IMAGE EVALUATION TEST TARGET {MT-3) t ^ // .r illustration, or to add examples ; presuming that such, where necessary, will be furnished in the Lecture-room or by the Tutor. University College, Toronto, April I, 1S5S. ai pi ch re or m ce of iin tic CO] "Wl to ini sui the otl wh ket CHAPTER I. DEFINITIONS AND PRINCIPLES. 1. A material particle in a portion of u.attor occupying Mater.,, an uulehnitely small space; or, a geometrical point endowed "'^'^•'" with the properties of matter. All bodies may be geomotrically conceived as made up of particles. ^ 2. Whoa the cUstance between two particles remains un- Fore, changed during any period of time, they are relatively at rest and wo conceive that they will contuu.e so unless one or bot.1 bo acted on by some cause to which we give the name of force. e """ The state of rest or motion of a particle can only be con- ceived of m Illation to others, but it is convenient to speak of It absolutely ius being at n.st or in motion, reference being .mderstood to ourselves (or some particles in a known rela tion to ourselves), and changes of rest or motion are to be considered ^ produced by forces acting on the particle alone. 3. When a particle at rest is set in motion by a force it ''''"ti.n mil begin to move in a particular liiie, which we may define Smitud, to be the directixm or Hits of action of the force. The motion "' ' ^"'" might be just prevented and the particle kept at rest, bv a suitable force applied in an opposite direction. In this case the two forces are said t« balance or counterbalance each other ; and the magnitudes of two forces are said to be equal Enn.. when each would separately counterbalance the same force. '"^*'" vL ^''''^^' "^^^l ^^'''^^^ "«*i^g o'^ any system of particles statics. keep them at rest, the forces are said to counterbalance, or KLof \v-ii,'ia. 1 J 11 it of <rw is nicasurod nutafricallv with refcnMico to th(i unit liy the niiiidxT of such unit*< (acting; siniullniicousiy at a pnint and opposite to the force) which it will countei'l)alaneH, its inat,'iiitude is said U* b<^ n. This Kupposoa w to he a wliolu munlwr, nnd wo can nlwayH take a uiiit-fiiivi; of such ina^'nituilo tliat it sliiill liu hd ; then, wlicn any •ithiT fori't" is Uiki'U us tin: unit, tliu inagiiituiln of unr (irit.'iii:il force will 111' cxiiressfil hy tho rjitio (wlii'thor a whole auniher era fraction) whitli its iMiigjutU'lc hears tn tliat of tlie force assiuued iis luiit. In if( iieral. (lie torni /n'l'ssnrc may lie u.setlier prf8- brasa care- id all othtT • 1 from til'' *, ''a ciiliii- lieit, wlu'ii 1 that i\\o go l>y tlif liner, tlic Mscalleil M. F'roru ilio (Iffliiitiofis laid down, it will 1w o1)S('rvcil that three oleiiieiits enter into k-\\>v\ force; (1), its point nf rtpplicntion, or tlio particlo on which it acts; (L»), its direc tiun : (3), its ina^'nitudo. When these nro known, the force IS fiillv detcnninod. 1). The t'ollowiii',' is the funrlaniontril law, deduced fr(, f^.\j>crinient. on whicli (he sciencp of Statics is l)as(Ml : '" Kxin'ritiioil l;il I iw !f tvo f>.,p((il forcdi >icf ns-jwvt'u'fl if on ftm pari in, n, ir/n'r/* are rifulh/ cmuieckil, in the lim-joininy them hut la oppa.sifr. directions, they will counterhalumc. Hence, either of these forces may ho transferreil to tli. '»ther particle, presorviiit,' the same direction, without alteia tion of its statical effect ; or : " A force 7)11(1/ he suppnficil to net ,it ANY polvt in its omn line of action, Ike ww j>oint of application heiwj rifuVij cov uPA^tcd xrith the forma' one;" and in this latter f.rm the lasv is frojuently stated. I". Th(( following ctjnsequenccs may he noted: WHion a pressure is coiniuunicated l)y means of a straiirht ngid rod in diivction of its length, the ])tvssure is wholly eflcctive in this / point of tlie string there aiv acting along it, in oppo site directions, two forces equal to the former, and either of ^hose is called the tension of the string, which is thus uniform thr.mghout its whole length. The same is true when the =!tring (if it bo peifectly ilexihle) is stretched over a smooth. surfa,ce. A .'^moolh surface i.s (.ue which can cxuri; a pressure at any point of it, only in (lireetiou of the noiiual at that point. .1 striiii; K- ri iiri>- fll'lltrd tiv RtraiKlit " 'mill 1 1 . Sinco the throo olomontH whicli Hcrvo to dettjrminft a jtnvssun' iim in tlitur iiaturo i(l(>ntk-til with tlioso which deter- iiiinoii Htriii;^ht lino — nainoly, niiignitinh', diroction, an.* point thntiigh which it is to ho dniwn — it foHown that a utraight lino may proporly ho taken as tho reproHontativo of a prew- 8ur«. When, howover, a lino Ali is so takon, it will b*; iindorstood that tho prosHuro acts in tho diroction from A towards li; if writton JiA, then from // towards A. Fro- (|ii(!ntly also, tho words " reprosentod hy " will 1)0 omitted, and wo shall use " the force AJi " to indicate tho force ropro- Honted in magiiitiido and direction by tho line AJi, acting in a direction from A towards Ji. 12. We now proceed to state tho two problems of Statics which aloiio will be hero investigated. (I). Tho conditions of equilibrium for any set of forces acting on the same particlo. (2). Tho conditions of equilibrium when forces act on a rigid system of particles which has a fixed axis round which it can turn freely, tho forces acting i)erpcudicularly to this ,-ixis. CHAPTKR II. FORCES ACTING AT A POINT. \f lie ZtlT" T "*""^*— 'y -' a particle at rest, n.n„.t,o„ „ tr tLe pa tide bogxn to move, its motion will commence in a "^^^'"'^ defu.te .Ixrection, an.l might be just prevented by a single force of suitable magnitude appli.l in an opposite directiot This force would then counterbalance the original set of forces, and a force e,ual and opposite to it wotld produce he same statical etfect as the fi,-.t set of forces, and 3 therefore termed their resulkmL it aVr.!!'"''' ''^''" '"^ '"' '^ ^^'^'^ ^^'^"S '^^ '^ point keep tcrbalancng all the rest, a force equal and opposite to any one of them is the resultant of all the others. 15. Hence also the condition, in order that forces acting , , at a point may keej. it at rest, is that the magnitude of '^^^^^ 16. When forces act in the same line and direction on . a point, heir resultant acts in the same direction, and its '=-''* mag^utude is equal to the sum of their magnitudes. If some of the forces be acting in the opposite direction, the mag- niU.de their resultant will be the difference between the Uin ti ?r^'1" '' *'^" ^^*"^° "^ *'- -« ^--tion and m the other, and xt will act in the same direction a. those forces whose sum is the greater. We can, however indicate oppositeness of direction by attaching to the magnitudes of the forces the algebraic signs .- and - : so that, any one force being considered positive, all forces in 10 that (liroc'tion will nlso Ih> coiisidoroil jKmtlir, l)ut forces in the opposite direction will bo considered necjatlw. The above results may then Vo combined into the fol lowing : Th.*;!- resul The resi'/fanl of any ,w/ of forcaa act in f/ on a point h>. the Sdinc lnii\ i/^ th'' (Uihhrdic stnn of ttio, lorces. i,v.n.liti.iii of 17. licnc- idsd llie condition tliat the point may be kept •e I'epresented in mag nitiide and din'ctinn by two straight lines drawn from a point, and the paj'allvlogram, of which the.se lin(>s are adj.i cent sides, be et)m[iU ted. that diagonal which passes through tiic point will rcjirescnt in magnitude and direction th<- resultant of the two forces. P'^iTiiii ,,f ^ "^ ^'"^ lows .1.1., A fi bo di-awn representing in magni- '•'•>"''••""• ^j^^n^^^^m^d^^i^^ui^ tude and direction two tbrces acting at .1 ; and let y/. 7 be the number- tlcnoting the magni , , ^^^_i_— i^_-.^M^^siis.^.^._MM-_>_ tudes of the foi'ces. f)iinani(^l s lir.Kil. Divide -1.1,, into /> < (pial parts inthopoint- .1,, .b„ .1.;, and AL into n uipial jiarts in the points 11, (', 1>, : liicn LMCli of tliese cipiul }iarts will icpresent in jnagnitudi; the unit force. Through these points, draw line-- ■•'M t forces ill I pohit h>, ij be kei>r 11 •parallel to tho ori.iriiml lines, comph'ting the pfimllelo.trram • 'iud suppose all the lettered points of the figure ri and ,> acting at A to the same forces acting at Z„ iii parallel directions to the former and this without alteration of tl,.-ir stati.'al etF.rt. Heiuv Z„ nuist be a point in the direction of the resultant of the original forces at .1; that is, AL^ u the direction of the resultant, which proves the principle enuuciated, so far as Hie direction of the resultant is coucerued. I 19 Mdgnitnde Let AB, AC, represent the two forces acting at A. Com- Aut plete the parallelogram AC DB. Then AD is the direction in which the resultant acts, and we have now to prove that AD represents also its magnitude. In DA produced back- wards take AE to represent this magnitude, so that a force represented by AE \n\\ bo Cfpial and opposite to the resultant ; and the three forces represented by AB, AC, AE, wUl keep the point A at rest, and each one of them is equal and opposite to the result- ant of the other two. Complete the parallelogram A EEC ; then, A F h the direction of the resultant of the forces A E, A C, and is therefore opposite to AB. Hence, FAB is a stmight line, and therefore FACD a parallelogram. Hence, the lines AE and AD are equal, being each equal to EC : but AE was taken to represent in magnitude the resultant of AB, A C, and consequently AD also represents it in magnitude. Q.E.D. UtiiioltitioD •f a forci*. 20. Conversely, a force acting at a point can be resolved into an equivalent pair of forces at that point in an infinite number of ways ; for, taking a line drawn from a point to represent the force, and constructuig on it as diagonal ani/ parallelogram, the two adjacent sides terminating at this point will represent an equivalent pair of forces. If this pair consist of two forces acting in perpendicular directions, each of them is called the ejective part of the original force in this direction. 13 Thus, if ff he the force at A, represented hy AD, and it ho resolved into two forces in per- pendicular directions- -namely, A" along AB and Y along A G ; then A' and Y are the effective parts of li resolved along AB and AC respectively. Com- pleting the rectangle ACJJB, A' and Z \vill be respectively represented by AB, AC, and, calling the angle BAD, 0, we have from the right-angled triangle BAD, X=:Rco3 0, Y=R sin 0. 21. Hence, to find the effective part of a force in any Rule for. given direction, or, more briefly, to resolve a force in any given direction, m^dtiply its magnitude hy the cosine of tfie angle contained between its direction and the given direction: and to resolve a force perpendicularly to a given direction, multiply it hy the sine of this angle. 22. So also from the same figure we obtain the resultant (/?) of two perpendicular forces (X, Y), and the angle {0) which its direction makes with one of them (A); for R' = X^+ P; and, tan 0^^. 23. When any number of forces act at a point, their whole effect in any direction will be the; algebraic sum of the so{»a- rate resolved forces in this direction, which will evidently therefore be equal to their resultant resolved in the same direction. Hence also the algebraic sum of the separate resolved forces in direction of the resultant is the resultant itself, and the correspond- ing sum in a direction parpendieular to the lesultant is zero. 24. To find the magnitude and direction of the residfant ncsuitant of of any forces acting at a point, tlteir directions being all in "qc plane, '" one plane. AUd 14 Taking any two directions at riglit angles to cacli otlior, let each foree be resolved into its components in these direc- tions. Let the algebraic sum of these resolved parts iu the one direction be A', and in the other 1 . Then the whole set of forces is efpiivalont to the two X, Y. }lence, if Ii be the residtant of the whole set, and there- fore also of the two X, V, ami the angle it makes with the direction of X, the equations in § 22 give li'^X'+Y\ tanO = Y X' ( ri'llti'DSof (.quilibfiuiii. Avhich determine the res,V)' lami's iui'uiulag. Polygon of for«i!». 16 and also the equivalent exprensions — R'^P' + Q^+2 FQ cos {P,Q) P'--^Q' + R' + 2QIicos(Q,R) Q^ = R» + P' + 2 RP COB {R,P) ■ 30. Polygon of forces. 1/ forces acting on a point be represented in magnitude and direction by the sides of a polygon, taken in order, they will keep the point at rest. For if ABC DBF be the polygon, the forces AB, BC, have for their resultant AC ; and the resultant of this and CD \s AD ; and so on till we come to the last side, which is equal and opposite to the resultant of all the previous ones. Hence the proposition, as well as its converse, is estab- lished. 31. In this way, the resultant of any number of forces at a point can be constructed geometrically ; for, having drawn consecutive lines, so that, taken in order, they are parallel to, in the same direction with, and proportional in magni- tude to the forces ; the lino drawn to complete the polygon will represent in magnitude and in reversed direction the resultant required. It may be noticed that the polygon referred to need not be a plane one, neither are re-entering angles or crossed sides excluded, nor does a change of the order in which the forces are taken affect the result. CHAPTER in. FORCES iK ONE PLANE ACTING ON A SYSTEM OF RtC.IDLY CONNECTED POINTS, WHICH CAN TURN FREELY ABOUT A FIXED POINT IN THE PLANE. 32. Tioo intersecting forces act on a rigid system, in the Comiitiun nr same plane with a fixed point romid which tloe system can wl"''' tn.:"" t'tm. tWd forces ini'ct. Let be the fixed point ; P, Q, two forces in the same plane with O.tlieir direc tions intei-sectiiig in A, at which point, rigidly connected with 0, they may be supposed to act. Then if R be the re- sultant of P, Q, in order that the point A and the whole system with which it is rigidly connected , _ may be kept at rest, it is necessary and sufficient that the direction of R shall pass through the fixed point 0: that is, ^0 must be the direction of i?. Draw OB, OC perpendicular to the directions of P Then resolving the forces at ^ in a direction perpendicular to ^0, we have (§21, 23): P sin 0^5 -(? sin 0^(7 = 0, and therefore P. OB-Q. 00 = 0, or, P.OB^Q.OC. B \l 18 Moment dc- flneU. nfJ. Tlio ])ro(luot P. OJl, wliidi is tlio jn-ndiict of tlio rmm- l»or fxin't'ssini,' tlio inii<,'iiit>i(l(' of tlio force, ami the lcii;,'th of tlic ))f'r|KMiD found from this expression. 35. Ihco forces act in parallel directions on a ritjid system. J^^.