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VVRfWHOIfr^ QC 133 CS" •» < r t JOHN LOVELL & SON, Printers, Montreal. 0ONTE)NTS. ' Mass, Density, &c. paob 1. The Balance 1 2. Specific Gravities 6 3. The Specific Gravity Bottle 8 4. JcUy'8 Spring Balance 10 5. Mohr*ii Specific Gravity Balance 13 6. Nicholson's Hydrometer 14 Distance, &c, 7. The Micrometer Calipers 16 8. The NTernier Caliper 18 9. Micrometric Measurement of Distances 19 Area, &o. 10. The Planinieter 21 11. The Mechanical Integrator 22 Volume. 12. Measures q^ Volume 24 Motion, Forge, &o. 13. Atwood's Machine (E) 27 14. Atwood's Machine (P) 32 16. The Inclined Plane , 37 16. Centrifugal Force 41 17. Statical Friction 43 18. Mechanical Equivalent of Heat 44 Oscillations, &o. 19. Torsion and Moments of Inertia 44 20. Maxwell's Vibration Needle 49 21. Bifilar Suspension 49 22. Reversion Pendulum 50 23. Simple Pendulum 54 T Contents. Maouinbs. paob 24. Simple Machines... 58 Pneumatics, &o. 25. Mercury Barometer 60 26. Aneroid Barometer 62 27. Boyle's Law 64 28. Wet Bulb Hygrometer 68 29. Stereometer ( Volumenometei ) 72 Api'KNDix I.— Theory of the Planimeter 73 •• II. — Theory of the Mechanical Integrator 79 1 1 MATHEMATICAL LABORATORY, I. THE BALANCE. Experiment. — To use the balance in determining the mass of a given object. Apparatus. — A balance. Box of weights. Camel's hair brush. A brass weight marked . . . grains. Method /. — Level the case if necessary (probably it is sufficiently level). Dust the pans with the brush. Bring the pans to rest, turn the key gently and set the beam swinging. The pointer would move backwards and for- wards along the scale for a long time ; do not wait for it to come to rest in order to determine the position of rest ; find this by observing the extreme positions of the pointer on the scale. Read the scale from left to right, counting each whole division as lo. Take an even number of conse- cutive readings on one side of the scale and an odd number on the other (say 2 and 3, or 3 and 4) ; find the mean of each set and then the mean of the two results ; this will give the position of rest of the pointer. It is best that the pointer should move over about half (the middle half) of the scale. Arrest the beam when the pointer is near the middle of the scale in order to avoid jarring the beam. The beam must always be at rest when any change is made in the body to be weighed, or in the weights, or in the position of the rider. a MATHEMATICAL LABORATORY. Again, free the beam and repeat your work. Take as the position of rest the mean of the several results. Enter the results thus :— Experiment No. i. I. First trial. Readings on left. Readings on right. Mean . . . Mean . . . Mean of both ... Second trial. Readings on left Readings on right. Mean . . . Mean ... Mean of both . . . Mean of the two trials ... //. — To find the sensitiveness. Place a rider at the division i on the beam. The mass of a rider is i centigramme. Hence, putting the rider at the division i is equivalent to putting one milligramme (.001 gramme) in the pan. Set the beam oscillating and again find the position of rest. The difference between this result and that of I. gives the sensitiveness of the bal- ance for I milligramme when the pans are empty. At least two or three determinations should be made. The sensitiveness will probably not be quite the same ^hen the pans are loaded, but the result you have now obtained will be of use in checking your subsequent weighings. THE BALANCE, Enter results thus : — Experiment No. i. II, Enter readings as in I. Position of rest is now ... " •' " was ... .'. sensitiveness for i milligrarome s . . . Ill, — To weigh the given body (the . . . grain weight) and hence to calculate the number of grains in a gramme. Place the body in one of the pans (say the left) and the gramme weights in the other. Always move the weights with the forceps, never with the fingers. After putting a weight in the pan turn the key until the pointer y«x/ begins to move. The direction will show whether the weight is too large or too small. Continue to adjust the weights until you are within a few milligrammes of the true resuh. The final weighing must be made with the rider, the case being closed. Determine the position of rest with the rider in one position, then shift the rider along the beam, and again find the position of rest. One of these two positions should be on one side and the other on the other side of the posi- tion of rest when the pans are empty. From these results find the number of milligrammes to be added to the weights in the pan. Enter results thus : — Experiment No. i. III. Position of rest with rider at . . . is . . . (( (( « « (( (( t . . l9 • • . pans empty is . . . MA THE MA TIC A L LA BORA TOR V. I .'. number of milHgrammes to add to weights in pan Weights in pan . . . Total weight . . . Hence i gramme « —- * ■ . . . grains. /^. — To weigh the given object (a piece of wood marked i), the result to be corrected for the buoyancy of the air. Find the apparent weight as in III. Each side is buoyed up by the weight of the air displaced by the bodies in the pans. Let the apparent weight be ff^ grammes. Then the weight of the given body is very nearly IV grammes. IV Hence its volume is — c. c. nearly, s being the number of grammes in a c. c. of tne wood. (This may be assumed as .56). Hence for the air displaced by the body the cor- W rection is + (.0012) — , .0012 grammes being assumed as the weight of i c. c. of air. Similarly the correction for the IV air displaced by the brass weights is — (.0012) — where Ji is the number of grammes in i c. c. of the brass, which may be taken as 8.4. Enter results thus : — Experiment No. i. IV. Apparent weight ... Correction for air displaced by the body, + . . . " . " " " weights "... Total correction ... Corrected, weight ... (Results of parts I, II, III, IV of this Experiment to be returned together. Each studem to sign his name.) SPECIFIC GRAVITIES, 5 2. SPECIFIC GRAVITIES. ExpiRiMENT. — To find the specific gravity (1) of a body which sinks in water, (II) of a body which floats in water, (III) of a liquid. Apparatus. — An ordinary balance arranged to weigh a body when immersed in water. Distilled water. Silk thread. Copper or other wire, Met,.jd I. — Place the body (brass weight marked 2) in the balance pan, and weigh it in air. Call this weight W\, Attach the body, by means of a light string or wire, to the hook above the pan, and weigh it again. Let the weight now be W^. Keeping the body still attached to the hook, bring underneath the vessel of distilled water, having pre- viously exhausted the air from the water. See that no air bubbles remain attached to the body when thus immersed in water. It may be found that the beam will not oscillate freely when the body is in the water; if so, adjust the weights so as to bring the pointer to the position of rest when the pans are empty, without depending upon the method of oscillations. Let the weight in the water be W^^ Then W%^ Wz is the weight of the displaced water. Hence the specific gravity of the body is m- ^3 All weighings must be corrected for displacement of air whenever such correction is appreciable. The IV2 differs from the JVi by the weight ot the string or wire. IVs must be corrected for the weight of water displaced by the string or wire. The specific gravity of silk is about I, that of iron about 7^, that of copper about 9. 6 MA THEM A TICAL LA BORA TOR K. The body and the water should be of the same temper- ature as the surrounding air. Let this be t^ C. The above value of the specific gravity is that of the body when referred to water at the temperature /<^. To reduce it to what it would be if referred to water at 4" C. we must mult' ly by the specific gravity of water at t°. Let this be calleu p. Then the specific gravity of the body at the temperature t^ is * The water should be freed from air by boiling, or by means of an air pump. • Find the specific gravity of the given brass weight, marked 2, by two determinations, {a) hanging it up by a silk cord, and ib) by a copper wire. Enter results as follows : — Experiment No. 2. I. (a) With silk cord. Observed. [ Corrections if any. Corrected values. I , ' / = . . . Specific gravity = ' J ' x ... = ... (Enter b in same way.) SPECIFIC GRAVITIES. ■ 11. — When the body is lighter than water it must be pulled down into the water by means of a sinker. Weigh the body (marked 3) first in the scale pan (W^i). Attach it and the sinker ( 2) to the balance hook, the sinker being below the given body. Weigh the two ivith the sinker in water ( IV2)' Add more water so as to cover the given body and weigh again ( W^s). Then the specific gravity of the body at the observed temperature when referred to water at 4^ is IV^ IV2 - ^K3 X P where p is the specific gravity of water at the temperature of the experiment. * Find the specific gravity of the given body using silk thread. Enter results as under I. ///. — To find the specific gravity of a liquid, take a heavy body (" sinker ") which will sink in the given liquid as well as in water. Attacli the sinker to the hook of the balance and weigh it in air (IVi), Next weigh it in water ( IV2) . Dry the sinkei and weigh it in the given liquid (IVi). Then IVi - IV3 and IVi - IV2 are the weights of equal volumes of the given liquid and water. Hence, the specific gravity of the given liquid at the temperature of the experiment, referred to water at 4° is IVi - ATa ^ where P is the specific gravity of water at the temperature of the experiment. 