve while still pulling. '''^■'^' 28. Air Pressure. — If A and B (Fig. 13) are two pieces of wood, what are the direc- tions of the two forces acting on them to hold them together ? PRESS USE. 23 To pull A and B apart, forces greater than the pressure exerted by the clamp must be used. Figure 1-i shows a " vacuum-tipped arrow " pressed against a board, to expel the air from beneath the rubber. What force would be required to pull the arrow from the board ? How can it be taken off without using this force ? Dip the end of an open tube into a dish of water (Fig. 15). Why does not the air, pressing down on the sur- face of the water in the dish, force the •water up higher in the tube? Notice that the air presses down in the tube. What pressures are balanced ? Lessen the pressure in the tube, by removing some of the air, with the mouth. The out- side air pressure on the surface of the water in the dish is no longer bal- anced, and causes the water to move up the tube. On sucking a liquid through a straw, what makes the liquid rise ? How high can air pressure raise Fig. 1 5. a liquid ? Fig. 14. (^ 29. How Air Pressure acts. — Make a large hole in the bottom of a tin can. Place the can on the plate of an air- pump, using tallow to make an air-tight joint. Put an apple on the top of the can (Fig. 16). The apple, being pushed by equal pressures of air above and below, is not moved. Exhaust the air from the " can. This removes one of these pressures •and leaves the other free to act. What pressure is unbalanced ? Tie a sheet of rubber over the mouth of the can, and try the effect of exhausting the air. Fig. 16. 24 PRINCIPLES OF PHYSICS. If the two springs shown in Fig. 17 press on an apple from above and below with equal force, the apple will not be pushed down into the can. But when the lower spring is removed (Fig. 18), the pressure above is unbalanced and is free to act. In various forms of cash carriers, air is exhausted from one end of a pipe. Explain how an object is then made to traverse the pipe. A pneumatic railway was operated on this principle for a short time in Glasgow. In preserving fruit in glass jars, the liquid, at the time of sealing, fills the jar; when cool, the liquid contracts, and a space is left at the top of the jar. What is there in this space ? What enters, if an opening is made under the rubber ring ? Why does the top then unscrew more easily ? Fig. 17. Fig. IS. 30. Elasticity. — Such substances as steel, glass, wood, india rubber, and paper, when bent or compressed, tend to return to their original form. This is due to a quality called elasticity. Wet clay, putty, and such materials are not elastic. A billiard ball dropped on a steel floor will rebound to practically the height from which it is dropped. If both substances were perfectly elastic, it would rebound the entire distance. Is air elastic ? Squeeze a tennis ball, or push down and re- lease the piston of a bicycle pump, while preventing the air from escaping. A pneumatic tire is inflated by pumping in from three to five times as much air as the tire would hold if open to the air. Does the pressure exerted by the air in the tire increase as more air is pumped in ? What can you say about the effect that a decrease of volume has on the pressure of a gas ? Remember how much air is pressed into a tire. What makes the water come out of a so-called siphon of carbonated water, when the valve is opened ? What pressure is unbalanced ? PBESSUBE. 25 31. Volume and Pressure of Gas. — Fasten the stem of a toy balloon air tight into an open end of a so-called one-eighth inch gas valve, V, Fig. 19. The other end of the valve screws into a bit of pipe, T, the lower end of which is to be fixed in a wooden base, B. Open the valve, V, blow air into the balloon, close the valve, screw on the base, place under the receiver of an air- pump, and exhaust the air. The tube P connects with the air-pump. Why did not the pressure of air in the balloon cause expansion be- fore exhaustion ? Admit air to the receiver, and explain the action of the balloon. A piece of the inner tube of a bicycle tire may take the place of the toy balloon and stand. The tube should be half filled with air and tied tightly at the ends. It may be kept off the pump-plate by raising it on a pasteboard box. The statement regarding volume and pressure of gas is called, after its discoverer, Boyle's Law. Fig. 19. 32. Measurement of Air Pressure. — When a bottle full of water is inverted in a shallow dish of water, the air pressure on the surface of the water in the dish holds up the water in the bottle. To measure this air pressure, we have only to find out how high a column of water it will hold up. This could be done, if the bottle were long enough to let the water rise as high as the air pressure could push it. Mercury, being heavier than water, is more convenient to use for this pur- pose, as the column need not be so high. Mercury weighs 13.6 times as much as water ; therefore the column held up by the air pressure is 13.6 times as short. 26 PRINCIPLES OF PHYSICS. 33. The Mercury Column. — Fill with mercuiy a heavy glass tube, 36 or more inches long, closed at one end (A, Fig. 20). The bore of the tube should be two or three millimeters. Hold a finger over the open end; invert the tube (as B), and place the end in a dish of mercury. Release the mer- cury in the tube, and let the column come to rest (C). Measure the height (h) of the mercury above the surface of the liquid in the dish. Record, with date and hour. Wliat holds the mercury up in the tube ? Why is it not held up higher ? What is there in the tube above the mer- cury? Before answering, tip the tube as in Fig. 21, and explain what happens. In moving the tube to this position, does the mercury rise ? Place a stick hori- zontally behind the tube in C, Fig. 20, at the level of the mercury ; incline the tube, and determine whether the mercury rises or falls. Attach paper to the back of the tube, so that it extends two inches above and below the level of the mercury. Mark lines for inches and tenths, and read and record the height twice each day. Fig. 2 1 . ZI * B S a Fig. 22. 34. To vary the Pressure on the surface of the mercury in the dish, insert the same or a similar glass tube into a rubber stopper, as in Fig. 22. Slide the stopper, S, a little toward the closed end of the tube. Fill the tube and the rubber pipe, E, with mercury. Put mercury in the jar, J. PBESSUBE. 27 Invert the glass tube, holding the mouth of the rubber pipe upward, and place it in the jar. Press in the stopper, and read the height of the mercury. Increase the pressure on the surface of the mercury, by blowing at T or by the use of a bicycle pump. Suck air from T, or use an air-pump. T may be attached to the air-pump in connection with the receiver in which the rubber balloon was placed (Fig. 19), and the experiment may be repeated. In place of a balloon, a long test-tube, tt, half filled with water and inverted in a Fig. 23. dish of water, can be used (Fig. 23). The test-tube should have a paper scale attached, or be marked with a cross-pencil. Read the height of the mercury and the length of the air column in the test-tube. The pressure on the air in the test- tube is nearly the same as that on the mercury, being only less in proportion as the column of water is shorter and lighter than the mercury column. Exhaust the air till the height of the mercury is reduced one-half, and measure the volume of air in the test-tube. See how far the mercury can be made to fall. Notice the action of the air in the test-tube. Allow air to enter, and note how far the water rises in the test-tube. Was all the air exhausted from the receiver ? What deter- mines this ? Did the mercury fall to the level of the mercury 28 PRINCIPLES OF PHYSICS. in the jar? Place a burned-out incandescent lamp, tip down, in a dish of mercury or water, and file off the end of the tip. Does the liquid entirely fill the bulb ? 35. The Barometer. — A space where there is no air — for instance, the space above the mercury in the glass tube in the last experiment — is called a vacuum. A^glass tube over thirty inches long, closed at one end, filled with mercury, and inverted in an open dish of mercury measures, by the height of the column of mercury, the pressure of the air. This in- strument is called a barometer (i.e. pressure measurer). When we say that one point on a mountain is 1000 feet higher than another, do we mean that the road up the mountain is 1000 feet long ? The height of the mercury column in the barometer, or, as is said for short, the height of the barometer, at the sea level averages about 30 inches, or 76 cm. Would the barometer stand higher or lower on Mt. Washington ? In the bottom of a mine ? At the level of the Dead Sea ? In ascending from the sea level, there is less and less air above to press down on the surface of mercury in the barome- ter. The mercury column therefore falls. On Pike's Peak or Mt. Blanc, the height of the mercury column is about one-half the height at sea level. The fall is one inch for 945 feet of ascent. What would the barometer read at Denver, 5000 feet above the sea level ? What is the difference between the barometer readings at the base and the top of the Eiffel Tower, 1000 feet high ? What is a balloon doing when a barometer in the car falls ? In a barometer, cross-section one square cen- timeter, how many cubic centimeters of mercury are in the tube at the sea level ? Mercury weighs 13.6 times as much as water ; what is the weight of the mercury column in the tube ? What, then, is the average pressure of air in grams per square centimeter, at the sea level ? How high is the column in a water barometer ? If a mercury barometer falls one inch, how much would a water barometer fall ? PRESSURE. 29 The Weather Bureau reports give as the height of the barometer at a place above the level of the sea, not the actual reading, but what would be the reading at that place if a shaft were dug down to the level of the sea and the barometer low- ered there and then read. This can be calculated when the height of the place above the sea level is known. © 36. Water lifted by Air Pressure. — In the apparatus shown in Fig. 24, why does not the pressure of the air on the water- in the dish force water up the tube ? Eaise the piston, thus reducing the air pressure in the tube. The air pressure on the water in the dish is no longer balanced, and forces water up the tube. If the piston was started from the surface of the water, and lifted, how far would the water rise, if the ap- paratus were in New York ? How far would mercury rise ? How far would oil rise, density = .85 ? glycerine, density = 1.26 ? sulphuric acid, density = 1.8 ? If the piston be raised so high that the water is no longer driven up, what would there be between the level of water in the tube and the bottom of the piston ? If this experiment were tried also at Denver, would the column be a different height from the one in New York? If the experiment were tried on different days at the same place, would the liquid always rise to the same height ? Hi Fig. 24. 37. Lifting Pump. — Water usually contains some dissolved air, which expands when the pressure on the water is reduced. How would the use of boiled water (containing no air) affect the height to which the water column could be raised ? In the tube (Fig. 24), the liquid falls again as soon as the piston is pushed down. Make a hole through the piston ; put a trap 30 PRINCIPLES OF PHYSICS. or valve on the top of the hole. Fix firmly in the bottom of the tube a plug of wood, or cork, having a hole covered on the upper part by a valve opening upwards. Water can go up through the valves, but not down. Fit the base of a student lamp chimney with a plug having a half-inch hole, covered by a piece of thin leather nearly as large as the plug (Fig. 25). The leather is held by a tack at one side. The pis- ton is made of a single piece of wood, pierced by as large a hole as its diame- ter will permit. It should be made a trifle smaller than the bore of the chim- ney, and be wound with soft white string till it is a loose fit. Place the base of the chimney in a dish of water, and work the piston. The apparatus operates as a lifting pump. What forces the water through the lower valve ? How high can that force drive it ? Study the pump, and make diagrams showing the position of the valves when the piston is ascending, and again when it is de- scending. How high can the water be raised when above the upper valve ? In case water is to be pumped up 200 feet, where must the upper valve be, at the highest part of the stroke ? Owing to the leakage of valves and to dissolved air in the water, instead of 34 feet, 20 feet is the practical limit for the distance of the piston above the water in a well. What would this height be if mercury were pumped ? If oil, density = .8, were pumped ? While the model pump is full of water, raise it from the dish, and continue pumping. Explain why bubbles of air enter the lower valve. With metal pipe and tight-fitting valves and piston, the apparatus would be a primitive form of air- pump. Fig. 25. PBESSUBE. 31 Fig. 26. 38. Cartesian Diver. — Nearly fill a small glass bottle -with water ; invert quickly in a dish of water. If the bottle does not sink at a slight blow, and slowly rise to the surface, remove the bottle, and add or take out water, as needed. Cover the mouth of the bottle with the finger, and re- move to a narrow jar. Fill the jar with water (Fig. 26), and tie sheet rubber over the top. If the jar is narrow, the palm of the hand may take the place of the rubber. Press on the rubber, and note what the bottle does, a.nd any change of level of water in it. Remove the pressure. How can the bottle and the air in it be made to displace less water? By what is a body in water buoyed up ? Is the buoyancy of the bottle increased or decreased by compressing the air in it ? ' This appa- ratus is called a CaHesian diver. Place a bottle containing a Cartesian diver under the re- ceiver of an air-pump. Exhaust the air a little, and then admit air. Repeat, exhausting more air each time. Explain the change of level of the diver, and the rising and sinking. Try a piece of hard ■wood in place of the diver. What change in density is there ? 39. Force-pump. — If the piston- rod of a lifting pump is made to pass through a tight-fitting cover, and an opening anywhere above the piston is connected with a pipe, water can be forced or driven by the piston to a height depending only on the force applied to the piston and on the However, in force-pumps it is usual Fig. 27. strength of the pump. 32 PBINCIPLES OF PHYSICS. to make the piston solid, and place the upper valve at the side in the tube through which the water leaves the pump, as shown in Figure 27, page 31. Make diagrams of a force-pump, with the piston ascending, and with the piston descending. What makes the water rise through the lower valve ? How high could the lower valve be above the level of the liquid in the reservoir, in pumping mercury? water? oil, density = .8 ? 40. Siphon. — Eeread section 36, and then look at the appa- ratus shown in Fig. 24, page 29. Suppose the straight tube in Fig. 24 were replaced by a bent tube with a weighted piston, as shown in Fig. 28. The pressure exerted by the air in the tube is in part removed by the weighted piston. Turn the tube up until a little watet is over the piston ; then replace the tube, and pull the piston down. The water acts as a pack- ing, and even if the piston does not fit, the water leaks slower than air. Work the piston up and down a few times, or turn the tube up and fill it with water. For all practical purposes, the piston is now made of water. This piston of water is drawn down by its weight, just as the first piston was, and in the same way it lessens the pressure of air in the tube. The atmospheric pressure on the water in the dish is no longer balanced, and drives the water up the short arm. How high can the bend be above the level of the water in the tank, when the barometer is 76 cm. ? How high can the bend be, if mercury is siphoned ? If the long arm be placed in the tank, what happens ? Which would exert more pressure, the liquid in the long or in the short arm ? How high does water rise in a siphon placed empty in a tank ? How should the length of the short arm be measured ? What ffl Fig. 28- PBESSUBE. 83 is the effect of making an opening at the bend ? at a point in the long arm at the level of the water in the tank ? Can water be siphoned over a hill 100 feet high ? What is the effect of lengthening the long arm ? 41. Experiments with Siphons. — Make a siphon in the form represented in Fig. 29, using a rubber tube, and raise and lower Fig. 29. "•^ Fig. 30. the opening. Partly close the opening, and try to find out how high the water spurts from different levels. Raise and lower one of a pair of tanks, connected by a siphon, as represented in Fig. 30. Remove the siphon. Does it empty itself ? Why ? Under what conditions will a siphon fail to work ? Consider the comparative lengths of the short and the long arms, the pressure of air, and the possibility of leaks in the bends. 42. Intermittent Siphon. — Remove the bottom of a bottle; put a siphon through the stopper. Place it where water will drip into it from a faucet or tank, as in Fig. 31. Explain what happens. Read the encyclopedia on " Tantalus cup " and " Intermittent springs." How could a sewer be flushed at regular intervals ? Fig. 31, CHAPTER III. LIQUIDS AND GASES. Exercise 8. SPECIFIC GRAVITY OF A LIQUID. BALANCING COLUMNS. Apparatus : Two glass tubes, connected by rubber tubing with a three-way tube (preferably of metal), to the third- arm of which is attached a short rubber tube, closed by a plug ; dish of water ; dish of some saline mixture. The plug may be a glass tube closed at one end. The glass tube A (Fig. 32) dips into pure water ; the glass tube B dips into a saline mixture, such as common salt or sulphate of copper. Remove the plug, G, and draw the air from the three-way tube, T, until the liquids rise nearly to the top of A and B. Pinch tight the short rubber tube near the open end, and insert the plug G. Neglecting the effect of capUlarity (section 78, page 64), measure the height at which each liquid stands in the tube above the level in the dish. Record these lengths on a diagram in the note-book. The liquid in B is as many times denser than the water in A as the length of the liquid in B is con- tained times in the length of the water in A. Suppose the length of the water in ^ = 70 cm., and of the liquid in B = 50 cm. Then the pi^ tala liquid in B is ^, or 1.4, times as heavy as E=.3 1^^ water, and 1.4 is the specific gravity of the liquid in B. Its density is also 1.4 grams per cubic centimeter. Find, in this way, the specific gravity of kerosene, mercury, and brine. 34 LIQUIDS AND GASES. 35 43. The Specific Gravity of a Number of Liquids can be measured at one time by the apparatus shown in Tig. 33. Short lengths of rubber tubing are slipped over the side branches of T, and hold tubes dipping into various solutions, some of which are suggested in the figure. Exhaust the air from T and insert the plug G. Make a diagram, and record the name of the liquid and the height in each tube.^ „__G T 1 nnnnn^ m nnni Fi&33. Problems. 1. If the liquid in B, Fig. 32, page 34, were a column of acid 50 cm. high, and the water in A were 60 cm. high, what would be the specific gravity of the acid ? ) Si- 2. A pump is starting to draw liquids through two pipes; one liquid is water, the other is mercury. Which column is the higher? How many times higher? 3. What is the specific gravity of a liquid, if a column of it (B, Fig. 82) stands 25 cm. above the level of the liquid in the dish, while the water column is 22 cm. high ? iThe measurement of each height may be made by a different pupil, who will report the reading to the class. Subsequently, every pupil computes the specific gravities of all the liquids. ^6 PRINCIPLES OF PHYSICS. 44. Volume and Displacement. — If a lump weighs five grams in air and three grams in water, the volume is two cubic centi- meters. The lump loses two grams of weight when weighed in water, because it pushes aside, or displaces,, two grams of water. Since one gram of water takes up one cubic centimeter of space, the size, or volume, of the lump must be two cubic cen- timeters. Suppose the same lump is weighed in oil ; the loss of weight is less than in water. The lump displaces two cubic centimeters in oil ; but oil is lighter than water, and the two cubic centimeters of oil displaced weigh less than two grams. It is not buoyed up as much ; that is, it does not lose as much weight in oil as in water. A match, made heavy by a small pin, floats in water and sinks in oil. 45. Specific Gravity of a Liquid by Immersion of a Solid. — Weigh a lump in air, then in water, and then in the liquid of which we wish to find the specific gravity. Suppose the loss of weight in water is 40 g. ; then the volume of the lump is 40 cc. In oil this lump displaces 40 cc, but loses only 32 g. of weight ; therefore, 40 cc. of oil must weigh 32 g. ; and 1 cc. of oil weighs |f = .8 g. What we have done is to divide the loss of weight in the liquid to be measured by the loss of weight in water. Problems. 1. A piece of rock weighs 200 g. in air, 130 g. in water. What is the specific gravity of a liquid in which it weighs 120 g. ? 2. A lump of metal weighs 22 g. in air, 16 g. in water, and 17 g. in another liquid. What is the density of the other liquid ? 3. What is the size of a lump of aluminum that weighs 52 g. in air and 32 g. in water? How much room does the lump take up? How much of any liquid would it displace? How many grams of oil, density = .8, would it displace? of acid, density = 1.5? How much weight would it apparently lose in these liquids? What is its weight in these liquids? LIQUIDS AND GASES. 37 46. Boyle's Law. — If the escape of air from a bicycle pump is prevented, the air within is pushed into a smaller space, or is compressed, by forcing in the piston. The greater the pressure applied to the piston, the smaller becomes the volume of air. When the pressure is removed, the air expands, and the piston returns to its former position. The volume, or size, of a mass of gas decreases as the pressure on it increases (see section 31, page 25). This fact, from the name of its discoverer, is called Boyle's Law. V c 11 .j_ Exercise 9. (a) FBESSUB£S GBEATEB THAN AN ATMOSFHEBE. Apparatus: Two glass tubes, one of which {A) must be of uniform "bore, and may be closed by a clamp and a rubber washer; thick-walled rubber tubing ; mercury. The glass tubes are attached to boards, which slide in grooves on an upright support, and are held by clamps in any position. Close A (Fig. 34) at c, by the clamp and rubber washer. Connect A with -ff, which is open at the upper end, by means of the thick-walled rubber tubing, B. Case I. — While in the po- sition indicated in I, Fig. 34, pour mercury in at R, and ■^ ■" l"r~~li^ open c so that air may escape. Close c. Read the length of the air column in A, making sure to read from c to the top of the curved surface of the mercury. Call this length the volume, V. Bead the height of the barometer in the room, and record. As x and y (I, Fig. 34) are at the same level, the air in A is under the same pressure as that in the room, and this pressure is indicated y. II. by the barometer. Fig. 34. 38 PBINCIPLES OF PHYSICS. Case //. — Raise R 20 em. or more (II, Fig. 34). The volume of the air in A becomes smaller, and this new volume, V^, is the distance from c to X. The air in A is under greater pressure, for it is submit- ted not only to the pressure of the air in the room, which presses on the mercury in R, but also to the column of mercury between the levels X and y. Call this difference of level, h. Case in. — Raise R (Fig. 34) higher than in Case II, and repeat the readings. Continue till the greatest possible difference exists between x and y. Arrange the observations as in the following table: — Barometer B h Pressure B + h = P Volume F PxV Case I . . . Case II . . . Case III . . Case IV Case V . . B is the height of the barometer in centimeters, and if the experi- ment is performed quickly, — that is, within an hour, — the air press- ure is not likely to change much, and the barometer reading will be the same for all cases ; h is the difference of level of mercury in the two arms; B + h is the total pressure, P, on the air in A. Multiply the total pressure, P, by the corresponding volume F, and put the result in the column PV. Considering the results in the last column, how do the first two or three figures in each result appear ? 47. A Constant. — When the computations in each one of a series of experiments give practically the same result in all cases, the result is called a constant. Many of the tables in the Appendix are made up of physical constants. LIQUIDS AND GASES. 39 Exercise 9. (6) PRESSUBES LESS THAN AN ATMOSPHEEE. Return to Case I, page 37. Lower R or raise A (Fig. 35), pressure on the air in A is now less than that of the atmosphere. Measure the air column in A and the distance, h, between the levels of mercury, x and y. Repeat several times, till the greatest possible difference of level is obtained. Arrange the results in a table, remembering that the pressure column is B — /i ; h is the difference between the levels of the mercury columns, and must be subtracted from the barometer reading to get the pressure of the air in A. The "T h .ty. ^} 48. Formula. — Starting with a given amount of gas, and keeping the temperature un- changed, the pressure, -P, multiplied by the volume at that pressure, always equals a con- stant. More briefly, P xV= K, K standing for some number. Changing the pressure to Pi, the volume changes ; call this new volume Vi- Then, as we have seen in the experi- ment, P^ xVi— K. These products are equal to each other. Pressure now times the volume now = anotlier pressure times the new volume. This can be abbreviated as P x F= Pi X Fi. Ftg. 35. Problems. 1. The volume of a balloon is 400 oc, the pressure on it is that of the atmosphere, barometer = 76 cm. What will the volume become when the pressure is 38 cm.? .4ns. V■^ = 800. In this example, 400 cc. = V, 76 cm. = P, 38 cm. = Pi. Substituting, PxF=Pi X Fi, 76 X 400 = 38 X Fi, 38 Fi = 30400 ; therefore Fi = 800. 40 PRINCIPLES OF PHYSICS. 2. One hundred cubic feet of air under a pressure of one atmos- piiere has the pressure increased to 20 atmospheres. What does the volume become? Ans. V\ = 5 cubic feet. 1 X 100 = 20 X Vi. 3. The volume of a mass of gas is 150 cc, the barometer stands at 76 cm. What is the barometer when the volume is 100 cc. ? A ns. Pj = 114 cm. 4. P = iO cc, V = 110 CO., Pi = 100 cm., Fj = what? '- - ' 5. P = what, when V = 24, if P^ = 200 when F; = 96? i ' "'' 6. A tank is filled with air, compressed to 1800 pounds to the square inch. The volume of the tank is 2 cubic feet. The air, before compression, was under a pressure of 15 pounds to the square inch (the average pressure at the sea level). How many cubic feet of air were compressed in the tank ? 7. In a certain compressed-air system for operating railway switches, the air is under a pressure of 95 pounds per square inch. What volume of air at this pressure will have a volume of 190 cubic feet, when subjected to merely the atmospheric pressure of 15 pounds per square inch ? Arts. 80 cubic feet. 8. For operating calcium lights, oxygen is sold in steel cylinders under a pressure one hundred times as great as that of the atmos- phere. When one-half of the gas has escaped, what is the pressure in a cylinder? 9. If the cylinder in the last problem holds five cubic feet, and if it should burst, how much space would the oxygen then fill? If, after half the gas has been used, the remainder is allowed to escape into a rubber bag, what is the volume of the gas in the bag ? 10, The Pintsch system of car-lighting employs oil gas ; cylinders having a capacity of four cubic feet are used, filled with gas under a pressure of 13 atmospheres. How much gas do they hold when full? What is the pressure on the cylinders ? 49. Experiment for Air Pressure Greater than an Atmosphere. — In the year 1772, when Eobert Boyle was studying the "spring of the air," flexible rubber tubing had not been LIQUIDS AND GASES. 41 Fig. 36. 11 / invented; so he had to use an apparatus made entirely of glass, like Fig. 36. This was a bent tube, closed at the short end, supported on a stand. \ / Unlike the apparatus used in Exercise 9, page 37, it cannot well be used for pressure less than the atmosphere. To repeat the experiment as Boyle did it, there is needed, in addition, two or three pounds of mercury, and a wooden base, with raised edges, to catch any mercury that is spilled. Rings cut from a rubber tube may be used to mark the levels of the mercury. Pour in mercury till the tube is filled above the bend. The level need not be the same in both arms. Slip on rubber rings at the level of the mercury in each arm. Add more mercury, till the long arm is one-third or one-half full. Mark the new levels of the mercury in both arms by rings, as before. Then nearly fill the long arm with mercury. Pour out the mercury, and lay the tube down flat on a table. Put a block or box against the bend, as shown in Fig. 37. Make a diagram like Fig. 37 in the note-book. Measure the distances h^, h^, hg, — the heights of the mercury in the long arm above the bend, — and «!, ttj, Oj, — the heights of the mercury in the short arm above the bend ; c is the total height of the short arm. The volume of the air in the first case is c — fflj. The pressure is the height of the barometer plus the amount that h^ is higher than aj ; the total pressure, then, is barometer -|- (/ii — Ui). This total pressure, P, on the air in the short Ui 'i^^ Fig. 3 7. h. ""f 42 PRINCIPLES OF PHYSICS. arm is the air pressure as read by the barometer plus the un- balanced mercury column in the long arm (/ij — Oi). Record thus : — Barometbk h~a p Total Pressuke Bab. + (*.-a) r Volume (c-a) P-x V Case I Case II Case III Before removing the rubber rings, compute the values for the last column in the table, and if the products are widely different, repeat the measurements. 50. Limitations of Boyle's Law. — Boyle's Law is approxi- mately true for all gases at temperatures far above the point where they can be liquefied. As the temperature is reduced, a gas is compressed at a greater rate than might be expected, and near the point where the gas becomes a liquid, the con- densation is very much more rapid than Boyle's Law would indicate. Problems. a. A day when the barometer reads 75 cm., mercury is poured into an apparatus like Fig. 37 : A, = 6 cm., and Oj = 4 era. How much greater is the pressure on the air in the short arm than the atmos- pheric pressure? Ans. 2 cm. b. What is the total pressure? Ans. 75-1-2 cm. c. If c = 20 cm., what is the volume of the an in the short arm? Ans. 20 -4 = 16cm. d. Suppose mercury is poured in the long arm till it stands 50 cm. above that in the short arm. The total pressure is now 75 H- 50 cm. or 125 cm. What is the volume in the short arm ? P X r = P, X Vy 76 X 16 = 125 X Vy Find F,. LIQUIDS AND GASES. 43 Exercise 10. DEKSITT OF AIE. Apparatus : A sensitive platform scale ; a two-liter bottle, fitted with a rubber stopper that is bored to admit a gas valve with a corrugated tip (Fig. 38) i the valve and stopper made air-tight with tallow or glycerine ; a cradle (Fig. 39) to hold the bottle on the sensitive platform scale, i Find the capacity of the bottle, by weighing it when empty, and again when filled with water. The number of grams of water it holds is its volume in cubic centimeters. Mark this number on the bottle. Exhaust the air from the bottle, and record the amount of air exhausted. If all the air could be exhausted from the bottle, /^ 7^ ^^^ capacity of the bottle would show the number l^-ry f^ ' of cubic centimeters of air exhausted. But since an ordinary air-pump must leave an appreciable amount of air in the bottle, and any process that Fig. 3.9, would exhaust practically all the air would be slow and tedious, the following appa- ratus is used to measure the amount of air drawn from the bottle. The bottle is connected with the air-pump, by the tube B, as shown in Fig. 40. 7" is a three-way connection ; H, a glass tube one meter long, resting in a dish of mer- cury. The rubber tubes, A and B, are of sufficient length to aUow the bottle and pump to be in convenient positions on the floor or on a table or Fig. 40. 1 From a number of platform scales, the most sensitive may be selected by loading them in turn with weights of a kilogram or more in each pan. Balance 44 PRINCIPLES OF PHYSICS. shelf. C is a bottle that acts as a trap to catch mercury, in case the tube H is accidentally raised during the exhaustion. P connects with the air-pump. The trap C is not needed, provided the gauge is 90 or 100 cm. deep, and not more than half filled with mercury. Instead of C and H, a U-shaped tube, used as a pressure or vacuum gauge (M, Fig. 214, page 242) is convenient. Exhaust the air so far as possible. Read the barometer and the height of the mercury in H,^ and calculate the amount of air taken from the bottle. Suppose the barometer reading was 76 cm., the gauge reading in H, 70 cm. All the air was not exhausted, or the mercury in H would have risen to 76 cm. ; ^ of the air was pumped out. Find the number of cubic centimeters of air pumped out, by multiplying the capacity of the bottle — for instance, 2200 cc. — by Jg. This gives 2026 cc. The accuracy of the weighing is not sufficient to make the 6, or perhaps even the 26, of any account. Call the amount of the air 2020 cc. Close the valve, disconnect the bottle, and weigh it. Open the the pans by adding bits of wood or chalk, if needed ; then note what distance the sliding weight must be moved to make the pointer move one division while the pans are swinging a little. For this experiment a scale should be found that indicates a change of one-tenth gram in weight. 1 A large pump exhausts more quickly. Put the bottle on the scale and counterpoise, having the sliding weight near the four-gram or five-gram mark. Without changing the weights, attach the bottle to A, and let one pupil operate the pump with strokes the full length of the cylinder, two others measure the height of the mercury column in H, and another turn the valve, V, at the moment of reading. One of the two pupils measuring the height in H keeps the meter stick exactly at the level of the mercury in the dish. Place the bottle on the scales, and move the sliding weight till exact balance is obtained. Record the weight as X (the weight of bottle and valve) -|- the reading of the sliding weight in grams and tenths of a gram. Admit air, and get the balance by moving the sliding weight. Record as X-1- slider reading. Read the barometer and thermometer. The exercise is rapidly repeated by a large number of pupils, the bottle, as soon as the air is admitted and the weighing made, being given to another pupil. Divide the difference in the weighings — an amount between one and four grams — by the amount of air, and record the weight of one cubic centimeter of air at the observed barom- eter and thermometer readings. It would prove instructive to fill the bottle with coal gas, hydrogen, or carbonic acid gas. The first two gases should be passed into the inverted and unstoppered bottle by a tube reaching to the bottom. With carbonic acid gas, the bottle must be filled mouth up. Pass the gas for fifteen minutes. LIQUIDS AND GASES. 45 valve, and record the increase of weight. Find the weight of one cubic centimeter of air. 51. The Weight of Air. — Weigh an incandescent lamp; one having a broken filament is as useful as a new one. Then, file till air is admitted, letting the filings fall in the scale-pan. Weigh the lamp and the filings. The experiment is made less tedious if most of the tip is filed away before the first weigh- ing is made. Of course the filings made before the first weigh- ing are thrown away. Problems. 1. In Exercise 10, page 43, if the barometer reads 74 cm., and the gauge reads 72 cm., what proportion of the air has been pumped from the bottle? 2. What is the reading of the gauge when half the air is ex- hausted ? 3. If one cubic centimeter of dry air, at 20° Centigrade, barometer = 76 cm., weighs .0012 g., what is the weight of one liter of air 'I 4. A pneumatic tire has a capacity of 4200 cc. ; what weight of air does it hold? (Use data of third problem.) 5. If air is pumped in to 45 pounds to the square inch (that is, 3 atmospheres in addition to the one atmosphere already in it), what additional weight is there in the tire? 6. Does a bicycle weigh more or less when the tires are blown up hard? 7. One cubic centimeter of mercury weighs 13.06 g., one cubic centimeter of air, as in Problem 3, weighs .0012 g. ; how many times heavier than air is mercury ? 8. When the air pressure holds up a column of mercury in the barometer 76 c, what is the pressure per square centimeter ? Figure the weight of a column of mercury 76 cm. long and 1 cm. square. 9. A body is buoyed up by the weight of the air it displaces. Find the volume of 140 g. of zinc (density = 7). How many cubic centi- meters of air does it displace? What is the weight of that air? How much lighter does the zinc appear to be in air than in a vacuum ? 46 PRINCIPLES OF PHYSICS. 10. How long is a column of air that weighs as much as a column of mercui-y 76 cm. long, 1 cm. square ? Divide the result of Problem 8 by the weight of 1 oc. of air. This is the height of the so-called "homogeneous atmosphere." If the air did not become thinner at higher altitudes, it would not extend so far as it really does, and the height — of course an imaginary one — would be about five miles. 11. If a lump of gold (density = 19.3) is balanced against a lump of aluminum (density = 2.7), which occupies the greater amount of space, and which is buoyed up the more? What would happen if the scales were put under the receiver of an air-pump and the air exhausted ? 12. Which displaces the more air, a pound of feathers or a pound of gold ? In which case would you get the greater amount of feathers iQ a pound, when weighed, balanced against iron weights, in air or in a vacuum? 13. From the following data, calculate the density of air : — Capacity of bottle, 2000 cc. ; Barometer, 75 c. ; Height of mercury in gauge, 72 o. ; Weight of bottle, air exhausted, X 4 1.9 g. ; Weight of bottle, air admitted, X + 4.1 g. 14. How large is a bottle that holds .5 g. of air ? 15. A toy balloon has a volume of 400 ec. ; how much is it buoyed up by the air it displaces? How much can the balloon and contained gas weigh and the balloon still float ? 16. Some one has proposed an air-ship or balloon of thin sheet steel, made air-tight. The air is supposed to be exhausted. If the volume is 600,000,000 cc, what is the buoyant effect? How much can the balloon and its load weigh, and the balloon yet float? What would be the tendency of the air pressure on the thin covering? 17. A Whitehead torpedo weighs, in air, 522,000 g. ; in water it weighs 250 g. ; how large is it ? If the propeller does not operate, will it sink or float? On the longest run possible, 32,000 g. of air escapes to drive the little engine ; the torpedo is 32,000 g. lighter than when launched. If it misses its mark, will it sink or float? How many cubic centimeters will float out of water? LIQUIDS AND GASES. 47 82. Air-pumps. — The ordinary lifting pump (Fig. 25, page 30) acts as an air-pump, before the water in the well or cis- tern is drawn up to the piston. The first few strokes pump air, removing the downward pressure on the water in the pipe. This allows the pressure of the air on the water out- side the pipe, being no longer balanced, to force water up the pipe, past the valves. When such a pump is made especially for pumping air, the piston is packed with leather, string saturated with tallow, or metallic rings fitting in grooves in the piston. These rings spring out and prevent leakage. The valves are often made of oiled silk, whi&h is much lighter and more air-tight than leather. It is impossible to pump all, or even nearly all, the air out of a bottle or globe with this kind of a pump. When the piston descends, the air, being compressed, lifts the upper valve and passes out. When the piston is raised again, the air in the receiver expands and fills both the receiver and the pump cylinder. Suppose C, in Fig. 41, page 48, holds one liter of air, and R four liters. Then the four liters of air in R, after one stroke, occupy five liters of space, and the entire amount of air in C — one liter, or one-fifth of the original total amount of air — is removed at the first stroke, and four- fifths remain behind. The second stroke removes one-fifth of the remainder. 53. Rate of Exhaustion. — Draw a line, AB, of any length, in the note-book. Take away one-fifth of this, and make- another line the length of the re- ^ ^ mainder. Repeat this ten or more times. Notice that the actual amount removed decreases each time, and that even after a hundred or more of these diminutions, an appreciable amount remains. In an air-pump, the leakage of air into the pump increases as the air is pumped out. After a certain number of strokes, air leaks in as fast as it is pumped out. 48 PRINCIPLES OF PHYSICS. 54. Exhausting by Air-pumps. — By a three-way tube, connect enclosed barometer, B, Fig. 41, with, the tube leading to the receiver, li. The height of B measures the pressure of the air in the receiver, B. Taking full strokes with the pump, notice the relative amount B falls at the end of each stroke. If B falls one-fifth of its first reading, one-fifth of the air has been removed. Try a small receiver at R, or, if necessary, discon- nect R a,t D and let the pump exhaust the air out of the con- necting pipes and barometer bottle merely, so that the volumes of the cylinder, C, and the space to be exhausted are equal. This will be the case when B falls to half its height in one Fig, 41. stroke. At the next stroke, one-half of the remaining air, Or one-fourth of the original amount, is removed. Air-pumps of immense size are used to remove the air from the " condenser " of condensing steam engines, such as are used now on all large steamers. The vacuum gauge on one of these engines is a bent, flattened tube, fastened at one end. The other end moves a pointer, or needle, over a scale. Increasing press- ure inside the tube tends to straighten it. On decreasing the pressure, the spring of the tube makes it bend, or curl up again. Close a piece of thin-walled rubber tubing at one end with a cork or plug of wood. Connect the open end with a water main. Lay the pipe in a sharp curve, and turn on the water. What does the pressure do to the tube? LIQUIDS AND GASES. 