„ ,,,r,iii,i forces an' T>i't P. Q 1)0 tlio two foivoia ; 0, nny jioiiit in tlioir ])lane. t.) a s'in'gU' Dniw ()(, B j)or- IH-ndioular to the tbrce.s. At 11, t\ I '4H'L^ ^^^'^ I'quiil anil opposite forces T\ these will in no way atloct the system. Let A* be the ro- „,,„,,,,., ^^ mltant of P, T, tl'em. Lotiug at B ; and 5' that of Q, T, at C. Tlieu the directions of R, IS will in general meet : let them do so in A, and suppose them to act at this point. They can now be here resolved into their original com 20 */\\nnc mug- potients, P, T; Q,T\ of which the forces T, beinj? equal and f'liei'r Hum Opposite, may bo romovetl altogether, leaving the forcen P, (^ acting in a direction parallel to their original direction, anti combining into a single force (P + (J). Again, the moment of this single force (P + Q) al)Out O is e<]ual to tlus algebraic sum of those of its conn>onent» P and *S' (S 31); and the moment of A' is equal to the sum of the moments of its components, namely, P and T ; and so is that of (|uili1n-iiiin \n tiot jtoHsiMo ; In tlu* latter, ciiiiililM'iuiii will siilisiHt it' tli<< rt'suitiiut ititlioi' Im> /.no or piiHH tliroiii,'h tin* tixi'd point, iiml »'iicli of thtrso supiioNi- tiouH will nmko its niouMUit iil>out this jtoiut viiiuhIi, and th<>t'««t'oi'u also tho iil^'*>hruic hiiiii of the nioniiMLtn of all tiiu forces, to which siuu it has been hIiuwu to ho c/' the moments of all the forces about the Jixcd point III lift ridiitth. This principle will always ho ipiotcd by the nani<' of " tho vanishini,' of moments." 3!>. Tlio sumo priiiuiplt; iii.'iy easily lie seen ti> apiily when tlu' rigid l)(iily i« civiKilileof t'iriiiiigiil)ii\it ii tixuil straiglit line or ax is, iunl the forces ivru nut nil in one pliiau lint are lu'i'iii'inlicuiiii' to tiiis axi.-i. The mninent of each force being taken almut tlia*"; point of the axis whieii is cut liy a pei'pi'inlieular plane containing the force, wc cuu statu the uonditiou of c([iiilil>riuui in the form : Tlie alufhralc tiuin of the momi'iits of the Join .s altuiU the Jixtd nx'm mii.it f- f/ntiii oJ'J'oirvn, viz.: — by drawing lines from a point in dircciions of the two axes and proportional in length to the magnitudes of the moments, and completing the parallelogram of ■vvliiuh these are adjacent sides ; then the «liagonal represents the direction of the resultant axis and the nuxgnitude of the moment abcnit it. Hence any set of forces acting on such a body are e it at rest is that this resultant moment shall vanish, and (as in § 20) the necessary and sulHcicnt conditions that this shall be the case are that the alijebraic i>iiiiif I HUppOMi- iiiisli, and of all thd 1. uilihridin fCS It.IlOIlt II' I )f"tll I'll till' rigiil sis, iiiiil till.' ;(i this jixi-t. of tlic axis It-f, WU L'illl (' Jixed nxh licli it w.vfl in itself, point arc ii'sultant jiiiriiUi'Ut- ludoas of k'8 of the these are on of the Henee to a single it at rest §20) the e case are It To recapitulate the condilioini of e(iuililirium ; (1.) Wliiii forces act lit II point and keep it at rest, the resultant force nnist vanish, and thcrcfori^ thf nlijihroir kiiiiii nf tliv j'mri.i rfHith'rd along thrre iniitualhj pifjiindU'ular llinn mitut Hepamkbj raiiUh. i (2.) When forces act on a ri^id liody which can tui'ii freely ahoiit a lixed point, the resultant nn>nient must vanisli, and therefore Ifn' nliji'liriiir siinix of fin viuiin n/s (if thr j'nrciH alniut thru., mutanlli/ pi r- jwiidii'iilitr (ijr.i iiiii.it Mepui'itttlj/ niiil.ifi. (.'{.) When forces act on a riu'id hody which is perfectly free, and kceji it at rest, tiie iirecedinj,' conditions must lioth he fnllilled ; that is, the forces must i»u such that (I) if they acti:il at a jioiiit retaining tlicir directions and magnitudes, tiiey would keep it at rest ; ami (2) if a point of the hody were tixed and the limly free to turn ahout it, tiiey would keep it at rest. Thus involving on the whole six necessary and snilicicnt conditions for e([uililirium. (4). When any system of rigid liod'-s, acting on each other hy contact or in otlur ways, is in ecinililu'iuin under impressed forces, the in'inciple laid down in Newton's Tliinl f.n'ir n/ Afuflini (hereafter (juotcil) asserts that the actions of these hodies are mutual, eipial, and opposite ; and therefore by forming the conditions of enuili- brium for each body se|>aratily we obtain enough ens to elimi- nate their mutual forces and leave tiie resulting conditions among the impressed forces. Although this is the comjdote solution of all the statical problems which arise in the cimsidetation of rigid systems, it must ))e noted that in the practical apidication to the cases which occur in nature, these residts can only be regarded as a first approximation, because no such thing as a rigiil body is known to exist. All natural bodies are nun'e or less yielding, and this circumstance gives rise to prob- lems of extreme ditlicnlty and complexity, in the solution of which not much progress has yet been maile.] CHAPTER IV. CENTRE OF PARALLEL FORCES, AND OF GRAVrTY. iirMiit^iiit of 40. It lias 1)0011 shown that two parallol forcw (not formiut: two |^i':llli'l . . . I. mis lias a coujilo) actiiig on a rigid system, lia\ o fov rosultant a single force in tlio sanio plane ; its dirootion being parallel to that of iiioiiiont the two, its magnitude boing the algebraic sum of their mag- luiKliiuiar nitudes. and its inouient alxmt any {)oint in this plane Ix'ing the algebraic sum of their moments about this point. Also, the moment of a force about a line to which its direc- tion is perpendicular has been detined to b(> the product of the number expressing the magnitude of the force, by the }>erpeiulicular distance botwceii its ilircction and the line. 41. It will now bo shown that the moment of this rosultant of two parallel forces, about anij Hue {»erpeiidicular to their direction, is e(|ual to the algebraic sum of the moments of the two forces about this lino. ri|iial to the Sinn (if the Supposing th(> forces /', Q, to be acting in the same direction perpendicularly to I lie piano of the tiguro and mooting this plane in the j>oinls Ji, ('; and their re- sultant A' (which - P + Q, and acts panillel to and in the same plane with them), to Mieot the ))laiie of the ligun^ in A. Let Inic Ik^ any line in this jilano, and draw to it the ])orpondiculars Aa, JJh, Tc. Then, if JIAC bo parallel to hue, Aa, Jib, (\; are all (Mpial. and the pro})ositioii is manifestly true, for Ji\ Aa - ( /' + Q) Aa _ P. Bb + Q. Cc. m [{AVITY. (not fonuing Itaiit a single lli>l to that of of tlu'ir mag- s piano Ix'ing [>oint. Iiioli its iliroc- lie pnulnot of foioi', by the id the line. this resultant •ulai- to their .^ moments of , to 1)0 acting 'ndieiilarly to meeting this anil their re- el acts panillel ith them), to in A. Let ', and draw to h, Co. 'riien, (f, Jil), (\; are 1 is manifestly c. But ifBAC be not parallel to bac, let them meet in 0. Then, being a point in the plane of the forees, the moment of a' round is e.pial to tlie sum of those of P and Q round it, and therefore A". OA - P. on + Q. oc. But by similar triangles 'OA M ~0B Cc OC and therefore iiioniciits .i( t)ic fi)Ui'.s. P. Aa^P.Bh uQ.Cc, Nvhieli }.roves the proposition for this case, and the proof holds gootl also feu- the eases where the forces or their moments are m opposite directions, having due regard to algebraic sign- 4l>. An;, paralJel force,, actiwj on a ruihl s,,sf.'m, are Anv ,.,„„.„., e,fher re^h.etble to a con/>/e or e/se to a sin,,Je resultant force t^^^^tl ione/> net, ,n a p,mdM dhrction. Its nnujnitnJe be!n,, the '.".i^M,"' ah,ebro,c stun of the nim/nifmtes of aU the forees. oil its """''^"" moment obont on,, assnnnul line perpen>lie>,b,'r to their direc- tion, benn, ainol to the ahjebraic sam of the moments of all the Jorees about this line. Taking any two of the forces (which do not form a eouph-) w., hud their resultant, which acts in a iKiraUel direction Its magnitude being the algebraic sum of their magnitudes' uiul lis moment, about any assumed lim- perpendicu'lai to the direction of the forees. being e.pial to tlie algebrai.. sum of their .Moments about this line : ..ombining this. vsultaut with any third force to form a new resultant, aii.l this a-aiu with a fourth, a.ul so on as in 5$ ;57, we arrive at last v\i\un- at a couph. or a sin»lo resultant fore.* acting in a pa.-all,.| ,|i,vc tion, its magnitude being e,p,al to the al-..l,raic sum of the ■uagmtiules of all the forces, and its moment about the :issumed line In-ing o.pial to the algebraic sum of all their uioments about this line. Hi n 26 Centre of pimltel forces. 43. The centre of panUhl forces. When given parallel forces, aclinrj at given points of a rigidlij connected Sjjsteni, are reducible to a single resultant, its direction passes through a point whose position is invaria- ble with regard to the points of the system, tohatever be the direction of the forces. Tdke any two of tlio forces P, Q, acting it tli<; given points Ji, C Join JW and lot ;hi;ir rcsnltant Ji cnt it in A. Then, the moment of Ji about any point in the plane HUiig ('(jnal to the algcldviic sum of the nonients of P, Q, about this [)oiut; let thcsc! moments bo taken about ^1. Tlio moment of Ji about A is zero ; lence, dniwing bAc per[)enilicular to the lirection of the forces, and, l)y similar triangles, An Ah P. Ab - Q. Ac - 0, , and therefore, AG Ac P Hence JiC, which is given, is cut in the point A in a ratio which is independent of the direction of the forces with I'egard to IJ C, and the position of A is therefore given with regard to Ji and C Now, taking any third force, acting at D, wo may combine it with the resultant of P and Q, and the point in which the new resvdtant cuts AD will be given in position with regard to A and JJ or to A, B and C. And thus we may go on till we arrive at the final resultant. Hence, the projiositiou as enunciated is true. This point is called the centre of parallel forces. 27 44. If this point — the centre of i)ar{illel forces — in a given The syst«n system be rigidly connected with the system and Hupjioitcd tmn.ii or fixed, the system will be kept at rest, and will remain so "^^ when the forces are turned about their points of action into any other direction. It will also still be at rest if it bo turned about this point into any other position, the forces acting always at the same points of the .system and being always parallel to each other, though their directions may be varied at pleasure. The pressure supported l)y this fixed centre is evidently the algel)raic sum of the forces, and the a]gel)r,iic sum of their moments about any line through this point vanishes. Cfiitivdf Kv;ivily. 45. The centre of ynivitij. When the only forces acting on a system are the weights of the several particles of that system, if we suppose the vertically-downward directions in which th(!se weights act to be parallel to each other, and the weight of any ]>articlo to be independent of its position ; then, since the forees all act in the same direction, they have a single; resultant which is eejual to their sum, that is, to the weight of the whole system, and acts vertically downwards thnxigh ihe centre of jxtrallel forces, which is in this cas*; called the centre of (jraclt.ij. [Neither of the suppositions above nuiosing it to be ^s•itllout weight, and the i,,' rulkiui' whole weight to l)e collecti^d at its centre of gravity and there '' ' to act — this point, however, being rigidly connected with the syste'm. We may also, without alteration of the statical effect, con- ceive the system to be geometrically divided into any nuniber of systems, and the weight of each of these to be collected at ■1*11 28 its own centre of gravity ar.< i tliore act, these? partial centres of gravity being rigidly connected with each other and the system. . ' i^ystein 47. Also, if the centre of gravity of a system be supported iiiioiit in all or fixed, the system will balance about this point in all posi- I osi louH . tJQjjy under the sole action of the weights of the parts of the system, these being I'igidly connected with each other and the centre of gravity, and this is sometimes made the defini- tion of the centre of (jravity. Ilmv I'lninJ. of iiuuiform liuUy. 48. The position of the centre of gravity relative to a given system will be tleterniinecl from the consideration, that, placing the system so that any given line in it shall be horizontal, and e<[uating the moment of the whole weight collected at the centre of gravity with the moments of the several weights of the particles about this line, the distance of the centre of gravity from the vertical plane passing through this line will be found. Taking thus three planes in succes- sion intersecting in a point, the distances of the centre of gravity from each of these planes can be found, and its position therefore determined. 49. Since the position of the centre of gravity in the sys- tem depends on the relative and not the absolute weights of its ])arts, this position will not be affected by increasing or diminishing pro])ortionally these weights. 50. If a rigid body be of uniform density : that is, if the weight of a given volume of its substance be tlie same in every ptirt of the body ; then, if there be a line about which the form of the body is symmeti'ical, the centre of gravity will be in that line ; and if there be two such lines, the centre of giavity will be their intersection. Thus the centre of gravity of a circle or sphere is the centre ; of a parallelogram or parallelo[>iped, the intersection of its diagonals ; of a regular prism or cylinder, the middle j)oint of its axis. 51. If a uniform body balance in every position about a line, the centre of gravity will lie in that line ; and if about two such lines separately, it will be their intersection. Thus a triangular area will biilance about a line drawn from one angle to bisect the opposite side, for the triangle can be generated by a line moving parallel to one side, and the small area gene- 28 fated at any stage of its motion will balance about tlio lino Of a tnaii which bisects it. Hence the centre of gravity of a triangular '' area is the intersection of lines drawn from the angles to bisect the opposite sides, and this intersection is at a distance from the angle of two-thirds of the bisecter drawn from it. For let ABC be the triangle, and BD, C^})i8ect AC, AB, and meet in i.T. Then G ia the centre of gravity. Join ED, which is parallel to BO (Eucl. B, VI. 2), BO BG Then -- = — , by similar triangles BGC, EGD GD ED , by similar triangles AGB, ADE _CA 'ad' 2 Hence BO, being double of GD, is — BD. o The same point O is also the centre of gravity of three equal bodies placed at the points A, B, C. Cor, In this way can the centre of gravity of any poly- ofanypoiy gorial area be found ; for, dividing the figure into triangles ^'^"'^ the weight of each of these may be supposed collected at its own centre of gravity, and the centre of gravity of the whole figure will be that of these weights, considered as lieavy particles situated in those points. The method of finding this latter will be treated in the following article. i 52. To find the centre of gravity of a system of particles or all in one plane. .nny heav* jxirticlfs ii> OIll.' plilll'' Let Ox, Oy be two perpendicular lines in this plane, with iregard to which the positions of the particles are known. Let P be the place of one of the particles^ w its weight. no Draw PX, PM porppndioulai' to O.t', Otj, vospectivoly, and denote PM by .