8 MA THEM A TICAL LABOR A TOR Y. Find the specific gravity of the given liquid, using silk thread to suspend the sinker. Enter results as under I. (Each student to sign his name.) I 3. THE SPECIFIC GRAVITY BOTTLE. Experiment. — To find, using a specific gravity bat- tle, the specific gravity (I) of a liquid, (II) of a solid in fragments. Apparatus. — A balance. A specific gravity bottle. Distilled water. Method I. — To find the specific gravity of a liquid. Weigh the bottle ( IV{). The specific gravity of the glass may be assumed to be 3. Hence, find the correction to the weighings for the air displaced by the glass. (It may not be appreciable.) Fill the bottle with water; insert the stopper, wipe the outside dry and weigh {W^). Empty the water, fill with the given liquid and weigh (PK3). Then the sp. gr. ^ is The given liquid and the water are assumed to be of the same temperature, and the above fraction represents the specific gravity referred to water at the temperature of the experiment. To reduce to water at 4° C. multiply by p the specific gravity of the water at the temperature of the experiment. f ■f THE SPECIFIC GRAVITY BOTTLE. Enter results thus : — Experiment No. 3. I. Wx^ CO ff^3 = . . . , Temperature = ... p for this temperature = . . . Wz- Wx rrected u tt lue = //.—To find the specific gravity of small fragments of a solid. ' Weigh the fragments (IVx ). Fill the bottle with water, wipe dry and weigh (IV2). Put the fragments in the bottle, fill up with water if necessary, wipe dry and weigh { PV3 ). Then IVi s = Enter results thus : — EXIERIMENT No. 3. II, lVi= ... , corrected value = . . . li'2 = . . . JV2 - J^3 corrected for air displaced by difference of (brass weights «... Temperature «... p for this temperature = . . . ' s — (Each student to sign his name.) w 10 MATHEMATICAL LABORATORY. 4- JOLLY'S SPRING BALANCE. Experiment. — To find (I) the weight of a small body, (II) the specific gravity of a small body, (III) the specific gravity of a liquid. Apparatus. — A scale of millimetres is etched on a ver- sical strip of mirror glass in front of which hangs a long spiral spring carrying two pans. A white bead is strung on the wire which connects the upper pan to the spring; the top of the bead corresponds to the pointer of an ordin- ary balance, and in reading its position on the scale the line of sight must pass over the top of the bead and the top of its image in the mirror. In this way any error of parallax in reading the scale is avoided. The small plat- form, adjustable as to height, is intended to carry a vessel of distilled water, or of other liquid, in specific gravity de- terminations. The spring is intended to support weights up to 6 grammes, but a more delicate one carrying up lo 3 grammes, and a less delicate one carrying up to lo grammes, are provided, as is also a small glass sinker, which may be suspended in the place of the two pans. Method /. — To find the weight of a given small object. Place it in the upper pan. Give a little support to the pan at first and let it down gently. Read the scale when the spring comes to rest, or deduce the position of equilibrium from the oscillations along the scale. Now remove the body and in its place put weights from the box until the bead comes to the same position as at first. Enter results thus : — Experiment 4. I. Scale reading in millimetres when body is in pan Weight of body in grains JOLLY'S SPRING BALANCE. II \ II. — Within certain limits the extension of the spring is nearly proportional to the force which produces it. To verify this, read the scale when the pans are empty. Next place in the upper pan any small weight from the box, say 10 grains, and again read the scale. Add another weight and take the new reading. Enter results thus : — Experiment No. 4. II. First scale reading for weight . . . grains. (( (( — > — , 40 40 40 ' i. e., .025, .050, .075. Turning the bar over we find a scale of centimetres. The Vernier reads to -^ih ofacen- 500 timetre. Hence the Vernier readings are to be doubled ;and called thousandths of a centimetre. Method. — Loosen the clamping screw and slide the movable jaw until it comes up to the fixed jaw. The Ver- nier should now read ; if it does not, the error must be inoted, and applied to subsequent readings as an index correction. Now separate the jaws until the distance be- tween them is a little more than the distance to be measured. Clamp the sliding jaw, and turn the adjusting screw until the jaws just include the given object. Read the scales on one side. Turn the caliper over and read the other scales. Repeat the work two or three times, and if the readings differ take the mean. To measure an internal diameter, close the jaws and with a micrometer caliper, or otherwise, measure the breadth of the MICROMETRIC MEASUREMENT OF DISTANCES. 19 jaws, ([t is .3000 in. = .7620 cm.) This must be added to the readings of the scales when the jaws are opened to correspond to the internal diameter. The given cylinder is in three parts. Measure the three diameters in order, beginning with the smallest. Find also the length of each part, and the internal diameter. Enter results thus : — Experiment No. 8. External diameter (i) (3) Length of (i) <( <( (< <( <( (( Inches. " " (3) Internal diameter . . . (Each student to sign his name.) Centimetres. 9. MICROMETRIC MEASUREMENT OF DISTANCES. flfb . Experiment — To find ihe distance (not exceeding one metre) between two lines on a metal bar. Apparatus — Two microscopes with micrometer eye- pieces. A standard metre scale. Method — The microscopes are placed so that the centres of their fields are near the marks on the given bar, <«0 MA THEMA TICAL LA BORA TOR Y, and are then securely clamped. During the remainder of the experiment the microscope stands and tubes must not be disturbed. The parallel crosshairs of each micro- scope are then moved so as to be equally distant from the corresponding line on the bar. Read the divided circles, and repeat the operation several times, taking the mean of the several settings. Remove the bar, and, without disturbing the microscopes, replace it by the standard metre, adjusting the latter until its divisions are in focus. Move the metre scale until a millimetre division is near the cross- hairs of one of the microscopes, while the crosshairs of the other are at the same time near one of the fine divisions of the scale. Now set the crosshairs on the nearest con- venient divisions of the scale and take the readings, being careful to note the whole number of turns (if any) of the screw. It is also necessary to determine the value of one divi- sion of each of the screw heads by moving the crosshairs over the divisions of the standard scale. Enter results thus : — Experiment No. 9. i Given distance = ... cm. Error of metre scale for this distance ^ , , . cm. .*. corrected distance = . . . cm. Each micrometer division = . . . cm. microscope A. B. li <( <( (( (Each student to sign his name.) e )f >_ g THE PLANIMETER. 10. THE PLANIMETER. 21 Experiment. — To measure a plane area. Apparatus. — Amsler's Polar Planimeter. This instru- ment consists of two metal bars jointed together. The extremity of one bar is provided with a needle point which remains fixed, while the tracing point at one end of the other bar is carried all round the perimeter of the given area. Near the end of the latter bar is a little celluloid wheel whose circumference is divided into loo parts ; with a vernier, tenths of ihese parts are read. Complete revol- utions of the wheel are recorded on a metal circle which is turned by a worm on the axis of the first wheel. It may be proved that the given area is proportional to the dis- tance recorded by the wheel.* The area is also propor- tional to the distance of the tracing point from the joint. This distance can be altered, and hence the wheel madv to record in different units of area. Method. — Let the sliding bar be set at the line marked " 100 sq. cm". Read the recording circles after the tracing point has been moved to a point on the curve. Keep the needle point fixed and move the tracing point round the curve from left to right, i.e., in the direction in which the hand of a clock goes round the dial. When the point has gone entirely round, read the circles again. (The reading of the metal circle must be increased by lo if it is less than it was at first.) The difference of the readings multiplied by loowill give the area of the curve in square centimetres. For example, suppose that the metal circle read 5+ and the celluloid circle 672; then the first reading recorded * For the theory of the Planimeter see Appendix I. II ii m MA THEM A TICAL LA BORA TOR Y. is 5.672. After the motion, let the metal circle read 7 + and the celluloid circle 535 ; the second recorded reading is 7-535- Then 7.535 - 5.672 = 1.863. .