49 55. Degree of Exhaustion. — By attaching a U-pressure tube or an enclosed barometer to the receiver of an air-pump, as shown in Fig. 41, and exhausting the air, the number of inches that the column B drops is indicated on the scale. If all the air is pumped out, the gauge reads 30 inches. Since the water con- densed in the condenser is at a temperature of about 50° Centi- grade, the vapor exerts a certain amount of pressure (see Exer- cise 26, page 196). Considering this steam pressure, 27 inches or more is a good vacuum ; this means that the vacuum is as good as when |^ of the air is exhausted from a vessel or bottle. When the enclosed barometer (Fig. 41) is used to determine the degree of exhaustion, all connections should be made with thick- walled tubing to prevent collapse. Blow into A to determine what pressure can be exerted by the lungs. Then suck the air out ; as much as half a vacuum can be obtained by some persons. 56. A Liquid Piston of water or mercury offers many advan- tages. Figure 42 is a rough diagram of the form invented by Sprengel. Connect the tube ^ to a faucet, and turn on the water. The water, in passing by the side tube, breaks up into drops. Each drop acts as a piston, and takes some air from B be- hind. The tube connected to C should be a foot or more long. Air and water mixed will pass out of D into the dish, and water or mercury will rise in E, showing that air is pumped from E. This form of pump is much used for exhausting the air from the bulb of incandescent lamps. For this purpose it is made in the form shown in Ifig. 43. A millionth, or less, of Fig. 42. Fig. 43. 50 PRINCIPLES OF PHYSICS. the original ajr is left in the bulb before it is sealed off. Geissler tubes require a little less exhaustion, and Crookes and Eontgen-ray tubes require still more. 57. Geissler's Mercury Pump. — If a barometer tube, more than 30 inches long, is filled with mercury (sec- tion 33, page 26) and inverted in a dish of mer- cury, the space above the mercury in the tube is nearly a perfect vacuum. If the mercury was boiled in the barometer tube before inverting, there is in it a little mercury vapor, but almost no air. If a barometer tube is made with a bulb at the closed end {B, Fig. 44), the bulb may be sealed off after the tube is filled and inverted. This will leave an almost perfect vacuum in the bulb. Fig. 44. 58. Practical Forms of Geissler's Pump. — In the Boyle's law apparatus (Fig. 34, page 37), loosen the cap over the tube on the left-hand slide. liaise the open tube till the mercury drives out all the air and begins to run out under the cap. Screw down the cap, and lower the right-hand tube till the level of the mercury in it is 80 cm. or more below the cap. The space under the cap is a good vacuum. An air-pump made on this principle is shown in Fig. 45. L is an incandescent lamp; A and B are valves. Close A, open B, and raise C; air is driven out of R. Close B, open A, and lower C; mercury runs out of B, and the air of Fig.45. LIQUIDS AND GASES. 51 the incandescent lamp L expands and part flows to B. Close A, open B, and raise C; air from B is expelled. Repeat, till the air in L is practically all exhausted. Then turn a current on the filament in L, thereby heating it and driving off the gases in its pores, and seal the bulb off, while working the pump. Crookes and Eontgen-ray tubes are often exhausted on pumps of this type. 59. Exhausting by Condensation. — An interesting, though not very practical, way of obtaining a good _4 ^ vacuum is to condense all the air in the glass /^~ \ A ^ bulb, B, Fig. 46, by pouring liquid hydrogen ^^— ^ V_y over it. The air condenses as a solid on the ^'s- '^^• inside of B, completely leaving A. The small tube between A and B is then sealed off. 60. Pressure in a Liquid due to its Weight. — Like all other substances, liquids exert a downward pressure, due to their weight, or apparent attraction by the earth. It is not down- ward pressure, however, that bursts a dam, or makes water spurt up through a leak in the bottom of a boat, but a sideways or upward pressure caused by the downward pressure. 61. Downward Pressure. — Tie sheet rubber over one end of a glass tube (a student lamp-chimney, for instance), and hold the open end up. Fill the tube a quarter full of water ; notice the sag of the rubber. Pour in more and more water, till the tube is full. Any increase of pressure is shown by the in- creased bulging of the rubber. Does the pressure increase with the depth of the water in the tube ? The same experi- ment may be tried by holding the hand over the bottom of a lamp-chimney while water is being poured in at the top. 62. Sideways and Upward Pressure. — Unless there were enormous sideways pressure exerted by a liquid in a deep 52 PRINCIPLES OF PHYSICS. tank, there would be no need of strong hoops. The upward pressure in a liquid is shown by pushing down into a dish of water a glass tube with thin rubber tied over the lower end. Any movement of the rubber may easily be made apparent by a straw reaching from the rubber to the top. Press a tin can, having a hole in the bottom, '^' ■ down into a dish of water (Fig. 47). Does the water exert upward pressure ? fr Exercise 11. PKESSURE IN A LIQUID DUE TO ITS WEIGHT. Apparatus : A small-bore glass thistle tube, the mouth covered with sheet rubber, and connected by rubber tubing to a small-bore glass tube contain- ing a drop of colored ink ; a large battery jar full of water ; a student lamp-chimney, stopped at one end with a cork. The water in the jar should be at the temperature of the room. Case I. — Push the thistle tube, G (Fig. 48), to different depths in the jar of water. (The pressure on the rubber j across the mouth of the thistle tube is indi- cated by the position of the drop of ink in the level tube, /. What effect does the depth of water have on the amount of pressure ? Case II. — Try the experiment, using the chimney, filled with water, in place of the jar. Case III. — Replace the apparatus by an- other, in which the gauge at G can be turned in any direction without raising or lowering the rubber. Immerse the gauge and turn in differ- ent directions. Does the drop of ink in / indicate any difference of pressure in different directions? Case IV. — Place the closed end of the lamp- chimney above the face of the gauge, as in Fig. 49, noticing any variations of pressure. Note also that when the face of the gauge is under the closed end of the lamp-chimney, as at A, the depth of the water immediately over it is much Fig. 49. less than when it is at B. C^ Fig. 48. LIQUIDS AND GASES. 53 Case V. — Fill the chimney with water, and invert it over the face of the gauge, as in Fig. 50. Case VI. — With the form of gauge used in Case I, push the gauge up into the chimney, which is inverted and full of water (Fig. 51), noting any variations of pressure. At what point is the fe §? j 1 =-s=^^:^ "^ :^^^ Fig. 50. Fig. 51. pressure inside the chimney the same as in the open air ? At what point is the pressure less ? How long would the chimney have to be, so that the gauge could be pushed up to a point where there would be no pressure? Where would that point be if the liquid were mercury? 63. Relation of Pressure to Depth. — These experiments show that pressure increases with the depth of the liquid ; that at a given point below the surface of the liquid open to the air, the pressure is the same in all directions, and is the same whether a solid or a hollow vessel, as in Case IV, or a column of enclosed liquid, as in Case V, is above the gauge. The pressure of the liquid at its surface (2/, Pig. 61) is zero, and the pressure inside the tube of water at the same level is also zero. For although the gauge inside the tube, when at the level L, is under a column of water, the depth of the gauge below the level L of the surface of the water open to the air (or the free surface, as it is called) is zero. The pressure is nothing at that level. By pushing the gauge up into the water above the level L, the pressure becomes less than at L. What is the pressure on the gauge at L anywhere in the room outside the dish ? A 54 PRINCIPLES OF PHYSICS. pressure of one atmosphere = 76 em. of mercury, more or less, as shown by the barometer. This pressure becomes less and less as the gauge is moved up in the tube. If the chimney reached fifty feet above the level L, would it stay filled with water ? 64. Average Depth. — A liquid exerts pressure due to its weight. As a liquid is free to move in every direction, unless hindered by the walls of the vessel, it exerts pressure in every direction. Figure 52 shows a can 1 cm. wide, 1 cm. broad, and 6 cm. long. It holds 6 cc. of water, or 6 g. The pressure on the bottom equals 6 g. ■ The pressure at the level A is 1 gram on 1 sq. cm. of surface. What is the pressure at B? At C? AtZ>? KtE? At F? Consider 1 sq. cm. of the side, as ;S^. The pressure at S is zero; at A, 1 gram per square centimeter. The average pressure is zero plus one divided by two, or .5 g. The press- ure on one square centimeter of the side, as AB, is 1 + 2 "^ = 1.5 g. The usual way of expressing this A— >m B — -^ ;r^= C E— ^E F — =^ - Fig. 52. is to say that the pressure per square centimeter equals the average depth. Ou the section AB just mentioned, the depth of u4 is 1 cm., the depth of B is 2 cm. The average depth is 1 + 2 = 1.5 cm., and the average pressure over the section AB 2 is 1.5 g. What is the average depth of the other sections? "What is the pressure on each? 65. Calculation of Pressures. — In all problems relating to the pressure of liquids due to the weight of the liquid, the following should be kept in mind : — The total p-essure in grams on a Iwrizontal surface equals the area in square centimeters times the depth in centimeters. Tlie total pressure in grams on a vertical or slant surface equals the area in square centimeters times the average depth in centimeters. LIQUIDS AND GASES. 55 These rules hold for water, the weight of which is one gram per cubic centimeter, or nearly that. If mercury is the liquid, the pressure is 13.6 times as great as for water. Problems. L ti. Find the pressure on the sides of a box 5 cm. by 8 cm. by 4 cm. deep. The average pressure = the average depth = "^ = 2 g. per square centimeter. The area of one small side, 5 cm. x 4 cm. = 20 sq. cm. The total pressure on the side = the area x the average depth = 20 X 2 = 40 g. b. Find, in the same way, the pressure on a large side. Ans. 64 g. c. Find the total pressure on the bottom, ^ns. 5 x 8 x 4 = 160 g. 2. In the top of a square box (Fig. 53), a square tube 2 cm. by 2 cm. by 6 cm. high is attached. Th# box is open where the tube is attached, so that pouring water in the tube fills the box and the tube. The box is 8 cm. by 10 cm. by 5 cm. deep. a. Find the average pressure per square centi- meter on the side A. '^'S' ^^' The depth of the top of .A is 6 cm., that is, the top edge of .4 is 6 cm. below the level of the water in the tube. The depth of the bottom of A is 6 + 5 = 11 cm. The average depth is "'" =8.5 cm. The average pressure per square centimeter on A is 8.5 g. b. Find the total pressure on A by multiplying the area of A by the average depth, 8.5 cm. Ans. 8 x 5 x 8.5 = .340 g. c. Find the total pressure on B. Ans. 425 g. d: Find the average depth of D. Ans. 3 c. e. Find the total pressure on side D. Ans. 36 g. / Find the area of the top, C, of the box. g. Find the area of the bottom of the box. h. Find the pressure per square centimeter on the bottom, under the tube. 56 PRINCIPLES OF PHYSICS. i. On what does pressure in a liquid depend ? Ans. Depth below the free surface. j. Is the pressure at any part of the bottom any more or less than right under the tube ? k. What is the pressure on the bottom ? Ans. Area 80 x depth 11 = 880 g. I. What is the upward pressure on C ? Ans. Area 76 x depth 6 = 456 g. m. Find the 'difference between the pressures on the bottom and on a. 71. What is the total weight of water in the tube and can ? A71S. 424 g. 3. Figure 54 is a tank 6 cm. by 6 cm. by 4 cm. deep. Out of the top rises a tube 20 cm. long. a. Find the pressure per square centimeter at the bottom. Ans. 24 g. b. What is the pressure per square centimeter at the top of the box ?. c. Does the diameter of the tube hav» any effect on the pressure on the base ? 4. What are the pressures in Problem 3, if the liquid is kerosene, density = .8 ? 5. A cubical box, 10 cm. on an edge, has a pipe, filled Fig. 54. yfiiii water, rising 80 cm. from the top. What is the pressure per square centimeter on the bottom ? What is the total pressure on the bottom? What is the average pressure on a side of the box ? What is the total pressure on a side of the box ? 6. Find the pressure per square centimeter on the side of a canal 50 cm. below the surface of the water. 7. What is the pressure, near the keel, on a vessel drawing 600 cm.? 8. What is the pressure at the bottom of an ocean 750,000 cm. deep? Neglect the increase in density of the water as the depth increases. Water is slightly compressible. 9. A submarine boat sinks 1200 cm. below the surface of the water; what is the pressure per square centimeter on the sides of the boat? LIQUIDS AND GASES.. 5J 10. The air in a diver's suit is pumped in under pressure at the top, and escapes from the lower part of the suit. The pressure of the air in the suit must equal that of the water outside. What is the pressure 1500 cm. below the surface of the water ? 11. The pressure of water in a street main or in a pipe in a house service depends on the depth below the surface of the water in the reservoii", or the difference in level. The size of the reservoir has no influence on the pressure. If the water pressure at a certain point is 2000 g. per square centimeter, how much higher is the surface of the water in the reservoir ? 12. If one atmosphere of pressure sustains a column of mercury 76 cm. high, or a column of water 1033 cm., what is the pressure of an atmosphere, in grams, per square centimeter ? 13. An atmosphere equals 14.7, or nearly 15, pounds per square inch, and this pressure holds up or balances a column of water about 34 feet high. (To find the exact amount, reduce 1033 cm. to feet.) What is the pressure in pounds per square inch 34 feet under the surface of a lake ? 68 feet under the surface ? 14. What is the difference of level between a place where the water pressure is 45 pounds per square inch and the level of the reservoir ? 15. If the water pressure in a town is 170 pounds, how high is the reservoir above the town ? 66. Pressure and Weight. — -The bases of the dishes A, B, and G (Fig. 55) are the same size, and. the dishes have the same depth. Which holds the most water ? Which the least ? Dis- regarding the weights Pig' 55 of the dishes, vrhich weighs the most when filled with water? The depth being the same in all three, how does the pressure per square centi- meter on the bases compare? Since the areas of all three bases are equal, how do the total pressures on the bases com- pare ? Not considering the weights of the dishes themselves, the pressure that B, when filled with water, exerts on the table, exactly equals the pressure of the water on the base. \^^ LJ A 58 PRINCIPLES OF PHYSICS. Fig. 56. A, full of water, weighs more and C less than the pressure on the bases of the dishes. This paradox seems clear, if we notice that the slanting sides of A help to hold up some of the water, and the sides of C hold down some, as it were. 67. Pressure and Depth. — Why does not the water in a kettle drive the water out of the nose ? Pressure depends on depth only. In Eig. 66, the weight of the large amount of water in the large arm D does not force the water out of the small arm F, because the height ED equals the height EF. The pressure at F to the right, due to the column of water DE, is exactly equal to the pressure to the left, due to the column FF. These pressures balance. " Water seeks its level," because a higher level at one point means a greater depth and a greater pressure in all directions under the point of a higher level. This greater pressure causes a flow of liquid till the surface of the liquid is at the same level everywhere. 68. A Body buoyed up in a Liquid. — Experiments have shown that a body is buoyed up by a force equal to the weight of the liquid it displaces. A cube, C (Fig. 67), is 4 cm. on each edge. The vol- ume is 64 cc. The loss of weight in water is 64 g. Why is the cube buoyed up ? The top of the cube is 6 cm. below the surface ; the pressure per square centimeter at that depth is 6 g. ; the area of the top of C is 16 sq. cm. The total downward pressure, SA, is 6 X 16 = 96 g. The area of the bot- tom of C is also 16 sq. cm. ; but the depth is now SD, which is 10 cm., and the pressure per square centi- meter is 10 g. The total upward pressure on the bottom is 10 X 16 = 160 g. The difference between the downward Fig. -57. LIQUIDS AND GASES. 59 pressure on the top, 96 g., and the upward pressure on the bottom, 160 g., is 160 - 96 = 64 g. ; 64 cc. is the volume of the cube. Work out the figure, if the cube is 20 cm. under water, SA is 20 cm., and SD is 24 cm. Does the loss of weight of an immersed body depend on the depth below the surface ? 69. A Body buoyed up in Air. — A body in air is buoyed up by the weight of the air it displaces. A cubic centimeter of air at ordinary temperature at the sea level weighs about .0012 g. How much heavier would a cube 100 cm. on an edge be in a vacuum than in air ? A balloon of silk or a soap-bubble, filled with hydrogen or coal gas, rises because the weight of the balloon or bubble is less than the weight of the air displaced. The air pressure increases the further we go toward the centre of the earth (see section 35, page 28). At the sea level, the pressure of the atmosphere per square centimeter is 1033, or about 1000 grams. As water is about 800 times as heavy as air, it is necessary to descend 800 cm. in air to get the same increase in pressure that would be obtained by descending one centimeter in water. A barometer, if taken from the level of the bottom of a balloon to that of the top, registers the difference in pressure that makes the balloon float. Problems. 1. A vertical tube, DB (Fig. 58), 30 cm. long, is conuected with the side of a tank. 5 is 5 cm. above C, and ^ C is 25 cm. The tank and pipe are filled with water, and open to the air at D. Pressure depends on depth below the free surface. The free surface here is D. Find the pressure per square centimeter at the level of A ; also at B and at C. How far is C below the opening at Z» ? Ans. Pressure at 4, 10 g. ; at B, 30 g. ; at C, 35 g. C is 35 cm. below D. 2. Suppose DB (Fig. 58) is 30 cm. ; AB, 15 em. BC, 20 cm. Find the pressure per square centimeter Fig. 58. 60 PRINCIPLES OF PHYSICS. at the level of ^ ; of B; of C. Would these pressures be increased if the tube DB had a greater diameter ? 3. EF (Fig. 59) is 20 cm. ; HF, 30 cm. ; and FG, 10 cm. a. What are the pressures at the level of H, E, F, and G 'I h. If the tank and tube are filled with water, will the pressure at H he greater or less than at E ? c. If, after the can and tube are filled, a sheet of rubber is tied over E, will the rubber bulge in or out ? d. If an opening is made in H, would water run P in or out of El e. What holds the water upio the level of H"? f. li E were 50 feet below H, what would happen Fig. 59. ■ to the water? What pressure would the top at H have to resist, and from what source would the pressure come? 4. The story is told of a little schoolboy in Holland who thrust his arm into a small opening in a dike and stopped the flow of water, thus preventing the opening from growing larger and causing an inundation. Suppose the distance below the level of the North Sea was 150 cm. ; what pressure per square centimeter did he have to withstand? How could a little boy hold back the whole North Sea? 5. Compare the difficulty in stopping the flow of water from a large and from a small opening in a water-pipe or faucet. Why are large pipes made thicker than smaU'ones? 6. Why does a bicycle tire need reenforcement with strong cloth to withstand the same pressure that the rubber tube connecting the pump easUy stands ? 70. Forced Pressure of Liquids and Gases. — The pressure exerted by liquids and gases is not always due to their weight. Water driven by the piston of a force pump or syringe, and air pumped into a bicycle tire, owe but a small fraction of their pressures to their own weight. The pressure on the piston is transmitted in every direction. Stretch a rubber band by two pins or matches placed inside the band. What shape does the band take ? Use three or four, or any number of pins, all pushing out equally. How many must be used to make the band assume the form of a LIQUIDS AND GASES. 61 .circle? A flexible tube under pressure from the inside always takes such a form that a cross-section is a circle. Look at a rubber hose carrying water under pressure. Close one end of a thin-walled rubber tube, and blow into the other end. A rubber band is a small section cut from a large, thin rubber tube. In how many directions must pressure be applied from the inside of the band outward to make it take a circular shape ? Why are boilers made round, instead of square ? 71. Distribution of Pressure. — From a study of the experi- ments mentioned above, or by pressing on any part of a bicycle tire, one is led to believe that pressure in a liquid or a gas is exerted equally in all directions. A soap-bubble is almost a perfect sphere. Any greater pressure of the air inside, in one direction more than another, would make the bubble bulge out at some part. The following experiment shows that the film of a soap-bubble exerts pressure : — Blow a soap-bubble with a large pipe. Hold the mouth of the pipe toward a candle-flame. The air inside the bubble is under pressure, and blows out the flame. Blow into a tube having a diameter of \ inch ; I hold one finger over the other end of the tube. Blow in at c. Fig. 60, and try to keep the piston D in. (The piston fits much closer than the dia- h i \ ^^ D gram represents.) Exactly the same pressure per square centimeter was applied to the finger pig. ^ closing the small tube and to the piston D, but the total pressure on D was greater, because there are in it more square centimeters for the pressure to act on. , 72. Hydrostatic Bellows. — Lay an inner tube of a bicycle tire on the table in the form of a U. Place a large drawing- board on the tube, and blow in the opening of the tire. See how great a weight can be raised. A person can stand on the board and lift himself. This apparatus is known as the Hydrostatic bellows. 62 PBINCIPLES OF PHYSICS. W Fig. 61, Set the apparatus (Fig. 61) on a stand, and find the largest weight that can be lifted by the piston D, by blowing in the tube c. Calculate the area of the piston. Find what the pressure of the air is, by testing with the en- closed form of barometer (Fig. 41, page 48), or by using a U-shaped mercury gauge (M, Fig. 214, page 242). Eemember that each centimeter the mercury is made to rise measures a pressure of 13.6 g. per square centimeter. How nearly does the computed total pressure on the piston com- pare with the weight raised ? Of course the weight of the piston itself and the friction cause some appar- ent loss. Attach a stout rubber tube to c, and connect with a bicycle pump. Which has the greater total pressure, the piston D or the piston of the pump ? Which has the greater area ? 73. Efficiency of Pistons. — • Connect c, Fig. 62, by a heavy rubber tube to a small bicycle pump, or, preferably, to the pump described in Section 34, page 27. In the former case, see that the connection is made to the opening on the pump marked "exit." The pump is then used as a force-pump. How great a weight can be lifted by the piston D? There is, of course, a large amount of friction in both pistons; and since the effective lifting power of D will not be as many times greater than the force down on A as the sur- face of D is times greater than that of A, the pressure on A should be tested by a spring balance, *S. Suppose D is twenty-five times as large as A. Then any pressure ex- erted on A — for instance, 10 pounds — is transmitted to D, and the total pressure on Z) is 25 x 10 pounds = 250 pounds. However, any force applied to the handle of A, as registered by S, is not all effective. A little Fig. 62. LIQUIDS AND GASES. 63 is lost in friction between A and the sides of the pump. Again, at D, there is a further loss from the same cause. A pressure of 10 pounds on the handle A, instead of balancing 250 pounds at D, may balance 200 pounds. The efficiency of the machine is — - = .80, or 80 per cent. If a liquid, such as water, alcohol, or oil, is used in the experiment (Fig. 62), in place of air or steam, the friction of each piston is from one to five per cent of the force on it. In all problems, disregard friction for the present. 74. Elevators. — Pistons, such as D, Fig. 62, are made several feet in diameter and twenty or more feet long. The piston D is sometimes used as shown in the diagram, to raise loads from one story of a building to another. More often the piston, by means of connecting pulleys and ropes, raises or lowers a car rapidly, while the piston itself has a slow movement. Water is almost always used to move the piston D. Figure 62, then, is the simplest form of the hydraulic elevator. 75. Hydraulic Press. — Set the cylinder and piston, D, in a rectangular frame of iron (Fig. 63). Place a cork, E, on D, or on pieces of wood or metal resting on D. The top of the cork should nearly or quite reach the iron frame. Connect a rubber tube to c and blow into it. Replace the cork by an English walnut, and try to crack the nut. A bicycle pump con- nected with the tube gives higher pressure. This (Fig. 63) is the simplest form of a hy- draulic press, which is used to .compress cot- ^^ g3_ ton into compact bales, to press metal into various shapes with dies, to punch holes in steel plates for steam boilers, and, in fact, for any work where enormous press- ures are required, and where the slow motion of the piston is no disadvantage. Properly, the term hydraulic should be ap- plied to this press only when water is used in the cylinder. 64 PRINCIPLES OF PHYSICS. Problems. 1. The area of D, Fig. 62, is 50 square inches ; the area of ^ is 1 square inch. The pressure on A is 12 pounds ; what is the total press- ure on Z) ? 2. What is the pressure per square inch on Z) ? 3. How many times larger must D he than A, so that a pressure of 20 pounds on A will lift or balance 240 pounds on D ? 4. If 60 pounds on D balances 2 pounds on A, what relation does the size of D bear to that of yl ? 5. If the water in a street main has a pressure of 40 pounds per square inch, what pressure is D capable of exerting, if the area of D is 100 square inches? 76. Surface Films of Liquids. — Dip a glass rod in water. Examine the shape of the surface of the liquid next to the glass rod and the sides of the jar (Fig. 64). Lift the rod out of the water. How much water clings to the rod? Try the same experiment, using mercury in place of water. Try tubes of small bore instead of the rod. Try a still smaller glass tube, made by heating one of the f^'s- 64. small tubes a few inches from the end, and drawing it out quickly. Color the water with ink, to make it visible. The water rises higher in the smaller tube. 77. Experiment. — Draw out a fine tube, C, from a larger tube, A (Fig. 65). Hold a match near X. As soon as the fine tube bends, change the po- sition of the tube till a complete bend, like B, is obtained. Pour mer- cury into B. Notice the depression in the small tube. 78. Capillarity. — Water wets the surface of glass; they attract one another strongly. Mercury does not wet glass ; LIQUIDS AND GASES. 65 the surface of mercury in a narrow tube is convex. As these phenomena are noticeable in the case of small tubes, the name capillarity (from the Latin, capillus, 'a hair') is used. The barometer column of mercury is slightly depressed, by capil- larity. Water, oil, etc., fill the spaces between the threads of cloth or wicking. If a few drops of water are added to a tumbler level full, this force of capillarity prevents overflow. Tarnished metals are not wet by mercury. Show, by a drawing, the shape of the surface of mercury into -which a piece of tarnished metal has been put. Zinc, or any other metal, freshly cleaned by dipping in acid, can be wet by mer- cury. Show the form of surface between zinc and mercury. Problems. 1. Assuming that a 34-ft. head of water gives a pressure of 15 pounds to the square inch, or one atmosphere, as it is called, how many atmospheres of pressure are there 1700 feet below the surface of a lake? Ans. 50, in addition to the pressure of the air itself on the surface of the lake. 2. If 200 cc. of air at the level of the lake are under a pressure of one atmosphere, what would the volume become 1700 feet below the surface, where the total pressure is 51 atmospheres ? 3. The water pressure in a city water main is 80 pounds per square inch ; the diameter of the piston of an elevator is 10 inches, and the area of the piston is 78 square inches. How heavy a load can the elevator lift under pressure from the water main, disregarding fric- tion? Taking the friction loss as 30 per cent, what load can be lifted? 4. What must be the area of the piston of an elevator or hydraulic jack to raise one end of a car weighing 10 tons, the pressure of water being 200 pounds to the square inch ? (The jack lifts half the weight of the car, or 5 tons. This equals 10,000 pounds. How many square inches, each having a pressure of 200 pounds, must the piston have to give a total force of 10,000 pounds ?) CHAPTER IV. rOEOES. Exercise 12. COMPOSITION OF FOKCES. Apparatus : Three 2000-gram spring balances ; adjustable clamps ; fish-line. Tie two spring balances, A and B, Fig. 66, to opposite ends of a piece of flsh-liue. Hold B in the hand, or fasten it to a clamp. Pull A till one thousand or more grams are registered on the two balances. If the balances are accurate, they will both register the same pull in grams. The pulls are in opposite directions, and are exactly equal. LI Tie a loop in the end of the line, and through the loop J pass another piece of line (Fig. 67) . Connect the balances C and D. Pull C and D till A registers the same pull as before. To hold the balances in place, tie stout strings to their rings. Hold the strings by the cams of clamps that slip over the edge of the table. Adjust the pull of A by loosen- o iBg the cam and moving A by Fig. 66. the string attached to the ring, till it reads the same as before. Then rotate the cam to fasten the string. Record the readings. The result of the pulls of C and D is exactly the same as if the single force B, directly opposite in direction to A, replaced C and D. The single force B is called the resultant of [B the forces C and D, and the forces C Fig. 67. and D are called components. The sum of the forces C and D is greater than the single force B. To see if this is always so, make the angle between C and D as E D COMPOSITION OF FORCES. 67 large as possible. The limit is reached (Fig. 68, I) when the force which has been registered at A cannot be obtained by C and D pulled out to the limit of the scale. Make diagrams of the general directions of the forces in Fig. 68, recording the readings of the balances. It will be noticed that the sum of the component forces C III. IV. and D is always greater than the resultant or single force that can be substituted for them. The sum of the components decreases as the angle between them becomes smaller, till their sum, when the angle between the forces is very small, as in Fig. 68, IV, is only slightly greater than the resultant. If two boys pull on a sled, one boy toward the east and the other toward the north, the sled will move in a direction somewhat between east and north, — in the direction of the resultant of the two forces applied. If the boy pulling north pulls hard, and the other boy pulls gently in an easterly direction, the sled will move a little east of north, but nearly north ; for the resultant is always somewhere between the two components, and its direction is nearer to that of the greater component. Arrange the apparatus about as shown in Fig. 69. See that no balance registers less than 1000 g. Place a sheet of paper under the lines, with its centre near E. Tap the balances, or jar the table slightly, so that the effect of friction is eliminated. Press the line down at E and at ^ . If the line does not almost touch the paper, put a note- book underneath the paper. With a sharp- pointed pencil, held vertical, mark along on both sides of the string near A and E. The dotted lines (Fig. 69) indicate where the fine pencil marks are to be made; they are, of Fig. 69. 68 PRINCIPLES OF PHYSICS. course, closer to the fish-line than is shown in the figure. Mark EC and ED in the same way. On AE record the pull of the balance E. On the other lines, record the pulls of the balances attached to them. Remove the paper. Make a fine mark, freehand, between the two parallel lines at E, and also at A. Draw a line through the marks just made, using a ruler. This line will represent the direction of AE. In the same way locate and draw CE and DE. If they meet in a point, the work has been well done. The three lines represent the three forces in direction only. To make the lines represent the pull of each balance, mark off spaces, centimeterSj for instance, on each line, making a space for each 200 g. of pull. If the pull oi AE is 1400 g., then AE is 7 cm. long. Erase the ends of each line not marked off into spaces. Repeat the exercise, varying as much as possible the angles between the forces. Have the balances register as nearly as possible the full amount of the scale. 79. Parallelogram of Forces. — In Pig. 70, AE represents 1400 g., and DE 1600 g. Make a parallelogram by drawing a line through D parallel ' to EA, and another line through A parallel to ED. These new lines cross at F. Draw the diag- onal EF. Is the direction of EF opposite to EC? Measure the numbers of centimeters in EF, and call each centimeter 200 g. ; what is the force represented by EF ? Is it equal to CE? Exact correspondence may not be obtained, through errors in the balances and slight inaccuracies in the work. ^ To draw a Line parallel to Another. — To draw a line through z parallel to xy (Fig. I), place a card with its edge coinciding with the line xy (Fig. II), and bring a ruler close up to the end of the card. Now, holding the ruler in place, the edge of the card is always parallel to xy, no matter how far up or down the ruler the card is moved. Slide the card up, till its edge touches z. Draw a line along the edge of the card, passing through z. This line is parallel to xy. Fig. I. Fig. II. COMPOSITION OF FORCES. 69 Fig. 70. A and D have been taken as the components. The force EF, which could replace A and D, is the resultant of those forces. CE, the force which balances A and D, is called the equilibrant. How do the resultant and equilibrant compare in direction ? how in magnitude? If A and O are considered compo- nents, ED is the equili- brant. Complete the par- allelogram, having AE and GE as two sides. Its diag- onal represents the result- ant of A and C Compare it with DE. Take O and D as the components, com- plete the parallelogram, and study the diagonal, as before. A boat is rowed across a river at the rate of four miles an hour; the current is two miles an hour. Find the path taken by the boat, as follows : Draw two lines, representing the river bank, of any length, four centimeters apart. Draw a perpen- dicular between them, to show the course the boat would have taken had there been no current. Measure down two centi- meters on one of the lines representing the bank. Draw the diagonal. What is the path of the boat ? If the wind blows ten miles an hour from the east, and a bicyclist rides north ten miles an hour, from what direction does the wind seem to him to come ? (From the northeast.) With what apparent velocity does it blow ? Does a flag on a moving boat ever indicate the real direction of the wind ? If the wind is blowing from the north, a yacht, in order to go north, sails first to the northwest and then tacks and sails northeast. Why does the wind apparently shift during the' tacking? 70 PRINCIPLES OF PHTSICS. 80. Direction and Amount of Resultant. — A force of 4 g. pulls north (JV) and another of 3 g. pulls east (E) .on the object (Fig. 71, I). What single force can be substituted for these, and what is its direction ? Draw a line north and another east from 0. Use arrow-heads to show the direction N I. ->-B -I \—^E II. of the forces. Make ON four units long and OE three units (II). The units may be any convenient length, one centi- meter, for instance. Complete the parallelogram (III). Draw the diagonal OF, starting from (IV). This diag- onal represents the resultant in direction and amount. Find the number of units of length in the diagonal. It is 6. When the angle between the components is a right angle, the amount of the resultant is easily computed. FE—0N=i, OE = 3. In a right triangle, the square of the hypothenuse OF equals the sum of the squares of the other sides, OE and FE. 4^ = 16 3'' = _9 26 The sqviare root of 25 is 5, and 5 is the number of units of force in the resultant OF. Problems. 1. Find, by drawing, the resultant of a force north of 4 pounds and one east of 2 pounds. COMPOSITION OF FOBCES. 71 2. Find ths same by computation. ^ns. iJ =Vl8 = 4.2 +. 3. Find the resultant of forces 6 north and 5 southeast. Use a diagram. What force would balance these two ? Should it be called a resultant or an equilibrant? 4. Compute the resultant of forces 5 and 12 at right angles. 5. Forces of 300 and 5C0 pounds can be brought to bear on a stump. What is the direction of these forces, to make the resultant as large as possible ? As small as possible ? 6. Find, by drawing, the resultant of forces 5 and 3 at an angle of 30° ; 45° ; 60° ; 90° ; 120°- In which case is the resultant the greatest ? How great can the resultant of these foi'ces be ? How small ? If the angle between the forces is 180°, — that is, if they pull directly oppo- site, — • what is the resultant ? 7. If an arc light is suspended over the centre of a street by a wire rope attached to two posts, one on each side of the street, explain why the rope will break long before it can be pulled tight enough to be straight. Could the rope, without any weight attached, be pulled straight? Try the experiment, using string. Sight along the string. 81. Three or More Forces. ^ — -It will be noticed that, in all the cases of the composition of forces, the two components and the equilibrant meet at a point ; that is to say, they are not par- allel. The resultant can also be found of three or more forces meeting at a point. If a dozen horses, each, hitched by a separate rope, draw a light building on rollers, no two horses pull in exactly the same direction; yet the twelve forces have a resultant — a single force that could replace the twelve forces. Sup- ^_^ pose three horses, A, B, and C (Fig. 72), pull the house H. To find the resultant of these forces, think of only A and B. Fig. 72. Cover up the line C for the moment, and find the resultant of A and B; call it R. Eub out A and B, and uncover C There are now two forces tending to move H, one force C, the other R. Find the resultant of R and G. This is the resultant of, or the single force that could be sub- 72 PEINCIPLES OF PHYSICS. Fig. 73. stituted for, A, B, and C. In a similar manner any number of forces may be compounded. 82. Resolution of Forces. — A single force can be replaced by, or resolved into, any number of forces, just as a single force can replace two or more forces acting on a point, or as one strong boy can take the place of a number of boys pulling a double runner. ^ A and O (Fig. 76) are two posts ; a force, R, is applied to the rope connecting them. This one force R causes two forces, one, AH, pulling A to the right ; the other, HC, pulling C to the left. When one force, like R, causes two or more forces, we say that R is resolved into the other forces. 83. To Resolve a Force. — Resolve a force of 3 pounds into two forces, one of 5 pounds, making an angle of 30° with the given force. Draw the line HR (Fig. 74, I), represent- ing, by three divisions, the force of three pounds. As the next step, make an angle of 30° at H, as shown in II, Fig. 74. Draw HC, five divisions. Since the resultant is always between the forces, the other com- ponent must be on the other side of the resultant. Connect R and C; draw DR parallel to HC, and HD parallel to CR (Fig. 74, III). HD is the other component. The divisions used in laying off HR and HO must be all of the same length See how many divisions can be found in HD. The number of these divisions will be the number of pounds in the com- ponent HD. Fig. 74. COMPOSITION OF FORCES. 73 Problems. '^ 1. Resolve a force of 5 pounds into two forces, one of 3 pounds, making an angle of 30° with the other ; what is the other force 7 2. Resolve a force of 5 pounds into two forces, one, at an angle of 45° with the 5-pound force, having an intensity of 1 pound. 3. Resolve a force of 4 pounds into two forces, one being 6 pounds and making an angle of 90° with the given force ; find the other. 4. A force of 3 pounds is resolved into two forces ; one is 5 pounds, making an angle of 60° with the given force ; what is the other V 84. Direction of Force. — A strip AB (Fig. 75), fastened on the table, represents a track rail. C is a ruler, driven along Xc D Fig. 75. rF -^E Fig. 76'. by a force applied at F in the direction of the arrow. C can move only in the direction of its length. A sharp point on the end of F prevents its slipping on C. Vary the experiment, as in Fig. 76, where a string, F, is shown attached to a pin at C. The force has the same direc- tion as before. C may be a car or a boat. Sideways motion in the direction CD is prevented by the flange of the wheels or by the keel. Supposing there is no friction between C and AB, or that it is very small, the force F is resolved into two components. One is CE, as is shown by the fact that the boat or car moves in that direction. The other component is CD, perpendicular to AB. Test this by laying the ruler C (Fig. 77) on a sheet of paper. Push a Fig. 77 74 PRINCIPLED OF PHYSICS. pencil in the direction indicated by F. Prevent any sideways movement of the pencil by the finger at G. Provided there is little friction between the pencil and O, the motion of C will be in direction of CD, pei-pendicular to the surface of C. The pressure on a surface where there is no friction is always perpendicular to the surface. 