<;, PN Ity //. Suppose tlie ))lane of tho fiijfun! to 1)0 liorizontal ; tliou Ox is a lino poipcndicular to the direction of the wcijrhts, and thci'cforo (^ .'}()) tlio nioiiiont iil)out Ox of the whole weight colU'cted at this centre of gravity is ecpial to the algoliraic sum of the separate moments ahout it. ITenee if W l)e the wholt* weight, mikI the distance of the centre of gravity fi'oni Ox bo denoted hy //7 we hiive and // ^ W where I" denotes the idgel)raic snm of all the products corros- ])onding to that within thi^ bracket. Also, if a luon.ent be reckoned positive when /* is above the line Ox, it will plainly be negative when P is Ixdow the line, and the diffia-iMice in sign of the moments will therefoi'e at once be indicated by considering a ij positive when drawn uj)wards from Ox, and negative when downwards. Similarly, by taking moments round 0//, if ~^be the dis- tance of the centre of gravity from Oy, we liave where x will be considered positive when drawn to th(> right of Oy, negative when to the left. The distances of the centre of gra\it y from these two lines beinff thus found, and the directions in which these distances are drawnbeing i nd icatod by the signs with which they arc; affec- ted, the ])osition of the centre of gravity is fully determined. p -X Y P M +y -^y N' N -y -y -X +x p r p 81 Cor. If ihe pui-ticlpH all lie in tlio same line, take this for Of imrticies ' ' III a struiKht Ox. Tlion, cviny v/ lieiuj:; z(u-o, »/ is so also, and tho ceutvo l'''«- of gi'a\ity is in Ox, its distance from boing given l»y ^' ir The following independent pi'oof of tliis may be noted : 53. r, that of a particle on the otlier side of O will be negative, and the difference in algebraic sign of the moments will therefore at once lie indicated by considering the x'h of the particles to be positive or negative according as they lie on one or the other side of (>. 5-t. When a rvj'id body rests suspended from or snpprn'tcd n.'uvy i^dy hij a fixed point, and acted on onJy by its weight, the vertical fnriy, its line drawn throutjh the centre of yravity will pass through the p-'ivltyis point of suspension or support ; and, coiwersely. ]!i!o\i'\,rh.;- Iriwtlu' point For the weight of the body may be sui)posed collected at Bum. its centre of gravity, and tliere to act vertically downwards ; and the necessary and suffici(!nt condition of equilibrium is that its moment about the fixed point must vanish, which requires that its direction shall pass through this point. Tfm Cor. 1. Wlieii the contre of gravity is vertically Under tlio point of Hn]))»ort or suspension, if the body be slightly disturbed from rest, the moment of the weight will tend to bring it buck again to its original jmsition. The eqnilibriuni is therefore said to bo stable. When the centre of gravity is Vertically above, the contrary takes place, and the equilibrium is unstable. Cor. 2. This affords a practical method of finding the centre of gravity of any phmo area. Thus, susj)ending it freely from any one j)oint, trace on it, when at rest, the direction of the vertical passing through this point : then, taking any other point (not in this line) for a new ])oint of susiiension, trace also the vertical through it. The intersec- tion of the two lines thus drawn is the centre of the gravity required. Jidiiy riiacod 55. Wken a rinid body, havina a plane base, is placed till iiiuio, wUk tins in contact with a Jixea horizontal plane, and is 1.1 (ill I Dvei. acted on only by its oion weight, it will stand or fall accord ing as the vertical through its centre of gravity passes within or without the base. By the base is here meant the figure included by a string stretched completely round the outside of the plane section of the body which is in contact with the horizontal plane. If the body fall over, it must begin to turn round some tangent to the curve formed by this string, and the moment of the AVeight, stipposed collected at the centre of gravity, must tend about this tangent in a direction from the inside towards the outside of the area of the base, and the vertical through the centre of gi-avity will pass outside the base. Also, when this vertical passes outside, the body must fall over ; but if this vertical pass within the base, the moment of the weight about every tangent to the string tends in direction from the outside towards the inside, and the body cannot fall over. CHAPTEK V. THE MECHANICAL POWERS. 56. The science of Mechanics was (as its name implies) The . , : no macliiim in old times restricted to the employment of force on machines, and it is still usual to treat of the simple machines, or mechanical powers as they ai-e sometimes called, under six classes, namely — the lover, the wheel and axle, the pulr leys, the inclined plane, the screw, and the wedge. Of these, the wedge will not be here considered, as , in its pmc- tical application the investigation on the principles of the foregoing chapters would be of small utility. When a power P sustains on any one of these machines a MechauUiai weight W, the ratio W : Pis called the mec/ianical advantage defined. " of the machine ; and the machine is said to gain or lose ad- vantage according as this ratio is gi'eater or less than unity. In the following investigations, bodies will be supposed rigid, sur- faces smooth, strings perfectly flexible and of insensible size, and the parts of the machine to be without weight, unless otherwise specified. 57. T/ie lever. straight A straight lever is a rod capable of turning freely in one i«ver. plane about a point in itself which is fixed. This fixed point Archimedes is called the fulcrum. and UaVinci. Case 1. — The weight W at one end of the lever supported Fig. i. Ijy a weight P at the other end. liAC the lever ; A the fulcrum. Draw bAc horizontal, and therefore perpendicular to the direction of the weights. e ^9^ M Then liy the Viinitiliiiii.' ut' luomoiits, J\AI,- \r.Ac~i), m J\A(,.]r.Ac. AB AC But, by siiuilai' ti-iiuiglos, -77- = -jy ; and tlicroforo P. AH .W.AC. Cor. 1. The |>rcHsuix> on till) fulcrum is the woiglit (/'+ If';, iictin*' voitifiillv ilownwiirds. Cor. 2. SiiicH' tlu! rolution /*. AB - If. AC does not invoh »• tlic angle at which the h'vor is inclined to the horizon, it lol- I0W8 that if the ]i/v r Ik; at rest in any one position (ex(C|it a vc^-tical one), on licinj^ turned into any other position it will still Ik; at rest. i.ij. ., Cask II. — The jiower I' and the weight IKivcting in o|ij»o- sito (but jnirallel) dii-ections, iuid the weight nearer to the fulcrum than the power. Using the wune construction and reasoning as in tin- Cortuer case, we have heie also P.ABW.AC. Cor. ] . Tlie pi-essure on the fulcrum is hei^e W - P, rtctin;: vertically downwards. The second corollary also holds. KiM .; Case III. — The jiower P and the weight W acting in opftositL! but pinillel din;ctions, and the jwwer nearer to the fulcrum than the weight. As before, we have P. AB^ W. AC. Cor. 1. The pressure on the fulcrum is P- TF and acts veiliadly upwards. Tlie second corollary aJ&o holds. Midi idv. f'8. In all these cases, the mechanical advantage ( ,, ) i*^ AB AC 7- or the ratio of the arms of the power and weight, (n 35 r!ii,sii()Metl ciiUoctid at tlie centre of gravity (/ w(hicli will he the l"'^"l'i' '^^ niiiliUe jioint if the lever he uniform). bet the vertical through G meet the horizontal Ahc in ;/. Then, \>y the vaniahing f(i). Proceeding in this way, we come at last to the weight sup- ported at A„ = 2" P, and this is the attached weight. Hence, W = 2" P and the mechanical advanta,HMiB'^ fffll'*Vy»Lll l.^UWU And the whole power required will be the sum of these ; therefore, w, w„ Third sys- tem, FiL', !i. P=-^ w +....+ n 2" w + — , or 2" ^ The weight of the pulleys, therefore, lessens the advantage of the machine. Cor. If the weight of each pulley be the same («>), then jr = 2'*P-(2'^-^-l- 2'*-2 + ...+ l)«; = 2"^P-(2''- l)w. 68. Third si/stem of pulleys. Each pulley hangs by a separate string, which is attached to a bar or block carrying the weight, and the free portions of all the strings are pai'allel, and therefoi-e vertical. This is the second system turned upside down, the weight becoming a fixture, and the beam to which the strings are attached becoming a movable bar carrying a weight, and the mechanical advantage miglit be inferred from the preceding. The pressure supported by the beam in the second system is the sum of the tensions of the strings, that is, P+2P + 2''P+ ... to n terms, = (2»-l) P, and this becomes tlit- weight W in the third system. Therefore, W={2''-l)P. The last pulley {An ), however, becomes fixed, so that the number of movable pulleys is only ()i-l). Making n the number of movable pulleys, we have jr=(2'^+^- i)p. The following is an independent investigation for this case. Let .^u -^2, ^3, ... A,^, 1)6 the pulleys, n being their num- ber exclusive of the last one A, which is fixed, and « m the number of .strings. 41 iiii- Bi, Bj, B3, ... -B„+i, the points at which the respective strings are attached to the straight bar which carries the weight W. Number the strings 1, 2, 3, ... according to the pulley over which each passes. The tension of each separate string is the same throughout. The weight supported is the sum of the pressures of the strings at Bi, B^, B3. ... The tension of (1) is P, and this is the pressure at Zf,. The weight supported at A2 is double the tension of (1) and = 2 P, and this is therefore the tension of (2) and the pressure at £2- The weight supported at A3 is double the tension of (2) and = 2(2P) = 2 P; and this is therefore the tension of (3) and the pressure at By Proceeding in this way, we obtain the tension of the {n + \)th string and pressure at 5„+i = 2"' P. Taking the sum of all these pressures, ^=^+2^+2^^ + + 2"^ -(2'' + '-l)F, iind the mechanical advantage is 2 - 1. Cor. The weights of the pulleys may be taken into account by Puii. observing that each may be considered as a power acting by means '""" of the string from which it hangs, and supporting a weight on the system of movable pulleys above it. Let io^,W2,iOq,...w^, be the weights of the pulleys, blocks included. The weight supported by w^on (n-\) movable pulleys is (2" - 1) h'j. •ys sii)'- illlc.tVV. II II II II Also tOj on (n - 2) w^ on Pon » -,n-l (2''-'-l)w.^. (2' (2 -IK. n+l 1)/'. i'iilil|i;nvil. Spiiiii.^li Baituii h'iu. 111. 42 The whole weight W (including that of the bar) is the sum of these ; therefore, ]V = P{2"-+^ -1) +iv^ (2"--\) +w., (2'»-^-l) +...+H',J2-l). The weight of the pulleys therefore increases the advantage of the machine. Cor. 1 . If the weight of each pulley be the same (lo), then, W = P{2"-+^ -I) +io(2"+2"~'^ +...+2-n) = P(2"+l-l) +i«(2"+^-2-H). If we put P—0, we have ir=w(2"+^-2-H), which is the weight that would be supported by the pulleys alone. Cor. 2. The point of the bar to which the weight should bo attached, in order that the bar may be horizontal, will be the ccnlri' of imrallel forces for the tensions of the strings and the weight of the bar. If we neglect the weight of the pulleys and the bar, this point will remain the same in a system, whatever be the power ; if, however, the weight of bar and pulleys be considered, it will be dif- ferent for different powers. G9. Taking the same number a of movable pulleys in each system, the respective mechanical advantages are 2h, 2 , .j/i+i _ j^ _^j^j these numbers are in ascending order of mag- nitude. Hence the mechanical advantage of tlie third system is greater than that of the second, and of the second than of the first when there is more than one pulley. 70. The following combination of pulleys may be noticed. It is called the Spanish Burton. T'he tension of the string to which P is attached is the same throughout and = P. That of the other string is also tlie same throughout and = 2 /'. Therefore W= 4 P. If we take the weights of the ])ulleys A, B into account, wehavo ir+.B = 4i'+^. m dif- jacli iced. , the also )unt, 71. The inclined I dane. indm i Tlii.s is a plane fixed at a certain angle (called its iticVnia- tion) to the horizon, and on it a heavy particle i.s supported by a force applied and the reaction of the plane. Since the plane is smooth, its reaction is exerted in a normal direction ; also the weight of the particle acts vertically : therefore, if a vertical plane be drawn through the particle and the normal to the inclined plane, since the plane thus drawn contains the directions of those two forces acting on the particle, the third force or power must also act in this plane. Let the figure represent this plane ; A B, the section of Ki.:^. 1 1 the inclined plane; AC, horizontal. The angle BAG is the inclination, a (suppo.se). Let P, the power, act at an angle to AB, and let B be the reaction of the jdane exerted perpendicularly to AB. W the weight of the particle acting vertically downwards. The particle is then kept at rest by the three forces F, B, W. Taking the resolved parts of these along AB, that of P is P cos ■ of A' is ; of W is W cos (90° - «) := W sin a. Hence by the " vanishing of the resultant," PcoaO- irsin« = 0, wliich gi^es the mechanical advantage \-\ - (-. — )• Cor. 1. For a given inclination, the mechanical advantage is greatest when cos is greatest; that is, when 0^0, and the foi'ce acts parallel to the plane. For a force acting at a given angle to planes of difTei'tnu inclinations, the mechanical advantage increases as the incli nation diminishes. m-^<9m 1 — Tirriiifr-"^'-'^-"-""'"*^''*^ 44 Cor. 2. Resolving the forces horizontally, we have P cos (tf 4= a) - ^ sin a = 0. Also resolving them perpendicularly to the direction of F, W cos (0 + a) - R cos = 0. These two equations give H in terms of P or W. Or these relations might at once have been asserted from the " triangle of forces" (§ 29) : for this gives W R sin a cos cos ( = 0, or at once by resolving the forces along A B, observing that P is the tension of the string. P - IF sin o = 0. W and the mechanical advantage — = cosecant of the inclination. am a and is the li BC (vertical) be called the height of the plane, AB its BC length : then, since sin a = ^T^i '^^^ have --4^-> Olf, P _ height W~ length" the its 46 Again, resolving forces perijendicularly to the plane, we have R ~ IT cos a = 0, AC or and R = IF cos a = W R base AB' W length Hence the power, weiglit and pressure on the plane are proportional to the height, length and base of the plane. Cor. This latter result is at once seen from the "triangle of forces;" for, drawing CiV perpendicular to AB, the sides of the triangle BCN, taken in order, are parallel to the directions of the forces, and therefore represent them in magnitude; and the triangle ABC is similar to BCN. 7 3. The screw. Screw. The screw is a circular cylinder, on the surface of which Pig. u. runs a protuberant spiral thread, whose inclination to the axis of the cylinder is everywhere the same. This thread works freely in a fixed block, wherein has been cut a corres- jKjnding groove. The power is applied perpendicularly to a rigid arm which passes perpendicularly through the axis of the cylinder and is rigidly attached to it, and the weight is supported on the cylinder (whose axis is here supposed to be vertical), and may be supposed to act in the direction of this axis. 74. The complement of the invariable inclination of the thread to the axis, or (the axis being vertical) the inclina- tion to the horizontal line which touches the cylinder at the l)oint, is called the pitch of the screw. If a right-angled F'f-' ' • triangle BA C be drawn, having the base A C equal to the circumference of the cylinder, and the angle BAC equal to the i)itch of the screw (a), and this triangle be wrapped smoothly on the cylinder, its hypothenuse will mark on the cylinder the course of the thread, and by superposing similar i-i'tm li^;. 