*. the area is 186.3 square centimetres. In this case each unit of revolution recorded corres- ponds to 100 sq. cm., but if the sliding bar be set at the mark " 10 sq. in." each unit will correspond to 10 square inches. If it be set at 200 sq.' %" = 1' , each unit will cor' respond to 200 sq. feet on a diagram constructed on a scale of ^ of an inch to the foot. The area should be traced out several times and the mean taken for the final result. Enter results thus : — Experiment No. 10. I. Area of given circle. First trial, area = 100 (...-...) Second " " = 100 (..,-...) &c. Mean . . . sq. cm. W3 = . . . (Each student to sign his name.) liM T2. MEASURES OF VOLUME. Experiment. — To verify certain measures of capacity. Apparatus. — A half-litre flask. A pint flask. A burette with glass float. A thermometer. A balance, water, etc. A litre is the volume of i kilogram of water at 4° C, the weighing being corrected for displaced air. It was intended to be, and is very nearly indeed, equal to one cubic decimetre. A gallon is the volume of 10 pounds of water at 62^ Fah., the weighing being by means of brass weights, and not corrected for displaced air. The pint is one-eighth of a gallon, or the volume of 20 ounces of water. One twen- tieth of a pint is called a fluid ounce; each fluid ounce is divided into 8 fluid drams, and each fluid dram into 60 minims. Hence, 8 x 60 or 480 minims of water weigh one ounce or 437.5 grains. Thus the graduations of measures of capacity depend upon weighing. n MEASURES OF VOLUME, 25 Method I. — Taking rather more than a pint of distilled water, exhaust the air from it by means of the air pump. See that the half-litre flask is quite dry, both inside and outside, and weigh it in balance E or F. Fill it with water up to the mark, wipe the outside if necessary, and weigh again. The difference of the two weighings, corrected for displaced air, is the weight of the water. Dividing by the specific gravity of the water at the temperature of that used in the experiment, we have the weight of the flaskful of water at 4°. This should, of course, be 500 grammes. Enter results thus : — Experiment No. 12. I. Apparent weight of empty flask = . . . grammes. " " " ' ,sk and water = . . . ** Diflerence = . . . Correction for displaced air = . . . Corrected difference = . . . Temperature of the water = . . . ° C. Spec, gravity of water at this temp. = . . . .'.weight of flaskful of water at 4^ = ' \ ' = . .*. volume of flask =...... ccm. grammes. //. — Proceed similarly with the pint flask, using grain weiglits. The weighings are not to be corrected for dis- placed air; and, as the temperature will be about 62^, no temperature correction will be required. Enter results thus : — Experiment No. 12. II. Weight of empty flask " " filled " .'. weight of water = . . . ounces. — • • • grains. u 26 MA THEMA TJCAL LA BORA TOR V. III. — To verify the graduations of the burette. Nearly fill the tube with water. The mark on the float will serve as an index in reading the scale. Draw off water until the mark is below the top of the scale. Then draw off about lo cc. of water in a vessel which has pre- viously been weighed, and find the weight of the water. Read the scale in cubic centimetres and tenths, and find the weight in grammes and tenths. Continue to draw off about 10 cc. at a time until the float reaches the bottom of the scale, weighing each time. As the results are approximate only, it will not be necessary to correct for temperature or displaced air. Enter results thus : — ' Experiment No. 12. III. 1 i' I Scale Readings. Difference. Weight of water. I. . cc. and «... . cc cc. grammes 2. . (( • 1 • • • • • • t • It 3- • << (( 4, . (( .... • «< (( 5. • irence is 10 grammes. The 105 on IV,- balances 105 on IVi, but the whole 220 will be set in motion by the attraction of the earth on the 10 grammes. Find /i the time of falling through any distance s. The acceleration is 2S A T WOOD'S MA CHINE, II Now move the remaining 5 gramme disc to the left. The sum of the masses is the same as before, but their difference is twice as great. Find h the new time of falh'ng through s. The acceleration is now 2S If this is about twice as 2S great as — ^-^ we shall have shown that the acceleration is proportional to the force when the mass remairs the same, or that the rate of change of momentum is pro^ ortional to the force which produces it. Enter results thus : Experiment No. 13. III. 2S s = . . . cm. ■—; t\ = . . . sec. • • • 2S .*. first acceleration X 2 = . . . second " = . . . JV. — The value of ^. Let the acceleration produced by the weight of 10 grammes acting on 220 be called ai. Then^, the acceleration produced by the weight of 220 grammes acting on 220 grammes will be 22^1. Similarly g = i laz, where a^ is the acceleration produced by the weight of 20 grammes acting on 220 grammes. The resulting value of ^ is too small, since the friction retards the motion. The true value ofg is about 981. 081 - 22rtl . , ^ . Hence, is the fraction of the weight of 10 981 ** grammes, which goes to overcome the inertia of the pulleys, the friction, and resistance of the atmosphere. This fraction may similarly be determined from 02. 3* A/ A Til EM A TICAL LA HO R A TOR \\ Enter results thus: — Experiment No. 13. IV. 981 - 22«i 98 [ - II'<2 "^^ Z ~- • • • ,, 981 981 (Each student to sign his name.) 14. ATVVOOD'S MACHINE. Experiment. — To verify the first and second laws of motion. Apparatus. — Atwood's Machine (with wooden scale), electric attachments and water clock. The machine consists of a light aluminium pulley whose axis rests on friction wheels also of aluminium. Over the pulley passes a silk thread carrying two weights ; the one on the left we shall call IVi and the other W^. The mass of W} is 240 and that of VV,. 225 grammes. On the weights may b2 placed brass discs and bars of 5 and 10 grammes mass. Two rings are attached to the scale ; one of these serves the purpose of stopping a bar placed on the top of the miss Wi while the mass itself moves onward through the ring ; the other (the " break-circuit " ring) is arranged so that Wi in passing through it breaks the electric circuit. The binding screws at the top of the switch-board are connected by wires behind the board (as indicated by the dotted lines) with the swinging arm S and thecantact pieces B^ C, A>, to which S may be moved. /iJWOOD'S MACHINE, "1'/' / i < A/ ' ' h When 6" is moved to B the circuit is completed, but the machine is in no way affected. When it is moved to C the current passes through the electro-magnet of the machine and also that of the water clock ; hence the platform which supports IVi drops at the same instant that the water clock is opened. When S is moved to D the weights are arrested. (The arrangement of the wires should be carefully studied before commencing the ex])eriment.) Method!. — Taking as the unit of time the time in which 10 c. c. of water falls into the graduated vessel, we must first find the equivalent of this in seconds. Balance the weights (unless they are already balanced), and set the water running slowly into the cistern. Place under the cistern a flask which will contain a considerable quantity of water (say 500 c. c). Move S to B. Then move to C, noting carefully the time. When the flask is filled to the mark, move ^ quickly back to A, at the same time noting the time. From this find the time in. which 10 c. c. will fall. ir 34 MA THEMA TICAL LABOR A TOR Y, Enter results thus : — ExPERlMtNT No. 14. I. Time of . , . c.c. . . . sec, first trial. ""..." . . . " , second trial. " **...*' . . . " , third trial. Mean of times « .'. 10 c.c. is equivalent to . . , sec. //. — To verify the first law of motion . Place the ordinary ring at any position near the top of the scale (say 50 cm.) and the break-circuit ring a little (say 20 cm.) below it. The weights being still balanced, bring up Wi above the upper ring and place on it a bar (say the larger one), and then pull Wi up carefully to the top of the scale, adjust the platform and let IVi come down gently on it. At this point in the experiment and in all similar cases brifig W^to rest. See that the contact points of the break-circuit ring are together. Move S to B and then to C. Do not remove the hand from S; watch the weight and as soon as it breaks the circuit by passing through the lower ring^ immediately move S to D, This will stop the motion. Take hold of the string on the right, move .S back to A^ then pull IVi up again and arrange as before. Now read the water clock and repeat the work. let the mean of the two water limes be t\ , 10 q.c. beiii^ taken for unit of time. Now lower the break-circuit ring any o - veil' ■;? amount, say 50 cm,, and proceed as before, getting t2. Again lower by the same amount as before, getting tz ; and continue to lower until Wi has been placed in say 5 posi- tions. Then find h-t^^ ti- ts j /s - ^2 , t^-ti. If these are equal we shall have shown that a body will, when acted on by no force, move over equal distances in equal times. AT WOOD'S MACHINE. . Observe that all distances on the scale are to be con- sidered with reference to the bottom of Wi. Enter results thus : — Experiment No. 14. II. Distance s of upper ring from o of scale . . . cm. d by which lower