85. A Keel is a thin board, held vertically, that prevents the sideways movement of a boat. It answers the same purpose ■g. ri ^^ *^® flange of a car-wheel. A keel is usually attached to the centre of a boat. Occasionally it is held some distance to I the side, as in Fig. 78. The sail and the keel form a wedge, like HBO, Fig. 79. Fig. 78. 86. An Ice-boat may sail Faster than the Wind. — The fric- tion between polished steel and ice is small. It is easily computed from the measurements of the height and horizontal distance (section 96, page 81) of an icy hill on which a sled just keeps sliding. DGH, Fig. 79, is a wedge; HG is longer than RD. See that the edges are rubbed smooth and polished with graphite. Press against the wedge with a rounded piece of metal, W, in a direction at right angles to A and B. W, in moving the distance HD, drives the wedge the distance HO. If HD = 3, and H0= 4, then the forward motion of the ice- boat is four miles for every three miles the wind blows in the direction DH. Though there is some loss, the boat will go faster than the wind. However, in sailing before the wind, — that is, in the direction in which the wind blows, — the boat moves a little slower than the wind. COMPOSITION OF FORCES. 75 87. Why a Boat sails into the Wind. — The pressure of the wind on a slanting sail is not exerted in the direction in which the wind blows, but nearly perpendicu- lar to the sail. Suppose a force, W, Fig. 80, is applied by a pointed stick that does not slip along the sail, S; -^* the boat will move sternwards, or from A to B. But if a breeze strikes the sail in the direction W, the boat moves forward. The action is somewhat the same as if a lot of rubber balls were thrown in the direction W and in bounding off from the sail caused a pressure on it per- pendicular to the surface. The fact that the wind pressure is perpendicular to the sail is perhaps more clearly seen in the case of the old-fashioned kite. The wind blowing horizontally strikes the inclined surface of the kite a glancing blow. If the pressure on the kite is in the direction of the wind, then the kite-string must pull in exactly the opposite direction to the wind; that is, horizontally. This is not so ; every boy knows that the string is very nearly perpendicular to the surface of the kite. The pressure exerted by the wind must be oppo- site in direction to the pull of the string, and therefore must be per- pendicular to the surface. The wind blowing in the direction WE, Pig. 81, causes at any point, E, a pressure, ED, nearly perpendicular to the sail. This force ED is re- solved into two forces, one, EG, which moves the boat very slowly sideways, and another, EH, which drives the boat forward. In the case of the ice-boat, there is no slip nor any sideways movement toward G. CHAPTER V. PEIOTION. 88. Surface Resistance. — We all know that a sled draws easier over snow than over gravel or an iron rail, and that a loaded sled pulls harder than an empty one. The resistance caused by one surface sliding over another and tending to stop the body is called Friction. Bxercise 13. (a) COEFFICIENT OF FRICTION. - First Method. Apparatus : Board having a smooth surface ; wooden box, to the bottom ol which three small pieces of wood or metal are glued, like runners of a sled ; a 2000-gram spring balance, and various weights. Sandpaper the sur- face of the board and round the three projections on which the box slides. Weigh the box, containing a certain load. Place the board in a horizontal position (Fig. 82). Find how many grams' pull is re- quired to keep the box slowly sliding. The friction on starting is greater, but this is to be en- tirely neglected ; what is to be measured is the friction resistance while the box is moving. The box will stick a little in places where the friction is greatest, but several trials should be made and the average taken as the friction resistance. It is easier to read the balance if the board is drawn under the box than if the box is pulled over the board. The pressure between the surfaces of the board and box equals the weight of the box and its contents. Find how much of the pull reg- istered by the spring was required for each gram of pressure between the surfaces. For instance, if the weight is 400 g. and the pull to make it slide is 100 g., then the puU for 1 g. is ^ as much, |^= 0.25. 76 Fig. 82. FRICTION. 77 Increase or decrease the load in the box. Does the pull, or force, required to make the box slide vary? Make a statement showing what effect increased pressure has on the friction. Does the friction increase as the weight or pressure between the surfaces is increased ? Compute the friction caused by each gram of weight for each case ; in other words, find the force required to make one gram slide. Use as small a spring balance as will register the load without the pointer striking the bottom of the scale. Vary the weight again, and repeat. 89. Fonnula for Coefficient of Friction. — The force required to make one gram slide on a horizontal surface is called the co- efficient of friction. It is found by dividing the force required to make the body slide by the weight. In shorter form, Coefficient of friction = • Weight Find the coefficient of friction between metal and wooden surfaces ; between rubber and leather, using both the smooth and the rough sides of the leather. Try the effect of unplaned surfaces. Rub together two pieces of wood, — matches, for instance. Cover them with blacklead (rubbing with a soft lead pencil will answer) ; rub them together again, and notice that the friction is less. Oil, graphite, tallow, etc., used to reduce friction, are called lubricants. 90. Effect of the Load. — The values obtained for the coeffi- cient of friction will be found to vary somewhat. Yet the coefficient of friction is nearly the same for light and heavy loads. The following law is ,approxiniately true : The coeffi- cient of friction is independent of the load. What is meant by this ? Does increase of load in a sleigh increase the friction ? Does it increase the coefficient of friction ? If the pull indicated by the spring balance in Fig. 82 is 200 g., and the load, W, is 1000 g., what is the coefficient of friction ? (^^ = .20.) What is the force required to make a load of only one gram slide ? (.20.) 78 PRINCIPLES OF PHYSICS. 91. Effect of the Surface. — If the coefficient of friction, that is, the force required to make one gram slide, is the same for large and for small loads, it follows that the coefficient of friction is practically independent of the size of the sur- faces in contact. Suppose two blocks, each weighing one gram, have the same surface. The same force is required to move them, whether they are side by side or piled one on top of the other. The coefficient of friction is therefore nearly the same, whether the sliding surface is large or small. The usual way of stating this is: TJie coefficient of friction is inde- pendent of the surface. 92. Exceptions. — These two laws are really two ways of stating one law. They hold only while the surfaces are un- changed as the body slides along. . Very heavy loads may cut and roughen the surface and so increase the friction. A sled- runner as thin as a knife would cut into the ice and draw very hard. The sliding surfaces should be large enough to prevent one cutting into the other ; else the friction will be increased. 93. Advantages of Friction. — Friction is useful at times. Without it, standing on the side of a hill would be impossible ; a locomotive could not draw a train, nor could the train, once in motion, be stopped by putting on the brakes ; a bicycle would slip sideways, and its wheels, though made to turn, would not send it ahead ; a knot in a string would slip. In most cases, however, friction is a disadvantage, and the work done in overcoming it is all wasted. Find the force needed to make a loaded car slide on a board when the wheels are wedged or blocked so that they cannot turn. Let the wheels turn freely, and again find the force. Friction is reduced by making the load smaller, by using hard, smooth surfaces, and by flooding them with oil. The use of wheels diminishes the loss due to friction. A carriage wheel thirty-six inches in diameter, for example, has an axle one inch FRICTION. 79 in diameter, on which it turns. The circumference of the wheel, and consequently the carriage itself, travels thirty-six times as far in one revolution as the bearing on the axle, where the rubbing friction occurs ; so that the force of friction has to be overcome through only -^ the distance ; and, besides, the bearing and axles are made smooth and oiled. Problems. In the following problems, consider that the surfaces are horizontal. 1. A pull of 40 pounds is requh-ed to make 200 pounds slide. What force is needed to make a weight of 1 pound slide, and what is the coefficient of friction ? ,( 40 oa Ans. — = .20. 200 2. The coefficient of friction between two metal surfaces, well lubricated, is sometimes as low as .03. What force must be applied to make a one-pound weight slide ? What force to make 600 pounds slide? Ans. .03 lb. ; 600 x .03 = 18 lbs. 3. If the weight is doubled, what is the effect on the friction? What is the effect on the coefficient of friction? ,4ns. About double; almost none. 4. Compute the coefficient of friction between two pieces of wood, the pull being 44 pounds, and the pressure between them 88 pounds. 5. Compute the coefficient of friction, if a pull of 6 pounds is needed to make a body weighing 30 pounds slide. 6. Cast-iron on cast-iron has a coefficient of friction of .15. What force must be used to overcome the friction of a 32-pound lump? 7. If two horses pull 3 tons of coal on a sled, how many horses should be attached to pull 6 tons ? 8. What is the effect of making the runners twice as wide? 94. Effect of Speed. — The coefficient of friction becomes less as the speed is increased. A locomotive engineer, in stopping a train at full speed, turns sufficient air pressure into the cylin- ders under each car to drive the piston and to force the brake- shoe firmly against the wheels. As the train slows down, the coefficient of friction increases, and the wheels are sometimes gripped fast by the brake-shoe and made to slide on the track. 80 PRINCIPLES OF PHYSICS. To lessen the pressure and to keep the wheels from sliding, the engineer turns off the high pressure and then turns on a lower pressure. He is careful not to grip the wheels so hard that they stop turning and slide on the rails ; for as soon as the wheels begin to slide, the friction between them and the rail is lessened, and the train cannot be stopped so quickly. The wheels are held on the track by the weight of the train ; but the brake-shoes can be forced against the wheels with a much greater pressure. In case this is done, the wheels stop sliding against the brake-shoe and slide along the rails. Exercise 13. (6) COEFFICIENT OF FRICTION. - Second Method. Apparatus : The same as for Exercise 13, page 76. Raise one end of the board till the loaded box, W (Fig. 83), slides down slowly after it is once started. Measure the -F" height, h, and the horizon- tal distance, d. Divide h by d. Change the weights in the box. In each case ^. „ Fig. 83. vary the height, h, till the box just continues to slide after it is once started by the hand. Record as follows : — h d h d Case I. Box empty Case ir. Box lightly loaded Case III. Box with added weight Case IV. Box heavily loaded Case V. Box lightly loaded Case VI. Box empty FRICTION. 81 Let each case be an average of several trials. The values in the last column will be found to vary a little. How do they compare with the values for coefficient of friction, as found by the First Method (page 76) ? The value - is the coefficient of friction. Is it d the same for light and heavy loads ? 95. The Coefficient of Friction is the force, or pull, required to make a body slide when the pressure between the surfaces is one gram or one pound. Of course, the force and the press- ure must be measured ir^the same unit. ^ n tc • I. j^ c ■ i.- Force required to make body slide Coemcient of friction = -i 1 -^ ; Pressure between surfaces or, letting F stand for force and P for pressure, the formula becomes _ F P 96. Resolution of the Force. — Study Fig. 84. The incline AB is changed till the weight keeps sliding at a uniform velocity. The force F driving the body down the slant is then just equal to the friction. The weight of the body, W, is a force acting on it and pulling it down. This force W may be resolved, or split up, into two forces, — P, 82 PRINCIPLES OF PHYSICS. the pressure perpendicular to the surface, and F, the force tending to make the body slide down the incline. (See sec- tion 82.) The pressure P on the slant is n always less than the weight, and becomes less and less as the slant becomes greater. On a horizontal plane the pressure equals the weight. When the slant becomes straight up and down, as AB, Fig. 85, the e^-(^S50 pressure of W perpendicular to AB is zero. The perpendicular pressure P and the Fig. 85. force F (Fig. 84), tending to make the " body slide down, can be measured directly. Attach spring balances F and P to the weight W, as shown in Fig. 86. Do not let W touch the incline. There is now no fric- tion, and jP just balances the force, or component, necessary to drive W down the slant, and P balances the perpendicular pressure that would be exerted against the in- cline. Start with FW horizontal. F reads zero, and P equals the '^' ^^' weight. Lower P, keeping the strings at right angles; P becomes less and F increases. 97. Computation of the CoeflScient of Friction. —In Fig. 84 (page 81) we do not know the lengths of F and P directly. Study the little triangles made by F, P, and W. Make and letter a card or paper model of Fig. 84. Cut out the little triangles. Apply one little triangle to the other, and notice that they are exactly equal in every respect. Apply the angles, or corners, of the little triangle to the large triangle ABC, and find which corners fit. The little triangles are models of the large one. Measure the short side, DF, of the little tri- FRICTION. 83 angle, and the side P. DF stands for the force necessary to overcome friction and make the body slide. P stands for the pressure between the surfaces. Write out as follows : — Force to make body slide DF Pressure between surfaces ~ P Find what the value is, and call it the coefficient of friction. Find the side of the large triangle corresponding to DF; it will be found to be BC, or h (the height of the incline). Then find the side of the large triangle corresponding to P; it is AC, or d. Measure h and d, and find the value of h divided by d ; that is, -• As might be expected, this gives the same result as dividing DF by P. It is much more easy and accurate to measure, in the first place, the height h and the length d, than to cut out the little triangles and measure them. The coefficient of friction between the surfaces is found by adjusting the slant till the body just keeps moving when started, and then measuring the height'^ and the distance d, and dividing h by (Z. Problems. 1. If, in Fig. 84, A = 4, d = 20, and the body just slides, find the coefficient of friction. What force is required to pull the body along the same surface in a horizontal plane, if the body weighs 1 pound ? If it weighs 200 pounds ? 2. If BC (Fig. 84) = 120 feet \ AC = 1200 feet, what must be the coefficient of friction between the weight and the incline, so that the body will not slide? Ans. A little more than ^■^. 3. If the coefficient of friction between two surfaces equals .3, and a body slides on the incline, what must be the height and the hori- zontal length of the incline? Ans. Height, 3; horizontal length, 10. 4. Find the coefficients of friction, when the dimensions of the incline are: — a. Height = 40, horizontal length = 800. Tj. Height = 60, horizontal length = 1200. c. Height = 12, horizontal length = 20. d. Height = 1, horizontal length = 40. 84 PRINVIPLES OF PHYSICS. 5. There is an incline of which the horizontal length is 30. Pind the height in each of the following cases, if the body just slides : — a. When the coefficient of friction is ^. 6. When the coefficient of friction is .1. c. When the coefficient of friction is .4. 98. Variations of the Laws. — In the case of smooth surfaces, especially of metal, increasing the weight or pressure between the surfaces increases the friction, so that doubling the press- ure pretty nearly doubles the force required to make it slide. But the friction caused by a one-gram pressure is about the same whether the load is light or heavy, just as we might say the fare per passenger on a train is the same, whether one or one hundred travel. In actual experiments, one often finds that the coefficients of friction vary with the load, and may cite the instance of a heavily loaded sled coasting further than a light sled. The laws of friction are only approximate. In Exercise 13, the empty box required a steeper slant than the loaded box. Still, the laws that the coefficient of friction is independent of the weight and area of sliding surface are sufficiently accurate to be very useful. Suppose the coefficient does vary from .14 to .16 by changing the load ; the mechanical engineer considers these practically the same, for the difference is much less than would be caused if the surface were roughened or if oil were applied. The laws and rules that engineers use are often only approximate. CHAPTER VI. PAEALLEL FOEOES. 99. The Resultant of two forces meeting at a point and mak- ing a large angle is small (Exercise 12, page 66). As the angle is reduced — that is, as the forces become more nearly parallel — the resultant increases. When the forces are par- allel, and pull in the same direction, the resultant force that could replace them is equal to their sum. Exercise 14. (a) PAEALLEL FORCES NOT MEETING AT A POINT. - First Method. Apparatus: Stick or meter rule; 2000-gram balances; fish-line; adjustable clamps to hold balances in position. Tia a string, 6 inches or more long, to the ring of each balance. To the hooks, tie pieces of string having loops to slide over the stick, AB, Fig. 87. The strings, C C, C, C are held at any point by clamps that slide on the edge of the table. Make the pull on E 1500 to 2000 grams. Adjust the pulls on ^ and B till all three strings are parallel to each other and perpendicu- lar to the stick AB. Over- come friction by tapping AB vrhile making adjust- ments of the balances. Try several cases, vary- ing the distances AE and EB. Make diagrams ; re- R E Fig. 87. ?c A i 85 86 PRINCIPLES OF PHYSICS. cord the distances AE and EB, and the forces exerted at each point. Keep tlie following questions in mind, and try to determine the answers : How does the sum of A and B compare with E ? What must be the value of a single force, R, applied at E, to replace A and B and balance the equilibrant E ? The resultant R, therefore, always being opposite E, where is the resultant applied when A and B are equal? Nearer which force is the resultant applied when A and B are unequal? These questions may be answered by letting two boys, A and B, carry the ends of a stick on which is hung a basket. Where must the basket be hwag for A to carry one-half the load? one-fourth the load? 100. Conclusions. — In each case of equilibrium (Fig. 87), multiply the force at A by the distance AE ; also multiply the force at B by the distance BE. How do the products com- pare ? In a case where the force at A is three times that at B, how many times as long as AE is BE? See if the following conclusions can be drawn from the exercise : — The resultant of parallel forces in the same direction equals their sum. The resultant is always somewhere between the components. Tlie resultant is always nearest the greater force. If one component is a certain number of times the other, then the second component is just as many times farther on one side of the resultant. One component times its distance from the resultant equals the other component times its distance from the resultant. By the word distance is meant the shortest distance from the line of the force to the point where the resultant is ap- plied. Stating each case of equilibrium you have recorded in the note-book, show how nearly the conclusions given above apply. Parallel forces are studied more completely in the exercise on page 88, where the board, Fig. 89, may be thought of as a widened form of the stick AB in Fig. 87. PARALLEL FORCES. 87 Problems. 1. Two horses pull at A (Fig. 88), one horse pulls at B. Where on AB must the wagon be attached, or where must the resultant be ap- plied ? As the force A is twice the force B, a B must be twice as far from the resultant A « as A. Divide AB into three parts, and I I locate the resultant. Fig. 88. 2. Let AB, Fig. 88, be ten feet long, and the force at A equal 2, and at B equal 3. Where is the resultant applied? How large is it? 3. Where must a load be hung on a four-foot stick so that a boy at one end of the stick will carry one-third of the load ? i. Ji E (Fig. 87, page 85) = 15 pounds, and B = b pounds, how large is .4 ? . A and B together must equal 15 pounds. If BE equals 4 feet, how long is ^E? 5. A (Fig. 88) = 5 pounds, B = 7 pounds ; find where the resultant of these forces is applied. 5 -I- 7 = 12. Divide AB into 12 parts. Resultant is five divisions from which component ? 6. A bar is suspended by ropes, A and B, Fig. 88. The rope B is weak and likely to break under a load of 40 pounds. On what part of the bar is it safe for an acrobat weighing 120 pounds to hang? 7. When a steam roller weighing twelve tons has, gone one-fifth of the length of a bridge-span, what proportion of the weight is borne by the support at each end of the span ? 8. Suppose Fig. 88 represents a ladder used as a staging by painters. When a painter is on the ladder, the pull on rope A is 120 pounds, and on B is 20 pounds. What is the weight of the painter? How many feet from A is he, if ^ B = 12 feet ? In this problem, we will not consider the weight of the ladder, which is very small. 9. One end of a car is supported by a two-axled truck. The axles are four feet apart. The motor is geared to one axle, and it is de- sired that this axle carry three-fourths of the load, so that the motor wheels may not slip. On what part of the truck does the car rest? Make a diagram. PRINCIPLES OF PHYSICS. Exercise 14. C6) PAKALLEL FOKCES NOT MEETING AT A POINT. - Second Method. Apparatus : The exercise on page 85 may be profitably repeated, using the following apparatus; a square board, having forty-nine holes, in seven rows, one inch apart, with a border round the board, and extending a little below it ; fish-line, used as a string ; three 2000-grani spring balances ; clamps ; marbles ; wire pegs fitting the holes in the board. Part I. Lay the board on four or five marbles. Insert three pegs in holes in one line. Adjust the pull on the balances, holding, by- adjustable clamps, the strings attached to the rings Cj, Cj, Cg. Make the forces such that they are applied in parallel directions and along the lines of holes in the board, as shown in Fig. 89. Cut out a square of paper ruled with cross lines (Fig. 90). Make three points to represent the position of the pegs. Draw arrows showing the directions of the forces, and on each arrow mark the number of grams' pull that force has. Try several cases, changing the positions of A, E, and B, making them different distances apart, and changing the row of holes, but always keep- ing them in a straight line. Record each case on a separate piece of squared paper, and paste in note-book. Study the results. Disregarding the errors of the balances, the two components A and B, when added, are equal to the equilibrant E. Another way of stating this is : the sura of the north forces equals the sum of the south forces. Part IT. Taking any one of the cases of equilibrium in Part I., move one of the pegs to different holes along the line of the string, — that is, along the line in which the force acts, as, for instance, in Fig. 91, where the peg A has been moved from its position in Fig. 89. See if the board is in equilibrium, when the forces A, E, and B have E Fig. 90. PARALLEL FORCES. 89 the same intensity they had before. Try moving another peg along the line of the string attached to the peg. Suppose B is to be moved. Any point in the end row of holes may be tried. While anywhere in this row, B is always four spaces from the row of holes in which E is. Another way of saying this is : the line of direction of the force B is four spaces from the line of direction of the force E. It will be seen that a force has the same effect, if applied or attached at any point along the line on which the force acts. If the line A C were extended in each direction, the peg A could be placed anywhere along that line, and, if the three forces were unchanged, the board would still be in equilibrium, — that is, it would not move nor twist. < • . k •E • B Fig. 91. Fig. 92. A force, then, can be applied anywhere along the line of direction of the force, and always pro- duce the same effect. Part III. Try the effect of turning the board as in Fig. 92. Keep the strings parallel, by sliding the clamps along the edge of the table. Adjust the positions of the strings in the clamps, till the same forces as before are applied. Is the board in equilibrium? Move one peg out of the straight line AEB; equi- librium cannot be had with the same forces. Problems. 1. What changes can be made in the point of application of a force so as not to disturb the equilibrium of the case recorded in Exercise 14? What changes cannot be made ? In Fig. 91, the strings could be fastened to the board at any point in its length and yet not cause the board to move or the forces acting on A, E, and D to change. 2. Force A, Fig. 87, page 85, is 8 pounds; B is 4 pounds; what is force Et Ans. 12 lbs. 90 PRINCIPLES OF PHYSICS. 3. Nearer which force is £ applied? Ans. Nearer^. 4. Suppose A = 20 pounds and C = 34 pounds what does B equal? Ans. 34 -20 = 14 lbs. 5. How must the forces A and B compare, if the resultant R is halfway between them? 101. Translation and Rotation. — When a force is applied to a ball or a top, it sometimes is set moving without any twisting or spinning. It then has a motion of translation merely ; that is, a movement from one place to another, without turning or rotating. The ball or top can be set spinning or rotating with- out moving from one place to another. It is then said to have a motion of rotation. Both motions often exist at- the same time, as when a ball is pitched or a bullet sent from a rifle, — advancing and at the same time twisting. 102. Moments of a Force. — In the various cases we have been studying, the forces were in equilibrium. They were balanced in such a way that they tended neither to move the body along nor to make it rotate. The following ^^ experiments show the tendency of a force to make a body turn about a point. Fig. 93. Study the force required to lift the cover of a book. First, apply a force in the direction of A, Fig. 93 ; then of B. H is the turning-point. To raise the cover, the same moment of force is required, whether the force is applied in the direction A or B. A door is hinged at H (Fig. 94). By means of a spring balance, find how much jj C pull it takes to move the &^ door slowly, exerting the pull first at A, then at C, and afterward at B. ^''^- ^*- It takes the same moment in all cases, but the force required in a direction near to that of B is very large, and becomes PARALLEL FORCES. 91 necessarily larger as its direction is more nearly parallel to the door. By moment of a force is meant the tendency the force has to make a body turn or twist about a point. 103. Value of a Moment of a Force. — The value of the mo- ment, or the twisting power of a force, is found by multiplying the force by the distance of the force from the turning-point. The distance is always the shortest that can be measured from the line of the force to the turning-point. A certain moment is needed to make the cover (Fig. 93) turn about H ; but since the distance from H to the line B is very small, — that is, the leverage is so small, — a large force must be applied at B to have the same moment, or turn- ing efEect, as a much smaller force at A. Exercise 15. MOMENTS OF A FOBCE. Apparatus: A piece of wood about a foot long, one inch wide, and one-half inch thick, bored at intervals with holes that fit the pegs used in Exercise 14 ; a pin slightly smaller than these, to be used as a pivot ; a board that can be clamped to the table, in which this pin should be set ; spring balances and cord, as in Exercise 14. Pivot the stick by the middle hole on the pivot-pin, C, and exert two forces, as in Fig. 95, one of them, A, at right angles to the stick. Record the forces on a diagram representing the arrangement uised. The mo- ment of the force B must equal the moment of the force A, for the moments tend to make the stick turn in opposite directions, one just balancing the other ; but the force B is the greater. The shortest dis- tance from C to the line of the force A \s AC. Multiply that distance by the force exerted at A. Mul- tiply the force B by the shortest '^' "• 92 PBINCIPLES OF PHYSICS. distance from C to the line of the force B. This line is produced in Fig. 96. CD is the shortest distance from C (the turning-point) to the line of the force B. The shortest ■^ X ? distance is always a perpendicular one. Since a force can be applied at any point in the line of its direction without changing any condition, the force applied at B has exactly the same effect as if applied at D. Try other cases, with forces applied at dif- ferent points on the line ABC, and at different angles. In every case the moment of A equals numerically the moment of B. Find the mo- ments of A and B see how nearly they are equal. Pivot the stick at .4, as in Fig. 97. Find the moments of the forces D and E about the point A. Measure from the turning-points to the line of the force. As shown in Fig. 97, ADi& this distance for the force D. AE \a the distance from A to the line of the force E. Compute the moments. Are they numerically equal ? Fig. 97. 104. Positive and Negative Directions. — Take any case of equilibrium. Suppose, for example, that in Figure 89, page 88, the force ^ = 1800g, ^ = 1200g, 5 = 600g. Consider the point A fixed, as if a pin were driven through it into the table, so that the board revolves about A, if it revolves at all. Holding A, give a little extra pull to E. The board turns in the direction of the hands of a clock, or, as it is called, in a plus, or positive, direction. Force B will be found to make the board turn, or rotate, in the opposite, or anti-clockwise, direction, called a minus, or negative, direction. The moment of the force E about A = 1800 x 2 = 3600 The moment of the force B about A= 600 x — 6 = — 3600 The sum of the moments is zero. PARALLEL FOBCES. 93 Or, put a pin through a bit of cross-ruled paper, on which this case has been represented (as in Mg. 90). Hold the pin fast ; the pin is the turning-point. With the sharp point of a pencil give a push at E in the direction of the force E, as indicated by the arrow. 106. The Sum of the Moments. — When a body is in equilib- rium, that is, when it does not move, consider any point the turning-point, and the moments of all the forces about that point, added together, are equal to zero. This must be so, because if the sum were not equal to zero, the moments would not balance and the body would rotate. Next, consider E (Fig. 98) the turning-point. The moment of the force A is positive: 1200 x 2= 2400 The moment of the force B is negative : 600 x —4= — 2400 As before, the sum of the moments equals zero, and the board has no tendency to twist about the point E. ; Notice that the force acting on E does not tend to cause rotation about the point E\ 1200 eoo there is no moment to the force E; \ | why ? How far is the line of the ••i*a...l force E from the point E? la a. ^ -^ \ ^ similar manner, consider B the turn- isJo ing-point, and take the moments ^'e- 98- about B. Take a point outside of the board as the turning-point. To explain this, let us consider the point D (Fig. 98), although any other point, however remote, would serve as well. 7200 -7200 The sum of the moments about any point whatever, near or distant, would still equal zero. Moment of ^ = 1800 x 4 = Moment of ^ = 1200 x - 2 = - -2400 Moment of jB = 600 x - 8 = - -4800 94 PBINCIPLES OF PHYSICS. Problems. 1. What must be the force at A, Fig- 99, to balance the others? What is its dLrection ? IS [ A Forces^ and O, added together, equal forced. J ^ 2. In the preceding problem, find the force C at j4 by taking moments about B. ^x 2 = 12x3. A = what ? Next, find A by taking moments ^^ about C. ^ X 5 = 30 X 3. Fig. 99. In taking the moments about B, the force of 30 has no moment, because there is no distance between the turning- point and the line of the force B. 3. Find the force at C, Fig. 100. Take moments about A. 80 X 2 = C X 5. Find the force at A by taking moments about C. 80 X 3 = j1 X 5. Prove the work by adding A and C ; the sum • • i - should equal 80. A B C 4 - B C so S6 Fig. too. Fig. 101. 4. Find the forces A and C, Fig. 101. 5. A weightless beam is supported at B, Fig. 102. On the end A is hung a six-pound weight. How heavy a fish hung at C will balance ? -A , _ . -B What is the pressure on the sup- [ " J^ port B? 6. A meter stick is balanced at its Fig. 102. centre. On one end is hung a two- pound weight. How far on the other side of the balancing point must three pounds of tea be placed to balance the two-pound weight ? 7. One end of a 6-foot fishing-rod is held in the left hand ; the right hand grasps it 1.5 feet from the same end. Neglecting the weight of the rod, what force must be applied by the right hand to raise a one-pound fish from the water? How much must the left hand press down ? Add the downward push of the left hand and the weight PARALLEL FORCES. 96 of the fish. How does this sum compare with the pressure on the right hand? Ans. 4 lbs.; 3 lbs. 8. A stick, AB, floats east and west in the water. At A there is a force north of 20 pounds ; four feet from A there is a force south of 50 pounds. What force will keep the stick from moving? Where must that force be applied? 9. A crowbar, A C, Fig. 103, is pivoted at B ; what must be the downward pressure at A to raise a weight of 200 pounds at C? If the bar is pivoted at C and the weight placed at B, what must be the upward pull on Al If the bar is pivoted at A and the weight Fig. 1 03. is at C, what upward force must be exerted at B'! 10. A bat is 34 inches long ; the left hand holds one end fixed ; the right hand grasps the bat 10 inches from the end, and moves 14 feet a second. How fast does the free end of the bat move ? 106. The Lever. — In problem 9, the bar is called a lever, and is the simplest form of a machine. When the support is at B, Fig. 103, the point B does not move ; that point is called the fulcrum. When C is the support, that point does not move ; C is then the fulcrum. The fulcrum in a machine is a point that is considered not to move. We say considered, because, in the case of a fishing-pole, either hand may be held station- ary, and may, therefore, be the fulcrum. Both hands may also move at the same time ; but, for the purposes of calculating the forces, either point may be considered as stationary, and that is the point about which the moments are taken. Suppose, in Fig. 103, that the left hand at is stationary, and the right hand, grasping the bar at B, moves, the fish being attached to A. Which moves the faster, the right hand or the fish ? At which point is the greater pressure exerted ? A large force at B, moving a small distance, raises a small weight at A through a large distance. Use a pencil or pen-holder for the lever, and a bunch of keys for the weight. T^ X - F 96 PRINCIPLES OF PHYSICS. 107. Weight and Power. — A flat stick, or rale (Fig. 104) is supported on a pivot through F. At equal distances from F are pins, on which weights can be ■'"'' 'P hung. Put four equal weights one ~n space from the fulcrum, or balanc- ing point, F, and see if one weight, four spaces on the other side of the fulcrum, will balance them. W Pig I ()4_ Call the four weights, W; and call the distance of the point of sus- pension from F, the weight arm, w (in this case, one division of the stick). Call the other force P, and the distance of the force P from F, the power arm, p. Pour times one (that is, the weight times the weight arm) equals one times four (the power times the power arm). In shorter form this is written Wxw = Pxp, but is always read as weigM times weight arm = power times power arm. Eaise and lower the power arm, p (Fig. 104), touching it at the point where P is applied. Do the same with the weight arm, touching it where W is applied. The power travels four times as fast as the weight. To raise the weight, W, one inch, the power must move through a distance of four inches. In this case, a weight four times as great as the power is raised. 108. Power and Speed. — This principle is used in making machines to raise a small weight rapidly; for instance, to strike a blow with an axe or hammer at a higher speed than the hand can be moved. The power and the weight change places in such a case. The power applied, for example, one division from F, in moving one inch, causes the weight to move four inches — four times as fast. Place three weights two divisions from F, and find where PARALLEL FORCES. 97 two weights will balance. Try other cases, testing the law W xw = P xp. What is gained in potver is lost in speed, and what is gained in speed is lost in power is an old-fashioned way of stating this law. In the case shown in Fig. 104, there is a gain in power, — that is, the power moves a greater weight than itself, — but there is a loss in speed, for the weight moves slower than the power. When a machine of any kind, as, for instance, the lever in Pig. 104, is in motion, the power moves over as much greater distance than the weight as the power arm is longer than the weight arm. The formula Wxw = P xp can then be read as weight times weight distance = power times power distance; or, the weight multiplied by the distance through which it moves equals the power times the distance through which it moves. Always measure from the fulcrum to find the weight distance and the power distance. 109. Formula for Problems on Machines. — The same formula, or law, W xw = P xp, holds for all problems on machines. In solving questions, first decide what point is the fulcrum. Always measure from it to where the weight is applied, in order to get the length of the weight arm. Eind the length of the power arm in a like manner. In some problems, it is better, instead of finding the weight arm and power arm, to get the distances the weight and power move in the same time. Problems. 1. Suppose W, in Fig. 104, page 96, is 18 pounds, and is six divi- sions from F; P is two divisions from F. How large is the power? Which travels the greater distance ? 98 PRINCIPLES OF PHYSICS. 2. Where should the power and weight be applied so that the weight may move two-thirds as fast as the power? 3. A weight of 60 pounds is 4 inches on one side of the fulcrum ; what power must be applied 5 inches on the other side to lift the weight ? 4. A fisherman grasps the handle of a 9-foot fish-pole in his left hand ; his right hand is 1 foot from the same end. If he holds the left hand still and moves his right hand 1 inch, how far does the tip move ? Disregarding the weight of the pole, what weight at the end of the pole would a force of 4 pounds exerted by the right hand lift? If the right hand is held stationary and the left hand lowered 1 inch, how far does the point of the pole rise ? Ans. 9 in.; | lb.; 8 in. 5. In a pair of shears for cutting metal, the handle is always longer than the blade. Which moves the faster, the power or the weight ? Is there a gain in power or speed ? From the centre of the bolt on which the shears turn, to the handle where the power is applied, is 8 inches ; from the bolt to the end of the blade is 2 inches ; let a force of 5 pounds be applied to the handle, what is the force of the blade at the end? Ans. 20 lbs. 6. The handles of a pair of nippers for cutting wire are 10 inches long ; the cutting edge is i inch from the bolt, or pin. What force must be put on the handle to exert a force of 400 pounds on the edge? 7. At which end of a pair of scissors must a force be applied to cause a greater force at the other end ? 8. In a nut cracker, where is the fulcrum? If AC, Fig. 105, is 12 inches, and AB S inches, what pressure at B would be exerted by 6 pounds acting on CI What would be the pressure if B were 1 inch from A ? '^' ■ 9. If B is 4 inches from A, AC = Q inches^ and the power acting a,t B is 6 ounces, what is the pressure on C? 10. If a man shovelling snow holds the end of the handle in the left hand, which hand exerts the greater force ? CHAPTER VII. MAOHINES. -PULLEYS. 110. A Combination of Pulleys (Fig. 106) is a modified form of the lever. A rope is wound around the large pulley, and another around the small pulley. Pull on A. Which moves the faster, ^ or jB ? F is the fulcrum. The radius of the large circle is the power arm, if A is the power ; the weight is at B, and the weight arm is the radius of the little circle. A model of this pulley is made by slip- ping a spool, S, rig. 107, on a piece of wood or pencil, sharpened at both ends. A string is wound around the spool, and another around the pencil. Hold the ends of the pencil between the thumb and forefinger. the fulcrum. By means of the simple form of straight lever. Fig. 103, page 95, a weight can be lifted a short distance. Using pulleys, the dis- tance depends only on the length Fig. 1 07. of the ropes coiled around them. Fig. roe. FF is 111. Study of the Pulley as a Machine. — Fasten a car (ap- paratus of Exercise 16, page 109) to a table, so that the forward axle hangs over the edge. Let Fig. 106 represent the wheel of the car. Tie a stout linen string, A, to the wheel. For this purpose the rim is grooved, and through the edge of the wheel, to the groove, a hole is bored. Fasten another thread, B, to 99 100 PRINCIPLES OF PHYSICS. the shaft. Wind the strings a few times around in opposite directions. Put a 2000-gram balance at B, and a 250-gram balance at A. Fasten to the floor the ring of the balance joined to B. Apply different forces at A. Record the readings of both balances. Divide the readings of A by the correspond- ing readings of B. Divide the diameter of the wheel by the diameter of the axle. The radius is really the length of the arm of the lever, but the diameter is always twice the radius, and if the diameters of both are used, the result is the same as if the radii had been used instead. Move B (Fig. 106) one inch; how far does A move? On which rope would you apply a force to obtain a gain in speed ? a gain in power ? Try the effect of applying forces not par- allel. A given force at A always has the same moment, or turning effect ; for no matter in what direction A is pulled, the distance from the fulcrum to the line of the force is always the radius of the circle, and all the radii of a circle are equal. 