14. fi).. 1^. 4G triangles, the whole course of the thread may lie cnntinut'd. lie is then the distance between two contiguous threads, and we have tan BAC BO AC 1 > or tan a — distance between two contiguous threads circumference of cylinder 75. The screw is kept at i-est by the weight (W) which acts vertically, by the power (P) which acts horizontally, antl by the reactions of the groove on the thread at the various l)oints in contact. Since the thread is smooth, the reaction at each point of it is normal to the thread ; and the angle betw(;en the direc- tions of this normal and the axis, being the same as that between the thread and the horizontal tangent which are respectiv3ly jwrpendicular to them, is a, the jntch. If then we resolve this reaction at any point, B (supjiose). into two forces— one vertical, and the other horizontal aiid touching the cylinder — the former will be B cos a, and tlie latter B shi a. All these vei-tieal portions being parallel, will fonn a single vertical resultant whose magnitude is cos a I (/.'), and this must counterbalance the weight W, since all the other forces are horizontal. Hence cos « - (B) = W. (1) Again the horizontal portions of the ^'s tend to tuin the- cylinder about its axis, and since each acts in a horizontal direction touching the cylinder, the radius of the cylinder is itself the perpendicular distance between the axis and the direction in e-ach case. Hence the moment of one of these (.R sin fit) is B sin a X radius of cylindei", and the sum of tlicni all is sin « X nulius of cyliiulor x 1' (/»'), and this must l>e equal and opjiosite to the moment of tin* [lower, namely, jPx arm of i*. Hence sin « x nuHus of cylinder. I" (/»') - P x arm of J\ Dividing the sides of this ecjuality by those of the equality ( 1 ). sill a ,. „ ,. , 7' X arm of P X imlius oi cyluiuer = — cos « w sin « distance Ijetwwn threads Hence, since = tan « - -: — p— , • cos a circumierence ol cyliiuler ,- , . . 1 / nidius ^ ,. , , /* X arm of J' dist.bet.thi-eadHx (-^ ot cylinder) ,-,^ circuni. W hut radius 1 circum. 2;r ' and 2;: x arm of F is the circuniforcnce of the circle wiiicli the end of P'h arm would describe, and may be bi-iefly called the cii'cumferenee of P; hence P w (ILstance between contiguous thn^vds circumference of the power and this ratio inverted is the mechanical advantage. Cor. 1. Tlie mechanical advantage is inci-easeil Ijy diiiiin ishing the distance between the tlu'eads or incro keep open a rift in a body, it acts generally by means of friction, and not by a weight applied to it ; it is therefon* useless to proceed with its examination on the principles employed hitherto. VIRTUAL VELOCITIES. yiiniiii yy jf a machine at rest under a power Pand a weight W be put in motion, so, however, that its geometrical relations are unaltered, the space described by the point of applica- tion of the power, estimated always in the direction of the power, is called the virtual velocity of the power: and similarly for the weight. jviiK ipif of The principle of virtual velocities asserts that the product of P by P's virtual velocity is equal to that of W by JF's virtual velocity. 78. This principle is only a special application of a far more general one, which it is not here necessary to examine. We shall therefore only establish the principle, as above rdis 49 luct. stated, in a few of the more simple cases, by proving that it leads to the relations of equilibrium already found. The geometrical relations in a machine being such that the spaces described in different displacements are always proportional, it will be only necessary to prove the principle for a particular displacement, and we may select this as convenient. 79. The straight lever wider weights at its ends. Let the lever BAC be horizontal, and displace it round A into the position B^A C, the directions of P and W meet- ing BA C in b, c. Then B^h is P's virtual velocity, and C^c is IT'S. The principle then asserts that P. BJ)^ W. C,c; B,A Utraiglit lever. Pig. in. but by similar triangles, C^A B,h Aic' hence P. AB - W. AC, the condition found in § 57, 80. Wheel and axle. Suppose the machine to make one complete turn; then, the space descended by P is the circumference of the wheel ; and by W, that of the axle. The principle then asserts that P X circumference of wheel - W x circumference of axle, jmd the circumferences are as the radii : therefore, P X radius of wheel - W x radius of axle, the condition found in § 63. 81. Pulleys. In the single fixed pulley, the principle is obviously true. In the single movable pulley (fig. 6), let the pulley be raised through one inch, then W is raised through one inch, D Whcil ami axle. Fin r,. I'ulleyn. SiiiKl«' (luUey. I' < >l 50 iiiul Olio inch of each portion of tlifi string is H(!t froe ; there- fore, P ascends through two inches, and the principle asserts that Kig. 7. Si'oorifl systum. TliiKi sysU'iii. which is tlio condition in § 65. 82. In the fii-st system of pulleys, let the lower block lie raised one inch : then ]V is raised one inch, and one inch of each portion of the string at the block is set free; therefore, on the whoh;, 2>i inches are set free, and this is the space through which /'descends. The principle then asserts that the condition found in § 60. 83. In the second system of pulleys, let W bi d 1 incli : then A„ rises through 1 inch, and each pulley i. ^a through twice as much as the one below it, and P rises through twice as much as the top pulley ; therefore on the whole, P'b ascent will be 2" inches ; and the principle asserts that Px2"=Wxl, the condition found in § 67. 84. In the third system of pulleys let W and the bar be raised 1 inch : then each pulley flesceuds through this 1 inch and twice as much as the pulley above it, and P descends through 1 inch and twice as much as A^. The last pulley A^ descends through 1 : therefore, the last bnt one descends through 1 + 2x1 " two " three and so on. 1 + 2(1 + 2) or 1 + 2 + 'I' 1 + 2 (1 + 2 + 2'^) or 1 + 2 + 2^ + 2^ last OS Inclincil plaiic FiB I.'. 01 On tho whole, /^ " 1 + 2 -t- 2» + 2^ f . . . + 2", or 2" + ' - 1. And tho principlo asserts that Px(2"+'-l)= IKx 1, the condition found in § 68. 85. In tho inclined plane, the power acting parallel to the plane, (fig. 12), let W be at tho bottom of tho plane and be drawn up to the top. Tiien IFs vertical displacement is the height of tho plane, and P'h descent is its length. The prin- ciple asserts that P X length - W y. height, tlie condition found in § 72. 86. In the screw, let one complete turn be made. Then Screw the distance moved through by the end of P'a arm, esti- i\. 1 1 mated always in the direction of P, is the circumference of P; and the space descend l by W in tho distance between two threads. The principle then asserts that P X circumference of P = W x distance between two threads, the condition found in § 75. 87. Assuming the truth of this principle of virtual velo- mties, it may be conveniently employed to find the mechanical advantage in many machines — as examples, let us take Jiober- ral's balance, the differential axle, and Hunter's screio. b.ilanct Fig. i; 88. In RobervaVs balance the sides of a parallelogram are roUmvii connected by free joints with each other and with a vertical axis passing through the middle points of opposite sides ; so that the figure is symmetrical about this axis, and the other opposite sides are always vertical. The weights P, W are curried by arms fixed perpendicularly to these latter sides, which arms are therefore always horizontal. If tho machine, when at rest, be displaced, one of the weights ascends as much as the other descends, and they are therefore equal. ms , will 62 This result is independent of the particular points of the arms from which the weights depend, and in this lies the convenience of the machine sis a balance. Iiilli-rintial MXlc. Kij:. IS. iliiiiU'i's sortw. 89. In the differential axle (fig. 18), two axles of different sizes inui fixed together on the same axis, and the weight is supported on these by a pulley, whose string is coiled i-ound these axles in opposite directions. If P be raised by a com- plete turn of the mjtchine, W descends through a space ecjual to hiilf the quantity of string set free from the axles ; that is, through half the difference of the circumferences of the axles ; and, the circumferences being as the radii, we have P X radius of wheel = TF. J (difference of radii of axles). In the common wheel and axle, the power and wheel bein<» given, the mtxihanical advantage is increased by diminishing the radius of the axle ; but this diminution is practically limited by regard to the strength of the axle. In the above machine, the mechanical advantage may be increased in- definitely, by making the s»xles more nearly of equal si a;, without too much weakening them. If the axles were absolutely equal, the mechanical advantage would )Hi infinite, and it is ohvioub that any weight would be here supported witlKiut a power at ail. OO. In Hunters screw (fig. 19), the weight is supported on a smaller screw, which runs in a companion in the interior of a large screw, the latter passing through a fixed block and being acte. diffei-ence between distances of contiguous threiids in the two screws. 53 IS, the te one reight tweeu Lence, tances The mechanical adva. cage can therefore be indefinitely increased by making the distance between the threads more nearly the same in each screw. In the common screw, the advantage is increased by diminishing the distance between the threads, but the diminution is practically limited by regard to the strength of the thread. If each screw had the same distance of threads, tlie advantage would be infinite, and it is obvious that any weight would be sup- ported without a power at all, the outer screw rising just as much as tlie inner screw descends withhi it, so that the weight would be stationary. 91. When a power P is supporting a weight W on any machine, if the machine be set in motion, it will continue to move uniformly so long as its geometrical relations with the power and weight are unaltered ; and if s, S be the si)aces gone through by the power and weight in any time (that is, their virtual velocities), we have P \ s = W x S. Hence a given force acting through a given space for any time will lift the same weight only through a given space, whatever be the machine through which it acts ; and if the weight lifted be increased, in the same proportion will the space through which it is lifted be diminished. Also when a given power lifts a weight through a given space, the greater the weight, the greater in the same proportion is the space through which the power must act, and (the motion being unifoi-ni) the longer is the time employed. Hence the principle of virtual velocities is sometimes stated in the form, that " in f'very machine, wdiat is gained in power is lost in time." I I. In every iiiachiiic. what is giutu'fl ill IiDwrr is lost in time. 92. Hence also this product Px s or W x S may be con- worit.ioii sidered the loork done by the niachiuo, and is sometimes ojuncy.' termed its duty; while with reference to the power, the names of mechanical ejfftciemy and laboring force have been given. In this sense, although advantage may be gained by a machine, no efficiency is gained or (theoretically) lost, but it remains the same as if the power were ai)plied directly with- out the intervention of the machine. Hiirse- P'lWcr. 54 Practically, efficiency is always lost, owing to the various resist- ances due to the parts of the mdchine. 93. Among engineers the standard of efficiency in tlie com- parison of machines lias usually been taken to be a horse- power, wliich is represented by 33,000, a lb. and foot being the units employed, and the power being exerted for one uiinute of time, Tlius a horse in one minute is supposed to lift 33,000 lbs. through 1 foot, or 3,300 lbs. through 10 feet, or 330 lbs. through 100 feet, and so on. A machine is then said to be of so many horse-powers, whence the work done by it in any time can be calculated. FRICTION. '''''"•'• 94. Hitherto the surfaces of bodies in contact have been considei'ed smooth, and exerting on each other no pressure except in a normal direction. In nature, however, all sur- faces are more or less rough, and when one surface is pres- sing or moving upon another, a force is called into play which acts in a direction contrary to that of the motion, or to that in which motion would occur if the surfaces were smooth. This force is called friction. I'.lli'rt ilt'i iiiai'ljiiics. In machines, when a power is supporting a given weight, the mag- uitmle of the power, determined on the supposition of the smooth- ness of the machine, may be increased beyond this value without disturbing the equilibrium, until it is great enougli to overcome the friction together with the weight ; and on the other hand, may be tliminlshed till it is so small as with the aid of friction just to pre- vent the weight overcoming it. So also, with a given power, the weight may be increased or diminished within certain limits without disturbing the e(iuilibrium. Generally, when the power is on the point of raising the weight, friction acts to the disadvantage of the power ; but, when the power is just preventing the weiglit from descending, friction acts advantageously. When the equilibrium of a system depends on position, this position may with the aid of friction be varied within certain limits of the position determined on the supposition of smoothness, and the eciuilibrium be still m.iiu- t/es, and have endeavoured to be as concise as pos- necessaV"'' -P^-^tions or illustrations that have appeared University College, Toronto, April I, 1858. n P ai tl dc de til vet dei aiii be wil Miitioii kI a jioint CHAPTER I. THE MOTION OP A PARTICLE GEOMETRICALLY CONSIDERED. 