ring is lowered each time . . . cm. /i = . . . t'i. — t\ — n /2 = . t\ — . /5 = . h - i\ = Sum of these = Mean = . d sec. Speed V after falling distance s = —r - - . . . * above mean Time / of acquiring this speed = . . . sec. ///. — To show that the acceleration is constant, or ia other words, that the speed is proportional to the time during which the force acts. We have the speed v acquired in the time /. Pla je the upper ring in some otlier position, and as in II., find the speed Kand tim^i 7" of falling to the new position. (For this purpose the lower ring need be placed in two or three V T positions only.) Then — and— should be equal. Enter results thus : — V = / = V ExpERiMENr No. 14. Iir. T = T les. I'f 36 MA THE MA TICAL LAB OR A 7 OR V. I \ ! il IV. — In II. the whole mass attached to the two ends of the string(the bar included) is set in motion by the attrac- tion of the earth on the bar. To determine the effects of different forces when acting on the same mass we might take a five gramme disc from the right hand side and put it on the left. Tne whole mass set in motion would remain as in II, but the moving force would be twice as great. As in II. we might find tbe acceleration which shduld be twice that of II. It will be simpler, however, to proceed as follows : — Remove the bar. The weights are now balanced and their sum is 480 grammes. Change a 5-gramme disc from right to left. The sum is still 480 grammes, but the differ- ence is 10. Put the break circuit ring at any convenient reading j, say near the middle of the scale. Find /, the time 2S ' of falling through the distance s. The acceleration is — - Return the 5-gramme disc to the right and move the 10- gramme disc to the left. The sum of the masses is the same as before, but their difference is twice as great. Find /2, the new time of falling through s. The acceleration If this is about twice as great as IS now we /2=* ° h'' shall have shown that the acceleration is proportional to the force when the mass remains unchanged, or that the rate of change of momentum is proportional to the force which produces It. Enter results thus: — Experiment No. 14. IV. J = . . . cm. Z\ ^^. ..™-. . . sec. 2 S 2 S THE INCLINED PLANE. '. first acceleration x 2 = . . . second ** = . . . ar F. — The value of^. Let the acceleration produced by the weight of 10 grammes acting on 480 be called ai, Then^, the acceleration pro- duced by the weight of 480 grammes acting on 480 grammes, will be 48^1. Similarly ^ = 24^2, where 02 is the accelera- tion produced by the weight of 20 grammes acting on 480 grammes. The resulting value of g is too small since the friction, etc., retards the motion. The true value of g is about 981. Hence, 981 - 48^51 981 - 24- CENTRIFUGAL FORCE. ^t. .*. mean value of the acceleration a for the inclination . . . is . . . cm. per second in each second. Ratio to value in III. = -^-^ = . . . • • • Ratio of sines of angles of inclination = ~-* = . . , F.— Assuming the theoretical value of the acceleration to be -i ^ sin e, find both from III. and IV. the value of §'. Enter results thus : — From III. From IV. Mean, Hence, Experiment No. 15. V. ^ = . . . cm. per sec. in each second. „ ^ i( (( (( (( (( 6 ~ • • • ^= • ' ^= • . ft. " (( (( (( <( (< (( (( (( (Results of the five parts to be returned together.) (Each student to sign his name.) 16. CENTRIFUGAL FORCE. Experiment. — To investigate the stress in uniform cir- cular motion. Apparatus^ — At the centre of a horizontal rotatory table is a chamber containing mercury communicating with a vertical mercury column in a glass tube alongside a fixed scale. One side of the mercury chamber is a thin steel corrugaged diaphragm, to the centre of which is attached a radial brass rod, the ot her end of which is attached to a iT MA r II EM A TIC A L LAB OR A TOR Y. ■Sil! rocking arm fastened lo the re- -g table. When rota- tion takes place the stress be . the revolving mass and the steel diaphragm causes .ncrease in the capacity ol the mercury box and a corresponding fall in the mercury column. An overhead conical countershaft permits of large variations in the rate of turning of the table. We may thus investigate the connection between the mass, velocity and radius in circular motion. Method I. — The first thing to be done is the calibration of the glass tube. By a spring balance find the reading of the top of the mercury column for various forces applied at the end of the radial bar. Plot these on section paper^ the readings of the column being taken as abscissas and the corresponding forces as ordinates, and draw a curve through the points. //. — Keep the rale of turning unchanged. Also keep the position of the sliding body unchanged, but change the mass of this body by gradual addition or subtraction, being careful to clamp it securely after each change. Plot the mass thus added or subtracted, and also the corres- ponding forces, and draw a line through the points. ///. — Keep the rate of turning unchanged, and also the mass, but move the latter to various positions along the rod, being careful to clamp it securely in each position. Plot the various distances from the centre and the cor- responding forces, and draw a line through the points. IV. — Keep the mass and the distance from the centre unchanged, but vary the rate of turning. The number of revolutions per minute may be noted by means of a speed indicator. Plot these and the corresponding forces^ and draw a line through the points. \, t\ STATICAL FRICTION, 43 From an inspection of the diagrams obtained in II, III and IV. state the relation between the force and the mass, radius and speed. Enter results thus : — Experiment No. i6. (The four diagrams.) (Tlie general statement as to results.) 17. STATICAL FRICTION. Experiment.— To find the relation between the normal pressure and the corresponding statical friction. Apparatus. — A spring balance. Weights. Surfaces between which the friction is to be investigated. Method. — Keep a record of the various pressures between the rubbing surfaces and also of the force required to just overcome the friction in each case. This force must be applied very gradually and in a direction parallel to the surfaces under consideration. Then plot on section paper the corresponding pressures and frictions, the former being taken as abscissas and the latter as ordinates. Draw a straight line which will pass nearest to the various points thus obtained. Enter results thus : — Experiment No. 17. (The diagram.) (Each student to sign his name.) r 44 MATHEMATICAL LABORATORY, 1 8. MECHANICAL EQUIVALENT OF HEAT. Experiment. — To determine the mechanical equivalent of heat. Apparatus. — A steel cone containing mercury in which the bulb ot a thermometer dips is caused to revolve in a similar fixed cone. Means are provided for measuring the amount of work done in heating the cones and the mer- cury by a measurable amount. For correction for radia- tion, and for other details of the experiment, see Pickering's Physical Manipulation. 19. TORSION AND MOMENTS OF INERTIA. Part I. Experiment. — A metal bar is suspended by means of a wire, the upper end of which is clamped. The direction of the wire passes through the centre of gravity of the bar, which makes horizontal oscillations under the torsional reaction of the wire. It is required to find the moment of inertia of the bar. Let k be the couple which would twist the free end of the wire through a unit angle (one radian). The couple which twists it through the angle d is kd. When the wire is thus twisted it exerts, as an opposite reaction, a couple kd. Thus, when the bar is swinging horizontally and makes an angle Q with its position of equilibrium, the moments of the external force& about the axis of motion =-kd (minus since the couple tends to diminish^). Let /= the mo- ment of inertia of the oscillating body about the axis of v/\J3^wdJ'-L«>-v - -^ C^jv>JJLJm {ra--<-*X*^^ ff TO/iSION AND MOMENTS OF INERTIA. rotation, and a = the angular acceleration. '1 hen la^-kB (i) 45 or. a = - / e (a) Let r - the distance of any point in the moving body from the axis, and multiply both sides of (2) by r. Then r a^-l. vB In this equation r a is the tangential acceleration of the point, and r 6 is the displacement. Hence the motion of the point is a simple harmonic motion and the time of 4 . vibratio-P is tt ^ __. Calling this time / we have s / = TT /_ •./ = TT* (3) Two bodies of equal mass m are arranged to slide along the bar. Let /i = the moment of inertia of the bar alone about the axis of rotation, /2 that of each sliding mass about a vertical axis through its centre of gravity. Then a being the distance of this axis from the axis of rotation, the moment of inertia of each of the sliding masses with reference to the axis of rotation is A + ma^ . Method 1. — The experiment is to be conducted as fol- lows : — I. The sliding masses are placed at distance a from the axis of rotation, the time of vibration being 4. >wnrii^i*H*pi^pi^^a y\ 4i MATHEMATJCAL LABORATORY. 2, The sliding masses are next placed at a distance ^, the lime being tt, 3 They are entirely removed, the time being now /a. Then /j + 2/2 + awa^ - — i^ /i + 2/2 + 2W^2 = L /l = /6/32 TT^' (4) (5) (6) Subtracting (5) from (4), 2W (^2 - /^2 ) = --L (/i2 _ /.