112. Crank and Axle. — Sometimes one of the pulleys is replaced by a simple lever or crank. Suppose that in Fig. 108 all but one of the spokes of the wheel were removed. If a handle were attached to the one remaining spoke, at right an- gles to it, and to the plane in which the wheel revolved, the machine would be a crank and axle. Fig. 108 Problems. 1. If the handle of a clothes-wringer is 10 inches from the axis, the rubber roll is 2 inches in diameter, and a force of 30 pounds is exerted on the handle, what is the force on the surface of the rubber roll? Ans. 300 lbs. Notice that 2 inches is the diameter of the axle, and 10 inches the radius of the circle in which the crank moves. Either double 12, when MACHINES. — PULLEYS. 101 the power and weight arfli are as 20 to 2 ; or halve 2, in which case the two arms of the lever are as 10 to 1. The proportion is the same in both cases. That there is a gain in power in the wringer can be shown by trying to hold back with one hand a piece of cloth from passing through the rollers, while turning the crank with the other hand. If the rolls are screwed tightly together, as they are when wringing clothes, there is a large loss from friction. 2. The crank on a grindstone turned by hand is 12 inches long, and the diameter of the stone is 30 inches. When a force of 60 pounds is put on the crank at right angles to its radius, what force can be exerted on the rim of the stone by a knife that is being ground? J;iVfr( < /:; ' Ans. 48 lbs. 3. A crank and axle is used to raise buckets of .earth weighing 50 pounds. The diameter of the axle is 6 inches. What power must be applied to the crank, which is 15 inches long ? Eemember that the radius of the axle is 3 inches. 4. How many times larger than the axle must the large pulley in Fig. 106 be, so that 40 pounds at A can raise 50 pounds at -B ? 5. If a belt travelling 1000 feet a minute passes over an axle that is 2 inches in diameter, what must be the size of the emery wheel, so that points on its rim travel a mile a minute ? 6. The diameter of a grindstone is 24 inches. The crank is driven by foot power, and has a radius of 2 inches. How many times greater is the radius of the grindstone ? How many times faster does a point on the rim of the wheel travel than a point on the end of the crank? What force must be applied perpendicular to the crank, to give a force of 30 pounds on the rim of the stone ? 7. Why is a force that is applied as shown in Fig. 109, less effective than one at right angles to the crank? 8. The blade of a screw-driver is J inch wide ; the handle is 1 inch in diameter. Applying a force of 6 pounds to turn the handle, what is the force tending to turn the screw ? Ans. 24 lbs. 9. Why can a screw be driven harder, or the head even twisted off, by using a screw-driver in a bit brace ? 102 PBINCIPLES OF PHYSICS. 113. Movable Pulley. — When a load is supported by two or more strings, ropes, or chains, its weight is divided between them. Support the handle of a pail by two balances, A and B. Use at first a pail that has a plain wire handle. The sum of the pulls on the balances always equals the weight of the pail. Eaise B. As the pull on B in- creases, that on A decreases. Substitute a pail with a wooden handle strung on the wire (Fig. 110). The pulls on A and B are each about half the weight. The handle serves as a movable pulley. In this form of machine there is a gain in power. Eaise B one foot ; how far is the pail raised ? In what do you find a loss corresponding to the gain in power ? The size of the pulley has no effect on the power, except that there is slightly less loss by friction in a large pulley than in a small one. Fig. 110. 114. Fixed Pulley. — The axles of the car used in Exercise 16, page 109, answer well as pulleys, and have almost no fric- tion. Screw or clamp a car. A, to a block, and fasten the block to a table (as in Fig. 111). For the weight, W, use another car with a load, which may be tied in. Raise the car, and read the balance. Correction should be made for the error _. , , , Fig. III. due to using the bal- ance in an inverted position. This is determined by hanging it from another spring balance, and pulling down tiU the upper MACHINES. — PULLEYS. 103 balance shows that the force exerted is that which is to be used in the experiment. If the lower balance reads the same as the upper, there is no error ; if not, the amount of the dif- ference should be used in correcting the readings. (See exer- cise on errors in a spring balance, Appendix, page 636.) The arrangement in Fig. Ill comprises a fixed pulley only. Fixed pulleys change the direction of a force. Is there any gain or loss in power or in speed ? 115. Combinations of Pulleys. — Try the combinations shown in Figs. 112, 113, and 114. In Fig. 114 there are three strings pulling up. There is the same pull on every part of the ft 'I V y w w Fig. I I 2. Fig. 113. Fig. I 14. strings, since the pulleys all turn easily. Head the balances at A and P. A force of one pound applied as a power at P is in reality exerted three times on the weight. Therefore the weight, W, is three times the power. Move P an inch ; how far does TTmove ? Does the law P x p = TT X w, still hold ? In the combinations of both Figs. 113 and 114, two ropes 104 PBINCIPLES OF PHYSICS. support the movable pulley. W is twice P. The upper fixed pulley in Fig. 113 serves to change the direction of the string. The weight moved is as many times the power applied as the number of ropes attached to the weight. In practice, there is some loss from friction. Problems. 1. What power in Fig. 115 is re- quired to lift a weight of 120 pounds at Wi Count the number of ropes support- ing the weight. Atis. 40 lbs., and a little more, to overcome friction. 2. How many ropes support the weight in Fig. 116? How many times greater is W than P? How many times faster than W does P Y move ? r*-i 3. Sk,etch a system of puUeys jw\ where W is five times the power; 1 1 six times the power. ^' ' ' 4. How great a force must a hy- draulic piston apply at W (Fig. 116) to lift an elevator weighing 3000 pounds, attached to P ? How many times faster does the elevator move than the piston ? How long must the cylinder, in which the piston moves, be, if the elevator has to travel 120 feet ? What is the total downward force which the upper pulley exerts upon its support? Turn the diagram upside down, and count the number of ropes run- ning through the pulley in question. The pull on each rope is the same. 116. Gear of Bicycles. — The diameter of the rear, or driving, wheel of a bicycle is usually 28 inches ; occasionally 26 or 30. By the expression " 70 gear " is meant such a combination of gears' (pulleys having, teeth) that the rim of the rear wheel — and consequently the bicycle itself — goes as far in one revolu- MACHINES. — P ULLBYS. 105 tion of the pedals as a wheel 70 inches in diameter would go in one revolution. The small rear wheel of a bicycle goes round as many times more than the pedals as the gear on the crank-shaft is times larger, or has times more teeth, than the gear on the rear wheel. If the rear wheel has an 8-tooth gear and the crank-shaft a 24-tooth gear, then the rear wheel turns ^, or three times as fast. Problems. 1. Find the gear of a bicycle, if the diameter of the back wheel is 28 inches, the number of teeth on the same is 9, and if there are 27 teeth in the crank-shaft gear (called sprocket). Am. -V- x 28 = 84. 2. Find the gear of the wheel described in Problem 1, if the diameter of the rear wheel is increased to 30 inches. Ans. 90. 3. Measure the diameter of the rear wheel, count the number of teeth in the gears, or sprockets, and compute the gear of any bicycle at hand. (Mark one tooth with chalk, calling it number one, and count around to it.) 117. Speed Gearing. — Place a bicycle upside down, and support it firmly, with the rear wheel free to turn. Attach a 2000-gram balance to the rim of the rear wheel and another to the shaft of the pedal. Fasten or hold firm the first balance, and exert a force of 2000 g. perpendicular to the crank. Eecord the readings of the balances, the diameter of the rear wheel, the length of the crank, and the number of teeth in each sprocket. Calculate the gear of the wheel. This number is the diameter of a wheel that will go as far in one turn as the bicycle is sent by one turn of the crank. Half this diameter gives the radius. How many times greater is this radius than the length of the crank ? The gain in speed may be found by comparing the gears of the bicycle with the length of the crank ; for instance, if the gear is 70 and the crank 7 inches long, the bicycle travels five times as fast as the foot moves. For this gain in speed, what 106 PRINCIPLES OF PHYSICS. is the loss in power ? Why cannot you ride up as steep a hill as you can walk up ? How many times greater is the pull measured by the spring balance on the pedal shaft than the pull on the rim of the wheel ? It is this pull, or push, on the rim of the wheel that sends the bicycle along. Move the crank one inch, and measure how far the rim of the rear wheel travels. This is the gain in speed. 118. Pulleys and Belts. — Power is usually transmitted from one part of a factory to another by belts of leather. An engine or water wheel drives a long rod, called the main shaft, and on this there are pulleys of various sizes. Arrange several cars on blocks fastened to a board, so that the bodies of the cars are vertical. Belt a little engine, or a water or electric motor to the groove on the wheel of the end car, using a piece of white string for a belt. By a loop of string, connect the axle of this wheel with the groove on the wheel of the next car. In the same way, belt the axle of the second car to a wheel of the third car. On the last axle wind a few turns of string, and fasten a weight to the string. Start the motor. Which runs the faster, the motor or the weight ? Is there a gain in power or in speed ? Such a gearing as this is used in cutting or planing metal, which must be done at a slow speed, or the cutting tool becomes hot, and dulls. It is used also for lifting heavy weights. Remove the engine or motor, and turn the axle of the last car by hand. What part of the train of wheels turns fastest ? Emery wheels, circular saws, and wood-working tools must travel at a high rate of speed. Make a diagram showing how, by belts and pulleys, this speed can be obtained. CHAPTER VIII. WOEK. 119. Resistance. — To do work, a body must be kept in motion where there is resistance. Eesistance may be of sev- eral kinds : friction ; inertia, that is, the resistance an object offers when started or stopped; the resistance a substance offers when cut or broken ; some forms of resistance that are to be studied under Electricity and Heat ; and the resistance a body offers to being lifted. When a body moves against any of these forms of resistance, it does work. ( Does a spinning top do any work ? The point bores a hole in the floor, and the surface fans and sets the air in motion a little, and the top at last stops. In a vacuum, a top with a hardened peg would revolve a long time on a smooth surface ; or a grindstone could be mounted so that it would revolve an indefinite length of time ; but the grindstone would do no work. If a knife were pressed on the grindstone, the motion would be resisted and work done. 120. Unit of Work. — A body resists being lifted, because it is apparently attracted by the earth. One pound raised one foot is said to require one foot-pound of work. In the metric system, the unit of work often used is the work done in raising one gram one centimeter. Remember that work is measured by the force required to make a body move, or, more briefly, work equals force times distance, W=fd. 107 108 PRINCIPLES OF PBYSICS. 121. Relation of Force and Distance. — It is easier to roll a barrel of flour up an inclined plane from the sidewalk into a wagon tlian it is to lift the weight straight up. In rolling the barrel up the incline, the force applied is comparatively small, — smaller than that needed to lift the barrel, — but the distance is large. To lift the barrel up calls for more force, but the distance is much less than before. Then, since work equals force times distance, the amount of work done is the same in either case. In the experimental study of the inclined plank, or plane, by the use of a car, the friction is so much reduced by the wheels that it may be neglected. Problems. 1. How mueh work is done iu lifting 1 pound 2 feet? In lifting 3 pounds 5 feet? One-half a pound 30 feet? 2. How much work do you do in walking upstairs, 20 feet rise ? 3. How much work is required to lift a 500-pound hammer 30 feet? 4. If one pound = 454 g., and one foot = about 30 cm., how many gram-centimeters in a foot-pound ? 5. How much work is needed in gram-centimeters to lift 1 g. 1 cm.? To lift 5g. 4cm.? To lift 10 cc. of water 42 cm.? 6. How much work is done in pumping 40 liters of water 50 m. high? 7. What amount of work is required to raise a ton of coal 3000 feet from the bottom of a mine? 8. If in walking a person rises at each step -^^ foot, how much work is done in this way by a person weighing 150 pounds ? 9. How much work is done in pulling a 200-pound sled 40 feet along a level road ? Can this be found until the friction resistance is known ? If the pull to make the sled move is 22 pounds, how much work is done? 10. Find the force required to draw a 500-pound sled, with iron runners, over iron street-car rails, the coefficient of friction being .3. How mueh work is done in pulling the sled 20 feet? WORK. 109 Exercise 16. INCLINED PLANE. Apparatus: Board; car with a weight ; spring balance. Lay a board, A B, Fig. 117, in a slanting position from the table to the top of a box. Weigh a car with a load. The pull required to draw the car up the in- cline is measured by a ^-^'^ spring balance. The zero .--C^^^***^ f ~ point of the balance should be recorded for the slant used, and the correction, if appreciable, should be used in making the computations. To de- Fig. 1 1 7. termine whether friction is a large part of the resistance overcome in pulling up the car, record the pull when the car is moving down the incline. The average of the two pulls is the force that would be required to make the car go up the slant if there were no friction. As this friction is very small, it is best disregarded. Use as small a balance as possible ; if the pull is less than 250 g., use a 250 g. balance. If the car were a horse-car, the horses would have to walk from A to £ to draw the car the dis- tance AB, and the work done would be the distance AS times the amount of the puU. Calling AB the slant, the work done = slant times piiU. If W were a load of coal, it could have been wheeled from A to C with very little work, and then raised vertically to B. The work performed is the weight, W, times the distance moved or raised, which is CB. The work performed is therefore the weight times CB. Calling CB the height, h, the work performed is W x h. Compare the work done in pulling the car up the slant with the work performed in lifting the car straight up from C to B. Repeat the experiment with different loads and different slants. What would the pull become if the slant were increased? If it were increased so as to be almost vertical ? 122. The Mechanical Advantage in the Inclined Plane is that a small force, such as may be exerted by a man or a horse, can, 110 PRINCIPLES OF PHYSICS. if the slant is gradual, move a heavier load than can possibly be directly lifted by the same force. On the slant, while the pull is less than the weight of the load, the distance the load travels is greater than when it is lifted straight up. The work done in either way is the same; the choice of method is a mere matter of convenience. 123. Formula. — As the work is the same, whether the load be pulled up the slant or lifted up vertically, we can write that the pull times the slant distance = weight times the height ; or PxS=Wxh. Since slant distance equals power distance, and height equals weight distance, this formula may be written : — P xp=Wx w. Power X power distance = weight x weight distance is the law that applies to all machmes. In this equation, if any three quantities are known, the fourth can be found. If the slant = 30 feet, the height = 10 feet, and the load = 2400 pounds, what is the pull ? Writing down the formula, PS = Wh, then erasing S and writing 30 in its place, writing 10 for h and 2400 for W, we have P X 30 = 2400 X 10, 30 P= 2400, P= 800. Problems. 1. What is the slant length of a plane on which a 50-pound pull moves 800 pounds, when the height of the slant is 20 feet? 2. What weight can be moved on a rise of 1 foot in 25, with a pull of 40 pounds? WORK. Ill 3. On a rise of two feet in one hundred (called a two per cent grade), what pull must a locomotive exert to draw ten cars, each with a load weighing 25 tons, friction being disregarded ? 4. On a rise of three feet in one hundred (the steepest allowable on a really good road), what is the force needed to move a 1200-pound wagon up the hill, if we disregard friction ? 5. How great a load can, by a force of 120 pounds, be kept from rolling down a 12 per "cent grade (i.e. a fall of 12 feet in 100)? 6. If a mountain railway has a rise of one foot in three, what force is needed to move a car and passengers, weighing 10,000 pounds, up the track? 7. Find the height of an incline 1000 feet long on which an 80- pound pull moves 600 pounds. 8. An electric car on a hill is attached to a dynamometer (which is practically a huge spring balance). The dynamometer reads 900 pounds, and the car weighs 6000 pounds. What is the grade? What is the rise in 100 feet of slant? 124. The Wedge. — Instead of pulling a load up a slant, the slant may be pushed under the load. For the purpose of rais- ing the weight W, Fig. 118, an inclined plane, called a wedge, is forced under the weight. In this case, the power by suc- cessive blows is applied to drive the wedge in a dis- tance, AC. The work done is Power x AO, or, Power X horizontal distance. The load, be it a safe or a chimney or a lump of stone, is raised the distance BC. The work performed is Wx BC, or TF x h. The work accom- plished in raising the load must equal, disregarding friction, the work expended in driving in the wedge. Force xAC= Weight x BC. C Fig. 118. 112 PRINCIPLES OF PHYSICS. If a wedge is 10 inches long and 1 inch thick, and is driven by a blow of 300 pounds, what weight does it lift ? In prac- tice, something is lost in friction, and the weight lifted will be much less than the computed amount. If the wedge be used for splitting, it is usually ^ a double inclined plane (Fig. 119). The '-E power distance is AO, and the work dis- ^'^' ' ' '• tance is BE. That is, by forcing the wedge through a distance AC, a log is split open a distance BE. Problems. 1. In a wedge 12 inches long, what must be the thickness (BE, Fig. 119) for a blow of 150 pounds to cause a splitting force of 2000 pounds ? 2. On a grade of one in twelve, how much must a horse hold back to keep a 3000-pound wagon from running away ? 3. On what grade can a horse move a load of 2400 pounds, if he is capable of pulling 400 pounds ? 125. The Screw. — Examine and make a rough drawing of a bolt having ten turns, or threads, to the inch. Rub the point of a pencil on one side of the bolt. Lay the page of the note- book over this, and rub the paper with the side of a pencil. A tracing like Fig. 120 is left on the paper. Lay off one inch from AtoB, 'f' i i i i r i i i r ■^ I i and count the number of ridges be- { tween those two points. The first "*" Fig~i2o"~ ridge, in beginning at A, is not counted, just as in timing a pendulum for a minute, the observer does not count "one" at the beginning of the first swing. Make tracings of the threads of other screws, microm- eter calipers, iron clamps, etc. 126. The Power of the Screw as a machine comes from the very long and slim wedge that is wound around it as a thread. WORK. 113 Cut out a paper wedge thirty times as long as it is thick at one end ; wind this around a pencil. Try to print the mark of a medal on the page of a note-book, by pressing with the fingers. Force the medal in, using a clamp (Fig. 121). Make a tracing of the thread, and count the number of ridges in one inch. Suppose the number is ten; then, in one turn, the screw moves for- ward yij of an inch. Call this the weight, or work distance. If the diam- eter of the handle is 2 inches, and the power is applied at the ends of the handle, the power in one revolution acts over a distance that is a circle 2 inches in diameter. The power dis- tance, the circumference of this circle = 2 x 3^^ inches = 6^ inches, or about 6.3 inches. In case the power applied is 20 pounds, by substituting the formula, Wx w = P xp, TF X tV = 20 X 6.3, W= 1260 pounds, the force that would be applied by the screw in case there were no friction ; but more than three-quarters of the force is lost in overcoming friction. This loss is more or less useful, be- cause this friction keeps the screw from turning backward of itself. Fig. 121. 127. Power compared with Speed. — Set a nut, N, Fig. 122, in a board, C. Tie a 20-pound weight, W, to the lower end of the bolt, B. E is a, rod or wrench attached to the head of the bolt. The spring balance is so placed that DE = 3^ inches. This distance is taken for convenience in computation. The circle in which E turns = 3^ x 2 x 3^^ = 22 inches. Suppose 114 PRINCIPLES OF PHYSICS. the bolt has ten threads to the inch ; then the work distance in one revolution, or the height that W is raised, equals ^ inch. Kead the spring bal- ■^ ance as E is slowly- turned, keeping the string at E perpendic- ular to DS. The spring balance measures the power applied. Calcu- late what weight should be lifted without fric- tion, and compare with the weight actually- raised. In one revolu- tion, how far does the point E move ? (Ans. 22 inches.) In one revolution, how far does the weight rise ? (Ans. J^ inch.) How many times faster does the power move than the weight ? (Ans. 220.) Fig. 122. 128. Couples. — Two tugs, E and W, Fig. 123, press with equal force against a steamer, S, which is not working its own engines. E pushes at the bow toward the east, and W at the stern toward the west. The effect is to turn the steamer around without moving her to a different place. What would be the effect if she worked one propeller. A, to drive her backward, and the other, B, to drive her forward ? Lay a pencil on a smooth surface. Try to make it turn round on its short axis by applying one force. Apply two equal forces, in opposite direc- tions, at the ends of the pencil. Make a diagram of the forces. W -129. Two Equal Parallel Forces in opposite directions cannot have a resultant. If they are applied at the same point, as in WOBK. 115 a tug-of-war, there is no motion whatever. If they are applied some distance apart, as in Fig. 123, there is still no resultant ; the body does not move away, it rotates. In that case the two forces form a couple. A couple is two equal and parallel forces in opposite directions applied some distance apart. A couple tends only to make a body rotate. 130. Moment of a Couple. — At two points, 10 feet apart, on a beam are two forces, each 50 pounds, one north, the other south. Take A, B, or C, or any point in the line or out of it as the turning-point. The moments of so the force in any case will be 500. This is ^ the same as 50 x 10. The moment of a "S S- I couple = one force times the distance be- I tween the two forces. A couple cannot 50 be balanced by a single force. In stopping ^'^' ' ^*' a spinning top by holding a finger against its side two forces are applied, one where the finger touches, the other at the peg of the top. Imagine the top whirling in space, as the earth does. How would it respond to the touch of the finger applied as before. 131. A Single Force cannot balance a Couple, but a second couple can do so, if it tends to make the body rotate in the oppo- site direction. Take hold of the handle of a broom and twist it, to make it turn to the right with one hand, and to the left with the other. Eepeat, grasping the larger part of the broom with one hand. If the broom does not turn, the moment of the couples exerted by the hands are equal ; but the forces are by no means equal in the latter case. Two small forces, acting as a couple and applied a great distance apart, produce the same moment as larger forces applied a short distance apart. The forces of one couple may be parallel to or make any angle with the forces of the other couple, provided they are all in the same plane. In the following exercise, the forces of one couple are at right angles to the forces of the other couple. 116 PBINCIPLES OF PHYSICS. Exercise 17. COUPLES. Apparatus : Board used in Exercise 14, page 88 ; four 2000-gram, spring bal- ances ; fish-line ; clamps ; marbles ; pegs. Place the board on the marbles, insert pegs, and apply forces A, B, C, and D (Fig. 125) at such points that the couple of CD tends to rotate the board in a direction opposite to that given it by the couple ^ AB. Vary the tension on the bal- ances till the forces lie along lines of the board. Notice that C = D, C-<- ■F»- ->-D and A = B, as should be the case with couples. What is a couple ? Record forces and directions on • • cross-ruled paper. Take, in turn, each peg as the turning-point, and find the moments of the forces, remembering that an anti-clock- wise movement is negative. See if the sum of the moments equals zero. Force A has no moment about E. Why? In taking the moments about G, note that both C and B have no moment. Find also the moments about some point in which there is no peg. Try other cases, with the pegs in different positions. 132. Calculation of Couples. — A bracket, ABC, Fig. 126, is screwed to the wall at A and B, which are ten inches apart. Neglecting the weight of the I Fig. 125. bracket, if 360 pounds are placed at C, what is the hori- zontal pull on ^ ? Supposing the force exerted at A breaks the screw; the bracket falls, turning on B as a pivot. The moment of 360 pounds at C about the point B = 360 x 24. exerted by the screw is .4 x 10. 24 in Fig. 126. The moment of the force WORK. 117 3 ft 360 X 24 = ^ X 10. A = 864. If the screw cannot stand this strain, the bracket falls. Make a model of such a bracket, with B as the turning-point, and attach spring balances at A and O. Multiply force A by the distance AB, and see how this product compares with the product of the force C times the distance AC. A door, three feet wide (Fig. 127), is suspended by hinges, A and B, six feet apart. The door weighs 200 pounds. What is the force tending to pull the hinge A out from the wall ? There is a downward force on both hinges. This can have no direct effect in pulling hinge A away from, or pushing hinge B into, the wall. The door is symmetrical, and the downward force of its weight may be considered as applied at the centre of the door. Imagine a strong bolt passing through B, and study the tendency that the forces applied to the door have, to make it turn about B. In other words, take the mo- ments about B. The hinge A pulls toward the wall. The moment of this force equals ^ x 6. The moment of the weight of the door is 200 x 1.5. ^ x 6 = 200 X 1.6. How How far is the line of the force at A from the point B ? far is the force of 200 from the point B ? Make up several problems similar to the above, and solve them. Problems. 1. If a gate 12 feet long, weighing 300 pounds, is held up by hinges •3 feet apart, what is the horizontal force on the hinges? Ans. 600 lbs. 118 PRINCIPLES OF PHYSICS. 2. If a boy weighing 120 pounds stands on the very end of the gate, what additional pull is there on the upper hinge? 3. If the pull needed to keep a wagon moving uniformly on a level road equals .01 of its weight, what is the pull in drawing a wagon weighing 2000 pounds? How much work is done in moving it 600 feet? 4. A tool is pressed on a grindstone with a force of 20 pounds ; the coefficient of friction is .3. What is the friction force ? How much work is done in one revolution, if the circumference of the stone is 8 feet ? 5. In a hydrostatic press, if the distances moved by the pistons are as 1 to 800, how many times greater is the total pressure on the large piston? 6. Which of two bicycle pumps, one J inch in diameter, the other 1 inch in diameter, can pump up a bicycle tire the harder ? 7. Is the coefficient of friction large or small when a body is slippery ? Explain the use of sand on a rail, and rosin on a violin bow. 8. If a force of 10 pounds just moves one surface over another, the pressure between them being 100 pounds, what force would be required if the pressure were increased to 200 pounds? What force would be required if the coefficient of friction were 3 times as great, the pressure remaining 100 pounds ? 9. If the coefficient of friction is .25 between the driving wheels of a locomotive and the rail, what must be the weight of the locomo- tive to exert a pull of 10 tons ? 10. If the coefficient of friction is .25, make a diagram of an inclined plane down which the locomotive would start to slide, the wheels being prevented from turning. 11. In the case of equilibrium represented by Fig. 125, are the forces C aiid Z> large when C and D are far apart, as shown, or when they are near together ? CHAPTER IX. OENTEE OF GEAVITT." 133. Action of Gravitation. — The earth exerts its downward attraction on each little particle of a body, — a block of wood, for example. If the block is cut into small bits, each bit falls (in a vacuum) as rapidly as the whole block. The forces of the earth's attraction act downward in parallel lines. All these parallel forces may be replaced by one force, called the resultant — in this case the weight of the body. To make the block fall down as the earth makes it fall, and without making the block turn round in any direction, this force must be applied at a certain point, which we shall call the centre of gravity. 134. Centre of Mass, or Centre of Gravity. — Hold the card or board AB, Fig. 128, I, by a pin near the corner A. On the same pin hang a thread, to which a weight is at- tached. The centre of gravity is somewhere un- der the line AC. Mark a line on the card where the thread touches. Ee- move the pin and place it in some other part of the card — near the cor- ner Z>, for instance. Let the card and plumb line DC hang freely, as before. Mark the line DC on the card (Fig. 128, II). Where the two lines marked on the card cross, insert the pin. 119 120 PRINCIPLES OF PHYSICS. Turn the card to different positions. It remains in any posi- tion in which it is put. The card behaves just as if all its mass, or weight, were concentrated at the point G. This point is called the centre of mass, or centre of gravity. Set the card spinning. Does it shake the hand holding the pin ? Make a hole about a centimeter from O, and insert the pin. Does the card stay in any position in which it is put, or does it turn and the point G move as low as possible. Set the card spin- ning, and notice the shaking of the support. This shaking is caused by the card trying to revolve about its centre of gravity. A wheel or any revolving part of a machine shakes its bearings, unless it re- volves about its centre of gravity. Weights are often added to an unbalanced wheel to make its centre of gravity in line with the bearings. Find, in a similar manner, the centre of gravity of pieces of board shaped like A, B, and C, Fig. 129. It will be found to be outside of the body itself, in some cases. For instance, the centre of gravity of a ring is inside the ring. If fine wires are attached to the ring D, Fig. 130, in the directions taken by plumb lines hung as in Fig. 128, the centre of gravity will be at the point where they cross, and, if supported at that point, the ring will rest in any position in which it is placed. LTTJ 135. The Line of Direction. — When a body falls, although, under some conditions, it may turn and twist, still its centre of gravity goes down in a straight line toward the centre of the earth. This line is called the line of direction. It is the direction in which a plumb line hangs. In Fig. 128 when the card is suspended at A, AC is the line of direction ; when sus- pended at D, DO is the line of direction. CENTRE OF GBAVITT. 121 136. The Base. — Since a body acts as if all its weight were concentrated at the centre of gravity, ttere must be some sup- port under the centre of gravity to keep the body from falling. The base on which a body rests is all the space that would be included inside a string wound round all its supports. The base of a three-cornered stool is a triangle, and the three points give as good a support as a triangle of solid wood. What is the base of an ordinary chair ? of a tricycle ? of a barrel of flour standing on end ? of a person standing on one foot ? of a person standing on both feet ? of a bicycle ? 137. Equilibrium. — A body tumbles over when its centre of gravity is not over the base. Another way of saying this, is to say that a body falls when its line of direction passes outside of the base. A block of wood or a brick, lying on its side, is hard to tip over. If it is tipped a little and released, it returns to its place. 138. Stable Equilibrium. — Make a diagram of a block, show- ing the position of its centre of gravity. Tip the block, keep- ing the edge on the table. Notice that the centre of gravity rises as the block is tipped. The block is said to be in stable equilibrium, because, when it is tilted, the centre of gravity rises. Now, since the earth's attraction tends to bring the centre of gravity as low down as possible, a body, when re- leased, if it has not been tipped too far, at once falls back to its former position. Compare the amount of tipping necessary to make the block tumble over when it is lying on its side, with that required to make it tumble over when it is resting on its end. 139. Unstable Equilibrium. — A pencil balanced on its point, a man on stilts, a bicycle rider, have unstable equilibrium. The least disturbance tends to make them fall. Their centres of gravity are at the highest point possible above their bases, and the bases have no size, being merely points or lines. The 122 PRINCIPLES OF PHYSICS. Fig.TfJI. least disturbance tends to move the centre of gravity from over the base, and the body falls. By changing the point of support and placing it under the centre of gravity, bodies in unstable equilibrium are kept from falling. 140. Raising the Centre of Gravity. — Cut a block of wood in the form of ABCD, Fig. 131. Change the slant at which the block leans, or vary the height till the centre of gravity is just inside >F the base. A slight tip causes the block to tumble over. The block represents, in exaggerated form, the Leaning Tower at Pisa, Italy. Sup- port the block by a wire through a hole at S, and show that S is the centre of gravity. Put another story, EF, on the tower. This raises the centre of gravity, and the line of direction falls outside the base. Does the tower tip over? Find the point by which the tower can be supported so as to balance in any position. 141. The Most Stable Equilibrium of all is that in which the support is above the centre of gravity, as in a swing or a pendulum. To make a body that appears to balance on a needle-point, A, Fig. 132, fasten a wire to a cork, C; attach a weight, W, to the lower end of the wire. , Does the apparatus balance on the point A ? Where is the centre of gravity ? By disturbing W, is the centre of gravity raised or lowered ? Why will a pencil, weighted with a knife, as in Fig. 133, balance on the finger ? Fig. 133. CENTRE OF GRAVITY. 123 Such forms of the pendulum as a hammock or a swing are in stable equilibrium. Why? Show by a diagram how a bicycle could be balanced on a wire, after removing the tire, of course. Bdlt a bar of steel to the frame of the machine, and let the bar extend down and under the wire and carry a heavy weight. Instead of building roads in a rough country, freight is sometimes carried in boxes suspended from a car that runs on a single overhead wire. Fold a card in the form of a V. Invert, and suspend it from a wire. Imagine each side loaded with passengers or freight. The loss by friction is reduced by wheels running on the supporting wire. This form of track and cars has often been suggested. In what kind of equilibrium is it? 142. Centre of Buoyancy. — In boats, the centre of gravity becomes an important consideration. In most boats the centre of gravity is low down, as O, Fig. 134, because of the heavy lead keel. Imagine all the weight concentrated at G. The forces that buoy the boat have likewise a point, B, where they might be considered as concentrated, called the centre of buoy- ancy. This may be thought of as the point of support. This being above the centre of gravity, as in Fig. 134, the boat is in stable equilibrium, and will right itself if tipped so that the sails strike the water. '5 It is as if the weight O were hung from the point B; when swung to one side or the other, it falls again to the perpendicu- lar position, like a pendulum. A match, having a pin for a keel (Fig. 135), represents this type of boat. Increasing the distance BQ in Fig. 134, makes Fig. 134 Fig. 135. 124 PRINCIPLES OF PHYSICS. the boat more stable, or more "stiff." Loading the deck, sending men aloft, or taking out ballast or heavy freight from the bottom of the boat, raises the centre of gravity. What effect does this have on the stability of the boat ? The flat, or skimming-dish, type of boat (Fig. 136, I), is more common in shallow waters. A flat block of wood (Fig. 136, II) may represent this type. The centre of gravity may be at G, and the centre of buoyancy somewhere below G, perhaps at B. The centre of gravity is then practically above the support. The boat is in unstable equilibrium, and tips a little, as shown in III. The centre of gravity is unchanged in position at G. The centre of buoyancy shifts over to C, since more of that part of the boat falls below the water-line than before, and is buoyed up by the water. Re- member that a floating body is always buoyed up by a force equal to the weight of the water it displaces. The two forces, one at G pulling downward and the other at C pushing upward, form a couple that tends to right the boat. I. Bi ■Q- II. 143. Neutral Equilibrium. — Eoll a sphere in any direction. What com- mon objects are spheres ? Roll a barrel or a cylinder (a pencil, for instance) on its side. The centre of gravity is neither raised nor lowered. Ifo amount of push- ing tips It over. It simply revolves, and remains in the posi- tion in which it IS left. A cylinder or sphere has neutral equilibrium. If a body is supported at the centre of gravity in what kind of equilibrium is it? Why? What ela., of objects have neutral equilibrium? CENTRE OF GRAVITY. 125 144. Fall of the Centre of Gravity. — When a body falls, its centre of gravity falls. Fasten a weight, W, Fig. 137, to one side of a round berry box. Make a hole, C, through the centre of the top, and another at the centre of gravity, G. This last may be found by the method of Section 134. Place the box on an incline, in position A. The box appears to roll Fig. 137, up the incline. The centre of gravity does fall as the box moves from position A to position B. Make two little holes, C and D, one in each end of an egg. Blow out the contents. Stop up D with gummed paper. Drop in lead shot and a little liquid glue. Keep the egg in an upright posi- tion {A, Fig. 138) for twenty -four hours. Then try to put the egg down on its side; it at once stands upright again. The centre of gravity in B, Fig. 138, is not over the support S. The centre of gravity ^ falls, being nearer to the f^ =^^=^=:=^— — — ^— — , table in position A. \" S E, Fig. 139, represents a e curved piece of wood, carry- ^'^' ' ^'' ing a long straw, S. When the centre of gravity, O, falls, the straw stands upright. Measure the distances from the centres of gravity in A and B, Fig. 138, to the table. In which case is the distance the greater ? Make a diagram of an inkstand that will not tip over. FigiJ38. 126 PRINCIPLES OF PHYSICS. Exercise 18, CENTRE OF GKAVITT. Apparatus : 2000-g. balance ; objects weighing 500 g. to 2000 g. ; a lever of irregular shape, or a piece ol board loaded at one end by a piece of wood or a clamp. Weigh the lever. Attach a weight ( W, Fig. 140) by a string to a screw in the end of the lever. Balance the lever on a pencil, and mark the spot. A, on which it balances. Record the distance AB and B %-.^ig%i -r,-,r,-rrrrrrr, ,,,,,,,,„■ . j. --Aa B\ w '\ i"-"^ w Fig. 140. Fig. I4[. the amount of the weight W. The downward force of W tends to make the lever rotate in a direction opposite to the hands of a clock. To balance another force, the weight of the lever must act as if applied somewhere to the right of A (as at C, Fig. 141). Then W x AB = AC X weight of lever (call the amount L). W X AB = AC xL AC=^Y^. Balance the lever alone on the pencil, and record distance BD, Fig. 142. Compare this with AB + AC obtained above. For purposes of computation, where ^'^' ' ^^" may the weight of a lever, or of any body, be considered as concentrated ? At B, Fig. 143, hang a body, X, of which the weight is to be found; find the balancing-point E. Consider E as the fulcrum, or turning-point ; find the value XxBE = LxEC J f^i --30 30 diagram (Fig. 145), put- ting the numbers in their places. Calling E the ful- crum, or turning-point, the moment of 30 about that point is 30 x 15 = 450. The force 12 is 5. away from E, the moment is 12 x 5=60. The force Y is 19.5 away from E, the moment is 19.6 Y. As 12 Fig, 145. 128 PRINCIPLES OF PHYSICS. A'*- -*i<- 2 »J Y 10 14 Fig. 146. 'T<-ti— i; these last two tend to make the board turn to the right, their sum, 19.5 r+60 = 450 19.5 T= 390 r= 20. A lever weighing 14 pounds has its centre of gravity 3 feet from A (Fig. 146). At A is hung 10 pounds ; at B, two feet to the right of C, is hung 2 pounds. Find the centre of gravity of the whole. Suppose this point is at E, at a distance d to the right of C. Consider E the balancing- point. 