1. When the distance between two jwirticlea changes con- tinuously during an interval of time, they are relatively in motion. The position, and consequently the motion, of one partich' can only be conceived in relation to other particles, but it is convenient to speak of a particle absolutely as being at rest or in motion, reference being made to ourselves or to some points in known relation to ourselves, considering these as fixed, and referring all motion and change of motion to the particle itself. By a particle is here to be understood only a geometrical point. Uniform motion. 2. When a particle is moving in a fixed straight line, its inastraiviit motion is measui'ed by the change of its distance from a fixetl point in this line, and the rate of this change of distance at any instant of time is called the velocity of the particle at v»i"iiit^ towards which s has been estimated jMsitu'eli/, or in the opposite direction. Both these cases can be included in one formula by indicating oppositcness of direction of velocity by the algebraic signs + and - . Thus, fixing on one direc- tion from the fixed point towards which when measured the distances are to be considered positive, a velocity in this direction will be positive, and in the opposite dii-ection. negative. Hence, if a particle move during successive intervals of time with different uniform velocities, and a be the distance from the fixed point at the beginning of the time, s its diiitance at the end, then s =: « + 2' {vt), where - denotes the algebraic sum of all the products cor- responding to that within the brackets; and the particle will be on one or the other side of the fixed point at the end of the time according as s comes out from this expression positive or negative. Vi>lii('i|.y, + ami - Kistrtiice III' a inovirji; ]iiiiiit rioiii u ill its liiK! o liiiitioM nfliT .tii> liini'. 63 mis of Istancc s its ts cor- |le will ke entl lession 'I'lio whole .ytacfl (hscribed will, howovor, l>o tlio iiuinoi'ioal Slim of these products, disregarding algebraic signs. 4. When a partiele moves from a fixed point in a straight line with different velocities during successive eijual inter- vals of time, each vtdocity continuing uniform throughout its interval, the distance of the particle from the point at tlie end of the time is the product of the time by the arith- tnotic mean of all the velocities. l>y the arithmetic mean of a number of quantities is meant their a!t;ebraie sum divided by the number of tliem. For let t be the whole time; the duration of each interval ; n ''i. *'« ^3, . . . the successive velocities during the first, second, tliii-d . . . intervals. Then the required distance will be the algebraic sum of the si)ace described with these velocities ; that is, by § 2 ; Lt'iniiia. Dist.iiii f tniiii stint iiiK I'liiiii Ill'tlT il I'liiitiiiiii'iu SI'lil'H 111' iiniriiriii iiiiitioiis ill lllll Silllir liiif, V t t t - H- v., . - + fo . - + n ' n n or. -. t. Q. E. I). is till' I'll' duct III' th.' iiicaii vi'l'. city aiul tlu- time. Tlie case of any of the velocities l)eing in the opposite direi^tiou (and therefore accounted negative) is liere inchuled ; the resulting sign of the algebraic sum determining on which side of the fixed jioint the particle is at the end of the time. Accelerated motion. 5. When the velocity is not uniform, but changes during Varyins tlie motion, the velocity of the particle at any instant is ^*^^""^' inciusurcd Vjy the space which it would describe in a unit of im^suir.i time, if it were to move uniformly during that unit with iliis velocity. The rate of change of velocity at any instant (provided it Airri(i:iii"M l)t' continuous) is called the acceleration. If the change of velocity in a given time be always the uiiiidUM, same throughout the motion, the acceleration is said to be il'i'casimii uniform, and it is measured by this change of velocity in a giv«n time. 64 The chnngo of velocity may bo cither an incroaso or decrease, ami in the latter cohc the acceleration ia iu effect a retardation. The use (if both termB is, however, reiulered unnecessary by introducing the algebraic signs + and - ; for a decrease ia algebraically a lUijutict increase, and thus a retardation is a negative acceleration ; and when we speak of velocity being increased, added, or generated, we also include the cose of velocity lieing diminished, subtracted, or destroyed. The vcidcity Taking a second as the unit of time, the acceleration J\ I'lcrind'i'Jf when the motion is uniformly accelerated, is the change of a"rV'u",'iti:|re88ion, their sum n = . «. Hnnce, the mean of them - — , or the mean of the first any the same stages in revei-se order, its distance diminishing till the time ^, when it is again at its starting point. It then moves to the other side of the point, s becoming negative and being now given numerically by the formula -_/*<■ - (^/, and the whole span- W^ 1 „ described in the time t Yxnivf —. + - ft- - ut. " J 'i All'illliM ill- of till' same NewtDii's IIII'lllHll KiiS. I. 9. The following is another investigation of the above proposition, after Newton's manner. Draw any line AK representing on any scale the number expressing the time t, and divide ^1 A' into eipial piirts in the points Ji, C, D Draw .la pcirpendicular to AK, and take its magnitude on the sjxme scale to represent the number expressing u the initial velocity. In the same way take A'A' to i-epresent // + /if, the velocity at the end of the time. Draw nk parallel to AK, and at each of the points B, C, I) «lraw jjerpendiculars to AK, mooting ak' in h', c', d'. . . ., and com- plete the paitillelograms in the figure. Then, since kk' rejire- sent« ^(5, which is the change of velocity in the time AK\ by sinular triangles, dd' will represent the correspond in;; change in the time AD, and Dd' will rejiresent the velocity at the end of this time, and similarly for each of the lines lili, Cc Now, if tlie particle moved unifornih/ during any interval a-s CI) with the velocity Cc, which it has at the beginning of tliis interval, the space doscril>ed (§ 2) would bo n^piesented tally e in ae -,. \" • " 2/- Tes in wilt '11 s othri- giv.'U 67 iiumerically by the area of the inner parallelogram CW ; sd also if it moved uniform! ij during C'/>* with the velocity !)({'. which it has at the end of this interval, the space descrihcd would be represented by the area of the outer parallelogram Cd'. If, therefore, the particle moved uniformly throughout each interval on the former supposition, the whole space described would be the sum of the inner parallelograms; and if on the latter supi)Osition, it would be the sum of the outer parallelogi'ams ; and the space (a-) actually described lies numerically between the spaces described on these two su}>- jjositions. But as the number of intervals is increased, and the magnitude of each diminished, the two series of })arallelo- grams both approach nearer and neai'er to the quadrilateral area AKk'a, and this must therefore be the value of s. Henc«' s is represented by the parallelogram Ak and triangle akk'. that is nimierically by Kk X AK + ^ kk' X ak, and, therefore, above .1 K time ^ lie D tude on u th»* )resent ■aw (ik ♦haw \d com- re]>ri' tonilini: ,cl.|. Ki-. I. M..lM,ll n ';iiilfi|. '■MT-:Ar--<-9t ■V.miiart'.i-tiAJwi I ; i MctiiHl tliiMI ivsl with Mliiroriii ai'- iM-l'lMtinJl. 68 time yliV ( - 2AL) this distance vauislies. and the particle is ;i!,Min at tlio initial point, the wliolo space described being represented by twice the triangle A La. Afterwards the particle passes to the other side of the initial point, and its distance from it at the time AP is rep'esented by the area Xii'j/P, while the whole space that has been described in this time is rej)resented by the sum of the triangles ALd an. 10. When the particle moves from r(^st and its motion is unifonidy accelerated, wo have seen that the velocity and space described at any time from the beginning of motion are V ^ u + ft,s = ut + -jr. Squaring the first, we have v^ ^ u-" + 2uft + r e = w + 2/s. If the velocity wciv retarded, we should have V- = u' -2fs. Cor. This result might have been obtained without finding thu SLOond equation, for we have directly, from § 5, iuid from § 10 or 8, V - u =ff, - t {v + u) = s ; uua.. plying these etiualities, we have „i - i,« ^ 2 fs. The following geometric d proof may also be noticed : Let B be the initial point, where the velocity is u ; HO the .space described (.s) when the velocity is f. ■Ij no ial the ted the sp; 71 liBt A be the point from which the ijarticle, proceeding from rest under the same acceleration, would acquire the velocity u at B. Then (§11), «' = 2 /. AB. Also, since the whole motion may be taken to proceed from rest at A, we have (§11) v^:=2/. AC = 2f. {AB + AC) ^2f.AB+2f.AC = U^ + 2 f8. By a proper adaptation of the figure, this proof may be extended to all the cases included in the algebraic formula. 18, Give the initial velocity and the acceleration, to find the time when the particle will be at a given distance from tlio initial point. Here u,f, s are given to find t. Solving as a quadratic in t tlie equation s ~ut + -ff, we liuve 2 f ^ U + yu- + 2fs Tile significance of the double sign is here note-worthy. If/ he i)ositive, or the velocity be numerically accelerated, tine of the values of t is positive, and the other negative. The former is the solution required, but the latter can be interpreted thus : Suppose A the initial point, AP the dis- t;tnce s, and the velocity u at A to be in the direction AJ\ Then the positive value of t in tlie above gives the time of moving from A to P ; the negative value gives the time thit would have elapsed if the particle had moved from P towards A, with a retarded motion, passed through A to the other side of it, been reduced to rest, and again returned to .1. If/ be negative, then, writing -/ for/ the values of t lieoome r.i ; f II 72 If then u- > 2 fs, both values of t are real and positive, ami the particle will twice be at the same distance from the initial point, once during the recess from and again during the return towards it. If xC- = 2 fs, the two values bocomo the same, and tlu' distance in question is that where the particle monieutarily comes to rest. If it" <[ 2 fs, both values of t are imaginary, and the i>ar- ticle can never leach that distance. If, however, 8 be negative, both values are real, and one positive, the other negative, the latter referring to a i\n\o previous to the epoch from which we ai"e reckoning, when the particle, if it had been moving towards the initial point from the negative side, would have been at the assumeil distance. Component velocities. :..i]i|..Miiim 19, The position and motion of a particle moving uni- Ivi'l-Milics. ... . . formly in a straight line liave been determined by the dis- tance of the particle from a fixed point in the line, and by the change of this distance in a given time. Its position. however, might have been delined by its distances from two fixed lines, measured parallel to these lines. Thus : let Oj% Oi/ be two fixed lines, B the place of a particle moviir^ in the line .1 /i 0, and .1 a fixed point in this line, the distance from which determines the place of the particle. Let C be the place at which the particle would arrive after any time if it moved uniformly with the velocity it had at H, and complete the figure by drawing lines parallel to Ox, Oy. The position of B is known when B P, B Q, its distanc<>s from these fixed lines, are given; and C E, CD, or theii' equals, BD, BE, would be the changes of these distan:v^ if the piirticle arrived at C by moving uniformly. Now BC, which would be the change of distance iu a given time from the fixed point .1, measures the velocity of .Ki,< 1 ' 11 73 the parti{;Ie : and JW JiF -..o „; and \h...f "^""'^^ proportional to BC, sent tlJ ^^''' ^"^' «* diagonal will repre- """''''" sent tlie compoxent vFrnfirtrc • ,i ,■ . ' sides. vcLocnri,;, ,.a the directions of those Conversely. // Me coMPONKXT VELOCITIES in two directions be airen toe actucd vrlocitu null ha r i • . yi-itn, l»l drawiZT , "' ""'"'"""'' "»'' ''»■«""« InZlf T ■' "*" '""•««"%'•«» "/»/.« n. li'lle r.ii. 1 !. B' CHAPTER II. ApplicaHoii ll':5 2) as the basis of our conception of a force. E.xperience shows that whenever a quiescent body is set in motion, wo can trace the action of some cause e.vternal to the body ; thus, when a body is suftennl to drop to the earth, we assign its motion to a pressure exerted on it due to the earth itself, and which would have no existence if the earth did not exist. Also, there seems no reason why a particle, apart from any external force, should begin to move in one direc- tion rather than another. Again, when a particle is in motion there seems no reason why it should change the direction of its motion in one way rather than anothei', unless some force be acting upon it to determine such change; and in all ca.ses of any such change, we can always trace the action of some external force; as, for instance, when a stone is projected from the earth in any direction, the deflection of its motion from a straight line is produced by the aforesaid pressure due to the earth, which we know is always acting vertically downwards. If this pressure be counteracted by projecting the stono hori- tlircc liiiv- First law. No fiiiv, nctiiiK, 111.' parti. 1- citlici' remains A rest, or IllllVl'S III ,1 stnii>;lii liiii' * Lex. I. GorpuH omne persei^erare in statu sua quieKcendi vel movendi nnifontt'der in directum, nisi (/uatenus a ririhus iinpressis cogitiir statum ilium mutare. — Princ. Leg. Mot. 76 ivilli iinitKMii veldiity. t^-ilili'ii. 