^2 ) TT .-. /6 = (7) Then from (6) / /l2 - /22 (8) /2 may be found by substituting in (4) or (5). k will be in force-distance units, and h in mass-(dis- tance)2 units. If I2 were known at first, equation (5) would not be re- quired. k X length of wire is the couple required to twist unit length -of the wire through one radian. This is the modu- lus of torsion of the wire. The radius of gyration of the bar about the axis ' motion = V/i / mass of bar. TORSION AND MOMENTS Of INERT/A. 47 Enter results thus : — Experiment No. 19. I. a = . . . cm., b = , , ' cm,, m - . . . grammes. t\ = . . . sec, ti = . . ' sec, /^ = ... sec> .". k - , . . and /i~... The radius of gyration of bar = y fH = . • • cm. Part II. A table is similarly suspended. Let /i = the moment of inertia of the table alone about the axis of suspension, the time of vibration being ^1 . Let a body of known moment of inertia be placed on the table, so that the centre of gr^'ity is in the same axis of suspension, and let the time of vibration now be ^2 . Let t^ be the time of vibration when a body of unknown moment of inertia I3 is placed on the table with its centre of gravity in the line of the wire. Then Ji = T'2 /l./2= ^14^, Ii+h = Subtracting (i) from (2) k 7r2 n-2 (■) (3) h = )ra (<2 2 - «l 2 ). (4) and sublracling (i) from (3) k /,= « (/32 -h') (5) TT- 48 MATHEMATICAL LABORATORY. I \ ^' "72 2 - /I 2 k may be found from (4) and A from (i). (6) may be written (6) /.= /2 /2 2 - /3 2 _ -__ - 2 and ^ ^ may be found once for all. Then the moment of inertia of any body is of the form , where a and b are constants, and / is the time of vibration when the body is placed on the table. Find a and b for the given table and cylinder, and then find the moment of inertia of the given pulley. Enter results thus : — I! I i ' l> i Experiment No. 19. II. Mass of cylinder = . . . grammes, Diameter of cylinder = ... cm. .*. moment of inertia /2 = . . . /i = . . . sec, <2 = . . . sec. Hence a = .. , b = . . , Time of vibration, table and pulley, t = . . . sec. .'. Moment of inertia of pulley - at'^ - b (Each student to sign his name.) IL BIFILAR SUSPENSION. 49 20. MAXWELL'S VIBRATION NEEDLE. ExPERiMENr. — The value of k (Exp. 19) may be found with greater accuracy by using an apparatus specially de- signed for this purpose by Maxwell. The time of vibra- tion may be found with great accuracy by using a reading telescope and a chronograph. For details of the experi- ment see Glazebrooke and Shaw's Practical Physics. \. n n 2\. BIFILAR SUSPENSION. ExPERLMExr. — To find the moment of inertia of a rod. Apparatus.^ A heavy brass cylindrical rod which is hung up in a horizontal position by means of two equal vertical strings, and is caused to make small horizontal oscillations about its middle point. Apparatus is also required to measure the lengths of the strings, etc., the mass of the rod and the time of a vibration. Method. — See that the rod is properly suspended, i.e., that it is horizontal and that the strings are vertical and equal. Measure in centimetres the distance b between the parallel strings, also h^ the distance from the axis of the rod to the point of support of the string. The mass m of the rod is 5343 grammes. Cause the rod to oscillate about its middle point through a small angle and find in seconds the time i of a vibration. Then the moment of inertia of the rod about its axis of vibration is * nig ^- /2 4 n^ k where ^ = 981. * See Wright's MechnHict , p. 314. D 50 MATHEMATICAL LABORATORY. Enter results thus : — Experiment No. 21. b = . . . cm., /; = ... cm., t = . . . sec. /. moment of inertia = . . . (Each student to sign his name.) 22. REVERSION PENDULUM (KATER'S). Experiment. — To find the value of ^. Apparatus. — A metal bar carries a pair of cylinders and parallel knife edges at or near its extremities. One of the cylinders is hollow, the other solid ; hence, although geometrically symmetrical, the centre of gravity is not at the middle of the bar. Let the distances of the knife-edges from the centre of gravity be hi and //2 and the times of small vibrations about these knife-edges be h and h , respectively. Also take k as the radius of gyration of the pendulum about the axis through the centre of gravity parallel to the knife-edges. Then /2 = TT TT g^\ to eliminate k and solve for^. Squaring, 7r2 g (0 (2) ( ir subtracting, A'Ji VERSION PEND UL UM, tx 2 hi - h - /^2 = '^ {hx '^ - //2 2 ) S 51 (3) which is identical with 27r2 _ h 2 + /. 2 /i 2 _ I, 2 ^ ~ //I + //J h\ - hi (4) (3) will give the same value of g as (4) ; the latter, how- ever, shows by what arrangement the greatest accuracy may be secured. The times t\ and h may be found with great accuracy, but the exact position of the centre of gravity is not easily determined ; hence, /n and hi are uncertain. The denominator h\ + 7^2 of the first fraction on the right of (4) is the distance between the knife-edges, and hence may be measured even if the centre of gravity were entirely un- known. The denominator h\ - hi of the second fraction cannot be determined with very great accuracy. On account of the fact that one of the cylinders is much heavier than the other, h\ - hi is not small. If then fi and fi are very nearly equal, the second fraction will be very small, and hence this second term will have only a very slight effect on the value of ^. Thus the pendulum should be arranged so that ^1 and ti are as nearly equal as possible. The distance between the knife-edges may be found by Exp. 9 (or more accurately by means of a dividing engine). The approximate position of the centre of gravity may be determined by balancing the pendulum on a horizontal knife-edge . The time of vibration may be obtained by means of coincidences of vibration of the pendulum and that of the mean time clock, which beats seconds. Special HHMP 52 MAniEMATlCAL LABORATORY, II arrangements are provided for the purpose of obtaining these coincidences. If/ = the number of seconds between coincidences, and if the pendulum gains as compared with the clock, the time of vibration of the pendulum = p ^ ^ seconds. The value of/ adopted should be the mean of a large number of determinations. Find the value of ^, arranging the work as in the follow- ing example : — Example. y 26.97 ti - - — — = . 96425 sec. h = 27.97 25.67 = .96250 sec. 26.67 A 2 = .929773, /2- = .9264i5' /i 2 + ^.^ 2 = 1.856188, h 2 - /.J 2 ^ .003358, h\ + h^i = 92.681 cm., //I = 62.01 approx., 7/2 = 30.67 approx., /^i-/^2 = 31.34 approx. 27r2 1. 856188 .003358 •• 7~ "" 92.681 31-34 ' = .0200277 + .0001071, .'. g = Q83.34, Correction for arc* = .05 ** *' buoyancyf = .15 .*. corrected J? =980.54. Note I. — When the knife-edges, etc., are arranged so that t\ and h are equal, it is known that h\ Jh = k^ . Moreover, the knife-edges should be at equal distances from the \, * See Note 2. t See Note 3. RE VERSION PEND UL UM. 39 i. f le middle of the bar in order that the action of the atmos- phere may be the same in both positions of the pendulum. The required adjustment may be made as follows : The knife-edges being near the ends of the bar, observe ^i in (i) and measure /^i approximately, also assume an approx- imate value of ^ and calculate >i2 ^ which = — --^ h\ - . Let :x: = the required distance of each knife-edge from the middle of the bar and let a ~ the distance of the centre of gravity from the middle. Then (x -a) {x + a) = Ii^ The shifting of the knife-edges will change k slightly, hence the work should be repeated. Note 2. — The correction to the time for the arc of vibra- tion is approximately — eV ^^ where Q is the radian measure of the kngle of vibration, Now, g = __(/;i + k2 ), nearly ; . . ^ = - — , .'. dg = -- *__> the arc being the same in both positions ; or, — Q^ may be subtracted from each time. Note 3. — The correction for buoyancy is approximately g -L where 5\ and ^2 are the specific gravity of the air and metal respectively. Approximately, - = ' = .00015 ; .*. ^ ^ '^ h Find the value of ^, arranging the results as in the following example : Example. h = 104.316 cm., ?' = 2.99 cm., .'. /= 104.350 cm. / = 41.185 sec, from 200 coincidences. .-. /[ Correction for arc * " for buoyancy f 7r{p- P 0]^ = 980 .49 :± .04 t = •15 -£■ =^ 980 .86 JVi;fe I.— From t = 7rjl vve have s - — d^ dl dt J Z ,, e , .-. -- = 7- - 2 — , or ^p' = - dl ~2^ dt. The approximate values of j?', /and t are 1000, loo, and i respectively. .'. dg - \o dl - 2000 dt. Hence, an error of .cot of a second in determining the time of vibration is twice as bad as an error of i millimetre in measuring the length. * See Note a, Exp. 22. t " " 3, " 22. MATHEMATICAL LABORATORY. Note 2. — Differentiating^ ^- 1 [^-- ^ ~ ^ J , we liave dg= 1.2 dp + 9.5 dl for thi*! pendulum. For arc, dg= .0031 a- , where a is the total arc of vibration in centimetres. Note 3. — From t = dp /-I dp we have dt = - 7- — ^~T- - — 7 — approximately. (/ - 1)- 1600 - ^ ^ 'I'hus an error of i sec. in/ would cause an error of about sec. in the vaUie of t. 1600 Note 4. — To reduce the value of ^to what it would be at the sea level, we have, since g is inversely proportional to the square of the distance r from the earth's centre, d^ dr i r The increase is .