10 X (3 + d) + 14d = 2 (2 - d) 30-f-10d + 14d = 4-2d 26 d = - 26 d = -l. The negative value shows that the supposition that the balancing-point was to the right of C is incorrect, and that the balancing-point is one foot to the left of C. Assume that to be the point, and see if the sum of the moments about it is equal to zero. Another way to deal with such problems is to take the moments about one end, as A, remembering that at E there must be a force upward equal to the sum of 10, 14, and 2. 146. Bodies that are not Uniform in Cross-section. — An iron mast is made of three pieces, 20, 16, and 10 feet long. The iirst piece weighs 80 pounds, the second 60, and the third 40. Find the centre of gravity of the mast. As each sec- tion is uniform, its cen- tre of gravity is at its -20- -10- -16- 80 60 -10- 40 Fig. 147. CENTRE OF GRAVITY. 129 centre. To support the mast at a distance d to the right of A, there must be an upward force equal to the sum of the downward forces (80 + 60 + 40 = 180). The moment of this force is 180 x d. This tends to make it swing upward on A (Fig. 147). The moments of the separate forces tend to make the pole swing in the opposite direction around A. 80 X 10 = 800 60 X 28 = 1680 40 X 41 = 1640 4120 This must be equal to 180 d. 180 d = 4120 d = 22.8 +, the number of feet to the right of A. Problems. 1. Find the centre of gravity of a spindle composed of three sec- tions, the first 8 cm. long, weighing 25 g. ; the second, 6 cm. long, weighing 30 g. ; the third, 12 cm. long, weighing 20 g. Taking the moments about the left-hand end, we have 25 X 4 = 30 X 11 = 20 X 20 = Find the sum of the products. At the centre of gravity, a distance d from the end, a force of (25 -1-30-1- 20) must be applied. Find d. Ans. d = 11. 2. Three weights, 4 g., 12 g., and 10 g., are hung on a weightless stick 20 cm. long. The 4 g. and 10 g. weights are at the ends of the stick ; the 12 g. weight is 6 cm. from the 4 g. weight. Wh^e does the stick balance ? Taking the moments about the left end, where the force of 4 g. is applied, this force has no turning effect. (12 X 6) -I- (10 X 20)= (?X (4 -t- 12-1- 10). Ans. d = \OA. CHAPTER X. WEIGHT AND MASS. 147. Mass means amount of matter. If we wish to know, in a general way, how much there is in a box or package, there are three ways of finding out. Measuring the size and measur- ing the weight are two familiar ways. By measuring the size, the merchant knows how much molasses, grain, or ice cream he is selling to his customers ; in fact, almost all liquids and gasiss, and some solids, are measured and sold by volume, or bulk. Similarly, when we ask for a certain amount of sugar, the grocer measures the amount by weighing it. In each of these cases we are buying a certain amount of matter, or, as the scientist says, a certain mass. In addition to measuring the volume or weighing a body to find its mass, there is a third way. A wagon piled full of boxes looks heavy ; but if it starts easily when the horses pull it, the boxes are either empty or filled with something that contains a small amount of matter. An empty barrel is made to roll or stop by a slight push. The mass or amount of matter that a body contains can be measured by the resistance it offers to being started or stopped. 148. Comparison of Masses. — Select two cars that run easily. Arrange them as shown in Fig. 148. W is, a. grooved wheel, five inches in diaineter, turning on a machine screw in a piece of board. The board is clamped to the table. Fasten the end of a rubber cord to a screw in the centre of the upper edge of the box of each car. The whole length of the cord may be two feet or more. Pull the cars away from the wheel 130 w HEIGHT AND MASS. 131 W, thereby stretching the cord. If one end of the cord is stretched a little tighter than the other, the wheel revolves, distributing the tension evenly between the two ends of the cord. Load one car. Let both cars start together, and notice which car first reaches a mark, M, six inches or jt more from the wheel W. rig. 1 48. Stop the cars before they strike the wheel. Load more heavily the car that gets first to the mark M. When the two cars are loaded so that they both reach the line M at the same instant, weigh each car. Provided the wheels run loosely on their centres, it is unnecessary to tip the board to overcome friction, since the friction is small. A locomotive engineer can tell, by the speed with which his train starts up, if a car has been detached from the train ; for equal forces pushing for the same time on equal masses impart to them the same speed. 149. Change in Weight. — Bodies weigh less near the equator than they do near the poles of the earth. This is because the surface of the earth is a little farther from the centre at the equator than it is at the poles. Besides this, the centrifugal force (see section 169, p. 138), due to the rotation of the earth on its axis, tends to make a body at the equator fly from the earth, like mud from a revolving wheel. If the earth revolved seventeen times as fast as it does, its centrifugal force would be greater than the force of gravity, and bodies actually would fly from its surface. The variation in weight of a body carried from the equator to one of the poles is about one pound in two hundred. A mass weighing 199 pounds at the equator would weigh 200 pounds at either pole. This difference could be de- tected only by a spring balance of some sort, as weights would 132 PEINCIPLES OF PBYSICS. gain or lose, when carried from place to place, equally with the object weighed. 150. Weight as a Measure of Mass. — If sufficient weight, a mass of rock, for instance, were put on a spring balance to make it read 200 pounds at either the north or the south pole, and the balance, with its load, were carried toward the equator, the spring would gradually shorten, and finally read about 199 pounds. The mass of the rock would be unchanged. The springs of a wagon would be nearer together at the poles of the earth than in the neighborhood of the equator. The differ- ence, of course, would be slight, but it could be made evident by using some multiplying device. A box of candy weighs a pound, let us say. Take this box to the surface of the moon ; -there would be just as much candy, — that is, the same mass or amount of matter, — and it would do just as much sweetening, although its weight would be only one-sixth of what it was on the earth's surface, because the moon is a smaller body and attracts objects on its surface less powerfully. Take the same box to the sun's surface ; the amount of matter would remain the same, but the weight would be 27|- pounds. The sun is a much larger body than the earth, and there is more matter in it to attract an object on its surface. On the surface of Vesta, one of the smaller planets, the box would weigh only one-thirtieth of a pound. While the weight of this pound box of candy, which we will call a pound mass, varies with the attraction on this mass, being greater on the sun and less on the moon and smaller planets, yet the force required to set it moving in a horizontal direction, in a given time, at a certain speed, would be the same in each case. A regulation baseball weighs nine ounces. On the sun its weight would be 27^ x 9 ounces ; on Vesta, ^^ of 9 ounces. The weight varies ; still, the force required to pitch this ball with the same swiftness is the same in all cases, for the mass does not change. WEIGHT AND MASS. 133 Mass can be measured by weight, as long as weight is con- stant, or practically so, as on the earth. Outside of the influ- ence of the force of gravitation a body would have no weight, but its mass would remain unchanged. 151. Attractive Force of Different Planets. — The speed, or velocity, acquired by a freely falling body depends on the attraction exerted upon it. On the earth, the velocity of a falling body, while varying in diiferent places, at the end of one second is about 32 feet, or 980 cm., per second. The velocity at the end of one second's fall is 27^ times this on the sun's surface, or 880 feet per second, — the velocity of a rifle ball. The attraction of the moon for bodies on its surface is about one-sixth that at the surface of the earth. A falling body at the end of one second, on the moon, has a velocity of about 6 feet a second, and would fall 2^ feet in the first second. A person on the moon could jump six times as high as on the earth. On the surface, of Vesta the attraction is one-thirtieth of that on the earth's surface. In one second a falling body would drop about six inches, and have a velocity of one foot a second. Human beings cannot exist there, for its attraction is insufficient to retain an atmosphere for any length of time. If they could exist, they would be able to carry 30 times as many bricks or jump 30 times as high as on the earth. 152. Mass and Weight. — The mass of a body, then, — that is, the amount of matter which it contains, — does not change, but the weight of the body depends entirely on the particular planet, or part of the planet, in which the body happens to be. Although the same mass has different weights in different latitudes,, it always has the same weight at any particular latitude. It is on this account that we are able to use the convenient method of comparing masses by weighing them. Mass is the amount of matter a body contains. Weight is the force of attraction pulling down on this mass. The mass remains the same everywhere ; its weight varies. 134 PRINCIPLES OF PHYSICS. 153. Measurement of the Earth's Attraction. — The metric unit of mass is the amount of matter contained in a cubic centimeter of water at a temperature of 40° C. In practice, mass is measured by lumps of brass, iron, or any metal, which are made to conform to a standard unit — a lump of platinum kept by the government. Since the weight of any mass de- pends on the attractive force of the earth, the weight is meas- ured by measuring that attractive force. The greater this force is, the faster it will pull a body toward the earth. It is conveniently measured, therefore, in terms of the velocity of a falling body, and expressed by the velocity such a body acquires in a second. On the surface of the earth this is about 32 feet,' or 980 cm., varying from 978 cm. at the equator to 983 at the poles. Since it is troublesome to measure exactly the velocity a falling body acquires in one second, or even the distance it falls in one second, various ways have been devised to dilute, as it were, the force of gravity and make a body fall more slowly, so that its velocity can be accurately measured. Galileo increased the time of fall by making a ball roll down a hill or incline. 154. Measurement of Velocity.' — If the height BG (Fig. 149) is one foot, this one foot is the distance the body really falls in going the length of the hill AB. Then, if AB equals 10 feet, the body goes 10 feet in falling one foot, and gets up speed one-tenth Pig i^g ^"^ as quickly as if it were falling straight down. Let a heavy mar- ble or steel ball roll down the groove in a matched board for one second, and measure the distance it goes. Start it again, and let it roll two seconds. Notice that the distance gone over in two seconds, in starting at rest, is much more than two times the distance gone in the first second. 1 Change feet to centimeters, by multiplying by 12, to reduce to inches, and, then by 2.54, because there are 2 54 cm. in an inch. WEIGHT AND MASS. 135 This method, first used in the study of falling bodies by Galileo, does not give accurate results, because of the friction of the board and the energy needed to make the ball roll around. 155. Vibration of a Pendulum. — A much more accurate way of measuring the earth's attraction is to measure the length of a simple pendulum and the time of one vibration. A pendu- lum is a falling body (at least the bob, or weight, is), and it moves up and down a sort of double inclined plane. The path is slightly curved, and the weight slides down one hill and up the other, repeating this many times, since the friction of the support is almost nothing and the resistance of the air is slight, because the pendulum moves slowly. By counting the number of vibrations for a long time, the exact time of one vibration is calculated. From this and the length of the pendulum, the velocity acquired in one second by a freely falling ' body is estimated. (See section 455, page 399.) To show that the rate of vibration of a pendulum depends on the downward force acting on it, suspend a weight of about 60 g., B, Fig. 150, by a thread one- fourth of a meter long. Set it vibrating, and count the number of vibrations per minute. Attach a rubber thread, B, fastening it at A. Set B vibrating, and count the number of vibrations for one minute. In- crease the tension on R, if possible, to three times the weight of B, as shown by a spring balance at A. If R is very long, the pull exerted by it will be nearly parallel with the earth's pull. Suppose B weighs Y* 50 g., — that is, the earth's attraction, or downward pull, is 50 g. This was the downward pull on B when L its vibrations were counted without R attached. Make pjg , 50^ the pull of R, as shown by the spring balance, 150 g. Then the total pull on B is 200 g., or four times what it was at first. The pull of the elastic has the same effect on the R f)A 136 PRINCIPLES OF PHYSICS. pendulum as if the latter were swinging on a planet heavier than the earth. Find the number of vibrations in a minute. They should be twice as many as before. Increase the down- ward pull to nine times as much as at first, and the number of vibrations will be three times as great. A pendulum, if carried to different parts of the earth's sur- face, vibrates at different rates. A pendulum clock that keeps correct time at any one place gains time on being carried toward the pole, because the pendulum beats faster. It is easy to count the number of vibrations it makes and to meas- ure its length. In this way the force of the earth's attraction is computed. A pendulum clock that keeps exact time at the equator gains nearly four minutes a day if taken to one of the poles. The discovery that the earth's attraction varies in dif- ferent places was made in 1671, when a clock, taken from Paris to Cayenne, near the equator, lost over two minutes a day till its pendulum was shortened. 156. Mass considered apart from Weight. — If we could take a piece of rock so far off into space that the attraction of the earth or any planet would be reduced to little or nothing, we could then experi- ment on it as a mass of matter having no weight. Instead of doing this, the weight of a body, or its downward attraction, can be neutralized by floating it in water or suspending it by a long thread. Of course the water offers resistance to the motion of anything in it, while a very long suspend- ^ „ ing thread, for the purpose of the experi- ment, is no practical hindrance. Fig. 151. As long as the body A, Fig. 151, is moved a short distance in a horizontal direction, there is practically no resistance outside of A itself to stop it. Give a little pull to the string S, and notice how quickly A begins to start. Pay WEIGHT AND MASS. 137 no attention to the pendulous motion set up, because we are seeking only the force required to start A, or to set it in motion. Try to start A more quickly, and see how strong a string at B can be broken. 157. Setting a Body in Motion. — Suspend a weight. A, Fig. 152, by a string, C, that is but little more than strong enough to support A. Fasten a similar string, B, to A and to a stick, D. B is not long enough to reach to the floor. By pulling down gently on B, C will be broken, because the pull on G is the weight of A plus the pull given to B. Pall down quickly on B, using, if need be, the stick D to give a quick pull, and notice that B breaks. Eeplace B with a string strong enough to hold up several times the weight of A. If the downward pull on B is quick enough, B will be broken in every case. The reason for this is that C stretches a little without breaking, but A resists being set in motion, and this resistance causes the forces to accumulate, as it were, below A, and breaks B before it can act suifi- ciently on A to break C A bullet, if thrown by the A hands, shatters a window- pane, but if shot from a gun it is likely to cut a clean hole. The glass, being elas- tic, can give a little without Fig. I 53. ' breaking ; and as it resists being set in motion so quickly, the bullet cuts its way through before the glass some distance from it is bent far enough to Fig. 152. 138 PRINCIPLES OF PHYSICS. break. A candle can be shot through a board without splitting the board. A card may be snapped out from under a cent, because the resistance of the cent to being started quickly is greater than the friction that would make it move with the card. Place a flathead screw, S, Fig. 153, on a sheet of paper. Strike down sharply on the paper at A, thereby pulling out the paper with- out disturbing the screw. 158. The Resistance a Body offers to being set in Motion may be very little, if the starting is slow ; or it may be very great, many times the weight of the body, if the starting is quick enough. To stop a body quickly requires a greater force than to stop it slowly. The shorter the space of time in which a body is stopped, the. greater the force necessary to stop it. Sand driven by a blast of steam or air makes little impression on paper or on the flesh of the hand, for they yield a little and stop the flying grains of sand more slowly than does a piece of glass or hard steel. These substances are rapidly cut by the sand blast. 159. Centrifugal Force. — Tie a spool to the end of an elastic string and swing it in a circle. The spool pulls harder and harder on the string, stretching it more ^C^v^C *"'-' *^® faster it is swung. In Fig. 154, A represents the elastic string, and B the •^ spool. If A should break, B would go off in a straight line except for the down- ward pull of gravity. As long as the elastic holds, however, it pulls against Fig. 154, ^^^^ tendency and makes the spool swing in a circle. The fact that the spool does have an outward pull is shown by the stretching of the elastic. Every point of a revolving wheel has this same tendency to pull away from the centre. Flywheels, grindstones, and WEIGHT AND MASS. 139 emery wheels fly to pieces when turned too rapidly, because the speed makes the outward pull too strong for the stone or iron to resist ; that is, the parts of the wheel tend so strongly to move in a straight line that the iron or stone is not strong enough to pull them into the circular motion. In Fig. 154, BC is the path the body tends to take, and BD the path the body is made to take by the pull of the string toward A. 160. Inertia. — A body resists being set in motion, or being stopped, or being pulled out of a straight line in which it is moving and made to revolve in a circle. The resistance the body offers in these cases to being started or stopped or pulled out of its line of motion is said to be due to its inertia. If a three-pound mass and a one-pound mass are moving at the same velocity, then the three-pound mass has three times as much inertia as the one-pound mass. The one-pound mass, for instance, has the same inertia if moving at a certain speed, whether it is on the earth and has weight, or is far away and has no weight. It would have inertia anywhere and everywhere. CHAPTER XI. VELOCITY. 161. Average Speed. — A man walks 4 miles an hour for 5 hours. The distance, or space, he goes over is 20 miles, for 20 equals 4 times 6. At one part of the journey he may have gone at a faster speed, or velocity, than at another, but he averaged 4 miles an hour. 162'. Distance. — To find the distance, or space, passed over by a moving body, multiply the average velocity by the time. In a shorter form, this may be written Space = average velocity multiplied by time, or, letting s stand for space, or distance, v for average velocity, and t for time. Problems. 1. If a body moves at an average velocity of 4 cm. a second for 20 seconds, how far does it go? Ans. 80cm. 2. What is the average velocity of a railroad train that goes 480 miles in 420 minutes? Find the distance it goes in one minute. Ans. 1.14+ miles a minute. 3. How long does it take a steamer to go 6000 miles if it averages 22 miles an hour? Aiis. 272+ hours. 163. Average Velocity. — On a uniformly increasing rate of wages a man earns at first 50 cents a day, and later $1.00 a day. His average earnings per day are found by adding $0.50 and $1.00, and dividing by 2. This gives $0.75 as the average. 140 VELOCITY. 141 When a train is slackening speed its velocity is not uniform, but is growing slower and slower each successive second. For example, if its velocity at one moment is 8 feet a second, but at the end of several seconds is reduced to 4 feet a second, its average velocity is (8 + 4) divided by 2. 164. Formula for Average Velocity. — The average velocity is found by adding the velocity at the start and the velocity at the end of the observation and dividing by 2 : — . piop't _ velocity at end plus velocity at start As before, write v for average velocity ; write Vq for velocity at the beginning, or initial velocity, and v^ for velocity at the end of the observation, or final velocity. Putting these abbreviations for the rule above, we have the short way of writing it : — 2 Eead this as v = (v sub one + v sub naught) divided by 2 ; also as Average velocity = {final + initial velocity) divided by 2. In becoming familiar with any new formula, practise reciting and writing it in both the long way and the short way. Vo and Vi are sometimes read "-y-oh," "v-eye." Notice that the three different v's stand for different veloci- ties. To distinguish the three different v's, they might be printed of different colors or sizes, or, as here, have some dis- tinguishing mark added, v^ does not necessarily mean that the velocity is zero, although any one of the velocities may be zero. For instance, if a body starts from rest, the initial velocity is zero; if, at the end of the observation, the body comes to rest, the final velocity equals zero. 142 PRINCIPLES OF PHYSICS. Problems. 1. A train at the top of an incline is going at the rate of 30 miles an hour ; at the bottom it is going at the rate of 70 miles an hour. What is the average speed down the incline ? ^ 70 + 30 _ g^ 2 2. What is the average velocity of a sled dowp a part of a hill, if the velocity at the beginning is 4 feet a second and at the end is 20 feet a second? Ans. 12 feet a second. 3. If initial velocity (that is, velocity at the beginning of the obser- vation) is 30 feet a second, and the body comes to rest, what is the average velocity ? ii= — — — . Ans. 15. 4. Find the average velocity of a sled in passing over a patch of ground, if the velocity at the beginning was 30 feet a second and at the end was 6 feet a second. Ans. 18 feet a second. 165. Formula for Distance. — In the formula S = V X t, sub- stitute the expression for average velocity, 2 Then s=(^^y. This may be read ; Distance equals average velocity times time. Call this important Formula No. 1. A train is moving at a rate of 80 feet a second ; in the course of 40 seconds it is slowed down by the brakes to 20 f^t a second. How far does it go in the 40 seconds ? Substitute, in Formula No. 1, The initial velocity, Vo = 80 Final velocity, Vi = 20 « = 40 s= pQ + ^^ \ 40. s = 50 X 40 = 2000 feet. After the answer to the above problem is known, it may be worked to find the time, assuming that the time is unknown. VELOCITY. 143 Other problems may be made from this one by assuming that the initial velocity or the final velocity is unknown. Problems. 1. K a body, starting from rest, in 10 seconds is moving at the rate of 50 feet a second, how far does it go? Ans. 250 feet. 2. How long does it take a train moving 30 feet a second to stop in a distance of 400 feet? Ans. 26.6 seconds. 3. A body goes 80 feet in 5 seconds and then has a velocity of 10 feet a second ; what was the initial velocity? 4. A bullet is stopped in ^Jj second while penetrating two feet into a wooden block. What was the velocity of the bullet when it struck ? Make up problems similar to the above and solve them. Work them backward. 166. Acceleration. — If a body is moving, at one instant, 10 feet a second, and at another instant 30 feet a second, the gain in velocity is 30 - 10 = 20. Or, Gain in velocity equals the final velocity minus the initial velocity ; or, Gain in velocity equals v^ —Vo. But what we usually wish to know is the gain in velocity in one second. Suppose five seconds elapsed ; then the gain in velocity in one second is | of 20 = 4. Gain in velocity in one second = (final velocity— initial velocity) divided by the time. Gain in velocity per second = — -• The gain in velocity per second is called the acceleration, and the abbreviation for it is the letter a. The formula then becomes t. Gall this Formula No. 2. If the body is stopping, there is a loss of velocity. The gain in velocity in such a case is a minus quantity. 144 PRINCIPLES OF PHYSICS. Problems. 1. What is the gain in velocity per second, if the initial velocity equals 6 feet a second and the final velocity equals 21 feet a second, and the time is 5 seconds ? What is the acceleration ? 21-6 . o a = = Am. 3. 5 2. What is the acceleration of a train, if the initial velocity is 45 feet a second and the final velocity is 15 feet a second, the time being- 6 seconds? Ans. —5. 3. u„ = 80 cm. a second ; Ui = 24; < = 8; a = what? What is the gain in velocity per second ? Ans. — 7. 4. Find the final velocity if u,, = 26, t = 2, a = 12. Ans. 50. Make up a few problems, solve them, and also work them backward. 167. Combination of Formulas. — The two fundamental for- mulas for moving bodies first studied are -^^y (1) a = ^^. (2) Any problems involving distance, time, acceleration, initial velocity, and final velocity can be solved by these formulas. It is more convenient to combine the two formulas and obtain other formulas that have fewer letters. The work of obtain- ing these new formulas is a process of algebra. 168. Elimination of t. — This is done by any of the methods used in algebra, — substitution, addition, subtraction, etc.^ In this special case, as t is in the numerator of one formula and in the denominator of the other, multiply the two equations (1) and (2). 1 Let some pupils try one method, some try another. VELOCITY. 145 Cancel the t and multiply the parentheses together; then multiply both sides by 2. 2as—v^ — v^. Problems. 1. Initial velocity = 10. Final velocity = 30. Distance = 5. What is the acceleration ? 2x5X0 = (30)2 _ (10)2, „ 900 - 100 an o=— j^— = 80. 2. Initial velocity = 20 ; final velocity = 24 ; s = 8 ; a = what? Ans. 11. •^ 3. What must be the velocity of a ball, if it has an acceleration of 2 feet a second, and after 5 seconds has a velocity of 18 feet a second, and goes 65 feet ? Evidently, s = 65, a = 2, ui = 18. Ans. v^ = ?i. 4. A car starts up an incline at a velocity of 16 feet a second ; its acceleration is —2 feet a second. After going 39 feet, what is its velocity? Ans. 10 feet a second. 5. A cannon is 30 feet long; at the instant the powder explodes, the velocity of the projectile is, of course, zero ; the muzzle velocity is 2500 feet a second. What is the acceleration ? ^ns. About 104,000. s = 30, »o = 0, »i = 2.500. In this problem, the pressure of explo- sion, and consequently the gain in velocity inside the cannon, is assumed to be constant, though this is not strictly true. 6. From the data in the preceding problem, find how long it takes the projectile to leave the cannon after explosion. Use the formula s = ''' T^ "° t. Ans. t = yMtt seconds. 169. Elimination of Vq and Vj. — From the fundamental formulas , = ^^L±J^x« and a = '^^^:^, 2 t 146 PRINCIPLES OF PHYSICS. two others can be obtained: one by getting rid of, that is, eliminating, %; the other, by eliminating Vj. The latter formula is a particularly useful one. Try to eliminate Vi from the two equations above, and obtain the formula s = -^+Vo t. When solving the following problems, practise substitution in this formula. Problems. 1. A sled starts down a hill with a velocity of 3 feet a second ; the acceleration is 2 feet a second. Find the distance it goes in 5 seconds. = 25 + 15 = 40 feet. 2. A railway train moving 10 feet a second starts down an in- cline ; the increase in velocity is 1 foot per second. Find the distance it goes in 30 seconds. Ans. 750 feet. 3. The initial velocity is 10 cm. a second ; the acceleration is 10. Find the distance traversed in 3 seconds. 4. Call the initial velocity ; that is, the body starts from rest. Show that the formula becomes s = — 2 5. Use this formula to find the distance a body falls from rest in 3 seconds, when a = 32 feet. Ans. 144 feet. 6. Find the distance a weight falls in 4 seconds ; in 2 seconds ; in 21 seconds ; in 1 J seconds ; in 5 seconds. 7. Find how much time is required for a body to fall 64 feet, when a = 32 feet. Ans. 2 seconds. 8. How high is a tower, if a bullet takes 1^ seconds to fall to the ground? Ans. 36 feet. 9. How long would a body take to fall to the ground from a balloon 900 feet above the surface of the earth? Ans. 7 J seconds. 10. How long does it take for a weight to fall 16 feet? 1 foot'/ 4 feet? 400 feet? J foot? VELOCITY. 147 11. A baseball is thrown straight up in the air. A person stai'ts to count seconds as it begins to fall. lu 3 seconds it reaches the ground. How high up did the ball go ? 12. As an icicle melts, two drops of water fall, the first ^ of a second before the other. How far apart are they when the second one falls? Ans. ^ foot. 13. Using the metric system, acceleration = 980 cm. Find the number of centimeters a body falls in 6 seconds ; | second. 170. Acceleration of Falling Bodies. — The acceleration, or gain in velocity, made by a falling body in 1 second is about 32 + feet, or 980 cm., varying a little on different parts of the earth's surface. A falling body every second increases its speed by about 32 feet. The exact amount of increase in any place where experiments are conducted is well worth knowing, and is best obtained by finding the distance a body falls in a measured time, and then computing by the formula s = — . The difficulty is in measuring the time accurately. In sections 154 and 155 there are described several methods (the inclined plane, the pendulum, etc.) of so "diluting" the attraction of the earth and increasing the time of fall that an accurate measure of time can be made. From this the velocity gained per second on the acceleration is computed. 171. The Dyne. — Since the weight of a lump of any sub- stance varies in different parts of the world, the weight of a body — that is, the downward force it exerts because of the attraction of the earth upon it — cannot be used as an accurate standard in scientific work. Make a cube of wood, 1 cm. on an edge. If this does not weigh a gram, bore a small hole in it and put in one or more shot, as needed to bring the weight to 1 g. Close the hole with paraffin. Suspend the cube by a long thread. The block contains the same amount of matter as is contained in 148 PSINCIPLES OF PHYSICS. a cubic centimeter of water and occupies the same amount of space. The force that pushing for one second on this mass gives it a velocity of 1 cm. a second is the same the universe over. This force is called a dyne — a word coined by scientists from a Greek word like it, meaning force. 172. The Dyne compared with the Gram Force. — A dyne, then, is a force, that, acting on a gram mass for one second, gives it a velocity of 1 cm. a second. But this little block weighs a gram ; how, then, does the gram force compare with the dyne ? In one second, one dyne would give this one gram a velocity of 1 cm. a second. Suppose the block is dropped out of the window. At the end of one second it has a velocity of about 980 cm. a second, — a little less at the equator, a little more at the poles. The force of the earth's attraction on it — a force that we call a gram force, or a gram — is 980 times that of a dyne. There are from 978 to 981 dynes in a gram, accord- ing to the locality in which the gram force is measured. A dyne force is extremely small. The weight of a mosquito can be scarcely felt, but it is a dyne, or more. Problems. 1. What is the velocity acquired by a mass of 1 g. acted on by a force of 1 dyne for 3 seconds? Ans. 3 cm. per second. 2. What is the velocity acquired by a mass of 6 g. acted on by a force of 1 dyne for 1 second ? Ans. ^ cm. per second. 3. What is the velocity acquired by a mass of 4 g. acted on by a force of 12 dynes for 3 seconds? . 12 x 3 „ , ■' Ans. = 9 cm. per second. 4. Find the speed a 1-gram lump attains if a force of 1 dyne acts on it for 1 second ; for 4 seconds. A ns. 1 cm. per second ; 4 cm. per second. 5. Find the velocity a force of 6 dynes gives to a mass of 2 g. in 3 seconds. Ans. 9 cm. per second. VELOCITY. 149 173. Velocity in Terms of Force and Time. — The velocity is made greater by increasing the force. The greater the force applied to a boat, the quicker it starts ; the longer the force acts, the greater the velocity, or speed, attained. In these two vrays, then, by increasing the force and the time of applying that force, more velocity is acquired. But by increasing the mass to be moved, the velocity gained is made less. A heavy train gets up speed slowly. Velocity equals force multiplied by time and divided by mass ; ft or, vz=J— m In this formula, m stands for grams of mass, t for time in seconds, v for velocity in centimeters per second, and / for the force in dynes. Problems. 1. What velocity will a force of 8 dynes give to 2 g. in 12 seconds? ^ _ 8 x 12 ^ ^^j_ 48 cm. per second. 2 ^ 2. How fast will a bullet, weighing 10 g., go, after being pushed by a force of 20 dynes for 6 seconds? Ans. 12 cm. per second. 3. What force is required to give 100 g. a velocity of 20,000 cm. a second in sV of a second ? Substituting, 20,000=-^^^^; then /= 100,000,000. The answer is in dynes ; change this to grams by dividing by 980 or 1000. 4. How long must 20 dynes act on 80 g. to give a velocity of 120cm. a second? 120 = — . t = 480 seconds. 80 From each of the above questions make others, letting each quan- tity in turn be the unknown one. For instance, Problem 1 might become : How long does it take a force of 8 dynes to make 2 g. move with a velocity of 48 cm. per second ? 150 PRINCIPLES OF PHYSICS. In Problems 5 to 14, inclusive, below, find the value of the unknown quantity. V t m 5. 1 1 1 6. 6 1 1 7. 1 4 1 8. 1 1 2 9. 1 1 .5 10. 20 o 3 11. 10 .1 2000 12. 90 4 5 13. .2 8 60 14. 980 1 1 15. What force is always nearly 980 dynes ? 174. Conversion of Grams to Dynes. — If a force is given in grams, change to dynes by multiplying by 980 before substi- tuting in the formula. If the answer is required in grams, divide the number of dynes by 980. 175. Formula for Force. shown in section 166, is -The formula for acceleration, as a = ^'-^°. If the initial velocity is zero, that is, if the body starts from rest, ■Wo = 0, then v^, being zero, drops out, and the formula becomes a =— • Since there is now only one v in the for- VELOCITY. 151 mula, there is no need of any mark to distinguish it from any- other V. Therefore, instead of vi, let us write v. The for- mula then is t Multiply both sides by t ; then v = at. Substitute at, which is the value of v, in place of v in the following formula : — m then m Cancel the t'a : aU, m or /= ma, an important formula in considering projectiles, throwing and stopping a baseball, starting and stopping a railway train, etc. Problems. 1. Find the force in dynes that will give 40 g. an acceleration of 10cm. a second. /= 40 x 10. Ans. 400. 2. What force must be applied to a mass of 200 g. to cause the body to go 3 cm. per second faster each second ? Ans. 600. 3. What is the gain in velocity per second caused by a force of 24 dynes on 4 g. ? Ans. 6 cm. per second. 4. If the acceleration of a falling body is 980 cm. per second, how many dynes is the force that the earth exerts on a gram mass ? 5. An artesian well spouts a stream of water 25 feet high ; what is the velocity of the water at the mouth of the pipe? Acceleration = 32. Use the formula 2 as = v^ (section 168). Ans. u = 40 feet per second. 6. If the diameter of the jet of water of Problera 5 is 3'j inches, the cross-section is about 10 square inches. (Calculate it.) How many cubic inches of water will flow per second? Find the number of gallons of water per second by dividing by the number of cubic inches of water in a gallon. 152 PRINCIPLES OF PHYSICS. 176. Multiply the formula, section 168, page 145, 2 as =?)" by /= ma term by term.' Then 2asf=mav' J. mv' which is another important formula in studying projectiles, etc. 177. An Erg. — In section 119, work was defined as making a body move against resistance, and formulated as Force times distance through which the force acts. This is what is meant by fs. Therefore fs in the formula stands for work that has been done or can be done. The other side of the equation, exactly equal to fs, is the expression -- — This latter repre- sents the work a moving body can do when it stops. Since the force is measured in dynes and the distance in centi- meters, the work is measured in dyne-centimeters. A dyne- centimeter is usually called an erg. This is the work done by pushing or pulling a distance of 1 cm. with a force of 1 dyne. If there are 980 dynes in a gram force, how many dyne-centi- meters are there in a gram-centimeter ? How many ergs ? What energy does a 20-gram rifle ball have when moving 10 cm. a second ? Substituting in the formula -^, we get -^ = 1000 dyne-centimeters, or ergs. Square the velocity, multiply by the mass, and divide by 2. fs can be read as "force times distance through which the force acts." 1 To be convinced that this is possible, consider the two equations 2 = 2, 3 = 3. The equality is not destroyed by multiplying the first two and the last two terms together ; for then 6 = 6. VELOCITY. 163 Suppose the question is : What force must be used to stop the ball in 1 cm. ? Then, / x 1 = 1000, and / is 1000 dynes. If the body moved 5 cm. while stopping, then s = 5, and / x 5 = 1000 ; /= 200. The force required over the longer distance is only one-fifth as great as before. Problems. 1. Find the energy of : a. A 20-gram ballet going 60,000 cm. a second. b. A 300-gram baseball moving 1500 cm. a second. c. A 6,000,000-gram electric car moving 600 cm. a second. d. A 50-gram weight moving 5 cm. a second. e. A 500-gram hammer moving 600 cm. a second. 2. Find the force required to stop each body in Problem 1, in 1cm.; in 2 cm.; in 10 cm.; in 1000 cm.; in .001cm. Notice that the force required increases enormously if the body is stopped in a short space. Why is the blow of a hammer on a solid, unyielding body more severe than on a flimsy stick ? 3. Find the force required to stop each body in Problem 1, in 3cm.; in 4 cm.; in 500cm.; in .005cm. 4. a. What is the energy of a 60-gram arrow moving 200 cm. a second? Ans. 1,200,000 dyne-centimeters, or ergs. 6. What force is required to stop it in 1 cm. ? Ans. 1,200,000 dynes. c. What force is required to stop it in 30 cm. ? ^ns. 40,000 dynes. d. What force must the string have applied to the arrow to set it going at a velocity of 200 cm. a second, the string moving 30 em. ? Ans. 40,000 dynes. 5. a. A force acts in the barrel of a rifle 60 cm. long ; the bullet is 20 g., and its velocity on leaving the muzzle is 60,000 cm. a second. What is its energy? Ans. 36,000,000,000 ergs. b. What is the average force exerted by the powder on the bullet? Ans. /= 600,000,000 dynes. Since the barrel is 60 cm. long, s = 60, / x 60 = 36,000,000,000. Change the force to grams, by dividing the answer by 980. 154 PRINCIPLES OF PHYSICS. 6. A 300-gram ball is thrown with a velocity of 1200 cm. a second. Find: a. Its energy. b. How much work it can do in stopping. c. What the force is with which it presses on the catcher's hands when he stops it gradually in 20 cm. (The catcher moves his hands back.) 7. A bicycle and rider weigh 50,000 g. They move at a speed of 1000cm. a second. Find: a. The energy. Ans. 25,000,000,000 ergs. b. The force necessary to stop them in 1 meter. c. The force necessary to stop them in 10 cm. d. The force necessary to stop them in 10 meters. 8. How high up a hill could the same rider coast? Since the energy is 25,000,000,000 dyne-centimeters, and the gram is 980 limes as great as the dyne, the energy in gram-centimeters will be flij as much, or, roughly, 25,000,000, and this will raise 50,000 g., ^'^'W '^™-> 0'' 250 cm. 9. A 1-gram mass moves with a velocity of 4 cm. a second. Find: a. Its energy. b. The amount of force that will stop it in 1 cm. c. The amount of force that will stop it in 10 cm. d. The height to which this energy will lift it. 10. Find: a. The energy of 1 g. moving 1 cm. a second. b. The energy of 1 g. moving 2 cm. a second. c. The energy of 1 g. moving 3 cm. a second. d. What effect does doubling the velocity have on the energy? 11. By use of smokeless powder and better cannon, the velocity of projectiles has been increased during the last century three to five times. How much has the energy of the moving projectile been increased 1 12. a. How many times greater is the energy of a train moving 100 miles an hour than that of one moving 50 miles an hour? Ans. 4 times greater. b. If the force that the brakes can exert to stop each train is the VELOCITY. 155 same, how much farther will the first train go, with the brakes on, than the second? Ans. 4 times as far. c. State the objection to a railroad speed of 200 miles an hour. 13. Which has the greater energy, and which will strike the harder blow, a 100-pound shot having a velocity of 600 feet a second, or a 10-pound shot moving 2400 feet a second ? This illustrates the relative effectiveness of the artillery of the Revolu- tion and of the present day. 178. Reaction. — If a person sitting in a svi^ing throws a heavy weight in a horizontal direction, the swing starts to move in the opposite direction. If a man jumps from the bow of a light boat to a wharf, the boat moves backward a little. From a heavy boat, a person can jump ashore without causing any perceptible motion to the boat. If one were to jump from successively lighter boats, the same effort to jump would appear to affect him less and the boat more. A gun "kicks," or moves backward, at the same instant that the bullet is being driven forward. Newton states this in his third law of motion: To every action there is an equal and opposite reaction. 