'I z<)nt;illy aloniij a fixol piano, tlio path npproiiclies to a st!Mi;4lit lirus, with only sm-h tloviatioiis as may bo acconnto;l for by friction or irro/^nlarities in the piano, or from the stone nut boiny small (;nou<|h to be considored a particle. fHo also with regard to tho volocity of the particle: it (lo«'s not Hoom i»ossible to conceive any way in which its velocity conld iuereasc or decrease unless l>y the action of some ext«'r- nal cause, and in actual cases of variation of velocity, we can always trace the existence of such causes. Thus wlieii a store is thrown horizontally along tho ground, it gradually loses its velocity and stojjs, but here the friction of the ground and tho resistance of the air act as retarding causes, and we see th.at in ])roportion as the surface on wliieh tin- stone moves is smoother, as on a slieet of ice, tho longer and more uniform does the motion continue. This law is seiautimea termed the Law of Inertia, being uiiderHtiKxl to express tliat a material particle id incrl, and has no tendency ti'):i. always acts vertically, ami (tor not groat lioight.< ahove the Htirt'a';,!) is sciisihly uniform. Hore thou i\w rcMiuiretl comli ti<»iis arc! fuitillcd, and tli(5 result of oxpnrinioiit, wlion ilu;- allowance is niaaruteli/. Hence it follows that n equal forces acting simultaneously on a particle in its lino of niotion will produce n times the aeeeleration which one of the forces alone would jnoduco on the same i)article ; and, consequently, the acceleration pro- duced in a given particle is proportional to the magnitude of the force acting. It will be hereafter shown how this may be tested by com- paring the accelerations of a particle down inclined planes of different inclinations. Hence also the change of velocity in a given time is pro- |»ortional to the magnitude of the force, the particle acted on lieing the same. (3.) When a moving particle is acted on continuously by a a sin^ii uniform force which acts always in the same direction anly tliu.s : if wo rcsolx <■ tilt! uriyiiml vclouity of the piirtidi; into two coiiiiJOiu'iits, oim- in (linu'tion of tlio forco, tlu5 otlior |HT|K>iulicular to it, the latter remains imaltermi, and tlie formtir is chaiij^tMl l)y tlie force preciw^ly as if it alone were the actual velocity of tlie [lartiele. So that Tfu' chtDKji; of vcloi'.'Ujj pntiluced bif the forcv, in a (/ioi:u I'inu'. lit hi direction of the J'orce, and in projjorlioiuil to it in indijniludf. In tlii»< ciisu the patli of the particlu in no longer a straiglit line, liut ii eiirve, tlie tangent to wliieh at any point in the direetion of tlie pai- ticlu'8 motion tliure. In otlior words, tlio above expresses that the dynnmicdl effect of a force on a particle ig luhoffi/ independent of nnn motion which the particle man ham, and is the same as if it mere exerted on the particle oriyinally at rest. Thus, the vertical doscontof a body let fall from the mast head of ii sliip in motion is precisely the wime in all its eii- cumstanccs as if the ship were at rest. The principle tan also 1)0 tested by comparing tlu; results of calculation with observations on the niotion of a body projected oblitjuely to tin; horizon and acted on by gravity, due allowance teing made for the resistance; of the air. AiiviHirs (i) When sevenil forces act simultaneously, retaining iliiy'iilm''- always the same magnitudes and directions, on a partich^ liiir'tii'ir' originally at rest, the motion is uniformly accelerated in the i.ili'iv'at'ivst direction of the resultant of the forces, and the acceleration is that due to this resultant acting singly. Also the velocity generated after any time, being that due to this resultant, is also that which is compounded of the velocities due to the forces acting singly on the particle fi-om rest. < : in motion. Also if thc particle be in motion when the forces begin to act, its velocity and direction of motion aftei- any time will n I),! (U;tormiiiO(l I»y coinpoimdini; its on^inal velocity with the velocity duo to tlio msultaiit of the foiros, or with all those (hio to tho forces scpiiriitely. Or, Wfien force.it avt on the adiuo parti'ch; under nnt/ eircwii- utancca jn'oviled each Jorce he auij'onn and aJirai/s j>reHen'r the same direction, the chatuje of vclocifi/ in a (jirrn time ditr to each force ia in direction of that force, and is proportional tit it in niaynitude. \xi dm- )f the IVoiu lilin to le will ari'(liTiiti..ii is |>1>'I • tialliil 1,1 II,, l..|-,'c. M.i-^s <|r|:l..,l (5). It follows from the preoediiii; that if/* he the aeeole- ,,.,„ n,^ ration due to a force P actintf on a certain jtarticle, then J'j"i"! ,','," tho mtio P: f is invariahlo for this particle. This ratio is found, however, to ])o different in different jxirticles, and we thus discover a (juality which distinj;nishe.s one ])artieh' from another, of which this nttio will serve as a mcNisure. The name of taorsa is <,'iven to it, and one particle has (he same inasn as another when the sivme forci; pi-cxhiccs in each tlie same acceleration. The unit of mass is arbitiary and it is not necessary to fix it, but we shall take as the measure of mass the above ratio of tlie numbei's expre.s,sini:j a force and tho jicceleration it produces on the particle. Thus, if m be the mass of a particle, and /', /lus above, P — - m. f It ha,s been mentioned that the acceleration pr(xlucera if a particle at rest \w acted on by two impulses which, v«i«c separately connnunicated, would give the particle respec- ah.si tively such velocities ;ni would cause it to describe uniformly the sides AB, .4(7 of a parallelogram BACI), the particle will acquire from the impulses simultaneously communicated a velocity which will cause it to describe unitbrmly the diagonal A D in that time. of ities itic APPLICATIONS AND TKSTS OF THK SECOND LAW OF MOTION. Vciticsil motum liT 27. The vertical motion of a particle under the action o/* ^''"^'"-^ ■ gravity. Oaiii.o. Th(? acceleration of gravity (g) has already been stated to t; - '-' - be about 3'2-2 feet per second, and to be sensibly the same for all i)odies in the .same latitude and at nearly the same heiyht above the sea-level. ['.ac/i hige of iitur. — Hence, a{)plying the foi'mulas in § 10, Ave have, when the Fnnii ksi particle moves from rest, v=.gt = 32-2 X r.- s = \ gt- = IGl X «'• Thus the spaces describetl from rest after the lapse of 1, 2 3, ... seconds, are 16-1, G4:-4, Hi-D, feet; and the velo- cities acquii-ed are 32*2, 64"4, 96"6, feet per second. If tlui body do not fail from n-st, but be projected down. Pia.ir.trii down wards with a velocity u, then we have § 15 . 1 . V — n -i- yf, s ~ ut -\- _^ !• if it l>e projected vertically upwards with a velocity n, tlieii tlic velocity and distance from tlie point of ])rojecti(in '""r ut the time t arcr given 'y V — u — gt, s vt 1 2 . oI/<^ 82 and the particle is brought momentarily to rest after having ascended a height — - in the time -, and then descends again by the same steps in reverse ordei*, reaching its point of pro- 2m jection in the time — . 9 The resistance of the air and the raj^idity of the raotion render it difficult to test these results directly by experiment. Motion down an incliaeil Galileo. 28. Motion doiou a smooth inclined 2)lune. Let a be the inclination of the plane. The particle is acted on by two forces, namely, its own weight (W) acting vertically, and the reaction of the plane in a normal direction. If we resolve W into two forces, one perpendicidar to the plane, and the other ( W sin a) down- wards along the plane, the motion estimated along the plane will be due to this latter only. Hence, / being the accelera- tion along the plane, we have § 26 (5), f = Force -j- mass of particle = W sin a = g sm a 'J And with this value of/ the fonuulas of §§ 10 and 15 avail to determine fully the motion. I., 'U Vine i. Cor. For the same particle, on planes of different inclina- tions, the accelerations are as the sines of the inclinations, or, the length of the plane being given, as the luiiglits ; anil this is the test mentioned in § '1T^, {'!), sillowance being made in performing the experiment for the resistance of the air and imperfect smoothness. 29. Tlie velocity acqidred hy moving from rest down an inclined pliDW, is equal to thit acquired by falling freely down the height of the plane. 83 kclina- lations, ; and made the air hon an l/reehj For, J the acceleration down the plane is g sin a, and if v be the velocity acquired in moving down its length, §11, vi^ = 2 / X length := 2 gr sin a X length = 2 ^ X height ; and this is the same as if the body fell freely down this lieight. Cor. If the particle were projected down or up the plane with a velocity u, and v be the velocity after moving through any length of it, we should have in like manner, § 17, »2- u^ ±2 (J sin a x length --.,» u^ ±:2 (J X height, which is the same as if the particle were projected vertically downwards or uj)wards with a velocity u, and moved freely through the corresponding height. 30. The time of moving from rest at the highest point of a vertical circle down any chord {considered a smooth inclined plane) is the same as the time oj falling freely from rest down the vertical diameter; and so is also the time of moving Jrom rest down any chord to the lowest point. For, AB being the vertical diameter, the acceleration down fi; the chord AC is g co? HAC, and therefore, § 10, (time down A Of = 2 AC 2AB (J cos BAC (J which is the square of the time down A Ji. So also, the acceleration down C Ji is g cos C B A, and 2CB 2AB (time down C 5)'= ^,,^ ^ ij cos OB A the same as in the former case. i/ w 84 A particle I'rnjcctcd n .my (lirec- tinn un(i aitcd oil by gravity. Galileo. 3 1 . Mot ion of a jyrojectile. 'i'iiiip of JtiKht. Let a particle be projected from a point in the horizon, with a velocity v, and in a direction making an angle a with tiie liorizon. The force acting on it being its weight which always is directed vertically downwards, the motion of the pai ^icle will be in one vertical plane. If we resolve its velocity of projection into two : namely. V cos a horizontally, and v sin a vertically ; the former con- tinues unaltered, and the latter is retarded and accelerated by gravity precisely as if the particle had been ];)rojected ver- tically with this velocity. Hence, g being the acceleration l)y gravity, the velocity v sin a is destroyed by it in a time , (§ 10) ; at this moment the particle is moving hori- zontally, antl has reached its greatest elevation abtve the horizon. The velocity v sin a is again generated by gi-avity by the same steps in a reverse order, till on again reaching the horizon the velocity is the same in magnitude, and its direction is equally inclined to the horizon in an opjwsite direction, as at the point of projection. The path, therefore, consists of two equal and similar branches on each side of the greatest elevation. V gin ft The whole time of flight is therefore 2. , and during this time the horizontal distance described with the uuifornt velocity v cos « is (§ 2) ?' sin a il V cos a. or (irpati st lieigh' . sm « cos a , or il — sin 2 a. a (Trig. §38.) Hangf. and tliis is called the nuigc. The greatest elevation is the space due to the velocity V .sin a for the acceleration g : that is (§11), ( /• sin II )- .'locit.v 85 Again. \{ X he tlie horizontal distance of the particle from riotn atfany the point of projection at the time t, and 1/ its elevation above the horizon at that time, we have (§§ 2 & 15), X = V cos a. t, y = V sin a. t — \ 9 t^ \ which determine the place of the particle at any time. The path of the particle is the curve called by geometers a para- Imla. In comparing these results with observation, tlie resistance of the air has to be taken into account ; and for large bodies, or con- sidera])le velocities, tliese results are thereby rendered ijuite wide of tlie observed facta. 32. Third Law of Motion.* When one m'tterial partole acts on another, the second ixwWa^- I'xerts on the first an action equal in amount and opposite in direction to that which the first exerts on it. The actions here spoken of may he of various kinds ; such :is the mutual pressures between bodies in contact whether at rest or in motion ; or the action of one particle on another by means of a stretched string or a rigid rod ; or the action may be of the nature of attraction or repulsion ; or finally , V' = U, .•xdmM>:r » ' veldcititig and the two balls exchange volocities. Thus, if the second were at I'est, the first after impact would remain at rest, and the second would go on with the velocity of the first before impact. IMAGE EVALUATION TEST TARGET (MT-3) A 1.0 I.I IIM 112.5 * 2.0 1.8 1.25 1.4 1.6 ^ 6" ► Photographic Sciences Corporation ^ ^ \\ ^ /^ and 2 i/"5~, find tho angles at which tho forces are inclined to one another, 11. Two forces are in the ratio of 3:2; find the angle between their directions when the resultant is a mean pro portional between them. 12. If two forces acting at right angles to each other l>e in the proportion of 1 to i/T", and their resultant be 10 llxs., find the forces. n I' : 100 13. A force 3 y^~3~ has a component 6 in one direction ; find the other component, the direction of the components making an angle 120° with one another. 14. Three forces acting in the same plane keep a point at rest ; the angles between their directions are 135°, 120° and 105°; compare their magnitudes. 15. Three forces acting at a point are in equilibrium when they make angles of 60°, 135°, 165° with each other; find ratio of the forces. 16. Three forces represented by the numbers 3, 5 and 9, cannot under any circumstances produce equilibrium upon a point. 17. Three forces acting at a point keep it at rest : if one of them be reversed, find the direction and magnitude of the resultant. 18. In the parallelogram of forces, if the two sides of the parallelogram approach coincidence in the same or opposite directions, what is the magnitude of the resultant ultimately] 19. Resolve a given force into two others at right angles to each other, one of which shall be half the original force. 20. Two equal forces sustain each other by means of a string passing over a smooth tack ; prove that either force : pi'essure on tack : : J ; cosine of half the angle between the directions of the forces. 21. Three smooth pegs form a vertical equilateral triangle whose base is horizontal, and two equal weights (w, w) are connected by a string which passes over them ; find pressure on each peg. 22. Shew that a force E can always be resolved into two, R sec 0, making an angle with, and R tan 0, perpendicular to the direction of R. What is the whole efiect of R in the latter direction? 101 23. Shew how to find geometrically three pressures, which, acting at a point in given directions, will be equivalent to one given pressure acting at the same point. 24. Three forces represented by the numbei-s 1, 2, 3, act on a particle in directions parallel to the sides of an equilateral triangle taken in order ; determine their resultant. 25. Forces of 2, 3, 4 and 5 lbs. act on a point in direction of the sides of a square taken in order : find their resultant. 