*. 2 - //, where h is the height of the pendulum above sea level. This is equivalent to increasing the metric value of ^'^ by .000093 for each foot above sea level. Note 5. — The adjustment of the cylinder was made as follows : The time of vibration of the whole pendulum is t-~ il ^^^ ^^ ^ "^ ^'^'^ "^^ " "^ ^^^ ^'^ ^ "^ ^"^ ^'^ ^ (1) ^Ig I mi hi + W2 /i2 -v mz hz - tn\ h\ m being mass, k radius of gyration about its line of sus- pension, and /; the distance of this line from centre of I SIMPLE PENDULUM. 5? gravity ; the four parts of the numr. and denr. referring to (i) the ball, (2) the knife-edge, screw, etc., (3) the wire, (4) the cylinder. The ms are found in tlie ordinary way, also h\^ h^^ hz. For the k's we have k\ - = h\ - + |r- , r being radius of ball ; k-2, may be found experimentally by letting the knife- edge and screw vibrate on the plane of suspension and noting the time of a vibration ; kz'^'u approximately \ (length of wire)2 ; ^^ 2 = //^ 2 + ^^ (length of cylinder)2 + \ (radius of cylinder)'- . Thus everything is known except /'4 . This is to be determined so that m-2 ki - + ;//3 ^32 + m\ k\ - m\ 1i\ - (^) m^ h2 + niz hz - ?n\ /i^ f?iv /n and hence depends upon the solution of a quadratic equation. The right hand side = /n + ^ 7— • This was found by measurement to be 104.344; /^^ was taken as o, //3 = 50.5, mi - 3, mi = 1.57, ;;/4 -= 21.37, k-^ - - 2, /^a 2 = 3500, h- = /^'^ +H+T = ^^^^ +'28. The above equation is then (approximately) ;; 6 + 5 500+ (2 1.37/^4" + 6) _ 79.29 - 21.37 ^^^ ~ ^^^ ^'^' which is very nearly satisfied by /i^ := 1.2. The cylinder being thus placed we have r t = ^g h\ +1 r- IT V980.6 /^^°4-344 = 1.0248 58 MA THEM A TIC A L LA BORA TOR V. If the cylinder were omitted we should have, substituting in (i) and making pn = o, t = 1.0247 This would make g about .2 too great. This was verified by experiment. li m 24. SIMPLE MACHINES. Experiment. — When work is done on a machine, a part of the energy thus expended is always used up in over- coming friction ; another part may be stored up in the ma- chine, while a third part may be employed in doing work upon some outside body or bodies. It is required to de- termine these several parts in the case of certain simple machines, also the mechanical advantage or purchase of the machines, i.e., the ratio of the force overcome to the force applied, and the efficiency of the machines, i.e., the ratio of the useful work accomplished by the machines to the energy expended on them. Begin with the Differential Wneel and Axle. When by pulling on the stwng the wheel is caused to turn, friction is overcome, useful work is done in raising the weight attach- ed to the pulley, and by raising the pulley itself energy is stored up in the machine. The unit force employed in this experiment is the weight of one-quarter ounce and the unit distance may be taken as one centimetre. Note the heights of the weight attached to the pulley and that attached to the wheel ; take hold of the latter and pull it down through any distance, and again take the two heights. The weight I SIMPLE MACHINES, 59 attached to the pulley multiplied by the distance raised is the useful work done ; the weight of the pulley multiplied by the same distance is the work stored up in the machine, while the work done upon the machine is the distance moved by the weight attached to the wheel multiplied by the force required to pull the string, so as to cause the motion. To determine this, attach the scale pan (weight = 1.7 qr. oz.) and in it place weights until the motion be- gins. Find the decimal which the energy stored up in the machine is of the work done upon the machine ; let this decimal be called D. Next the decimal which the useful work is of the work done on the machine ; this is the effi- ciency, call it E. Then i — (Z> + E) is the part of the applied energy which is expended on friction ; call this F» Also find the mechanical advantage A, Similarly with the systems of pulleys, marked 11^., 11^., I. Also with the differential pulley. In this case the weight of the chain may be neglected, as the weights of the strings are in the other machines. The pulley to which the 56-lb. weight is attached weighs 3.3-lbs. The force required to pull the chain so as to cause motion may be found by means of a spring balance. In the case of the screw, apply the moving force hori- zontally, by means of a spring balance, to the end of the lever and perpendicular to the lever. I^et it be understood that the lever makes one complete revolution. The weight of t' e lever must be included with that of the screw I 60 A/ A THEM A TIC A L LA BORA I OR V. Enter results thus : — Experiment No. 24. IJ. Differential Wheel and Axle. Pulleys Ha. " I. Differential Pulley. Screw. £. F. (Each student to sign his name.) 25. THE MERCURY BAROMETER. Experiment. —To find the pressure of the atmosphere. Apparatus. — A Mercury Barometer with a scale of millimetres and inches, and an attached Thermometer, giving Centigrade and Fahrenheit readings. Method. — Read first the attached thermometer. By means of the screw at the bottom of the barometer raise or lower the mercury in the cistern until the surface just meets the ivory cone which forms the zero point of the scale. Tap the upper part of the tube slightly so that the mercury will take its proper position. The vernier and the sliding piece at the back of the tube which moves with the vernier must now be moved so that the plane joining their bottoms exactly grazes the top of the column of mer- cury. 'I HE MEKCVKV nAROMETEK, 6i Read both the scale of miHImetres and the scale of inches. To ensure as great accuracy as possible, the read- ings finally adopted should be the means derived from several settings of the cistern and vernier. The observed height must now be corrected so as to give the equivalent height at the freezing temperature of water. It is known by experiment that the length of a mercury column decreases by about .000181 of itself for each degree (centigrade) in the fall of temperature, and a brass tube (such as that to which the scale is attached) by about .000019 of itself. The difference of these is .000162. Hence the correction to reduce the observed height to what it would be at 0° centigrade is - (.000162) h millimetres where h is the observed height in millimetres (corrected for the index error mentioned below) and / the temperature (centigrade) of the attached thermometer. If the temperature be Fahrenheit we must take \ of .000162 to find the change for each degree. Hence the correction to reduce to 32" F. is - (.00009) (^ ~ 32) h inches where // is the observed height in inches (corrected for the index error mentioned below) and t is the temperature Fahrenheit. There are several other corrections to be allowed for in order that the true pressure of the atmosphere may be found. J (i) There may be a little air at the top of the mer- cury column and there is certainly some vapour of mer- cury. (2) The ivory point may not be correctly placed. 4a MA TIIEMA TICAL LA DORA TOR V. (3) Owing to capillary action between the mercury and the glass tube, the column is not so high as it would otherwise be. These errors are more or less uncertain, but are nearly constant, and hence are best allowed for by combining them into an "index error" determined by comparison with a standard barometer. The index correction of the laboratory barometer may be assumed to be 1.2 mm., or .047 in/ Enter results as follows : — EXPERLMENT No. 25. Attached thermometer Correction to thermometer Corrected temperature Observed height of barometer Index correction Correction to reduce to freezing temperature Corrected height . . . °F." + 2° ... in. + .047 " The final result should not have more than four decimal places in inches and two in millimetres. (Each student to sign his name.) 26. THE ANEROID BAROMETER. Experiment. — To read the aneroid barometer and by it to determine a height. Apparatus. — The aneroid barometer consists of a cylin- drical box made of thin elastic metal, from which the air 'Read the thermometer to the nearest degree. THE ANEROID BAROMETER, 63 has been exhausted. A change in the pressure of the at- mosphere causes the top of the box to rise or fall through a very small distance, this motion being magnified by means of levers and communicated to a hand which moves on a graduated dial. The dial has three double scales ; the interior graduations corresponding to inches of a m ;• cury baromster, and the outer to feet in a vertical height through which the barometer may be carried. The in- strument is compensated for changes in temperature. Method. — Read the scale of inches. Also read the mercury barometer and reduce to 32 F. From this sub- tract the reading of the aneroid. This gives the correction to apply to the aneroid readings for correct pressure at Z^ F. Carry the aneroid barometer to the highest floor of this building. Read the scale of feet. Ntxt carry it to the lowest floor and again read the scale. From this find the distance between the floors. Enter results as follows : — Experiment No. 26. I. Mercury barometer reduced to 32 F, Aneroid reading .