179. Momentum of a Body. — In every instance the lighter body has the greater velocity. If the gun and the bullet were of the same weight, the velocity of the gun and the bullet would be equal. It has been found that even when the bullet and the rifle have different velocities and different masses, the mass of the rifle times its velocity backward equals the mass of the bullet times its velocity. Suppose the bullet in a toy cannon weighs 1 gram and the cannon weighs 100 g. On firing the powder, the bullet has 100 times as much velocity forward as the cannon has in the opposite direction. The mass of the cannon times its velocity equals the mass of bullet times its velocity. Mass times velocity (or m times v, or mv) is called the momentum of a body. The word " momentum " expresses something about a moving body that is more imaginary than real, but it is a convenient term to use in some problems. 156 PBINCIPLES OF PHrSICS. For experiments in momenta, boats in water or cars on a track have too much friction, and. there is too much difficulty in measuring the velocity. Experiments can be made, how- ever, with bodies suspended as are A and B, Fig. 155. Sup- pose a spring on A, which is pressing against B, is let go. A and B will separate. The larger ball, B, acquires the less velocity, and swings a shorter dis- tance than A. In fact, we must assume, what is dif- ficult to prove, that the velocity the pendulum A ot B A^^ has at the lowest point of the swing is proportional ^ B to the length of the swing it makes. This is nearly Fig. 1 55. true, if the swing is not too great. If either A or B swings 20 cm. before or after reaching the lowest point of the swing, then the velocity at the lowest point is twice as great as if the swing has been 10 cm. Instead, there- fore, of trying to measure directly the velocities of the bodies A and B before and after they strike, we have only to find the distance each swings before and after they strike. Exercise 19. ACTION AND REACTION. Apparatus : Two ivory or wooden balls, one larger than the other ; meter rod on base-board; two swinging stops, fitted to board, and arranged so that by pulling a string both balls are released at the same instant ; a board 8 inches wide and 12 inches long, with bevelled edges, and some means of supporting it six or niore feet above the table ; linen thread for suspensions. Pass the thread through the holes \. (>\ \j \ in the balls and over the support at C and D (Fig. 156). Make the distance between C and D such that A and B just touch. The distance between C and D equals the radius of A plus the n radius of B. Adjust the height of A [~Y '^''~\U' and B so that their centres are on the | '''''''''' '''''™ same horizontal level. This can be Fig. 1 56, done by tying knots in the suspensions to shorten them. VELOCITY. 157 Case I. — Record in a note-book on a diagram (Fig. 157): 1st, the distance, BC, that B is drawn back; 2d, the distance, BF, that B goes after collision; 3d, the distance AE, that A goes after collision. Record the two points on the meter rod that are under the centres of A and B. Record these, as well as all other measurements, on a diagram like Fig. 157, in note-book. Draw B back 20 cm., holding it in position by the stop. Release, and try to slide a block of wood, holding a ? "^^S. ^ card, so that the centre of B at its point of ©" ©(jy * farthest swing to the left is just in line F* — V—AB with the card. Bring A to rest, and repeat pig, 1.57. the trial, changing the position of the card, if necessary. Record on diagram the position of B and the distance it swings from B to F. While doing this, pay no attention to the distance A swings. Then, in the same way, find how far A swings from A to E. Weigh A , to determine its mass in grams ; weigh B. Before collision. A, being at rest, has no momentum. The weight, or mass, of B times the distance, BC, that it swings may be called its momentum before striking A. This is the momentum before colli- sion. At that instant B gives up, as it were, a part of its momentum to A. Compute the momentum of A after collision ; then of B. Mul- tiply the mass of A by the distance it swings, and the mass of B by the distance it swings after collision. Add these two products to- gether, and see how nearly the total momentum before collision equals the total momentum after collision. As we have assumed that the velocity of a pendulum at the lowest point of its swing depends on the distance it swings before reaching the lowest point, the dis- tances we have recorded represent velocities. Representing the mass of A by the letter A and the mass of B by the letter B, does B X distance BC =(Bx distance BF) + (Ax distance AE), Momentum before striking = momentum after striking ? Case //. — Pull A back 30 cm., and let it strike B at rest. Make a diagram (Fig. 158), and record on it the , >,S^ X^ position from which A starts and the (^ ®Cv ^D distance AE. Record also the position F*—*i I '^ of B before collision and the positions ^^' ■ B and A after collision and the distances they move. On striking B, A not only gives up all its momentum, 158 PRINCIPLES OF PHYSICS. but more than it has, and goes in debt for that extra amount, and on this account swings backward with a velocity that carries it to F. The momentum before collision = ^ x distance A E. The momentum after collision = (B x distance BC) — {Ax distance AF). The sign of this last product is a minus one. Either for the reason stated before, or because A and B after collision have velocities in opposite directions, the momentum of one body must be a negative one. Case III. — Draw A and B back 15 to 25 cm., and release them at exactly the same instant. A and B need not be drawn back the same distance. Record on diagram like Fig. E A B C 159 the distances, EA and BC, that (b ^yO^ C^ -^ ^^^ ^ S° before striking together; j,_^ _J u-._Tf- T^ also the distances, AF and BK, they go Fig. 1 59. after rebounding. Mark the weights on the outlines of A and B in the dia- gram. If A is very much lighter than B and is not drawn btiok a greater distance, then B may not fly back after striking, but may continue on. Compute the momentum of each ball before striking. As they are moving in opposite directions, one value will be plus and the other minus. Call the momentum plus if the body moves to the right. Find the total momentum before collision by adding the momentum of A (a plus value) to the momentum of B (a minus value). If A = 100 g., B = 200 g., AE = 30 cm., BC = 20 cm., then the momen- tum of A is -I- 100 X 30 = 3000, the momentum of B is - 200 x 20 = - 4000, and the algebraic sum is — 1000. Compute the total momentum after collision. If A moves toward F, then its momentum is minus; if toward K, its momentum is pins. Add the two values just as you would in algebra, and call the result the momentum after collision. How do the momenta before and after collision compare ? While these should be the same, the errors in measurement may make them differ by several hundred units. Even that amount is not large compared with the momentum of one of the balls, for instance (200 x 20 = 4000). VELOCITY. 159 180. Inelastic Balls. — Wooden or ivory balls are more or less elastic. To make them inelastic (that is, so there is no bounce), put a band of putty around one ball. Figure 160 shows it around B. Draw back AtoE and release. Measure the distance the balls go after collision ; this distance "will be the same for both, since they pg, leo. will be held together by the putty. The total momentum before striking = mass A x distance AE. The total momentum after striking = (masses A+ B and putty) X distance BC. 181. Recoil. — If a person standing in the bow of a boat throws weights forward,' he will cause the boat to move back- ward. If he throws them backward from the stern, the boat will move forward. There is an instance reported of a man- of-war pursuing another and losing ground with every shot fired from her bow guns, while the other, firing from the stern, finally gained enough to enable her to escape. If a vessel, complete, weighs 60 tons (100,000 pounds), and a 100-pound shot leaves the muzzle of the gun with a velocity of 800 feet a second, how much is the speed of the ship increased or decreased. The process is : 100,000 D = 100 X 800 ; v=: .8 feet per second. The shot,-then, has an instantaneous tendency to decrease the speed of the vessel .8 feet per second; that is, if the ship were sailing .8 feet per second, the recoil of the shot would bring her to a standstill. Continued and rapid firing from a vessel's guns in the direction in which she sails would per- ceptibly retard her progress. Make examples similar to the above. 160 PRINCIPLES OF PHYSICS. Problems. 1. A toy cannon weighing 200 g. is suspended by a long string. On firing, a 10-gram bullet has, at the mouth of the cannon, a Velocity of 5000 cm. a second. What is the velocity of the cannon backward ? The formula is mv = m^Vj, or the mass times the velocity of one part (the bullet) = the mass times the velocity of the other part. 200 V = 5000 X 10. Ans. u = 250 cm. a second. 2. A cannon is mounted on a car; both together weigh 10,000 pounds. If the velocity backward is 2.5 feet a second just after firing, what is the velocity of the projectile weighing 20 pounds? Ans. 1250 feet a second. 3. A form of lifeboat is driven by the reaction caused by pumping water out of a pipe at the stern. If the boat weighs 20,000 pounds, and the pumps send 2500 pounds of water per minute sternward with a velocity of 60 feet a second, what is the velocity the boat would gain in 1 minute, supposing there is no resistance whatever to the movement of the boat? Ans. 1\ feet per second. There is, o£ course, much resistance ofiered by the water, etc., and the velocity given above would not be obtained. Sucli a boat has little machinery to break down, but is very inefficient, because it requires a large boiler and pump to drive the boat at a slow speed. 4. A ball weighing 150 g. is moving 20 cm. a second, and strikes another ball, weighing 200 g., at rest. The first ball flies back with a velocity of 4 cm. a second. Find the velocity of the second ball. Ans. 18 cm. a second. The smaller ball not only gives all the momentum it has, but gives up more and acquires velocity backward. Momentum before striking = momentum after striking. 150 X 20 = - (150 X 4) + 200 v. 5. A, Fig. 159, p. 158, weighs 50 g., and swings 30 cm. before col- lision ; B weighs 150 g., and swings 25 cm. before collision. After collision A swings back 35 era.; how far does B swing? Letting D represent the distance B swings, (50 X 30) - (150 X 25) = - (50 x 35) + 150 i) ; D = - 3.3 to the left. VELOCITY. 161 182. Momentum, starting from Zero. — The experiment of Exercise 19, page 156, maybe varied by using flat boards, each sus- pended by four strings or wires. Place a coiled spring, tied with a thread, between the boards. When the thread is burned, A and B swing apart (Fig. 161), the smaller mass. A, having the greater ve- p. locity. The momentum of -4 = the momentum of B. In this case the momentum before start- ing is zero, for neither pendulum was moving. 183. Examples of Reaction. — A powerful stream of water from a fire-hose causes such a reaction that the united strength of two men is needed to direct the stream. In a small hand- hose the reaction can be felt as the water is turned on sud- denly. Lawn sprinklers are driven by the reaction force given the pipe by the water as it acquires velocity in coming out of the nozzle. This principle can be easily demonstrated by means of a glass tube with a nibber tube attached to each end. Connect one rubber tube with a faucet and allow the glass tube to fill slowly with water. When the tube is full, opening the faucet suddenly will make it swing backward. CHAPTER XII. E 3^ ELASTIOITT. Exercise 20. STREICHmO. Apparatus : A coil of more or less elastic wire, such as No. 30 brass, iron, steel, bronze, or aluminum; clamps; spring balance; a light pointer a foot or more long; a scale; an upright board, with a vertical groove in which a clamp can slide freely. Fasten one clamp to the top of the board (A, Fig. 162). Double a piece of wire of any convenient length more than two meters, and attach the loop to the clamp. Pass the two free ends of the wire through two holes near one end of the pointer and through the clamps S and C. Fasten the clamp B to the wire, as shown in the figure. Fasten the clamp C to the wire in such a position that the tip of the pointer is nearly at the top of the scale. Below C attach a spring balance or a scale-pan, D. Adjust the clamp C in the groove. The pointer is now adjusted to act as a lever. B is its fulcrum, and if BS is 10 times BC, a vertical motion at C is magnified 10 times at S. Determine the error of the balance for the posi- tion in which it is to be used (see section 114, page 102), and use the error in correcting results. Record the reading of S with no pull on A C. With the balance D apply a 200-g. force to AC; read S. Lift D, and find the zero load reading of S again. Then apply 400 g., 600 g., 800 g., etc., remov- ing the load each time to get the reading for no load. Cease to increase the load when the 162 Fig. 162. ELASTICITY. 163 pointer does not retuin to the first reading. Measure the diameter of the wire.i "When the wire begins to stretch permanently, its elastic limit has been passed, and any further increase in the force applied merely increases the permanent stretching. This, of course, makes the wire grow smaller in cross-section, just as a piece of candy, when " pulled," grows longer, but smaller in diameter. Find the elastic limit of soft copper, aluminum, or brass wire. The wire can be made soft, if not already so, by drawing it quickly through a hot flame. Fasten one end of the wire to a clamp on the edge of a table ; to the other end of the wire attach a spring balance. Put a paper pointer on the wire near the balance, and mark its position, with chalk, on the table. Measure the diameter of the wire. Stretch the wire by successively increasing forces. After each stretching, release the wire and mark the zero point. Notice the amount of permanent stretching. Try the same experiment with fuse wire, which is made largely of lead or tin. It will be found to have very slight elasticity. From the data in this exercise, what has been determined about : 1. The eifect that doubling the load has on the stretching? 2. The eifect of hardness on elasticity ? 184. Experiments on Stretching. — Compare soft annealed iron wire and spring steel wire, or hard spring brass and annealed brass. Compute the amount the wire stretches under a given load, remembering that the pointer magnifies the stretching. Look in a table of the areas of circles in the Appendix, find the area of the cross-section of the wire used, and calculate how many of these wires together would equal 1 sq. cm. If you wanted to stretch a wire with a cross-section of 1 sq. cm. to the same extent you stretched the No. 30 wire, how much greater force would you have to use ? 185. Young's Modulus. — When comparing the behavior of two different wires, it is convenient to know how much a 1 In case there are several pieces of apparatus, put a different kind of wire in each. The pupils may, in rotation, test the stretching of different speci- mens in a short time. V \ A 164 PBINCIPLES OF PHYSICS. piece of 1 sq. cm. cross-section and 1 cm. long would stretch or compress if a force of 1 g. were applied at the ends A and B (Fig. 163). Another way of stating the problem is : What force must be applied to A and B to make it stretch so much as to double its length ? p. I ^ This is absolutely impossible, but the number of grams necessary to do it, if it were possible, can be calculated. The number is a large one, and varies for different kinds of metal ; it is known as Young's Modulus, or measure of elasticity. 186. Stretching. — We have found that doubling the force applied to a wire produces double the amount of stretching; and, evidently, doubling the area of cross-section of a wire decreases the amount of stretching by one-half. Further, a wire 2 m. long will stretch twice as much as a wire 1 m. long under the influence of the same pull, for each half of the longer wire will stretch as much as the shorter one. Wires of different materials stretch different amounts. 187. Breaking Strength. — The strength of materials used in houses, ships, bicycles, and bridges must be known, or they would be built so heavy as to be clumsy, or so light as to be dangerous. Test the breaking strength, or the force required to break pieces of twine, thread, fishline, and fine wire. Exercise 21. BREAKING STRENGTH. Apparatus : A testing-machine, consisting of an extensible frame, a wedge, spring balance of 2000 g. capacity, ratchet, and crank, arranged as shown in Fig. 164 ; wires of brass, copper, aluminum, and soft iron (No. 27 is a useful size). The wire, stretched by the crank and ratchet, moves the frame and registers on the spring balance the pull used. As the frame moves, the wedge W drops down and holds it, when the wire breaks, preventing the recoil of the balance. The rings xx, Fig. 164, are attached to the frame and are stationary. Select a piece of wire without sharp bends or kinks. Wind one end three times around the end of the frame in which the wedge W ELASTICITY. 166 (Fig. 164) fits. Fasten the other end to the axle of the ratchet. Insert the wedge, as shown ; set the pawl to engage the ratchet wheel. Slowly turn the handle. If the wedge does not follow the movement of the movable frame, put a small weight on top of it. When the wire breaks, the wedge holds the hook of the balance from flying back. Record the reading of the balance. Take hold of the movable frame and release and lift out the wedge. Let the frame go back slowly. Make several trials, using each time a fresh piece of wire. The average is approximately the breaking strength of the wire used. Measure the diameter of a fresh piece of wu'e with a micrometer caliper. Calculate the area of the cross-section. The rule is : Square the diameter, and multiply by \ ir, or by .785. For instance, if a wire }o-&< >o Fig. 164. measured .5 mm. (this may be written .05 cm., the square of this = .0025), thus multiplying by .785, we have as the area .00196, or about .002 sq. cm., or about 3^ of a square centimeter. Suppose the break- ing strength of a wire of this diameter is 12 kg. The strength of a wire having an area of 1 sq. cm. cross-section would be 500 times as great, or 6000 kg. To find the breaking strength per square centi- meter, divide the breaking strength of the wire used by the area of its 12 , .002' cross-section. = 6000. 188. Experiments on Breaking. — Compare the breaking strengths of other kinds of wire, strings, etc. Notice that violin strings are often stretched to the breaking point. For sounding great depths in the ocean, Lord Kelvin iirst used piano wire. Which do you find the stronger for pieces of the same length and weight, piano wire or string ? Recently piano wire has been used in kite flying. Compare the strength of 166 PRINCIPLES OF PHYSICS. steel wire with that of soft iron wire. Soften a piece of brass or hard-drawn copper wire, and compare its breaking strength with that of an unannealed piece. Measure the diameter before and after breaking. The wire may be annealed by heating it slightly with a candle. Some idea of the amount of permanent stretching, or elongation, is found by marking with an ink pencil several places 20 cm. or more apart. Measure the length between two of these marks between which the wire does not happen to break. 189. Strength of Fine Wire. — If of two wires of the same material, one has twice the diameter of the other and four times the cross-section, it might be expected to stand four times the pull before breaking. Such, however, is not always the case ; because, as a wire is drawn finer and finer, the metal becomes stronger. A rope of fine wire is stronger than a rod made of the same amount of metal. The separate strands of a wire rope should be capable of stretching a little before break- ing. Before the most tightly stretched strand receives a pull big enough to break it, it stretches a little, and the other strands receive their share of the load. In well made rope all strands are wound equally tight. Problems. 1. If No. 20 copper wire breaks at 28 pounds, how much should a No. 10 wive stand? As the diameters (see Table, page 539) are in the proportion of nearly 3 to 10, the amount of metal in the second is — , or , or 11 times as 32 9 great. The larger wire will break at somewhat less than 11 x 28 pounds. 2. A piece of No. 24 iron wire breaks at 22 pounds ; what should No. 30 wire break at ? 3. Compare the strength of two similar wires, one .1 of an inch and the other .3 of an inch in diameter. 4. How many times stronger is a rod .7 cm. diameter than a simi- lar one .2 cm. diameter? Ans. ^ = 12.1. ELASTICITY. 167 190. Factor of Safety. — In the construction of bridges, ma- chinei-y, etc., enough material must be used so that the strength at any spot is always much greater than the force that is to be applied there, greater even than any force that would cause it to be permanently stretched. Enough metal, wood, or other material is used to withstand a force from 4 times (in the case of metal) to 20 times (in the case of wood) as great as is ever to be applied. When a beam breaks at 4 times the work- ing load, we say that the factor of safety is 4. 191. Bending. — Support a match on two other matches, as in A, Fig. 165. Press on the centre with the butt of a lead- pencil, and* notice the bending. Try the effect of doubling the width by using two ■'-' a matches joined together, laid flatwise, as ^ "1 in B. Work a little glue into the opening between the two matches ; on drying, the ^2 ^^ ' G two act as one piece of wood. Try the Fig. I 65. bending of this, first on its side and then on its edge. Shave a match down to the shape of G. Try the bending of this on its side ; then on its edges. Try the effect of reducing by one-half the distance between the supports. Which is the stiffer, or resists the downward pressure more, a board on its side or on its edge? Why are iron or wood floor beams laid on their edges and not on their sides ? Set a bit of card 2 cm. wide and 6 cm. long on edge, on [q j(__ 0=3 ij^q supports. Apply a slight force to make it ■^ .■^ ^ bend downward. Notice that it bends somewhat '^' " sideways, and as a result sags down. Bend a strip of card to the shape a cross-section of which is like A, Fig. 166 ; test its stiffness. The horizontal part hinders the verti- cal part from bending sideways. Why is a bicycle frame that is built of tubing stiffer than one of solid rod of the same weight ? Compare the stiffness of a paper mailing tube, a straw, and a tin tube, and the same 168 PRINCIPLES OF PHYSICS. materials in flat section. Explain why a T-rail (B, Fig. 166) is less likely to bend than the same amount of steel in a fiat rail {C, Fig. 166). 192. A Support, or "Girder," forming part of a bridge or the floor of a building, is sometimes supported merely at the ends, as the match is in Fig. 165, A. In other cases, one or both ends are rigidly fastened, thus increasing the stiffness. The timbers supporting the floors of modern houses are deeper than those used in old-fashioned houses. In which are the floors the more springy ? The laws of bending are most easily studied with beams free to move at both ends. Exercise 22. (a) BENDING. Apparatus : A wooden rod somewhat more than a meter in length and about i/i inch square ; two triangular supports ; a micrometer screw (M, Fig. 167), so placed that it can be screwed down to touch a tack in the centre of the rod ; weights of from 100 to 500 grams ; any convenient arrangement for suspending weights from the centre of the rod. The tack and microm- eter screw are connected witli a battery, the circuit of which also includes a bell ; a sounder, lamp, or galvanometer may be used instead of a bell. Case I. — Screw down M until the fluttering action of the bell shows that contact is j ust made between the end of the mi- crometer screw and the tack. Read the micrometer, and call this the reading with no load. Put a 100-g. load, W, on the rod, and read the micrometer. Turn the micrometer screw back and remove the load. Fig. 167. Find the reading at zero load again. In the same way find the reading when there is a load of 200 g., of 300 g., of 400 g., of 500 g. In each case remove the weights and find the reading with no load, ELASTICITY. 169 always remembering first to turn the micrometer screw back far enough so that the tack will not touch it when the load is removed. If a square metal rod is substituted for wood, the tack becomes unnecessary, and V-shaped supports of metal are used, the binding post being on one of these. Arrange results in a table, as follows : — Load Reading with load Ebadino NO LOAD AVEBAGE NO LOAD BEADING Deflection Deflection for 100-gbam LOAD The average no-load reading is the average of readings without load, before and after the load has been added in each trial. Fig. 168. 193. Process of Bending. — To show what really takes place when a rod bends, lay a long rubber eraser on a page of the note-book and mark its outline with a sharp pencil. Bend the rubber and tie a string around it to hold it. Make another tracing in the note-book. Measure the length of the side C (Fig. 168) by laying a strip of paper on it. Lay off this distance in the book and see how it compares with the original length of the eraser. In the same way measure the length of the side D. Notice that the con- vex side is lengthened and the concave side is shortened. When a rod is bent, the forces at any point in the rod tend to pull the fibres apart in the upper half of the rod and to push them together, or compress them, in the lower half. The problem is somewhat like the horizontal forces on the hinges of a door (section 132, page 116), where the upper hinge is pulled apart and the lower hinge is compressed. If the door had hinges along its entire side, instead of one at the top and another at the bottom, the analogy would be still closer. 170 PRINCIPLES OF PHYSICS. Suppose a beam or stick (Fig. 169; has a small section cut out, and rods or hinges (J to 6) are inserted. If the rod is supported at the ends SS, and forces FF are applied to bend it downward, 4> 5, and 6 will be stretched, and resisting the stretching will tend to hold the lower parts of the stick together j 1, 2, and S will be compressed and will tend to push or keep apart the upper half of the stick, as the arrows indicate. The leverage of 1 and 6 is much greater than that of S and 4. In- creasing the thickness of a stick a little increases its stiffness a great deal. Consequently, if S and 4 were removed and placed near 1 and 6, the rod would be much stiffer. Does this suggest the reason why a tube is stiffer than a rod of the same weight ? Bxerclse 22. Fig. 169. (6) BENDING. Apparatus : The same as for Exercise 22 («), page 168. Case II. — Place the supports, SS (Fig. 169), 50 cm. apart; use weights of 500, 1000, 1500, and 2000 g., and make the readings as in Case I. Notice how much halving the length of the rod affects the amount of bending for the same load. If it becomes one-eighth as much, that is, makes the rod eight times as stiff, what relation does its increase bear to the decrease of length? How many 2's multiplied together make 8 ? Case HI. — Using weights of from 200 g. to 1000 g., test the stiifness of a stick (^4, Fig. 170) double the width of the one used in Case II. Case IV. — ■ Set the stick used in the preceding case on edge, as B, Fig. 170. Use weights of from 500 g. to 2000 g., and find what effect making the thickness twice as great has on the amount of bending, or stifiness. Fig. 170. ELASTICITY. 171 Compare the various results, and record the conclusions reached, trying to find what connection there is between : — 1. Increase of bending and change of load. 2. Increase of bending and change of length. 3. Increase of bending and change of width. 4. Increase of bending and change of thickness. 194. Formula for Bending. — Combining these conclusions, we have : — Bending or deflection equals load x (length)^ ^^^^^ (thickness)^ X width some number, which varies with the kind of rod used. D = — X some number, which varies with the kind of i^X w rod used. What has been studied in this exercise is the stiffness, and not the breaking strength, of a rod. The laws are not the same. Making a rod twice as thick makes it eight times as stiff, but only four times as strong, and doubling the length makes it one-half as strong. Problems. 1. A floor sags, or bends, ^ cm. with a load of 100 pounds. What would be the sag if loaded in the same spot with a ton weight ? Ans. 2 cm. 2. The tip of a stout fish-pole, held horizontally, bends 2 inches when a one-half pound downward pull is exerted on the tip. What is the load on the tip when the rod bends 7 inches ? Ans. 1| pounds. 3. A brook is bridged by a long plank, 10 inches from the surface of the water. If 20 pounds at the centre make it sag one inch, what is the greatest weight that can pass over the bridge without sinking it in the water? 4. How much would the plank in Problem 3 bend under a load of 20 pounds, if it were made twice as wide ? If made four and one-half times as wide? one-half as wide? Ans. \ inch ; | inch ; 2 inches. 5. The same bridge is made one-half as wide ; what effect does this have on the bending ? Ans. It bends twice as much. 172 PRINCIPLES OF PHYSICS. 6. How will the bending of two similar boards in a plank walk compare, if the cross supports are put 2 feet apart under one board and 5 feet apart under the other ? Ans. As 2" is to 5', or as 8 is to 125. 7. A beam 3 inches thick will bend how many times as much as one 6 inches thick, supposing that all other conditions are the same? as one 9 inches thick ? Ans. 8 times ; 27 times. 8. If a pine stick 1 inch thick, 2 inches wide, and 6 feet long bends .1 inch when a load of 20 pounds is suspended at its centre, how much would a plank of the same material, 3 inches thick, 8 inches wide, and 20 feet long, bend under a load of 400 pounds ? This can be solved in several ways. First, consider only the difference in the thickness, then in the length, etc. Another way is to write out the question as follows : — A stick 1 inch thick, 2 inches wide, 6 feet long, load 20, bends .1. A stick 3 inches thick, 8 inches wide, 20 feet long, load 400, bends how much ? The thickness of the second makes it bend ^ as much as the iirst. The width of the second makes it bend | as much. The length of the second makes it bend — = — = as much. ^ 6' 33 27 The load of the second makes it bend ^ as much; Therefore the bending of the second stick is the amount the first bends, .1, multiplied by all these numbers, or 27 8 27 20 9. If a beam 6 feet long, 3 inches broad, 2 inches thick, under a certain load bends 2J inches, how much would a similar beam, 30 feet long, 5 inches wide, and 8 inches thick, bend in sustaining a load one hundred times as great ? 195. Twisting. — Hold a match by the ends, and try to twist it. Cut away some of the wood, reducing the breadth and thickness by one-half, and again twist it. The following exer- cise shows how the twisting of a rod or beam is affected by increasing the load, the length, and the thickness. ELASTICITY. 173 Exercise 23. TWISnNG. Apparatus: A rod h inch in diameter, clamped firmly to a table at A, Fig. 171, and fastened at B to a J-inch rod ; pointers, C, D,E, F, and 6f, all of which, except E and F, must be equally distant from each other ; a grooved disk, 6 inches in diameter, at the end of the J-inch rod, to which disk are applied forces tending to twist the rod ; spring balances, arranged as shown in the figure, to measure the forces ; a graduated circle on a support behind each pointer, so that the pointer marks the zero point on the scale ; supports like Fig. 172 placed at intervals under the rods. Record the reading of each pointer at no load. Apply a load of 500 g. at each balance. Record the position of each pointer. The difference between the readings of C and D is the amount the part between C and D has twisted. How does the twist of the length CE compare with that of CD ? Remove the load and read the positions of the pointers. Apply forces of 1000, 1500, and 2000 g. to each balance. What effect does doubling the forces have on the amount of twist? For the same force, how many times as much twist takes place between C and D as in an equal length (FG) of the larger rod? Is it about sixteen times as ^ much? The larger rod has twice the diameter, and 16 is the fourth power of 2. Therefore the twisting of shafts ^'^' '^^' of the same material is greater in the smaller shaft and decreases as the fourth power of the diameters. 174 PRINCIPLES OF PHYSICS. 196. Conclusions. — From, the exercise, we see that — 1. Twisting increases with the length. 2. Twisting increases with the moment offeree applied. It is easy to see that forces of 100 g. 10 cm. apart have the same eifect as forces of 60 g. 20 cm. apart. The moment of the forces is the same. 3. Twisting increases as the diameter decreases. The increase is as the decrease of the fourth power of the diameter. Problems. 1. Compare the twist of a bicycle spoke 1 cm. long with that of one 30 cm. long. 2. How many times as much does a rod twist if the twisting moment is increased five times? 3. If a shaft 1 inch in diameter is replaced by one 3 inches in diameter, and the same forces are applied, how many times is the twisting angle decreased? Ans. 3* = 81. 4. How would. the amount of twist in two propeller shafts 3 inches and 20 inches in diameter compare, if the same twisting moment were applied to each ? CHAPTER XIII. HEAT. 197. Heat is a condition of a body. The hotter the body, the faster the little particles of which it is composed move back and forth. They do not move at all in the sense of going from one end of a rod to the other, but approach and recede without passing one another. The hotter a body is, the more violently they strike one another. Three dishes are filled with water, that in one as hot as can be borne by the hand, that in the second as warm as the room, and that in the third ice-cold. The water in the second, if left standing fifteen minutes, will be near enough to the tem- perature of the room. Put one hand in the first dish for a minute, and then in the second ; how does the water in the latter feel ? Put the other hand in the third dish, and after- ward in the second; how does the water in this dish feel to that hand ? The sense of touch, for several reasons, gives little idea about the heat of a substance. For instance, in the experi- ment the water in the first dish feels warm, because it gives heat to the hand; and when this hand, warmed by contact with the hot water, is placed in the second, heat is absorbed by the water from the hand, and the feeling of cold is produced. In the same way, when the hand that was chilled in the ice- cold water is placed in the second dish, it absorbs heat from the water, which consequently feels warm. A body that can give heat to another body is said to be of a higher temperature. A body is of a lower temperature when it can receive heat from another body. 175 176 PBINCIPLES OF PHYSICS. 198. Distinction between Temperature and Quantity of Heat. — If a red-hot nail is dropped into a bucket of boiling water, the nail is cooled, showing that it is of a higher temperature than the water. But the water actually contains more heat, — that is, a greater quantity of heat, — as is shown by the fact that it gives off more heat in cooling, and warms the room much more than the red-hot nail. 199. Conduction. — We speak of heat as moving, or travel- ling through a substance. If one end of an iron bar is put in the fire, in time the other end becomes warm. The heat is carried, or travels, through the iron. This is called conduction. The motion of the little particles of a body is very small in the case of a solid. The transfer of heat from one part of a body to another is like the movement of a bend along a rope. Give a quick swing to one end of a long clothes-line ; the dis- turbance will travel to the other end. Twist the ends of two wires, one copper and one iron, around the end of a glass rod or of a pipe-stem ; or, if the ends are of about the same diameter, they may be held together by a smaller wire, as in Fig. 173. The length of each piece should be four inches or more. Run the drip from a lighted candle on the full length of A, B, and C. Support on a ring-stand, and heat pjg 173 at D. After a few minutes the wax will cease melting. Note how far on each rod the wax has melted. Which apparently conducts heat the best ? Why can a glass-blower hold one end of a tube in his hand, whUe the other is held in the flame, for as long a time as he pleases ? Examine a "soldering iron." It is a lump of copper with an iron handle, which is less likely to break than a copper handle. What other advantage is there in the iron handle ? Why is it difficult to solder a large lump of copper ? BEAT. 177 200. Conductors of Heat. — If pieces of wood, lead, iron, and glass are left in a room for half an hour, — long enough for them all to reach the same temperature, that of the room itself, — they will seem to the touch to vary somewhat in, temperature. This is because one conducts heat away from the hand faster than another. A loaf of bread and the pan in which it is baked are of the same temperature just as they come from the oven ; but the pan feels hotter, because it con- ducts heat to the hand more rapidly than the bread does. In winter, why does an iron fence feel colder than a wooden one ? How can the hand bear the hot air from a baker's oven when the bricks are hot enough to cause a- burn? Wrap paper around a metal pipe or bolt, or attach a label to the bottom of an iron kettle, and try to burn the paper. Paper wrapped about a piece of wood and exposed to a flame scorches at once. Metal, being a better conductor, carries off the heat so fast that the paper is kept comparatively cool. Lead may be melted or water boiled in a box made by folding up the corners of a thin sheet of paper of good quality. This experi- ment is more difficult, but the paper may be kept from scorch- ing by shaking the box. Water can be almost entirely boiled away from over a lump of ice without melting the ice, if the ice is weighted at the bottom of a test-tube and the water heated above the weight. How does the conductivity of liquids for heat compare with that of solids ? Liquids and gases are poor conductors of heat. A long time would be needed to heat a dish of water, if the heat were applied at the top. 201. Experiment on Convection. — Nearly fill a beaker with water. Drop in a piece of the lead of a copying pencil, and apply a gentle heat. The water near the flame is warmed, expands, and thus becomes lighter and rises. Cold water flows down to take its place. Repeat the experiment; instead of colored lead, use sawdust, which also will show the direction of the currents of liquid. 178 PRINCIPLES OF PHYSICS. 202. Convection. — Fill a tube, bent as in Fig. 174, with col- ored water. Suspend it in a jar of clear water. Heat one side of -the tube, and observe the movement of the colored water. Soak paper or cloth in a solution of one part nitrate of potassium to twenty parts of water. Roll into the form of a taper. Ignite, and by aid of the smoke, study the air cur- rents in a room. Just as a crowd buying tickets is more quickly served by passing in line, instead of having the persons in the rear receive and send messages to the ticket-seller through the intervening crowd, so a liquid or gas is more quickly heated by having each particle in turn go to the heated surface and receive heat. And just as one person can pur- chase tickets for a number, so it is unnecessary for each par- ticle in every case to go to the heated surface, for those particles that do go become strongly heated, and, on rising, mix with and share their heat with other particles. This process of dis- tributing heat is called convection. Fig. 174. 203. Radiation. — If a hot object is held above the hand, the heat is felt, although the transference of heat is not by either conduction or convection. Hold the hand on a lighted incan- descent lamp ; then turn out the light. Could the glass have cooled as quickly as the heat disappeared? Was the heat the hand received transmitted by conduction through the glass ? Within the bulb of an incandescent lamp there is practically no air ; yet the heat comes from the filament across the empty space as the heat of the sun comes to the earth. This method of transmitting heat is called radiation, because the path is a radius, that is, a straight line. Scientists believe that all space is filled with a weightless, elastic ether, as it is called, which vibrates, and sends along heat and light vibrations. HEAT. 179 Try an incandescent lamp which is dim from use. Air has, perhaps, leaked in a little, and by convection heat is carried from the hot filament to the glass. Place an incandescent lamp under water, and turn on the current ; is the heat felt ? Turn out the lamp ; has the water been much warmed. 204. Effect of Surface on Radiation. — Fill a bright tin or nickelled brass can with water that has been heated to nearly 100° G} Place a thermometer in it, and read the tempera- ture every minute or half min- ute. Plot temperatures up- ward, and time in minutes to the right, and draw the curve, as in Fig. 175. Repeat, using a can painted black, and plot the curve. Which radiates heat more rapidly ? Should a stove be nickelled or black- ened? Should a coffee-pot, in which coffee is to be kept hot, be black or bright? Clouds retard radiation. Why does the earth cool less on a cloudy night than on a clear night ? In what kind of weather may frosts be expected? The conditions can be reversed, and the bright and blackened cans filled with cold water, and the temperatures taken regu- larly as before. For rapid work, place the cans in the sun- shine. Good absorbers are good radiators. Some heat is lost by convection when the cooling experiment is done in the air. Notice that the filament of an incandescent lamp cools less rapidly in the vacuum that exists when the lamp is new than in many lamps that have been burned a long time, and into which a little air has leaked. L minntea Fig. 175. 1 The water may be dipped from a large panful that has been heated before the beginning of the experiment. 180 PBINCIPLES OF PHYSICS. 