26. A particle placed at the centre of an octagon, is acted on by forces in directions tending to each of the angles of the figure, and of magnitudes, taken in order, 2, 4, 6, 8, 10, 12, 14, 16; find the magnitude and direction of their resultant. 27. Two equal forces act at a point ; through what angle must one of them be turned in order that their resultant may be txirned through a right angle 1 28. A segment of a circle being given, if two equal forces act along the sides of any angle of the segment, thei'e is a point in the circumference of the circle, about which, if fixed, there will be equilibrium, whichever angle be taken, and whatever be the magnitudes of the forces. 29. Three forces acting on a particle in one plane keep it at rest: of the six elements, viz., the magnitudes of the forces and the angles they make with one another, shew that in general three at least must be given to determine the rest. If the given parts are the angles, what is the only inference 1 30. If three forces act at a point, and be represented by three lines drawn from that point such that each of them would, if produced backwards, bisect the distance between the extremities of the other two, the point will be kept at rest. FT' 102 lit' 31. A,£,C and D are fixed points, and P, a moving point, is always acted on by forces represented in magnitude and direction by PA, PJi, PC, PD ; ii PQ always represent the resultant force on 7-*, shew that PQ always passes through a fixed j)oint 0, such that the i-atio of PO to OQ is constant. 32. Assuming tlie trutli of the parallelogram of forces for the magnitude of the resultant, prove it for the direction of the resultant. 33. Given li, P, a, in the formula P'^ = P^+Q^ + 2 PQ cos a, shew in what cases there will be two diffbrent values of Q which satisfy the problem ; and if these conditions be fulfilled, show that the arithmetic mean between the two values of Q is equal to the rectangular comi)onent of P in tlie direction opposite to that of Q. 34. If two forces acting at a point be represented in magnitude anil direction by the diagonals of a quadrilateral) their resultant will coincide with that of the forces repre- sented by two opposite sides. 35. A BCD is a parallelogi-am ; find the resultant of the forces represented by AD and AJi acting at A, and BA and AC acting at B, 3G. ABC is a triangle ; in BC take a point JJ such that IW = ^ BC. Shew that 3 AD will represent the resultant of forces represented by ^i^, 2 AC and CB respectively, all the forces being supposed to act at A. 37. ABC is a triangle; in CB take a point D such that CD is J^th of CB. Shew that i AD will represent the resultant of forces represented hy BC, 2 AB and 2 AC respectively; all the forces being supposed to act at A. 38. Z> is a point within a circle, and A any point on its circumference ; determine two chords, A li, A C, whicli, if taken to represent in magnitude and direction two forces acting at A, will have AD for their residtant. til at It the AG 103 39. In a triangle obtain a geometrical construction for the point from which, if strings sustaining weights propor- tional to the opiwsite sides pass over pulleys at the angular points of the triangular, the point will remain at rest. 40. is the centre of a circle about a triangle, on sides of which OJJ, OE, OF are perpendiculars : forces act along OD, OE, OF projwrtional to them in magnitude. Shew that if be kept at rest the triangle must be equilateral. ■tl. If 2» forces acting on a point lie represented in mag. nitude and direction by pairs of lines drawn from a point (not the centre) within a circle to the circumference in opposite directions, prove that there must always be a resultant who.se inclination to the diameter passing thi-ough 2 (Hin 2 0) I 1 X ^ , being the angle between «+ 2 (cos 2 0)' ® ° the point is tan' one of the forces and the diameter through the point. 42. If four forces act at a point and are represented in magnitude and direction l^y the sides of a (luadrilateral, either they keep the point at rest, or have a resultant which is double a force represented by one of the sides or one of the diagonals. 43. If six sides of a regular octagon taken in order repre- sent in direction and magnitude the forces acting on a point, Hud two ecjual forces which, acting at right angles to one another, will keep the point at rest. m its Ich, if lorcos it CHAPTER III. Since twice the area of a triangle is the product of the base by the perpendicular height, it follows that the moment of a force about a point may be represented by twice the area of the triangle whose base is the line representing the force, and vertex this point. Hence also, if three forces acting on a body are represented in magnitude and position by the sides of a triangle taken in order, their resultant moment about any point in the plane of the triangle is represented by the double of its area. A similar proposition holds when the forces are represented by the sides of a plane polygon, provided the sides do not cross. Forces acting in one plane on a particle will be in equili- brium if the algebi'aic sum of their moments vanish about two points in the plane which are not situated on a straight line through the particle. For let A be the particle, and B, C, two such points. Then since the sum of the moments vanishes about £, if there be a resultant, it must pass through B. Similarly it must pass through C. But it passes through A. Therefore it must pass through A, B, and C; which is impossible, since they are not in the same straight line. Hence the resultant vanishes. In problem 4 will be found an application of this principle. In general, when forces acting in one plane maintain equilibrium, by resolving them in two directions at right angles to each other, and equating to zero the algebraic sums of the separate sets of components, we obtain two equations — relations between the quantities, known and unknown, involved in the problem. A third relation is obtained by equating to zero the algebraic sum of their moments about any point in their plane. If the data be sufficient for the complete solution of the problem, and more than three unknowns be involved, additional equations 105 must be obtained from considering the geometrical relations of parts of the system. It must bo noted that if the two equations above spoken of are formed, and a third by taking moments about any point, no fresh independent relations are furnished by resolving the forces in other directions, or by taking moments about any other point. It will at once be seen that the statement in Italics at the end of § 35 applies to the case of any number of parallel forces in one plane. The magnitude of their resultant is found by taking their algebraic sum, and its position from the fact that its moment about any point in their plane is equal to the algebraic sum of their moments. ntain right ibraic two and on is their pa be ■more Itions 1. If a force be represented by a straight line, what is the geometrical representation of the moment ? If an inch be taken aa the unit of length, what is the geometrical representation of the unit of moment 1 When is a force said to have a negative moment ? 2. Three forces are represented in magnitude and position by the sides of a triangle taken in order ; shew that the moment of the forces about any point in the plane of the triangle is represented by twice the area of the triangle. 3. If the algebraic sum of the moments of forces acting in one plane on a particle, has the same value for two points in the plane, then the straight line which joins thes»- two points is parallel to the resultant force. 4. Forces act at the angular points of a triangle along the perpendiculars drawn from the angular points on the respectively opposite sides, each force being proportional to the side to which it is perpendicular. Shew, by taking moments round the angular points, that the forces are in equilibrium. 5. Shew that the proposition contained in § 34 must hold in the case of parallel forces without assuming the know- m 1 !■ li ''l\ ^ § '«i 106 l(!(lge of the magnitude or position of tlie resultant in this •). Three parallel forces acting on a rigid system keep it lit rest ; investigate the relations subsisting between their magnitudes, directions, and distances. 7. A BCD is a squai'c; a force of 3 lbs. acts from A to //, a force of 4 lbs. from B to C, and a force of 5 lbs. from (J to D ; if the centre of the square be fixed, find the force wliich, acting along AD, will keep the square in equilibrium. 8. A weight W is siispended freely from a fixed point by a perfectly flexible string ; find what horizontal force applietl to the string will draw the upper portion of it 30° out of the perpendicular. ■ y, A weightless rod AB, A being fixed and B sustaining ;i weight P, is supported by a string BC di-awn to a point C vertically above A, AC being equal to AB, and the angle A liC 30°. Find the tension of the string. 10. If a cord, whose length is 21, be fastened at two points lying in the same horizontal line, at a distance 2« from each other, and if a smooth ring upon the cord sustain ;i weight W, prove that the tension of the cord is Wl iVt' - a} ' 11. A weight P, suspended from one end of a lever with- out weight, is balanced by a weight of 1 lb. at the other end of the lever; when the fulcrum is removed through half tlie length of the lever, it requires 10 lbs. to balance P : find P. 12. Four weights, 1, 3, 7, 5, ai-e placed at equal distances on a straight rod. Find their resultant and the point through which it acts. 13. The arms of a weightless lever are inclined to each other : shew that the lever will be in equilibrium with 107 Stances point each with eijual weights suspended from its extremities, if the point midway between the extremities be vertically above or below the fixed point. 14. Two weights are suspended from the arms of a bent lever without weight, the arms being 10 ft. and 6 ft. long, lud inclined to the horizon at angles 30" and 45° respec- tively. Find the ratio of the weights. 15. Two weights, P and Q, are suspended, one at each (Mid of a bent lever, whoso arms (a, h,) ore inclined to each other at an angle a. If

ondicular from A on JiC. 2. Parallel forces P, Q, li act at the angular points A . Ji, C o{ a triangle ; show that the porpondicuhir distance of their centre from the side B C is ■J'tl' ! P+ Q + Ii 2 area of triangle ~BC • 3. Parallel forces P, Q, R act at the angular points A. B, C of a triangle, and their centre is at ; shew that p Q n area of BOC area of COA area oi AOB' 4. Equal like parallel forces act at five of the angulai j)oints of a regular hexagon ; determine the centre of par allel forces. 5. The circumference of a circle is divided into 7i c(jual parts, and equal like parallel forces act at all of the points of division except one ; find the centre of parallel force.s. 6. Two parallel forces being in the ratio of ^/iT + 1 to 1. shew that if one of them be reversed, the resultant will Ix- moved through a distance equal to that between the two forces. 7. If of three parallel forces acting in the same sense thf middle one is the resultant of the other two, prove that in whatever senses the forces may be severally acting, the line of action of the resultant of the three coincides with that of one or other of the forces. Ill 8. A rigid iHimlloloyram can turn about a flxod pcniit in one of its cliugonalH, and is acted on by four forccH alunj,' its sid«'8 (which also represent thcin respectively in magnitude), and each pair of parallel forces tend towards the same parts. Prove that either the parallelogram is at rest, or it can be kept so by a force whose moment round the fixed point is represented by four times the ni-ea of the triangle fornu'd by joining the fixed point to the ends of that diagonal in which it is not sitiuited. 9. A uniform rod 3 ft. long and weigliing 4 lbs. has ji weight of 2 lbs. placed at one end ; find the centre of gravity of the system. 10. Two weights of 5 and 7 lbs. are connectetl by a md. without weight, measuring G ft.; find their centre of gravity, and determine the change which will take place in its posi- tion if 1 lb. be added to the smaller weight. 11. A uniform beam, 18 ft. long, rests in equilibrium upon a fulcrum 2 ft. from one end, having n weight <^i' 5 lbs. at the end furthest from tlie fulcrum, and one of lid lbs. at the other. Find the weight of the beam. e ill jat in lint- liat of 12. A beam 30 ft. long, balances about a point at ont'- third of its length from the thicker end; but when a weight of 10 lbs. is suspended from the smaller end, the fulcrum must be movetl 2 ft. towards it in order to maintain eipiili- brium. Find the weight of the beam. 1 3. What particular case of the centre of parallel fin'Cfi. constitutes the centre of gravity? % Find the centre of gravity of a number of heavy |)artiuh's lying in a straight line, and shew that if the weight of each be increased by the same quantity, their centre of gravity will be shifted to the right or left according as it lay to tlie left or right of the centre of gravity of a number of particles occupying the same places, but all having the same weight. I ^1 112 14. Four particloH whoso weights aro as 1, 2, 3, 4, nre plac(Hl in onlor at coi-nors of a square; show tliat their centre of gravity lies on a lino through tho centre of the squai-e par- allel to two aidi^s, and at a distance from tho contro equal to one-fifth of side of square. 15. Find the centre of gravity of four equal particles plac(!d at the corners of a quadrilateml figure. 16. A rod AB, capable of turning about tho end A, rests with the end ^against a smooth vortical wall; construct for the direction of the reaction at A, and shew that it bears to tho weight of the rod the ratio of ^/'s' to 2, the rod making iiii angle of 45° with the vertical. 1 7. Wliat propositions are necessary to show that every lK)dy has one, and only ono, centi-e of gravity? 18. Prove that every system of bodies possesses one, and only one, centre of gravity. 19. Define stable and unstable equilibrium, and explain why an egg, when resting on its side, if canted towards an end, will return to its original position, but if, when on its end, it be canted towards a side, will not. 20. Three equal particles are placed at the middle points of the sides of a given triangle ; find their centre of gravity. 21. At the middle points of the sides of a triangle are placed three equal weights ; prove that the sum of the s«piares of their distances from their centre of gravity is i\ (»* + b^ + m C in lingle so ijlle are centre ^vas the ft. 117 57. If the sides of a triangle A BC be bisected in the points D, E, F, the centre of the circle inscribed in the tri- angle DEF will be the centre of gravity of the perimeter of the triangle ABC. 58. Shew that the distances of the centre of gravity of the perimeter of a triangle from the sides a, b, c, are as b+c c+a a+b ah' c ' 59. Prove by statical considerations (or otherwise) that the lines drawn through the middle points of the sides of a triangle parallel to the bisectors of the opposite angles meet in a point. If i* be this point for the triangle ABC, and be the centre of the inscribed circle, the three forces repre- sented by OA, OB, OC will have 2 OP for resultant. 