*. index correction of aneroid ... m. • • • II. Reading at highest floor " lowest " . • . distance between floors is (Each student to sign his name.) ... ft. • » • 64 MA THEM A TIC A L LAB OKA TOR V. t 27. BOYLE'S LAW. Experiment. — (i) To find the pressure of the atmos- phere, and (2) to verify Boyle's Law. Apparatus. — Two glass tubes connected by a flexible tube containing mercury are mounted on each side of a vertical scale. One tube is fixed in position, and is pro- vided with a tap near the upper end ; the other, which is open at the top, slides up and down the scale, and is balanced by a counterpoise to which it is attached by means of a string which passes over a pulley at the top of the scale. Method I. — The tap being open, slowly lower the coun- terpoise until the mercury just rises above the tap. Close the tap and raise the counterpoise. After raising it six or seven decimetres it will be noticed that the mercury is beginning to fall below the tap. Wait a short time for the mercury to come to rest, read in decimetres and decimals the heights of the tops of the columns on the two sides of the scale and subtract. This will give the length of the column of mercury which is supported by the pressure of the atmosphere. The work should be repeated at least two or three times. Enter results thus : — Experiment No. 27. I. Height of left- hand column. Height of right- hand column. Difference. First Trial.. . Second Trial. Third Trial. . • 1 1 BOYLE'S LAW. 65 The average of these resuhs gives . . . decimetres of mer- cury for ihe pressure of the atmosphere. //. To investigate the relation between^the pressure and volume of a given quantity of air. The counterpoise having been lowered and the tap opened, bring the mercury to about 6 on the scale, and close the tap. There is now enclosed in the left-hand tube a column of air about 4 decimetres in height and at a pressure already determined in Part I. of this experiment, and which we may call/>j . Read carefully the top of the mercury and subtract from 10.00, which may be assumed as the height of the top of the air column. Call this re- sult Fj, This may be taken as the volume of the air column, the volume of one decimetre of the tube being assumed as unit volume. Now lower the counterpoise about one decimetre and read the mercury columns on the two sides. The reading on the left, subtracted from 10, gives Fg , the new volume j the difference of the two readings, added to /j , gives p^ , the new pressure. Again lower the counterpoise about another decimetre, and proceed as before ; and continue until six or eight values of the volume and the corresponding pressures have been obtained. It will be found that the product of any volume by the corresponding pressure is very nearly constant, showing that the volume 's inversely proportional to the pressure. This is Boyle's Law. Having completed this, raise the counterpoise, open the tap, and enclose as before a column of air about one deci- metre in length and proceed as before, but in this case allow the air to expand instead of compressing it. E 66 MATHEMATICAL LABORATORY. Enter results thus : — Experiment No. 27. II. Pressures. Volumes. Product. For Compression. For Expansion. t i ..'.-,,-'>%•*■ iJV';'^'-^****;,*.-**. .-V-^ BOYLE'S LAW. IVRPI 68 MA THEM A TICAL LABOkA TOR Y. III. — To express the results by means of a curve. On a piece of section paper assume one centimetre to repre- sent, when taken horizontally, a unit of pressure; and, when taken vertically, a unit of volume. From any hori- zontal line cut off a length to represent the pressure of the air at any time in the experiment, and from the extremity a perpendicular line to represent the corresponding volume, thus obtaining a point A^ Represent in a similar way the other pressures and volumes, and through the points thus obtained draw a curve. In this way curves similar to AB and CD will be obtained. These curves are parts of hyperbolas, having the pressure and volume lines for asymptotes. Enter results thus : — Experiment No. 27. III. (Plot the curves.) (Each student to sign his name.) 28. THE WET-BULB HYGROMETER. (Double Interpolation.) Experiment.— To find the pressure of the aqueous vapour in the atmosphere, also the relative humidity and the dew point. Apparatus. — Two thermometers ; the bulb of one is dry, that of the other is covered with hnen which is kept moist by water which soaks up a wick. THE WET-BULB HYGROMETER. 69 Air at a given temperature will contain only a certain quantity of the vapour of water ; the higher the temperature, the greater the quantity it will contain. The rate at which water evaporates depends upon the ratio of the quantity of vapour actually present in the air to the total amount which the air is capable of holding. The smaller this ratio, the greater the relative dryness of the air and the more rapid the evaporation. Again, evaporation is always accompanied by a fall of temperature ; the more rapid the evaporation, the greater the fall of temperature. Thus the difference between the readings of the two thermo- meters will indicate the rate at which evaporation is going on from the wet bulb, and hence also the relative dryness or moistness of the atmosphere. When this difference is zero no evaporation is taking place, and the air is saturated with moisture. When in this condition a slight fall ot temperature will result in some of the moisture being con- densed into cloud or dew. Method. — Read the two thermometers, being careful to estimate first and quickly the number of tenths in the fraction of a degree on each scale. (If you remain near the thermometer a short time, the mercury will lise — hence the precaution just mentioned.) With the dry bulb tempe- rature and the difference of the two, enter the following table, which will give in millimetres of mercury the pressure of the vapour actually present in the atmosphere. 70 A/A Til EM A TIC A L LA BORA TOR V. "a c; Q Difference of Dry and Wet Bulbs. I 2 ; 3 i 4 ; 5 6 7 8 1. 10 II 4.6 3'7 2.9 2.1 1.3 1 J 4.9 4.0 3.2 2.4 1.6 0.8 f 5-3 4.4 3.4 ; ^-7 1.9 I.O 1 3 5-7 4.7 3.7 i 2.8 2.2 i 1-3 4 ; 6.1 5J 4.1 3.2 2.4! 1.6 0.8 • c ' 6.5 5-5 4-5 3.5 2.6' 1.8 1.0 1 m4 6 ! 7.C 5-9 4.9 3.9 2.9 2.0 I.I 2 7 1 7-5 6.4 5-3 4.3 Z'Z 1 2.3 I 4 0.4 C U 8 8.0 6.9 5.8 4-7J 3-7' 2.7 I 7 0.8 1 9 8.6 7.4 6.3 5«2; 4.J 3.1 2. 1 I.I 0.2 IC ' 9.2 8.0 6.8 5.7 4.6 3-5 2.5 1.5 0.5 II 9.8 8.6 7-4 62 5.1 4.0 2.9 1.9 0.9 12 iio-5 9.2 8.0 6.8| 5.6 45 3-4 2.3 1-3 13 ill. 2 9.8 8.6 7-3 6.2 5c 3-9 2.8 1-7 0.6, £ 14 In. 9 10.6 9.2 8.C 6.7 5-6 4.4 Z'Z 2.2 i.i 15 12.7 i'-3 99 8.6 7-4 6.1 5-f^ 3-8 2.7 1.6 0-5 E- 16 'M'-. 12. 1 10.7 9.3 8.C 6.8 5-5 4-3 3-2 2.1 1.0 17 J14.4 [3.0 II-5 lO.I 8.7 7.4 6.2 4.9 3.7 2.6 1-5 0.4 18 !i5-4 13.8 12.3, 10.9 9-5 8.1 6.8j 5-5 4-3 3-1 2.0 0.9 19 J16.4 14.7 13.2 II. 7 10.3 8.9 7-5, 6.2 4.9 3-7 2-5 1.4 20 17.4 '5-7 14.1; 12.6 II. I 9-7 8.3 6.9 5-6 4.3 3.1 1.9 21 18.5 .16.8 rS-i; 13-5 12.0! 10.5 9.0 7.6 6.3 5-0 3-7 2.5 22 .19-7 17.9 16.2 14.5 [2. 911. 4 9.9 8.4 7.0 5-7 4.4 31 23 20.9 19.0 17-3 15.6 13-9 12.3 10.8! 9.2 7.8 6.4 5-1 3.8 34 22.2 20.3 18.4 16.6 14.9 "^I'Z i''7; 10. 1 8.7 7.2 5.8 4.5 25 23.6 21.6 19.7: 17.8 16.0 14.3 12.7! II I 9.5 8.0 6.6 5-2 26 25.0 22.9 21.0 19.0 17.2 15-4 13.7 12*1 10.5 8.9 7-4 6.0 27 26.5 24.922.3, 20.3 18.4 16.6 14.8 13.1 II. 4 9.8 8.3 6.8 Example i. — Dry bulb 15°, wet bulb 11". Difference — ^o The table gives 7.4 as the pressure. Example ?. — Dry bulb 16". 3, wet bulb 11**. 7. Difference 4''.6. : .*. pressure - 8.0 - (.6 x 1.2)4- (.3 x .7) =7.51. THE WET-DULB HYGROMETER, 7« 15 8 The reading of the dry bulb is of course the temperature of the atmosphere, i.e., of the mixture of air and aqueous vapour in contact with the thermometer. Opposite this temperature and under the column headed o we find what the pressure of vapour would be if the air were saturated with moisture. For example, at 15° the saturation pres- sure is 12.7 ; at 16°. 