205. Effect of Heat on the Size of Substances. — Invert a flask , or test-tube, fitted with a long tube inserted in the stopper, and place the open end of the tube in a tumbler of colored water (Fig. 176). The water may be colored by scraping the lead of a copying pencil into it. Heat the flask with the hand ; cool the flask by blowing upon it. Heat the flask with a burner, and allow it to cool. Remove the stopper from the flask. Fill the flask with colored water, and replace stopper and tube, so that the liquid stands some distance up in the tube, as in Fig. 177. Heat the flask, but not to the boiling-point. Allow it to cool, marking the heights of the liquid in the tube, by a label or a Cross pencil. What happens to a dish full of cold water, when heated ? Which must be the lighter, cold or warm water? Of two liquids of different densities, which floats ? In summer, fish stay in the cooler parts of lakes ; why are they not oftener found near the surface at that season? Why does a heated liquid rise ? Repeat the experiment, using test-tubes of the same size, in place of the flask. Fig. 176, one filled with water, the other with a different liquid, — alcohol, for instance, — and determine if the rate of expansion is the same. Fig. 176. Fig. 177. 206. Maximum Density of Water. — One cubic centimeter of water weighs a gram exactly at the temperature at which water is most dense. To find this point, that of the maximum density of water, fill the metal can, C, Fig. 178, with water. The lower part of the can should be wrapped with cloth, to prevent absorption of heat from the room. Pack the basket, B, with HEAT. 181 ice and salt. Read the thermometers, Ti and T2, at frequent intervals, until both are constant. Which thermometer first shows a change? How low does it go? "Why does the cold water fall at first ? How cold can the water in a deep lake become at the bottom of the lake ? The water cooled by the ice and salt in the basket, B, con- tracts, becomes denser, and sinks. When, however, the water has cooled to about 4° C, it does not contiaue to contract and become denser on fur- ther cooliug, but instead it expands, grows lighter, and rises to the top. Do 100 cc. of water at 50° C. weigh more or less than 100 g. ? Do 100 cc. of ice-cold water weigh more or less than 100 g. ? . Which is the more buoyant, water at 3° C. or at O'.C. ? How cold will water in the bottom of a pitcher half-filled with ice become on a warm day ? In testing the freezing-point of a thermometer, what error would there be if the bulb were left in the water below the ice, vsrithout stirring ? Fig. I 78, Z> R 207. Coefficient of Expansion. — The metal rod, R, Fig. 179, — that of a ring-stand, for example, — is supported horizon- •tally; one end rests on the pin to which the pointer, PP, is attached by sealing- wax. The pin rests- on the glass plate, G, or on I any smooth, hard surface, which is levelled by boards or books ; the rod B does not touch the glass plate. Push the ring-stand, D, toward the pointer, PP, and record the direction of the movement of the pointer. Heat the rod ; Fig. 1 79. 182 PRINCIPLES OF PHYSICS. then let it cool. Drop some water on it. Were the rod to lengthen an amount equal to the circumference of the pin, the pointer would describe a complete circle. The pin may be measured by a micrometer caliper, the circumference computed, and the expansion of the rod calculated from the part of the circle through which the pointer moved. If the pointer moved through half a circle, the expansion was half the circumference of the pin. The numbers given in books to express expansion are called coefficients of expansion, and they mean the amount a rod one centimeter long would expand for one degree of increase in temperature. 208. Linear Expansion. — Metals do not all expand at the same rate. Lay a strip of copper, | inch wide, -^^ inch thick, and 12 inches long, over a similar strip of iron. Make a small hole in the two, every quarter of an ' L L IT inch, by driving an awl or sharp nail P, I gg through both. Put small tacks in the holes, and hammer down the points. Fig. 180 shows a portion of the two strips, after riveting. Hold one end with pliers, and heat. Copper expands more than iron, so the riveted strips will curve when heated. Wood swells when wet. A thin board, wet on one side, will warp. Measure the length of two opposite sides of an oblong rubber eraser. Cut two strips of paper, the exact length of the eraser, and gum them on opposite sides of it, attaching the paper only at one end. Bend the rubber, and notice what change there is in the length of the sides. 209. Examples of Expansion. — Metallic thermometers and some forms of heat regulators, or thermostats, are made of two metals riveted or soldered together. A balance wheel of a watch expands in warm weather, grows larger, swings slower, and the watch loses time ; but if another metal, which expands faster, be fastened to the rim, the halves of the rim, which are HEAT. 183 fastened only at one end, are made to curve in toward the centre of the wheel when the temperature is increased. A bal- ance wheel of two metals, skilfully designed, vibrates at the same rate, whether hot or cold. As the wheel grows warmer and expands, its tendency is to swing slower, like a lengthened pendulum. This is compensated for, because the unequal ex- pansion of the two metals of which the rim of the balance wheel is made, causes a part of the rim, which is free to move, to bend in toward the centre. This action tends to shorten the pendulum and make the wheel vibrate faster. 210. Micrometers. — As metals expand comparatively little, the increase of length due to a change of temperature must be magnified in some way, so as to be read easily'. Micrometers and levers have been used for this purpose. A micrometer meas- ures the increase in length in the same way that a micrometer- caliper measures thickness. Practice finding, by a micrometer, the diameter of wires, the thickness of paper, sheet metal, and glass plates. Set the micrometer for the following readings : 1 mm., .02 mm., 6 mm. 3.02 mm. An English binding-post may be studied as a rough model of a micrometer (Fig. 181). Count the number of threads in half an inch. Common binding- posts have thirty-two to the inch ; one thread to the millimeter is to be preferred. Hold the nut, N, and turn the head, H, one whole rev- olution. How far does the point A advance ? How many revolutions must H be turned to advance A one inch ? BDE, Fig. 182, is a wooden frame. A hole is made in E and a milled nut (like N, Fig. 181) is pressed in PI igj firmly. A circle of cardboard, C, is B A ffVWWU .r r 1 ■■< 1 VVWVL— f [oj. I'l'i'in'iy B 184 PRINCIPLES OF PHYSICS. glued to H. Mark any number of divisions on C, ten or one hundred, for instance. On E fasten a little scale of cardboard, S, which is marked off into thirty-seconds of an inch. Place S so that the zero on C, when close to S, is opposite one of the marks on S. Turn the screw till it touches B, and if the zero on C is not close to S, file away B. Practice measuring. If the circle G is divided into ten parts, turning C one division moves the point A one-tenth of one-thirty-second of an inch, or -j-^ of an inch. In measuring the expansion of a rod, we use a micrometer somewhat similar to Fig. 183. BB is removed to permit the rod to touch the end of the screw, and Z) is of a shape suitable to be clamped to a sup- ^, ,„, port. If the screw has ten Fig. 18*. '^ threads to a centimeter, that is, one to a millimeter, one whole turn of the screw moves the point forward or back one-tenth of a centimeter, or one millimeter. If the dial attached to the screw is divided into one hundred divisions, turning the screw so that the dial moves one division, moves the point y^ of a millimeter. Exercise 24. COEFFICIENT OF EXPANSION. Apparatus : A rod of glass, aluminum, brass, zinc, or iron, one-fourth inch in diameter and about 60 cm. long, pointed at the ends, and inserted through corks in the ends of a steam jacket, which rests on supports not shown in Fig. 184 ; a base, with metal upright near each end, one to hold a micrometer screw, M, the other to hold a metal stop, A ; a steam can (see Fig. 186, page 191) . The micrometer screw should have a friction rachet, or a handle, F, of small diameter, that will slip in the fingers when a certain pressure is applied. This handle acts as a friction slip, and prevents the micrometer from being screwed up harder at one time than another. Turn back M; slip a knife blade in front of M, to push the rod, R,to A. Read the tliermometer in the room. Screw up M till the HEAT. 185 friction-slip acts. Read the dial, and make a drawing of it in the position read. Note the mark on the scale, L, that coincides with the edge of the dial. Turn back M two full _„ fS X ,^1 turns. Connect S with the steam can by a thin rubber tube. Lead the drip from D by a tube to a dish. Read the barometer. ' The e^ M. ^ i-m-rq L Fig. 184. temperature of the rod is assumed to be that of the steam. Determine that from the barometer reading (section 226, page 201). For in- stance, if the barometer is 76 cm., the temperature of the steam not confined is 100° C. After steam has been coming from D for two minutes, turn the screw M up slowly, till contact is shown by the slipping of F. Read the dial, and turn M back a little. Repeat until the reading is constant, and make a drawihg of the dial, recording also the scale reading on L, which is marked in millimeters. Suppose that at the beginning the temperature = 20°, the reading on i is 4 mm., and the dial, where it touched L, reads 60 (that is, j'^ of a millimeter); and that at a temperature of 100°, L reads 5, and the dial 40. Then the difference in length of the rod was 5.40 mm. — 4.60 mm = .90 mm. This is .09 cm. We use centi- meters, since the coefficient of linear expansion is the amount that one centimeter of metal expands for 1° C. The coefficient of ex- pansion in this case equals .09 divided by 60, — since a one-centi- meter rod would expand one-sixtieth as much as the one used, — and divided also by 80, the rise in temperature, since the expansion for 1° would be one-eightieth as much as for 80°. Then Coefficient of expansion, that is, the amount a sec- qq tion of the rod 1 cm. long expands when warmed = gQ ^ qq = .000018 In this way find the coefficients of expansion of glass and of one or more metals. To repeat quickly the reading with the same rod, or to cool the jacket for another rod, attach the rubber tube connected with S (Fig. 184) to a funnel or faucet, and let cold water run through the jacket, around the rod, and through D into a dish, in which there is a 186 PRINCIPLES OF PHYSICS. thermometer. After two minutes read this thermometer. When the thermometer no longer falls, record the temperature, which is that of the rod. Turn up the micrometer screw, and record the reading. In this manner, several readings may be taken in an hour. > Problems. ' 1. A meter of brass at 20^ C. is how long at 30° C, if the coefficient of expansion of brass = .000018? 2. A copper wire stretches across a river 1000 feet. How much longer is the wire at 25° C. than when it was put up in "winter at — 10° C. ? Does the wire sag as much in winter as in summer? (Con- sult tables for coefficients, in Appendix.) 3. The outer rings of heavy cannon are shrunk on. When made, the rings are smaller than the tube or the other rings that they are intended to surround. How can they be made large enough to fit ? The sheets of iron of which a boiler is made are fastened with hot rivets. Do they become looser or tighter as they cool? Why is a space left between the ends of raUs on a steam railroad ? 4. How much does a bridge of iron, 1000 feet long, increase in length when warmed from - 10° to 20° C. ? This problem is done exactly as if the length were 1000 cm., and the result is read as centimeters or feet, according to the problem. 5. Which contracts the faster per degree, iron or glass ? If the leading-in wires of an incandescent lamp be made of iron, what hap- pens as the glass and iron cool down from the high temperature at which they must be sealed ? Find, in the table in the Appendix, a metal better suited and one less suited to this purpose. 6. In the left-hand pan of a delicate balance is a mass of 50 g. This is counterpoised by weights on the right-hand pan. The sun 1 falls on the left arm of the balance. Does the mass appear to weigh more or less than before ? 7. If, in turning a piece of iron, the metal is heated to 60° C, and measures 3 cm. in diameter, how much too small will it be, or how much will it shrink, when cooled to 20° C. ? If a ball is held in the micrometer (Fig. 183), why does the ball fall out when the bend of the micrometer is warmed? HEAT. 187 8. If the temperature of an iron steam pipe 600 feet long is raised from 10° C. to 120° C. when steam is admitted, how much does it lengthen ? 9. Calculate the coefficient of expansion of lead, if the length at 0° C. = 150 cm., at 90° C. = 150.37 cm. 10. If 10° C. is the average temperature of the three days before a road is laid of 40-foot iron rails, how much space should be left between the ends, that they may not touch in summer (temperature = 25° C. ? Consider the rails fastened firmly at their eastern ends, if the road runs in that direction, so that all the expansion appears at the western ends. 11. What will be the efiect of cold on a piano, if tightening a string or wire raises the pitch ? 211. Expansion of Rails. — When lengths of a mile or more of street car rails are welded together, as has been done on a few roads, no provision is made for expansion, and the rails do not expand and contract. Suppose a rail, when warmed a certain amount, expands one-tenth of an inch. As much force would be exerted by this rail in attempting to expand as would have to be applied to stretch it one-tenth of an inch without any change in temperature (see section 186, page 164). The force required to prevent an expansion of one-tenth of an inch, when the rail is heated, is as much as would have to be applied to stretch it one-t*enth of an inch, there being no change of temperature. The rails are nailed to cross-pieces, and are so firmly imbedded in the paving that the force tending to pro- duce expansion due to ordinary changes of temperature is more than counteracted. 212. Cubical Expansion. — In Exercise 24, page 184, the lengthening or linear expansion of a solid was measured. The rod, however, grew broader and thicker also. In the case of a liquid, confined like mercury in a thermometer, the expansion can take place (disregarding for a moment the ex- pansion of the glass) only lengthwise. As there is little or 188 PRINCIPLES OF PHYSICS. no increase or expansion in breadth and thickness, the entire increase must be indicated by movement in the direction of length. In considering how much greater the entire expansion (cubi- cal expansion) of a body in all directions is than the expansion in length merely, assume that a cube 10 cm. on an edge ex- pands 1 cm. in length, breadth, and thickness. Suppose ABO (I, Fig. 185) is the cube before expansion. The increase in length adds a piece 1 cm. thick, covering B (II). Adding the A / z /« / ^ J / / / / / a / n ni Fig. 185. ^ IT same amount to the width and thickness, the expanded cube appears as aba (III), where the increase in all directions, or the cubical expansion, is three times the increase in length, or the linear expansion. But we know that after a cube has expanded it is still a cube, and does not lack the little pieces needed to make ahc a perfect cube. Adding those little pieces (IV, Fig. 186), the cubical expansion is a little more than three times the linear. In the case of a solid, however, the expansion of a 10 cm. cube for 100° increase in temperature is about the thickness of two leaves of this book. Fit two sheets of paper on the faces, A, B, C, of a 10 cm. cube, such as the model cube iised in teach- ing the metric system, or on a box having nearly the same dimensions, and determine whether or not the little pieces are needed to complete the cube, as they were in abc, Fig. 185. In this case they are too small to measure. HEAT. 189 213. Cubical Expansion by Computation. — Let us compute the volume of a cube 1 cm. ou an edge. The volume is 1 X 1 X 1 = 1 cc. If warmed 100°, the increase in length is between j-^Vir ^'i*! nnrTT °f ^ centimeter. Of course, the cube has expanded equally in all directions. The dimensions now are 1.002 at the most. The volume = 1.002 x 1.002 x 1.002 = 1.006012008. All figures beyond the 6 must be disregarded, for they are too small to be of any account. Practically, the volume is 1.006, and the increase in volume, or the cubical expansion, equals .006, which is three times .002, the linear expansion. Suppose a cube 100 cm. on an edge increase to one 101 cm. Cube 100, that is, find 100 x 100 x 100. Cube 101, and show that the cubical increase is practically three times the linear increase. Problems. 1. How many times as much room does a lump of iron 10 cm. by 10 cm. by 50 cm. take up at 80° C. as at 20° C? 2. What metal would have a greater expansion than iron ? a less expansion ? 3. Give a reason, other than the great brittleness of glass, why, when suddenly heated, a thick piece of glass cracks more easily than a metal. 4. Why is a pendulum rod of wood better than one of brass ? 5. If a brass pendulum is 100 cm. long at 20° C, what is its length at35°C.? 6. Why do long lines of iron pipe screwed together have loops in them every few hundred feet? CHAPTER XIV. THERMOMETEES. 214. Measurement of Temperature. — As expansion, or in- crease in size, is the most common effect of heat, it may be used to measure temperature. For measuring most tempera- tures, solids expand too little and gases expand too much. Of liquids, water freezes and does not vary in volume uniformly ; alcohol, though not easily frozen and useful for determining low temperatures, boils at a lower temperature than water; mercury is generally used. 215. Construction of a Thermometer. — A flask and tube like that shown in Fig. 177, page 180, filled with mercury or alcohol, or even with water, could be used as a thermometer. But if the tube and flask are made of one piece of glass, the thermometer is more lasting and reliable. The flask, or bulb, can be made on the end of the glass tube. To make a model thermometer, first close the end of a tube. This is done by heating the end of the tube and touching the hot end with another. Continue heating, and pull gently on the pieces of glass ; the hottest part draws out fine, melts, and seals the tube. By repeatedly heating the tube and blowing into it, a large bulb can be made. While the bulb is still warm, put the open end of the tube into a dish of mercury. As the air in the bulb cools, the mercury rises and partly tills the bulb. Holding the open end up, re-heat the bulb till the mercury boils, and then replace the mouth of the tube in the mercury dish. After several trials a tube can be filled a little distance up the stem. When this is warmed, the mercury rises higher in the stem. The glass expands, too, but not so rapidly as J90 THERMOMETERS. 191 the mercury. Repeat the experiment described in section 205, Fig. 177, noticing what happens when the heat is first applied. A fall of the liquid in the tube shows that the glass expands, the flask becomes larger, and more liquid flows down into it. In a moment the liquid in the flask begins to warm, and, as the rate of expansion of the liquid is greater than that of the glass, it rises in the tube. 216. The Standard Temperatures generally used in fixing the scale of a thermometer are : (1) the freezing-point of water, or the melting-point of ice (though exactly- the same temperature, the melting-point is more convenient to use) ; (2) the boiling- point of pure water under the pressure of one atmosphere, that is, when the barometer reads 76 cm. ; (3) the melting-points of various chemical salts. Exercise 25. TESTING A THEEMOMETEK FOE 0° AND 100° C. Apparatus : Steam can (Fig. 186) ; mercury thermometer ; beaker or tin can. Place the thermometer to be marked or to be tested in a beaker or tin can full of snow or fine ice. Keep the snow or ice well up to the height of the mercury, and notice the reading when the mercury stops falling ; or, if the tem- perature be unmarked, make a little scratch with a file at the height of the mercury, and call thispoint " ice melts " (Fig. 187) . Remove the thermometer; warm it with the hand, and insert it in the long top of the steam can (Fig. 186). Stop the pipe in the side of the can, and allow steam to escape freely from the pipe in the long top. Read the thermometer when the mercury / — ^'f' \ ^^^ ceased to rise; also read f -* ^ ^ the barometer. If convenient, Fig. 186. push the thermometer down water boils ice melts n ice and salt Fig. .187. 192 PRINCIPLES OF PHYSICS. into the boiling water, and read the temperature. The boiling-point used as a standard temperature is that of the steam. On the unmarked thermometer mark with a file the level of the mercury, and call that the boiling-point (Fig. 187). If the bore of the tube is uniform, the space between the marks "ice melts" and "water boils " can be divided into any number of convenient equal distances, called degrees. On the centigrade ^ thermometer, " ice melts " is called zero, and "water boils," 100 degrees. The space between the two is divided into 100 parts, or degrees. These experiments serve to test the thermometer under a pressure of about one atmosphere. 217. The Zero Point is not the point of no heat; there are lower temperatures, as every one knovsrs. Fahrenheit chose the point reached by the mercury in ice and salt as the zero point, thinking that this gave the greatest possible cold. It was an unfortunate choice, because the melting-point of ice and salt, or ice and other substances, is difficult to determine. Lower temperatures than zero Fahrenheit are common. He divided the space between " ice melts " and " water boils " into 180° on his scale. There were thirty-two of these divisions in the space between the melting-point of ice and that of ice and salt. ! 750 ice boils 100 melts %nd salt Fig. I 88. 218. Fahrenheit and Centigrade. — The differ- ence between a Fahrenheit and a centigrade thermometer consists only in the marking of the scale, and both scales are sometimes made on one thermometer. If the boiling-point in the Fahrenheit scale is 180° above '<»ice melts " (Fig. 188), and if the melting-point of ice is 32° above that of ice and salt, — the zero of the Fahrenheit scale, — how many degrees from zero to the boiling-point ? (180° -I- 32° = 212°.) There being 180° from the melting-point (usually called the freezing-point) to the boil- ing-point in the Fahrenheit scale, and only 100 1 Centigrade means 100 steps, or divisions. THERMOMETEBS. 193 degrees in the centigrade scale, it follows that 180 Fahrenheit divisions equal 100 centigrade divisions. Divide both these numbers by 20 ; then 9 Fahrenheit degrees equal 5 centigrade degrees. If a Fahrenheit thermometer shows a rise of 18 de- grees, how many degrees does a centigrade thermometer rise ? A fall of 30 degrees on the centigrade scale would be registered by how many degrees Fahrenheit ? The minus sign applied to temperature means below zero. — 10° C. is read " minus 10 degrees centigrade," or " 10 degrees below zero centigrade," or " 10 degrees below the melting-point of ice." So, in the same way, — 6° F. means 6 degrees below the Fahrenheit- zero. Every Fahrenheit temperature is reckoned, not from the freezing-point, but from a point wrongly thought to be the greatest cold possible, — 32° lower. Starting from this low point, all Fahrenheit temperatures are 32° larger than they would be if the freezing-point were the zero. 219. To change a Fahrenheit Temperature to a Centigrade, first, subtract 32. The reason for this is evident, if it is remembered that, while there are 180 divisions from the freez- ing-point to the boiling-point on the Fahrenheit scale, the number of degrees marked on the boiling-point is 212, or 32 more than 180°. Suppose the temperature be 68° F. Subtract 32; 68-32 = 36. These are still Fahrenheit degrees, of which nine equal five centigrade degrees. Multiply by f ; 36 X I = 20° centigrade. Change to centigrade: 212° F. ; 0°F.; 150° F.; 50° F.; -10°F. ; 32° F. In changing from centigrade to Fahrenheit temperatures, exactly the reverse operations should be employed. For instance, in changing 20° centigrade to Fahrenheit, first, mul- tiply by f ; 20 X f = 36. Add 32 ; 36 -}- 32 = 68° F. Change 100° 6. to F. ; 0° C. ; 40° C. Change - 40° F. to C. ; 1000°F. to C. Change -40°C. to F.; 12° F. toC; -20°C. to F. 194 PRINCIPLES OF PHYSICS. 100- Cent. T 111 —i:.SlS water boils 220. Comparison of Fahrenheit and Centigrade Scales. — As- sume the space on a thermometer tube between the freezing and boiling points to be any con- venient distance, as four or five inches. Make, on one side of a line, the Fahrenheit scale, and on the other side, the centigrade scale (Fig. 189). On the Fahrenheit side measure down thirty-two spaces from " ice melts,'' and locate the zero point. There could be, . of course, any number of arbitrary scales. The Reaumur scale, still used a little, calls the freezing-point 0°, and the boiling-point 80°. ■ice melts. Fig. 189. 221 . How to vary the Boiling-point. — In studying the ther- mometer, we assumed that the barometer read 76 cm. when the mark indicating the boiling-point was made ; that is, that the pressure on the surface of the water was equal to one atmos- phere. The following experiment shows the effect, on the boiling-point, of variations of this pressure. Boil a little water vigorously in a test-tube, and while the water is still boiling, stop the test-tube with a one-hole rubber stopper, coated with glycerine and plugged with a glass rod or a closed glass tube. Place the test-tube, mouth down, in a dish of water (Fig. 190), so that if there is any leakage it will be of water, and not of air. Cool the test-tube by blowing on it, then by pouring cold water on it. What effect does this have on the water inside the tube ? Notice anything collecting on the inner surface of the test-tube. Pour on cold water until bubbles cease to form ; then note the temperature of the water in the test- tube. Take the test-tube out of the dish, Fig. 1,90. THEBMOMETEHS. 195 and raise and lower it quickly, to get the hammer-like sound of the water. What causes this ? What is the bubble made of that forms under the water when the tube is lowered quickly? Take the test-tube out of the dish of water, and remove the glass plug. What enters the tube ? What must have been the pressure on the water in the tube before the plug was removed? Try to get the water-hammer effect. Repeat the experiment from the beginning, but hold the mouth of the test-tube under water when removing the plug. When the water is first boiled the steam carries off the air from the tube. If the stopper fits tightly, or is held under water, no air can enter. At that stage the pressure of the steam is the same as that of the outside air. The effect of the cold water on the steam pressure may be determined by considering what happened when the plug was removed under water. The water-hammer effect comes from there being no air to cushion the water as it strikes the bottom of the test-tube. 222. Pressure of Steam. — If, instead of the plug, a glass tube 80 cm. long is inserted in the stopper, the variation of pressure inside the test-tube can be studied. Boil the water in the test-tube, insert the stopper, and continue boiling till steam comes from the end of the long tube. Then invert the test-tube, with the end of the long tube in a cup of mercury (Fig. 191). With the barometer reading compare the greatest height to which the mer- cury rises. Touch the test-tube from time to time to form an idea of its temperature. If the barom- eter reads 76 cm., and the mercury rises 60 cm. from the cup, then the pressure of steam in the test-tube (76 cm. — 60 em.) is 16 cm., or a pressure equal to that of a column of mercury 16 cm. high. Temperatures corresponding to pressure of steam may be read by a thermometer inserted in a modified form of this apparatus, as shown in Kg. 192. Fig. i9i. 196 PRINCIPLES OF PHYSICS. Exercise 26, in (a) TEMPEKATTJEES COEEESPONDING TO PEESSTOE OF STEAM.— First Method. Apparatus : A test-tube, held in a retort clamp ; stopper for test-tube per- forated with three holes, one for a glass plug, another for a thermometer, T, Fig. 192, and the third for a long bent tube, H, reach- ing down to a, cup of mercury. Put small pieces of flre-brick in the water to make it boil steadily. The test-tube may be covered with asbestos, except a section in which to watch the boiling. Boil the water in the test-tube, with the thermom- eter and the tube to the mercury cup in place. When the water has boiled for half a minute insert the glass plug and remove the lamp. Record, at intervals of one or two minutes, the reading of the thermometer and the height of the mercury column. In this method the tube, H (Fig. 192), is full of air, since steam is allowed to escape through the hole in which the plug fits, and is not forced through // to drive the air out. Owing to the volume of air in H, which expands and exerts some pressure, the boiling ceases before the temperature falls near zero. When the boiling ceases the experiment is finished. Record as follows : — >Flg. I 92. Temperatuek i. Height of Mercfey Gauge h. Height of Babometer i. Pressure in Test-Tube b -A. Calculate J — A in each set of readings, and plot on coordinate paper in any convenient form. For example, let spaces to the right represent pressure, and spaces upward temperature. The plot may THERMOMETERS. 197 well be on a larger scale than shown in Fig. 193. One point is shown by the cross. t 100° b 76 6-/j 76 S0° 70' Presswre in centimeters of mercwry Fig. 193. The point is on the 100° temperature line and on the 76° pressure line. Where these lines intersect is the point P. Locate the other points, and draw a curve connecting them. Practice reading off boiling-points for different pressures. Find the change in pressure from 100° down to 95°, for the nearest temperature recorded to 95°. Divide the change in pressure thus obtained by the number of degrees' fall in temperature. For example : — Prebsitre in Test-Tube Temperature 760 mm. 634 mm. 100° 95° The fall in pressure on the boiling water is 760 — 634 mm., i.e. 126 mm. for 5° change in temperature. For 1° change the fall in pressure is ^^, which equals 25 mm. In the same manner find the change of pressure per degree for the change of temperature between 95° and 90°, and so on for every five degrees down to the lowest temperature recorded. The pressure on the water in the test-tube, where A = 0, is the pres- sure as read on the barometer. What is the temperature ? What is the pressure of steam in the test-tube when the thermometer reads 85° ? Suppose the test-tube is cooled so that its temperature drops regularly, does the pressure fall regularly? This is the same as ask- ing if the mercury in H, Fig. 192, rises regularly. 198 PSINOIPLES OF PHYSICS. L_ Exercise 26. (6) TEMPERATUEES COERESPONDING TO PRESSUEE OF STEAM. — Second Method. Apparatus ' ; A glass tube, about 60 cm. long and 2 cm. in diameter ; a lubber stopper fitting this tube, in which is inserted a smaller tube of medium thick glass, 80 to 90 cm. long and 4 mm. in diameter, closed at one end and filled with mercury to within 1 cm. of the top; cup of mercury; varnished tin pan; thermometer; clamp; ring-stand; a funnel, attached to a third glass tube ; asbestos shield for the large glass tube. Close, with the finger, the open end of the small tube containing the mercury, and invert several times, to remove air bubbles. Fill to the top with water, and invert a few times. Cover the open end with a cork drilled half through, insert through the rubber stopper of the larger glass tube, as in Fig. 194, and place in a rack till needed. Before the labora- tory exercise is to be done, again fill the inner tube with water and invert till the ab- senceof airbubblesisassured, thus making a barometer, B, Fig. 195; then place, still inverted, in the cup of mer- cury. Set this in the var- nished tin pan, D. The thermometer, T, is hung on the stand, to which is attached the clamp, C, holding the outer tube. The shorter ring-stands of the chemical laboratory would rest on the table and the appa- ratus project down below, the cup and dish resting on a box on the floor. Place the funnel, F, in the apparatus, with its glass tube reaching to the bottom of the outer glass tube, or p|g 195^ N? 0=0 Fig. 194. 1 Three or four pupils can work with one piece of apparatus : one holds a meter stick at the level of the mercury in the dish ; another reads the height THERMOMETERS. 199 jacket, /. Pour in a little cold water and then, at once, a large amount of boiling water. At first, the cold water mixes a little with the hot, and prevents the glass from cracking ; but it is soon displaced by the hot water, the excess flowing into the pan, D, below. Nearly surround the jacket with an asbestos shield. Record many series of readings, taking the readings of the thermometer and mercury columns at the same instant. Continue to take readings every few minutes, until the temperature of the water in the jacket has fallen to that of the room. Ice water may be poured in, or fine ice added, and a few readings taken near zero. As the water cools, the steam generated in the barometer tube condenses and the pressure exerted by the steam decreases. The mercury in the barometer, B, is pressed down less and less by the steam vapor, and accordingly rises higher and higher. Record the temperature of the room and the height of a standard barometer. The more slowly the water cools, the less likely are the water and the water vapor in the barometer tube to be warmer than the thermome- ter. Record in note-book as follows : — Temperature t Height of Mercury h Barometer i Pressure of Water Vapor IN Barometer Tube b-h Plot the temperature and pressure of the water vapor, in colored pencil or ink, on the same paper used in the First Method for the first plot. The fall in pressure for one degree may be found as in the First Method. For pressures greater than one atmosphere, the special steam boiler used in the engine experiments, p. 257, and a U-shaped mercury gauge are convenient. Was there any difference in the readings of the standard barometer and the barometer, containing a little water, used in this exercise? of the mercury column : another reads the thermometer ; and still another records the observations. The experiment can he repeated several times in an hour, and each pupil take his turn in making the different observations. 200 PRINCIPLES OF PHYSICS. What happened to the water, when hot water was poured in the jacket? What was the pressure on the water at the top of the mer- cury column before the hot water was poured in the jacket? After it was poured in ? On what part of a mountain would water boil at the lowest temperature ? In what part of a mine ? The boiling-points of ether, alcohol, etc., under different pressures can be found in the same way as those of water. 223. Vacuum Pans and Digesters. — If water is boiled away from syrup in an open pan, the sugar that is left will not be granulated ; if water is boiled away from milk the milk will be cooked. How are granulated sugar and condensed milk made ? How are fruits dried quickly without cooking ? A vacuum pan is a closed kettle, from which the air or steam is removed by a pump. Milk heated in such a pan to 70° C. is not cooked, though the water boils away, leaving the milk condensed. In various manufacturing processes many substances must be raised to a temperature higher than 100° C. In some cases this must be done in a closed kettle which is called a digestor. The boiling point of the water or other liquid in it is raised above 100° C. as the liquid boils and its vapor exerts pressure. The utmost limit to which the pressure of steam in a boiler can be raised is reached when the boiling-point rises to the temperature at which iron begins to be red, and therefore weaker. 224. Boiling. — From the reading of the barometer tube con- taining a little water, in the preceding section, it is seen that the vapor of a liquid always exerts some pressure, — more at greater temperatures. When the vapor pressure of a liquid is greater than the pressure upon its surface, then the liquid under the surface begins to turn into vapor, bubbles of vapor form and rise, and the liquid boils. In the process of boiling, bubbles form in any part of the liquid, usually near the surface that is heated by a lamp or a fire. THERMOMETERS. 201 225. Evaporation. — Water that is not boiling disappears in time. Place under the receiver of an air-pump a dish of water that has just ceased boiling, and exhaust the air, but not rapidly enough to make the water boil. A dense fog fills the receiver, and condenses on the sides. This is the process of evaporation, which takes place only at the surface of a liquid, but at all temperatures, more rapidly at higher than at lower tempera- tures. Even snow and ice evaporate and waste away in cold weather. Evaporation is more rapid under low pressure. 226. Corrections for Pressure in Testing a Thermometer. — In Exercise 25 (page 191), the thermometer was tested under a pressure of one atmosphere, or 76 cm. of mercury. At or about this pressure, a variation of 2.7 cm. in the barometer reading causes a change of one degree in the boiling-point. Therefore, if the barometer reading was not exactly 76 cm. when the ther- mometer was tested, corrections must be made for the varia- tion. If the thermometer reads 100.2° in steam, when the barometer reading is 77.2 cm., the thermometer would have read lower with the barometer at 76 cm. As 2.7 cm. change in pressure causes 1° change in the boiling-point, and in this case the change is 77.2 cm. — 76 cm. = 1.2 cm., the thermome- 1.2 ter would read gy of a degree lower, or .4° lower, with the barometer at 76 em. The true 100° point is .4° lower than 100.2°, or 100.2° - .4°, which is 99.8°. At or near the boiling- point the thermometer reads .2° too low. To correct readings in this part of the scale, .2° should be added. The bore of the tube varies in size in different parts of the thermometer, and for very accurate work the true 50°, 25°, and 75° points need to be known. Following out the rule that 2.7 cm. change in the barometer causes a change of 1° in the boiling-point, a fall of 100° would seem to indicate a fall in the barometer of only 27 cm. But the barometer must fall 76 cm. to indicate no air pressure. 202 PRINCIPLES OF PHYSICS. Therefore the rule holds only at or near 100° C. Above 100°, a change of 1° in the boiling-point is caused by over 2.7 cm. change in pressure. Study the readings and the curve obtained in Fig. 193, page 197, and notice that farther down on the thermometer scale 1° change is caused by much less than 2.7 cm. fall in pressure. Suppose the lowest temperature reached by the thermometer in melting ice is — .3° ; then the true zero point is at — .3°. The thermometer in that part of the scale reads too low by .3°, and all readings near the zero mark must be increased by adding .3° to the reading. Problems. 1. At half an atmosphere (barometer 38 cm.), what is the boiling- point? (Consult curve.) 2. If a thermometer in steam reads 99° (barometer 76 cm.), what Is the true 100° point and correction ? 3. If the reading in melting ice is 2°, what is the correction ? 4. If the reading in melting ice is — .1°, what must be done to a low reading of the thermometer ? 5. Find the true 100° point, the correction, and the correction to be used for low temperatures in the following thermometers : — Reading in Steam Centimeter Reading OF Barometer Reading in Melting Ioe a 101.3° 78.4 .3° b 100.9° 75.2 -.1° c 98.2° 74.0 -.4° d 99.8° 76.8 .0° e 100.2° 76.0 .2° f 100.1° 75,8 -.2° 9 99.4° 74.4 .1° THERMOMETERS. 203 6. Unless exact points in other parts of the scale are obtained, the corrections to be applied in the middle of the scale may be taken as the average of the 100° and the zero corrections. If thermometer a reads 98°, what is the temperature ? If a reads - 5° ? If a reads 45° ? If c reads 90° ? If b reads 104° ? 227. Practical Working of a Thermometer. — When a ther- mometer is cooling from a high temperature, the glass of the bulb does not contract at once and the zero-point changes slightly. Re-determine the zero-point. The column of mer- cury in the stem should be heated to the same temperature as the bulb before a reading is taken. Find what difference there is in the reading when the bulb only is in steam and when the whole column of mercury is in steam. The stem of a thermometer is usually sealed with no air above the mercury. Invert an all-glass thermometer, tap the top gently on a soft board, and watch the thread of mercury run to the top. Which is the more sensitive — that is, which will show the greatest difference in length of the thread of mercury for a given change in temperature — a thermometer with a large bulb or one with a small bulb ? One with a small bore or one with a large bore ? 228. Melting-points. — The melting-point of ice, made by freezing pure water, is fairly constant, varying only a few thousandths of a degree. Solid ice, for instance, melts at a little lower temperature than slush formed by carefully stir- ring water that is slowly freezing. The melting-points of many crystalline substances other than ice are also fairly constant when the substances are pure. Of these, crystallized sodium sulphate melts at 32.6° C. ; sodium thiosulphate (the ' hypo ' of the photographer), at 48.1° 0. ; and barium hydrox- ide at 78° G. Many other points of the scale between 0° and 100° may be found and marked on a thermometer stem by using other salts. The following exercise shows a simple method of calibrating a thermometer from two other points : — 204 PRINCIPLES OF PHYSICS. Exercise 27.1 TESTING A THEEMOMETEE FOE POINTS BETWEEN 0° AND 100° C. Apparatus: Test-tubes, 1 inch by 6 inches; wide-mouthed bottles; cotton wool ; pans of hot water ; thermometer ; recrystallized sodium sulphate and sodium thiosulphate. These salts are prepared from the commercial salts by melting, in a graniteware pan, five or more pounds of the salt to be purified with one-third its volume of water. After a few hours' cooling, remove the crystals. Put these in another pan, melt as before, and recrys- tallize ; then put in glass jars. Expose a portion of the crystallized sodium sulphate to the air and dry it. Mix 10 parts of the crystallized salt, powdered as fine as granulated sugar, with 1 part of the dry salt. Fill a test-tube half full of the mixture. Heat over a Bunsen burner till the mixture begins to melt and appear like slush. Add about one-tenth as much of the unmelted mixture. Place the test-tube in a bottle lined with cotton wool. Set th& bottle in a pan of water at about 50° C. Wash carefully the bulb of a thermometer, wipe it dry, iijsert it in the test- tube, and stir the mixture with it. Record the reading. A large number of thermometers may be tested in turn in the same mixture. When the mixture is nearly melted, place the bottle in a bath of water a few degrees below 32.5° C. At this temperature the sodium sulphate recrystallizes. (The word freezes is used exclusively for the crystallizing of water as it turns into i(* or snow.) More thermometers may be tested as the substance crystallizes. The tem- perature remains constant during the melting and the solidifying, if some of both the dry and crystallized form of sodium sulphate are present. 