3 it is 13.5+ (.3 x .9) = 13.77- The ratio of the pressure of vapour actually present to the pressure at saturation gives the fraction which the vapour present is of the quantity at saturation. When this frac- tion is multiplied by 100 we have the percentage which the vapour present is of saturation, i.e., of the possible amount that might be present. This is called the relative humidity. Thus in Example i, the relative humidity = 100 ^12^ = 58.3, i.e... the vapour in the air is 58.3 per cent, of the possible amount at the temperature 15*^. The dew point is the temperature to which the air would have to be cooled before the moisture actually present would begin to condense. Look out in the column headed o the pressure already found for the moisture now in the air, and opposite if, in the first column, we find the dew point. In Example i, the dew point is C'^.S. Enter results as follows : — Experiment No. 28. Dry bulb Wet bulb Difference Pressure of vapour Saturation pressure at dry bulb temp. Relative humidity Dew point (Each student to sign his name.) 73 MA THEMA TICAL LA BORA TOR Y. 29. THE STEREOMETER. Experiment. — To find the volume and specific gravity of small quantities of sugar, powder, wool, etc., which cannot be conveniently weighed in water. An ordinary determination of specific gravity depends upon finding the weight of water displaced by the given body. By means of the stereometer the amount of air displaced by the given body may be found, and hence the the volume and specific gravity. For details see Stewart QX\diGt6's Practical Physics. APPENDIX I. THEORY OF THE PLANIMETER. w^ Fig. T, Fig. 74 MA Til EM A TIC A I. L A B OKA TOR Y. Imagine a straiglit line AB (Fig. i) to be provided with a wheel placed aiits midd'e point M so that its axis is in the direction of AB^ and that suitable graduations will record as in the case of the planimeter the number of revolutions and parts of a revolution which the wheel makes. Let ;/ stand for this number, i.e., for the change of reading of the circle between the time of starling and any subse- quent time. Let c = the length of the circumference of the wheel. Then en is the distance rolled through by a point in the circumference of the wheel. • Take l^ for the length of AB. Suppose ABf starting from the position in the diagram, to be an instant later at A\ B\ . This infinitesimal dis- placement may be considered as made up of a parallel translation to Ai E, and a rotation from Ai E to A\ Bi . In the first motion the wheel in moving from i)/ to ZT rolls through the distance MF and slides through FII\ during the second motion the wheel rolls through BMi ; hence MF+ HM\ - cfiy where n is the change of reading of the circle. The area which the line AB has swept out is made up oiAE and Ai E Lh . The area of AE is AB, MF, and that of AiEBi is AiE. lIMi (.'. HMi which may be regarded as perpendicular to A^E is \ EBy), Hence AB {MF + HM^) or ben is equal to the area ABB^Ay swept out by AB. Similarly if the line move forward into other positions the area which it sweeps out will continue to be the length of AB multiplied by the distance through which its middle point is displaced perpendicularly to the moving line, i.e., the distance over which the wheel rolls. In other words the area described up to any instant is bcn^ where n is the change of reading of the circle up to that instant. THEORY 01' THE PLAN I METER, 75 1 In order that this result may hold for all possible move- ments of the line, it is necessary to consider that the area, or a certain portion of it, is in certain cases negative. Let us agree to consider the area formed by all parts of the moving line which, as we look from A towards B^ are advancing towards the left as i-, and the area formed by those parts which are advancing towards the right as — . Then, whatever the motion of the line in the plane, the algebraic sum of the + and — areas swept out is equal to lyctiy where n is the final change of reading of the recording circle. Suppose, for example, that AD (Fig. 3) moves into the position A^ B^ by turning about C. Then in accordance with the distinction just made DCB\ is -!- andACAi is — , The sum of these (i.e , the arithmetical difference) is equal 'g- 3. to the product of AB by the perpendicular displacement MMi of its middle point. For the triangles DMi Bi and Ai Af\ E being equal, BCB^ ^ BE ^ ACAi orBCBi — ACAi = BE = AB.MMx = bm. Consider now the effect of putting the wheel at any point jy m AB (Fig. i). When ^^^ mowzs {0 A\ Bi the wheel will roll through IVR and GWi,. Hence, €n being the distance rolled over by the wheel, to get the area swept out by AB we must add b x IM\ to ben, Wi I being parallel to A\ E or AB. Draw (Fig. 2) a circle with IVM or IVi / as radius, and from its centre O draw OF and OPi parallel to AB and^i B\. Then PFi = lAfx . 76 J\/A Til EM A TIC A I. LAB OKA TOR Y. Hence, the area swept out by AB = ben + b, PP\ ; and if AB move into any jiew position Am Bi„ to which OPm is l)arallel, the resultant area swept out hy AB = ben i- b . arc FF,n . If AB move about, turn backward, and finally return to the position AB or parallel to it ; the area ben will require no correction, and the wheel will read exactly as if it were at M. But if ^^ mike a complete revolution and return to its first position, or parallel to ic^ we must correct ben by adding b x circumf. of the circle Fig. 2, i.e., by adding 2-bk where k - lV.\f, the distance of the wheel from the middle point of the moving line. Let any straight line move so that its extremities des- cribe any closed curves. Then in all cases the area swept out by the line will be equal to the arithmetical difference of the areas of the curves described by its extremities. • Fig. ^. When the areas are without one another, one will be described on the whole positively and the other on the whole negatively, while the area between the two which is swept out at ail will be swept out both positively and negatively. When they intersectj the common portion, in so far as it is swept at all, ivill be swept both positively and negatively ; the rest as before. When one curve lies , THEORY OF THE PLANIMETER. -j-j entirely outside the other, the portion of the latter which is swept at all will be swept both positively and negatively. Hence if, as in the planimeter, one end A of AB is constrained to move along a line (whether circular or straight) without tracing out any area, the area of the curve traced out by the othe. end ^ will be equal to ihe resultant area swept out by the line AB^ and hence will be^^« \{ AB return to its starting place without making a complete revolution. But if the fixed point of the planimetf.r lies inside the area described by the tracer B, the bar AB must make a complete revolution. Hence, the area des- cribed by B - ben + iirhk + circle described by A = hen + 2TTbk + ira^ = bc\ n + 2'iThk -I- ira^ Tc ] The second term in the brackets is a constant, and is en- graved on the bar. This number is then to be looked upon as a correction to n when the planimeter makes a complete revolution. In certain cases n may be negative. Let the distance of the wheel from the joint A be cal'ed d. Then (Fig. i) /^ - ~ - d. Hence area described by B = hcti + iirj I- - ii\ + 7r«2 = ben + TT {or + b'^ ~ ibd.) Place the planimeter so that the perpendicular from the fixed point to the sweeping bar passes through the wheel. Then the square of the distance from the fixed point to 78 MA THEM A TI CAL LA BORA TOR V. i I h the tracer B = a"- + I?- - ihd. .-. the area of the curve which B traces =/^c/^ + are i of circle described by the tracer when the planimeter is in the position just indicated. This circle is called the datum circle. (In describing this circle the wheel would slide and not revolve, and hence n would be o ) Let it be required to find If so that the area may be found in square centimetres by multiplying;/ by loo. We have l^cn = area = ioo« .-. dc = 100 100 or, = — centunetres, c where c is the circumference of the wheel in cm. -.s d = — mch es Similarly if^ = the circumference in inches, and the area is to be found in square inches by multiplying « by to. In general, if the area = ;/, Ifc is the unit area. pin 1 APPENDIX II. THEORY OF THE MECHANICAL INTEGRATOR. As we proceed from B to C by way of A (see figure), X changes from D to OE and \ ydx is the area DBA CE ; but, if we proceed from C to i5 by way of Pi , each element of area such as ydx is negative since dx is nega- tive, and hence f ydx is the area CEDB, but is negative. Hence, if we sum the elements such as ydx in the order of proceeding clockwise round the curve, the result = DBA CE-BDEC = B AFC, the area of the curve. Let A = this area, M = sum of the moments of the elements of the area with respect to OX, I = moment of inertia of the area, also with respect to OK. Then A - ydx, M^\ydx,^-=- ydx, J 2 2 J I^'^ydx J-^ ' j/^^. In the integrator one end of a sweeping bar traces a closed curve while ♦^he other end is constrained to describe a straight line OX. This part of the instrument is therefore a planimeter, and the area of *.he curve = bc^ni, where b is the length of the sweeping bai\ c^ the circumference of the wheel W^ , which the bar carries, and // ^ the change of reading of this wheel when the circuit vjj the curve has been made. The end of the bar which describes the ^o MA THE MA TICAL LAB OR A TOR Y, ' THEORY OF THE MECHANICAL INTEGRATOR. 8i straight line OX is also the centre of two arcs of radii 2a and 3^ which turn circles each of radius a, the circles carrying wheels W,^ and \V,^. When the axis of W^ makes an angle B with OX the axes of the other wheels make angles —-26 and 3^, respectively, with the same line. For jK substitute Z^ sin ^. Then A = d sin ^ {/x, I = 1/^^ J sin' Q dx. Now 2 sin ^ ^ = I - cos 2^ = I - sin ( ![ and sin 3 ^ = 3 sin ^ - 4 sin ' ^, 2^), or sin ' ^ = ^ sin ^ - - sin 3 Q. 4 4 .•.^=:^^^|[i-sin(^-26')] dx, = i^^[J^;c-fsin(--2(9) ^;c] «_l^^Jsin(!:-2^)^:x: since \dx^o for th.? complete circuit of the curve. /= V.' [(^ sin ^-' sin 3^) dx 3 •' 4 4 Also 8t MATHEMATICAL LABORATORY. = -<{»' sin dx _ -L sin 3 ^ dx. 4 J 12 J But by the theory of the planimeler the area A = bc^n^. .' . \ ixYi 6 dx = c^n^, />., when the axis of a wheel makes an angle ^ with OXyiQ have sin^ dx =c^nx* But the axes of the other wheels make angles -^ - 26 and 2,6 with OX. 2 s\w( —26^ dx = €2^2,, and sin 3^ .*. A = bc^n^y (IOC •— ^ Q ' 3 • 4 12 The maker has constructed the machine so that b ^ 2 decimetres, c^ ~ ~ dec, c = c-^ ^ ^ dec. 2 5 . • . -4 = «i , / « «1 - - «3 • 5