229. The 32.5° C. Point. — The melting-point of crystallized sodium sulphate is always the same (that is, constant), pro- vided the salt is pure. Just as the zero point of a thermome- ter is determined with great accuracy by noting its reading in melting ice, so the 32.5° C. point is determined by the reading of the thermometer in melting sodium sulphate. In a similar way, find the true 48.1° point, in a bath of melt- 1 This Exercise was outlined by Mr. J. B. Churchill, who discovered that many chemical compounds have definite melting-points. THERMOMETERS. 205 ing sodium thiosulphate crystals. Be careful to wash each thermometer before inserting it in a testing-bath, since very small impurities alter the melting-point. Barium hydroxide crystals have a melting-point of 78° C. 230. Effect of Pressure on the Melting-point of Ice. — A con- siderable pressure can be applied to ice without apparently changing the melting-point. Snow that is at the freezing-point is made into a snowball by pounding, and by continued pres- sure becomes a block of ice. Apply a heavy pressure to ice by a fine wire, to which a weight is hung. Support the ice on two ring-stands or boxes (Fig. 196). Over the ice pass a loop of fine wire, to which a weight equal to the breaking strength of the wire, is hung. The wire will not break, because the weight will be distributed between the two sides of the loop. When ^. ,„, ^ Fig. 196. the wire has passed completely through, notice that, while the ice has been melting a little all the time, the two pieces have frozen together solidly. Under the pres- sure of the wire the ice melted ; in doing so, it absorbed heat and lowered to a fraction of a degree below zero the temperature of the water formed. This water, escaping from under the wire, was no longer pressed upon, and froze at once. .Enormous pressures reduce the melting-point of ice very little, though enough to allow glaciers to flow and the snow to settle down and become ice under the pressure of more and more snow which falls on top. The ice-cap on Greenland and other Arctic lands is formed in this way from accumulations of snow. The ice slowly flows to the waller's edge and breaks off as huge icebergs. As they are formed of snow, what must be the melting-point of icebergs ? 231. Freezing-points. — The fresh water of rivers and lakes in winter freezes long before the salt water of the ocean. Salt 206 PRINCIPLES OF PBTSICS. melts the ice from a sidewalk, unless the weather is extremely cold. Pure water melts and freezes at 0° C. Almost all sub- stances in a pure state have a definite melting-point. Some substances, such as cast iron and platinum, have high melt- ing-points ; others, such as mercury, alcohol, and air, have low melting-points. Exercise 28. EFFECT OF DISSOLVED SUBSTANCES ON THE FREEZING-POINT OF WATEK. Apparatus: A beaker, or tin can; one-inch test-tubes; thei-mometer ; salt; snow or fine ice ; five per cent, seven per cent, and ten per cent solutions of common salt. Other salts may be tried. • Fill the can half full of a mixture of ice and salt. Test the zero point of the thermometer in ice. Find the lowest temperature of the ice and salt mixture. If necessary, make an opening in the ice and salt mixture and put in a test-tube containing 2 inches of a solution of salt. Insert the thermometer, and stir. Record the temperature every minute ; try to read to tenths of a degree. Remember that about once in ten readings the mercury column is likely to be on a whole number of degrees. Plot the results, as in Fig. 197. Vertical spaces represent degrees ; horizontal spaces, minutes. The curve may be somewhat like ABCD. AB shows the cooling down to the freezing- point. What is happening during the time from B toCi Why does the tem- perature again begin to fall at Ct What effect does the amount of the dis- solved substance have on the freezing- Fig. 197. point? 232. Dew-point. — In the warm, muggy days of summer, drops of water appear on the surface of a cold dish. This water does not come through the sides of the dish ; what is its source ? ^ Let each pupil make one set of observations of either pure water or one of the salt solutions. Compare the results. THERMOMETERS. 207 W D Thoroughly moisten a piece of muslin and suspend it from a board. Let the muslin reach well down into a test-tube of water, held in a wooden stand inside a glass jar {A, Fig. 198). The board, W, which should be well greased on its lower side, preveiits any change of air in the jar. Set up a similar apparatus, B, omitting the jar. Cover another jar, C, in which a dish D of calcium chloride or strong sulphuric acid has been placed. The cover, W, is made air-tight with tallow. After a time, the level of the water in the tube in A will gradually fall a little. Most of the water will disappear from B, because the air around it is continually changing and absorbing water. 233. Air saturated with Moisture. — Fill a bright calorimeter or tin can half full of ice water, or ice and salt, making it suffi- ciently cold so that moisture is deposited on it from the air in the room. Wipe off the moisture and hold the calorimeter in C, Fig. 198. No moisture is deposited. The air in G is dry. Then hold the calorimeter in A, first removing the muslin. Moisture will be deposited on the bright surface. Remove the calorimeter and allow it to warm so that moisture from the air in the room will not be deposited, then replace in A. Moisture will condense on the cold surface. The air in A is saturated with moisture, while the air of the room is only partly saturated. Air or an empty space can hold in an invisible form an amount of water depending on the temperature. The air is very seldom saturated with all the moisture it can hold. Dur- ing a long rain or fog, it becomes saturated ; in such an atmos- phere, water does not evaporate from wet clothing. The air in A is saturated, and, on being cooled, is unable to hold all the moisture. The air in C is dry, and would deposit no moisture, no matter to how low a temperature it might be cooled. 208 PRINCIPLES OF PHYSICS. Exercise 29. THE TEMPEEATTJKE AT WHICH MOISTURE IS DEPOSITED. Apparatus : A bright calorimeter ; a thermometer ; a bent strip of zinc. Fill the calorimeter, C, Fig. 199, half full of water. Suspend the thermometer, T, in it. Add ice water, or ice and salt, and stir continually by moving the bent strip of zinc, S, up and down in the water. Be careful not to breathe on C. Watch for the first sign of moisture deposited. An easy way to tell when moisture appears is to stand toward the light, and place a page of print facing C. When the reflection of the letters becomes blurred, the moisture has Fig. 199. begun to gather. Read the thermometer. At this point, stop adding ice, stir until the moisture disappears, and read the ther- mometer again. These two readings should not be more than a degree apart. The dew-point at the time of the experiment is the average of these two readings. Record the temperature of the room and of the outside air, and the condition of the weather. Copy in note-book the report of the nearest Weather Bureau station for the day on which the experiment is performed. On repeating the experiment, the temperature of the water in C should be reduced at once to within two degrees of the dew-point just obtained, and then the liquid cooled gradually till the moisture appears. It is instructive to take the dew-point in various parts of the building, in the cellar, and out of doors. These observations should be made about the same time, since the dew-point often varies greatly in a short time. 234. The Capacity of the Air for holding Moisture increases rapidly with, the temperature. At the freezing-point of water, a cubic foot of air, when saturated, holds about .1 g. of water, that is, about one drop. For each. 10° rise in temperature, the amount of water in a saturated space nearly doubles. At 40° C, one cubic foot of air can hold about 1.3 g., that is, about 1.3 cc, of water. A cubic foot of saturated air, if cooled to 0° C, deposits 1.3 — .1, or 1.2 g. of water. On cooling, the capacity of the air to hold moisture is decreased. Finally THERMOMETERS. 209 a point is reached where the air can hold only what moisture it has. It is then at the dew-point. Any further lowering of temperature causes some of the moisture to be deposited as rain, fog, or dew ; or if the temperature is below 0° C, as snow or hail. On the other hand, saturated air becomes dry by warming; not that it has any less moisture, but its capacity for holding moisture is increased and it absorbs, or evaporates, moisture from any source, — from vegetation, the surface of water, or damp cloth. During the process of evaporation, an im- mense amount of heat is absorbed. (See section 267, page 230.) 235. Wet-bulb and Dry-bulb Thermometers. — Suspend two thermometers, A and B, Fig. 200, some distance apart. Tie to the bulb of A some muslin in the form of a wick, long enough to reach into a dish of water, W. Before moistening the muslin, notice that both thermometers read alike. Fill the dish, W, and moisten the muslin with water that has been standing long enough to have the same tempera^ ture as the room. Record, each minute, the tem- perature of ^ and B, as long as there is any change. Then fan the thermometers, and continue the ob- "^ " servations till there is no change in either ther- '®' mometer. In the meantime, the dew-point should be taken by the method of Exercise 29 (page 208). The wet bulb is cooled by the evaporation of water from the muslin, and its temperature falls to some point between that of the air and the dew-point. The data in the table on page 539 of the Appendix are the results of experiments. The first column contains the temperatures of the dry bulb ; the second column, a number corresponding to each dry-bulb temperature. This number is used in multiplying the difference between the wet-bulb and dry-bulb reading, to give the number of degrees the dew-point is below the temperature of the air, as read by the dry bulb. 210 PRINCIPLES OF PHYSICS. When the dew-point is in the neighborhood of 0° C, the results of this method do not agree exactly with those of Exercise 29. Problems. 1. If the dry bulb reads 20° and the wet bulb 15°, what is the dew- point V 20 — 15 = 5° ; the wet bulb is 6° lower than the temperature of the air. The dew-point is still lower. Look in Appendix, page 539, for 20° ; next to it is the number 1.8. 5 x 1.8 = 9 ; therefore the dew-point is 9° below 20° C. ; that is, 20 - 9 = 11°, the dew-point. 2. Find the dew-point, when the thermometers read as follows : — Reading of Dry Bulb Reading of Wet Bulb a 20 17 b 20 10 c 20 4 d 30 22 e 15 12 f 10 9 22 22 236. Sensible Temperature. — The reading >of the wet bulb is practically the temperature that the air feels to us ; that is, it is the sensible temperature. Dry air feels cool, because of the cooling effect of evaporation of the perspiration from the skin. Set W, Fig. 200, page 209, in a glass jar that reaches above the bulb. Cover the top of the jar tightly, and half an hour later read A and B. The evaporation ceases as soon as the air in the jar becomes saturated. The bulb is no longer cooled, and registers the same as B does ; that is, it registers the tem- perature of the surrounding air. 237. Formation of Rain. — Air, by expansion, does work and becomes cooled (section 266, page 245). Suppose a mass of THERMOMETERS. 211 air rises ; it expands, and does work in pushing away the air around, and cools 1° C. for every 100 meters, or 300 feet, it rises. This is one reason why the air is cool at high eleva- tions. Should the cooling bring the temperature below the dew-point, a part of the moisture is condensed as rain or snow, and falls. Instead of trying to take a mass of gas up to a great height, the same result can be attained by removing the pressure by an air-pump. Leave a tumbler of warm water under the receiver of an air-pump for a few minutes; the air soon becomes saturated. Remove the tumbler and see that the receiver is dry on the inside. Exhaust the air rapidly ; the air expands, is cooled, and a dense fog appears. Open the inlet ; the air in the receiver is compressed and warmed, and the fog disappears. Problems. 1. Why is the dew-point on the ocean always nearly as high as the temperature of the air? 2. On a rainy day, why do wet clothes dry near a stove ? 3. Why should a thermometer be protected from rain if reliable indications of temperature are desired ? 4. Why does moisture form on a mirror when the mirror is breathed on? What would happen if the mirror were as warm as the breath ? 5. Does a high dew-point indicate a comfortable or an uncom- fortable day? CHAPTER XV. EVAPOEATION AND BOILING. 238. Dissolved Air in Water. — Set aside, in a warm place, two tumblers, one of freshly drawn water, and the other filled with water that has been boiled and cooled. From time to time look for the formation of bubbles. In which tumbler are there bubbles of air clinging to the sides of the glass ? Which has the pleasanter taste, boiled or unboiled water? Heat slowly a test-tube or flask of freshly drawn water; do not let it reach the boiling-point. Notice the small bubbles that form all through the liquid and on the sides of the test- tube. These bubbles are composed of the air that was dis- solved in the water. They rise slpwly, because they are small, and do not change much in size as they rise. 239. Evaporation. — As the temperature of the water in the test-tube approaches the boiling-point, hold a bright metal surface (C, Fig. 201) for an instant near the mouth of the test-tube. The metal surface may be a nickelled calorimeter in which there is cold water ; but it should not be cold enough to condense moisture from the air of the room. Instead of the metal sur- face, a plate of glass may be used. The moisture deposited comes from the surface of the liquid in the test-tube, and is noticed even at low temperatures, provided the metal or glass, C, is much colder than the water in the test-tube. This process of a liquid turning slowly into vapor is called 212 Fig. 20 1 . EVAPORATION AND BOILING. 213 evaporation. It takes place at all temperatures, but only at the surface of the liquid ; even in cold weather, snow wastes away by evaporation. It takes place more rapidly at high than at low temperatures. The effect of pressure is shown by placing a flask of warm water under the receiver of an air- pump and exhausting the air, but not carrying the exhaustion far enough to make the water boil. Evaporation takes place most rapidly at low pressures. 240. Boiling. — To study the phenomena of boiling, it is better to begin again with cold water. Review the phenom- ena noted in studying the boiling-point. Often, before the air bubbles have entirely escaped, one or more large bubbles form over the surface of the glass heated by the flame and at once collapse. There is a clicking, or singing, sound. As the water becomes warmer, these large bubbles, which are of steam, rise into the cold liquid above, and there condense, warming the liquid. Watch the level of the water in the test-tube as the steam bubbles are formed and condense. As a bubble of steam is formed it lifts the water. The bubble of steam condenses; the water falls, and, having nothing to cushion it, strikes a sharp blow, which gives rise to the click- ing, or singing, sound, and is, in fact, a water-hammer. The process going on is a steam heating plant on a small scale. Steam generated at the heated surface rises and is condensed by the colder liquid above. This upper layer of liquid soon becomes warmed to the boiling-point, and the bubbles of steam, not being condensed, escape into the air. The liquid is then said to boil. 241. Pressure of Steam Bubbles. — Watching the bubbles of steam at the stage in the heating when they immediately con- dense, what has a little bubble to do as it turns into a large bubble of steam ? It has to make room for itself. It, then, or its vapor, must exert enough pressure to lift the column of 214 PRINCIPLES OF PHYSICS. water above it and the air above that. In a shallow vessel the pressure caused by the depth of the water is comparatively small ; but the air pressure on the surface of the water at the sea level is, on the average, equal to that caused by a column of mercury 76 era. high, or 15 pounds to the square inch, or 1000 g. per square centimeter. If the atmospheric pressure is less, then the vapor pressure of the little drop of water that is about to turn into steam does not need to be so great, and therefore the water does not need to be heated to so high a temperature to cause it to have sufficient pressure to lift the water and the atmosphere above it and to expand into steam bubbles. At 100° C, — the boiling-point of water when the barometer is 76 cm., — 1 cc. of water forms about 1700 cc. of steam. Consult the results of the experiment on relation of pressure to the boiling-point (Exercise 26, page 198). Select from them one set of readings at a high temperature and another at a low temperature, and arrange as follows : — Barometer // Pressure Barometer-^ T 76.0 76.0 32.7 74.3 43.3 1.7 85° 20° In the first column is given the height of the barometer ; in the second column, the height of a barometer in which a little water floats on the mercury. The reason the mercury column in this second tube does not stand as high as the other is that the water vapor on top exerts some pressure. The pressure of this vapor increases with the temperature. At 100° C. (a tem- perature that we could give by supplying the jacket with steam or hot brine), the vapor pressure would drive all the mercury out of the tube and balance the atmospheric pressure without the help of a column of mercury. The third column EVAPORATION AND BOILING. 215 gives the amount of the depression caused by the pressure of the water vapor. This column shows the pressures exerted by water vapor at temperatures of 85° and 20° C. Other liquids give different results. Boiling takes place when the vapor pressure of a liquid is greater than the pressure on the liquid. In Exercise 26, page 196, it was suggested that fire-brick (many other substances would do as well) be put in the water to insure boiling. The temperature of water, especially if it has no air dissolved in it, rises a little above the boiling-point, and then the water boils violently and throws the liquid The liquid is said to bump. 242. Absorption of Heat in Evaporation and Solution. — Heat is absorbed when a liquid evaporates or a substance dissolves. The following is one way of measuring differences in temperature : — Through the rubber stoppers of two test-tubes insert a small- bore glass tube, bent in the shape indicated in Fig. 202, and carrying an index of colored water at /. The sensitiveness is increased by using large test- tubes with a very small-bore glass tube. Breathe on one test- tube ; hold the hand on one ; p apply water a little warmer than l^S the room. What happens at "^ first. Blow on one test-tube; does the movement of the index indicate that evaporation absorbs heat? Place the apparatus in two dishes of water, as in Fig. 202. When the index has ceased to move, stir salt or ammonium nitrate in A. Does the solution absorb heat ? Empty A and B, and refill with boiling water. Place burners under each, and as soon as the index is at rest, add salt to A. When boil- ing begins again, which is the warmer, A oi B7 Fig. 202. 216 PRINCIPLES OF PHYSICS. ^ Exercise 30. BOILING-POINTS OF WATER SOLUTIONS AND OF OTHEE LIQUIDS. Apparatus: Test-tubes 1 inch by 8 inches; thermometers reading to 120° C; ring-stands and clamps ; the test-tubes may be fitted with two-hole rubber stoppers. Fill the test-tubes one-third full, putting common grain alcohol (ethyl alcohol) into one, wood spirit (methyl alcohol) into a second, a strong solution of calcium chloride into a third, and pure water into several others. Suspend the test- tubes by ring-stand clamps, as in Fig. 203. On the base of each ring-stand put a number and the name of the substance in the test-tube. Put a thermometer in each test-tube, and boil the liquid. Record the tem- perature of each liquid when it boils. Add a little alcohol to one of the test-tubes of pure water ; sugar to another ; and salt to another. When the solutions boil, again read the thermometers in these tubes. Add more alcohol, sugar, and salt, and find the boiling-points again. A large amount of sugar, one-third or one-half the bulk of water, should be added to the tube of water in which the boiling-point of the sugar solution is to be determined. Hi Fig 203. 243. Distillation. — In the experiments on boiling and evap- oration it may have been noticed that while the water was sometimes cloudy, the condensed steam was clear. Into a boiler or kettle put water and either salt or sugar. Boil, and allow the steam to condense in a can or tumbler. A, Fig. 204, and to drip into B. A may be held with a cloth. Is the con- densed water either salt or sweet ? With the exception of liquids boiling below 100° C, any mixture of water on boiling gives off steam that condenses to pure water. Rain, being the condensation of water vapor evaporated from plants, bodies of water, and the soil, is also perfectly pure, excepting Fig. 204. EVAI^OBATION AND BOILING. 217 a few minute impurities it may wash out of the atmosphere in falling to the ground. Replace the metal rod of the linear expansion apparatus (page 185) by a long glass or metal tube, as in Fig. 205. Pass in cold water at B, and run a rubber tube from ^ to a sink or jar. The jacket of the condenser is supported in an inclined position by a clamp on a ring- stand. This forms a simple model of a still and condenser. By burning one pound Fig. 205. of coal, ten pounds of water, more or less, are turned into steam. The water into which the steam condenses is pure, and is fully as wholesome as spring or well water, though it does not taste so good. On long sea voyages, either a sufficient supply of fresh water must be carried or the salt water must be distilled. Every bit of space on a vessel is valuable. Is it more economical of space to carry fresh water or to carry coal to be used in a still ? 244. Difference between Evaporation and Boiling. — Define boiling. How does it differ from evaporation ? Any liquid, when boiling, turns into vapor by bubbles of steam rising from the hot surface of the boiler and by evaporation from the surface of the liquid. In a given time more water can be turned into steam by boiling than by evaporation. Engineers say that a pound of coal evaporates ten pounds of water, when they really mean boils and evaporates. The word evaporate has usually this meaning in mechanics. 245. Distillation of Alcohol, etc. — In the dipper used for heating the metal in the exercise on specific heat (page 221), put cider, or a mixture of alcohol and water, or a dilute solu- 218 PRINCIPLES OF PHYSICS. tion of molasses and water that has been allowed to ferment some, but not enough to turn the alcohol into vinegar. With a one-hole rubber stopper and flexible tubing connect the dipper with the condenser. Heat the water in the boiler nearly to boiling. Then, and not till then, put the apparatus in place, as in Fig. 206. Nearly close the opening S, and turn the gas down so low that the water in the boiler does not quite boil. See if the liquid dripping from •^'S' 206. ^;he glass tube in the condenser will burn. Alcohol, having a boiling-point of 78.6° C, boils before the water with which it is mixed. However, it carries with it a little water vapor, for water vaporizes rapidly at 80° C. without boiling, and a mixture of alcohol and water is condensed in the glass tube. Redistilling this product would increase the proportion of alcohol in the distillate. Crude petroleum is composed of many substances of different boiling-points. As the crude oil is slowly heated one sub- stance after another comes off in turn, each at a higher boiling- point, and is condensed in a tube, which for convenience is usually made into a coil and called a worm. Mercury mixes with lead, zinc, titi, and other metals. It is best purified by distillation. 246. A Heat Unit, or Calorie. — The amount of heat required to warm 1 g. of water 1° C. is called a heat unit or calorie. How many heat units are required to warm 1 g. of water 6° C. ? (6 heat units.) How much heat is absorbed by 5 g. of water in rising 6° in temperature ? (5 x 6 = 30 heat units.) How much heat is given out by 40 g. of water in cooling 25° C. ? How many heat units are needed to raise the tem- perature of 1 liter of water from 10° to 50° C. ? EVAPORATION AND BOILING. 219 247. Mixing Waters of Different Temperatures. — Find the exact size of a small tin box, to be used as a measure, holding 100 to 120 g. Weigh the box empty ; then fill with water and ■weigh. The difference in the two weights is the number of cubic centimeters the box holds. Find the temperature of a dish of hot water, and that of a dish of cold water. Fill the measure with the hot water, and empty into a calorimeter, which may be a large tin can or a glass beaker. Pour in a measure of the cold water. Stir with a thermometer, and record the temperature. How does the temperature of the mixture compare with that of the hot water ? with that of the cold water ? How many degrees did the temperature of the cold water rise ? How many degrees did the temperature of the hot water fall ? Find the number of heat units given out by the }iot water in cooling, by multiplying the weight by the fall in temperature. Find the heat units absorbed by the cold water. Repeat the experiment, using two measures of hot water to one of cold, or two measures of cold water to one of hot. In which case would the temperature be the higher ? Assuming that there is no loss or gain of heat from the outside, the amount of heat lost by the hot water must equal the heat gained by the cold water. Suppose 50 g. of water at 80° C. and 200 g. of water at 6° C. were stirred together, what was the temperature of the mixture ? Call this t. If the hot water cooled from 80° to t, or (80 — t) degrees, the heat given out was the weight, 50, times (80— t). If the cold water warmed up from 6° to t, or (t — 5), the heat absorbed was 200 (t — 5). The hot water must have lost as much heat as the cold water absorbed ; therefore 200 - 5) = 50 (80 - ty 200 t - 1000 = 4000 - 50 1 260 1 = 5000 t = 20°. 220 PRINCIPLES OF PHYSICS. Assume that Wj = weight of hot water, ti = temperature of hot water, w = weight of cold water, to = temperature of cold water. Then, using letters, the rule. Weight of cold water times rise in temperature = weight of hot water times fall in temperature, becomes , /* t\ it is w X (< — B ^0 yiL ^-B, placing the arrow J»£f ^ at different distances Fig. 237. irom the mirror. Draw the image of the arrow GH in the mirror EF. What would be the image of LM in the mirror IK? LIGHT. — REFLECTION. 269 302. Apparent Positions of Images. — In walking near a mirror a person sees his own image approach ; in what direc- tion is he going ? A man passing down an avenue sees his own image apparently coming into the avenue from a cross street ; how must a mirror be placed in a corner store to give this illusion ? What apparent effect on the depth of a store does a rear wall entirely of mirror-glass have? How can a store be made to look wider ? Stand before a mirror in which you can see your full height (the glass windows of a laboratory case may do). Let another person place a bit of paper on the mirror where the image of your forehead appears to be, and a second bit of paper at the apparent position of your feet. Measure the distance between the pieces of paper and compare this with your height. Move nearer to and farther away from the mirror, to see what effect distance has upon the height of your image. In case a large mirror is not available, use a smaller one, and find the smallest height of mirror in which the whole of your head can be seen. 303. Incident and Reflected Rays. —Let A, Fig. 238, be a point, and / be the image of that point in the mirror, MM. As light moves in straight lines, the ray AD (called the incident ray), after re- flection from the mirror, becomes DL (called the re- flected ray). Compare the distance AD+DL with the distance IL. Which would go the greater distance, a ball thrown from / to L, or one thrown from ^ to D and from D bounced, or reflected, to L? The -image, /, it will be seen, is apparently as far from L as the light has to go in its way to the mirror at D, and from D to L. Draw a perpen- 270 PBINCIPLES OF PHYSICS. dicular, DN (call it a normal, — normals are nothing more noi less than perpendiculars) ; measure the angle, LDN, and com- pare it with the angle, ADN. What is the law connecting the angle of incidence and the angle of reflection ? Test your conclusion in this way : Arrange the mirror and paper as in section 299, page 265. Draw a line, AB, Fig. 239, at any angle. Draw a line, BC, as the con- tinuation of the re- flection of AB. Mark the edge of the mir- ror on the paper. Draw a perpendicular, BN. What name has such a line in physics ? Cut with a sharp knife along AB and BC. Crease the paper at BN, and compare the angle of inci- dence, ABN, with the angle of reflection, NBC. Print a word — "school," for instance — in large letters on a card. Hold the card toward a mirror ; in the note-book draw the word as it appears in the mirror. Hold a picture toward a mirror and describe the image. Fig. 239. 304. The Brightness of a Reflection depends upon the angle at which the light strikes the mirror. Hold a candle over a cup of water and look almost perpen- dicularly down at the reflection. Holding the candle at the same distance from 'the cup, but almost on a level with the water, view the reflection. Decide in which case the image is the brighter. ^ M Mx Fig. 240. LIGHT. — BEFLECTION. 271 Hold a mirror, M, Fig. 240, vertically near a bright light, L (a ray of sunlight is preferable to a candle). Turn the mirror in the direction of the arrow, and follow the motion of the reflection as the mirror is turned through a right angle (90°) to the position Jfj. Which moves faster, the reflection or the mirror ? In turning the mirror through a right angle, what angle does the reflected ray go through ? 305. Location of an Image by Shadows. — Arrange a mirror as in section 299, page 265, using a candle for the object. Mark the edge of the mirror and the position of the candle on the paper on which they rest. At any points some distance apart, as A and B, Fig. 241, place tall pins, as nearly vertical as possi- ble. Look at the shadows the pin at B appears to cast. One, which may be called the direct shadow, goes to the mir- ror and is re- flected in the path BCD. Do not consider this at all, but trace on paper the other shadow, BE, which appears to be caused by a candle you see in the mirror. Trace a similar shadow for the pin at A. Remove the mirror, and continue the lines till they meet behind the mirror. Place a candle on this point, and notice if the real candle behind the mirror line casts the same shadow lines that the image candle cast before the mirror was removed. m^ Fig. 241. 306. Reflections in Two Mirrors. — Lay two pieces of mirror on a table, end to end. Slowly lift the outer ends. How many 272 PRINCIPLES OF PHYSICS. reflections of the face can be seen. Replace the mirrors, and lower or raise them slightly, until the image of a pencil held in various positions over and nearly parallel to the mirrors does not look bent at the edge where the two pieces of mirror join. Look along the surface of the mirrors, and see if they form, as it were, one flat mirror. Exercise 38. UIBBOBS AT BIGHT ANOIES. Apparatus: Two mirrors, each arranged on a block of wood, as in Exer- cise 37, and set at right angles on a sheet of paper, as shown in Fig. 242. Draw an arrow, A B, Fig. 242, between the mirrors. Place a pin at the head of the arrow, and by three sight-lines locate an image of the pin behind M. Then, in the same way, locate another image behind il/,. Locate a third im- age somewhere be- hind both. Draw lines at the edges of the mirrors. Put the pin at the other end of the arrow, and locate the three Fig 242. images. Having the position of the ends of the arrow, draw the three images of the arrow. The record of the exercise can be made as follows : — Lay down a sheet of paper in the note-book, with one corner near the binding. Place two mirrors, M and M^, on the edges at this corner, letting the mirrors touch ; remove the paper, and mark the edge of the mirrors. On this diagram make the record of this exercise. Write a word on a card. Hold it toward M, and describe the reflection ; then hold it toward M^ and describe the reflection. Finally, hold the card in the position of the arrow AB. The first two images were reversed, being reflected once only. The third image looks like the writing itself, and is reflected twice, first from one ^iv P^V y^^^ ■"^o. ■ ^^\m:>--''^ — 1^5^^-^^-^k^ ^H m— Z^ B , , . 1- M ^^H ^^^^E«=^-*^ ^ -^^^ =* -^_-^^^^^ ^"^-S^ ^— "i LIGHT. — REFLECTION. 273 mirror and then from the other. Push a card over the face of the mirror M, toward the intersection of the mirrors. What images are covered up when the card is entirely in front of M? When only partly in front of M '! Try the same with il/,. The third image is reflected from one mirror to the other, and then to the eye. Replace the arrow, AB,hy a. cork or spool, colored red on one side, blue on the other. (Instead of being colored, the cork may be marked A on one side and B on the other.) Let the red side face one mirror and the blue side the other. Bring the mirrors a little nearer together, making the angle less than a right angle. What does the third image appear to do ? Make the angle 60°, and count the images ; then 30°. Does the angle bear any relation to the number of images ? Notice that the images seen in one mirror are red, then blue, and so on. If the red face is toward M, the first image in M is red ; the next image is blue, and, as the blue side faces away from M, the light from the blue side must be reflected first from the other mirror, M^. How many images can you see in two large mirrors that are par- allel? Hold a card, on which the word " on " is written, between the mirrors, placed at different angles, and notice which images are reversed and which are direct, that is, which images spell " on," and which spell "no." 307. Images in Parallel Mirrors. — Keep in mind that an image in a plane mirror appears as far away from the observer as the light from the object has to travel before reach- ^ ing the eye, and find the position of images of an object between parallel mir- pi 243 ' rors. M and J/l, Fig. 243, are parallel mirrors 10 feet apart. Four feet from M stands a card, C, with the word " Moon " printed in red on the side toward M, and " Moon " printed in blue on the other side. The first image in If is 4 feet behind M; the first image in Ml is 6 feet behind My The second image in M is blue, and is not reversed. The light travels from C to Jfj 6 feet, and from Ml to M 10 feet more (as in Fig. 244, I) before reflection 274 PRINCIPLES OF PBTSICS. to the observer. The image, therefore, appears 10 + 6, or 16, feet behind M. In the same way, the light from C goes to M 4 feet, and thence, C after reflection, 10 feet to Mt (Fig. 244, II), before it is re- flected to the eye. The second image behind M^ appears 14 feet behind the mirror Mi. Locate two more images behind each mirror. The images in M, including the first, are, respectively, 4 feet, 16 feet, 24 feet, and 36 feet behind the mirror; those in Jf, are, respectively, 6 feet, 14 feet, 26 feet, and 34 feet behind the mirror. A man stands 10 feet away from and facing a mirror; a second mirror is 20 feet behind him, and enough out of the exact parallel so that he can see several images of himself. How far behind the mirrors do his- reflections appear ? Find three in each mirror. In which do the image of the man face him? Account for the fact that as a man walks toward a mirror his image approaches him. Place a photograph halfway between parallel mirrors, 4 feet apart, facing one of them ; locate four images of the picture side of the photograph. If two parallel sides of a room 24 feet wide are mirrors, and an arc light is hung 8 feet from one wall, where do the nearest two images seen in each mirror appear to be ? 308. Kaleidoscope. — Place two mirrors facing each other and meeting at an angle of 60°. Push a pencil between them. Put several coins, beads, or bits of colored paper between them, and note the appearance of the reflections. This forms a kaleidoscope, the construction of which is somewhat like Fig. 245. The mirrors meet at an angle of 60° (one-sixth of a circumference), or at an angle that is one-fifth, one-seventh, or any other even division of a circle. The end of a kaleido- LIGHT. — REFLECTION. 275 Fig. 245. scope is made of two layers of glass, with beads between them ; the outer layer is usually of ground glass. A covering is wrapped around the mirrors, covering the __ _-^ open space between the edges of the mir- rors. The observer looks in at the open end. On turning the apparatus the beads fall into different ar- rangements, which are reflected as sym- metrical figures. Place three mirrors as in Fig. 246, facing inward. Hold them in place by rubber bands, and examine the reflections of a bright object at the end B. Looking in at A, notice that a set of images is formed in each corner. What is the angle of the mirrors ? A triangular Fig. 246. 60° prism of the form AB is covered with paper, except the ends. On the end B make a figure or letter with ink. Holding the prism so that the end B is well lighted, look in at the end A, to the three corners of B, in turn. The inner surfaces of the glass sides of the prism act as mirrors, and the light is said to be internally reflected. It will be noticed that the image farthest from the object is faintest. This is because the last image has been reflected several times, and has lost light at each reflection. X CHAPTER XIX. LIGHT. — EEIEAOTIOir. 309. Refraction by "Water. — Hold a pointer or a pencil in a jar of water, perpendicular to the surface. Look down through the water at the part of the pencil under water. Slant the pencil a little ; then more and more. Notice that it looks as if it were bent or broken at the sur- ^,'' face of the water. ^ ^^■'' In a pan, close to one side, put a X^f' bright coin, 0, Fig. 247. Stand in - —<—=-/ such a position that looking along Pig 2^j BA the coin is barely out of sight. Keep the head in the same position while some one pours water into the dish until the coin be- comes visible. Obviously, the coin is not moved. To see where the change of direction of the light from the coin occurs, place in a battery jar. Fig. 248, an apparatus consisting of a vertical strip of board nailed to a heavy cross-piece, which rests on the edge of the jar. At B, which is about a centimeter below the lower edge of the cross-piece, with a double-pointed tack fasten a piece of wire solder, ABC. Straighten the part AB\ fasten A to the Fig. 248. board. Fill the jar with water to B. Look down from such a position that AB appears end on, or as a point only. Bend BG till AB and BO look like one straight line. Mark roughly the water-line, and remove the apparatus from the jar. The light travels in a straight line in water or in air ; but in passing from water to air the direction changes. 276 LIGHT. — REFRACTION. 277 Repeat the experiment once or twice, and in one trial let the part AB have a considerable slant (as in Fig. 248), 45° or more from the perpendicular. The more the ray of light repre- sented by AB slants in water, the more the ray is bent on leaving the water. Exercise 39. INDEX OF EEFEACTION OF WATEE. Apparatus : Battery jar ; pins ; sheet of paper ; centimeter rnle ; thin board nailed to a heavy cross-piece of wood. Stick a bright pin, A, Fig. 249, near the lower corner of the board, and put the board in the battery jar. Fill the jar nearly to the top with water, and mark the water-line by pins B and C. Add or take out water, or tip the jar by placing a few thicknesses of card under the bottom, till B and C are wet, but not covered by the water. Look into the jar on a level with the water. When prop- erly adjusted, the water just slants up on each side of the pin, without covering the top, as in Fig. 250, whei-e the pin B is shown touching the surface of the water, WW. Standing on one side, near C (Fig. 249), look down into the jar in a slanting direction at the pin, A. Keeping the head steady, put a pin (D) in the board a little above the water. jiii|iiiifiin i ' ' 1 n i Fig. 249. w If this pin does not cover the image of A, move the head un- W til it does. Put another pin, E, exactly in line with D and the image of A. Remove the board, wipe off the water, and hammer in the pins tightly; see that they are straight. Lay a ^-3-tO^ Fig. 250. 278 PBINCIPLES OF PHYSICS. Fig..25l. sheet of unsized white paper over the pins. Break the paper where the pins touch, and press it down to the board. Lay a ruler close to the pins BC, and draw a line with a sharp pencil, held vertically (Fig. 251). This line is the water surface. From the water surface the light went in a straight line to D and E. Lay a ruler against D and E, and draw a line from E through D to the water-line at F. The bending of the light took place on leaving the water at F; but in the water, from A to F, the path of the light from A was straight. Draw FA. The path of light from the pin A was to F, and thence along the line through D and E to the eye of the observer. Remove the paper and lay it flat on the table. Erect a normal, FN, Fig. 252. This is a line perpendicular to the water surface. With i^ as a centre, and any radius, draw a circle. From