60. A heavy uniform wire is bent into the form of an equilateral triangle, which is loaded at the angular points with weights in the proportion of 3:4:5; find the centre of gravity of the system, the sum of the weights being equal to the weight of the wire. 61. A. thin uniform wire is bent into the form of a tri- angle ABC, and heavy particles of weights P, Q, R are placed at the angular points; prove that, if the centre of gravity of the weights coincide with that of the wire, P : Q : R :: b + c: c + a : a + b. 62. A right-angled Isosceles triangle ABC, C being the right angle, is fixed at A, and BC ia kept horizontal by a horizontal force P applied at B. If W be the weight of the triangle, shew that P=^ W. Find also the direction of the resultant pressure at A. 63. A triangular lamina ABC at resi with the side BC downwards, is supported by a rod through the centre of the inscribed circle ; shew that if 5 C be the greatest side of the triangle, the equilibrium will be stable ; if least, unstable. w -'■) Hi ;■ ; I'; V'. u 1^ p^ iiiiii 118 64. A uniform wire is ormed into a triangle ABC, right-angled at C, and is suspended from C ; ia the angle which the side A C makes with the vertical ; shew that a + c tan = -r 6 + c 65. Three uniform rods form a triangle ABC, which is suspended at the middle point of BC with the vertex down- wards ; shew that the inclination of BC to the horizon is half the difference between the angles at B and C. 66. Three uniform rods of different materials are rigidly connected in the shape of a triangle, and suspended from one angle, the weights of the sides respectively opposite and adjacent to this angle being as 3 to 2 to 1. Shew that ui the position of rest, the perpendiculars from the other angles upon the vertical through the point of suspension are as 5 to 4. 67. A lamina of the form of an equilateral triangle, is hung up against a smooth vertical wall by means of a string attached to the middle point of one side, so as to have an end of this side which is uppermost, in contact with the wall ; determine the position of equilibrium. 68. ABC is a uniform triangular lamina in a vertical plane, right-angled at C, and capable of moving freely about an axis through G perpendicular to its plane. A weight w is attached at £; if 3w be the weight of the triangle, shew that it can be kept at rest, with the base BC horizontal, by a force 2w cosec B acting at any point in BA in dii-ection of its length. 69. A triangular lamina is poised on its centre of gravity ; shew that the balance will not be destroyed by circumscribing the lamina with a triangle formed by three straight wires of equal weight, the middle points of the wires being fastened to the angular points of the lamina. 70. A right-angled triangle is bung from a smooth peg by means of a string fastened to the ends of the hypothenuse 119 BC, and it is found that one or the other side is vertical according as the length of the string is ?i or l^ ; prove that /,' + Q = 10 BC\ 71. A heavy rod is movable freely about a fixed point in itself, and to one end of it is attached a string bearing a given weight, while to the other end is attached another string of given length carrying a smooth ring thx'ough which the former string passes. Find equations for determining the positions in which the rod will rest. 72. Particles of equal weight are placed at the n angular points of a polygon. Each particle is in turn omitted, and the centre of gravity of the remaining particles found; shew that if the points thus found ^successively be joined, a polygon similar to the original will be formed, and that the homologous sides will be in theratio of 1 : n - 1. 73. Shew how to place three unequal heavy particles on the circumference of a circle, so that their centre of gravity may coincide with the centre of the circle. 74. Perpendiculars are drawn from the angular points of a given triangle ABC upon the opposite sides, and another triangle is formed by joining the feet of these pei'pendiculars ; prove that, if ^, q,r be the distances of the centre of gravity of this triangle from the sides opposite to -4, ^ and C, p q r a'co3{B-C) ~ b''cos(C-A) t^cos{A-B) 75. Four equal particles are placed at the centres of the inscribed and escribed circles of a triangle ; shew that their centre of gravity is at the centre of the circumscribing circle. il! 11 CHAPTER V. Levers. 1. Two weights, 3 and 4, balance on the extremities of a lever 4 ft. long ; find the fulcrum. 2. In an ordinary lever of the first class the forces P and Q are inclined inwards, each at an angle of 45° to the lever. Shew that the resultant pressure on the fulcrum is \/F'^ + Q'- 3. AC, GB are the equal arms of a straight lever whose fulcrum is C; to C a heavy arm CD is fixed perpendicular to AB. Prove that if a weight be suspended to the extremity A, and the system be in equilibrium, the tangent of the inclination of CD to the vertical will be proportional to the weight. , • 4. A straight lever of uniform thickness, the length and weight of which are given, has two weights, P and ^, attached to its extremities, and is sustained partly by a fulcrum at a given point, and partly by a peg, the pressure on which is known : required the position of the peg. 5. If a uniform beam of length a and weight W be placed with one end on a fulcrum, and a weight 2 TF be placed at the other end, shew that, if a power P be applied perpendicularly to support the whole, the beam must be horizontal, and the distance of the point of application of P from the end on which is placed the weight cannot be more than — . 6. There are n weights, W^, W^ W^„ in geometrical progression, and W^, placed at A, one extremity of a lever, balances W^, placed at B, the other extremity. Prove that a weight equal to the first ?i - 1 weights, if placed at A, will balance a weight equal to the last n - 1, if placed at B. 121 ' + (^. [placed at the [ularly id tlie knd on ptrical (lever, that I, will 7. Two forces of 6 and 2 i/~W lbs. act at ends of a lever, and are inclined to the direction of the lever produced at angles of 30° and 60° respectively. Determine the position of the fulcrum, ard the resultant pressure on it. 8. The lever AC, without weight, turning about the fulcrum C, has two weights W, W, suspended from the extremity A and the middle point B respectively, and is kept at rest by the given weight P acting at A by means of a string passing over a tack at D; CD is horizontal and equal to AC; find the position of equilibrium. 9. A uniform bent lever, when supported at the angle, rests with the shorter arm horizontal; if the shorter arm were twice as long, it would rest with the other horizontal; compare the lengths of the arms, and find the angle between them. 10. A heavy unifonn rod ABC is bent at C, and there i*ests on a fulcrum, its arms making angles a, /3, with the vertical. If a weight equal to that of the shorter arm be attached to its extremity, these angles are reversed ; prove sin /3 = |/~3~ sin a, a being the angle the longer arm fii-st makes with the vertical. 11. Investigate the conditions of equilibrium of a bent lever, acted on by two foi'ces in any direction in the plane of rotation. 12. If a balance be false, having its arms in the ratio of 1.5 to 16, find how much per lb. a customer really pays for tea which is sold to him from the longer arm at 60 cents j)er lb. 13. A body, the weight of which is one lb., when placed in one scale of a false balance, appears to weigh 14 oz. ; find its apparent weight when placed in the other scale, the inaccuracy of the balance arising from the unequal lengths of the arms. What would be the apparent v/eight, if the inaccuracy arose from the unequal weights of the pans, the arms being equal 1 fi 122 14. If one of the arma of a balance is twice as long as the other, and a weight of one pound is put in the scale attached to the shorter arm, will a customer gain or lose by buying two pounds, weighing one pound in each scale? 15. Explain why, in the common letter balance, it is not necessary to place the weights in any particular position on the pan. 16. If Wi, Wj be the apparent weights of a body when placed successively at the two ends of a balance, what is its true weight (1) when the arms are unequal, (2) when one of the scales is loaded 1 17. If a substance be weighed in a balance having unequal arms, and in one scale appear to weigh a lbs., and in the other b lbs., shew that its true weight is ■y/'ab lbs. 18. In a false balance of which the arms are 10 and lO'l inches, the weight used for the 1 lb. is really less than 1 lb. by the ^xrkrrth part ; shew that a person will get just weight by buying 2 lbs. of a substance, and having 1 lb. weighed separately in each scale. 19. A grocer, knowing that his balance has arms of un- equal length, and wishing to act honestly, weighs (apparently) half of each of his sales in one scale, and half in the other; shew that he will not effect his object, and find his loss or gain per cent., if the lengths of the arms be as a to 6. 20. If a piece be bi'oken off from the longer arm of the Roman steelyard, how is the truth of the instrument affected? 21. In the common steelyard what change is made in the position of the point from which the graduations start, by increasing the movable weight? What change in the position of this point is made by in- creasing uniformly throughout it the density of the material of the beam ? 128 22. Having given the weight of a steelyard, 3 lbs., the movable weight, 4 lbs., the distances of the fulcrum and centre of gravity from the point of suspension of the sciile, 3 aud 5 inches respectively, graduate the instrument. 53. Tn the common steelyard, let A be the point from which the substance to be weighed (Q) is suspended, C the fulcrum, W the weight of the beam and hook, G the centre of gravity of these, N the point of suspension of the fixed weight, D the position of N when Q-0, then if ^ = W, prove that the rectangle DN.GC — rectangle A (J.CD. 24. If the common steelyard bo correctly constructed for a movable weight P, shew that it may be made a correctly constructed instrument for a movable weight nP by sus- pending at the centre of gravity of the steelyard a weight equal to n - 1 times the weight of the steelyard. 25. In the common steelyard, if the beam be uniform, and its weight — of the movable weight, and the fulcrum m, be — of the length of the beam from one end, shew that the greatest weight that can be weighed is 2m(n-l) + n-2 2^i times the movable weight, the body to be weighed being attached to the end of the beam. of snt [he by Pulleys. 1 . If the string to which the weight is attached be coiled in the usual manner round the axle, but the string by which the power is applied be nailed to a point in the rim of tlie wheel, find the position of equilibrium, the power and weight being equal. 2. In the wheel and axle if straps be used, coiled each on itself, and their thickness be not neglected, find whether the ratio of the power to the weight would be increased or 1' 124 ■ i I ml •iiininishcd as the weight waa raised, the straps being of the Hiiine thickness. 3. If there bo two strings at right angles to each other, and a single movable pulley, find the force which will sup- port a weight of |/"2" lbs. 4. Find the ratio of P to W in the single movable pulley, the strings making a given angle a with the horizon. 5. A man stands in a scale attached to a movable pulley, and a ropo having one end fixed passes under the pulley and then over a fixed pulley; find with what force the man must hold down the free end in order to support himself, the strings being parallel. 6. In the first system of pulleys, if there be ten strings to the lower block, what power will support a weight of 1000 lbs.? 7. In the second system, if a man support a weight equal to his own, and there be three pulleys, find his pressure on the floor on which he stands. * 8. In the second system of pulleys, draw a figure repre- senting the angles the ropes make with each other when the free portion to which the power is applied is held in an inclined direction, the weight of the tackle being neglected. 9. In the second system of pulleys, shew that if the weight of each of the puUevs is equal to the power, no mechanical advantage is gained or lost by the system. 10. In a system of three pulleys, if a weight of 5 lbs. be attached to the lowest, 4 lbs. to the next, and 3 lbs. to the next, find the power required for equilibrium. 11. In the second system of pulleys, the number of the pulleys being five, and the weight of each being one ounce, find the weight of the counterpoise. 125 be ho 12. If there be three pulleys of equal weight, find tin- weight of each in order that a weight of 5C lbs., atUichod to the lowest pulley, may be supported by a power of 7 lbs. 14 oz. 13. In the second system of pulleys, if the weight of eac-li pulley be w, shew that no mechanicul advantage is gained unless the power P be gi'eater than w. If W be the weight supported in this case, show that by introducing another j)ulley, an additional weight W-w will be supported. 14. Shew that 7 lbs. will support 49 lbs. on a system of thi"eo pulleys, each weighing 1 lb. What part of the power goes to support the 49 lbs., and what each of the pulleys '< 15. In the second system, if no W act, and the weights of the pulleys be taken into consideration, shew that a force less than the weight of the heaviest pulley will sustain all of them. 16. In the third system of pulleys, there being G pulleys, what weight can be supported by a weight of 1 2 lbs. 1 17. In the third system of pulleys, shew that a weight of three pounds will support twenty-five pounds on a system of two movable pulleys, each weighing one pound. 18. In the third system of pulleys, there being 8 equal heavy pulleys, find the ratio of the weight of one of the pulleys to the weight supported, in order that the latter may be supported by the weight of the pulleys alone. 19. A force of 6 lbs. supports a weight by means of .'^ . weightless pulleys arranged according to the third system. What force will be requisite to support the same weight by means of 8 pulleys arranged according to the first system, the weights of these pulleys forming an arithmetical pro- gression, of which the first term is I oz., and the common difierence ^ oz.1 126 20. In tho third syRtem of pulloyn, if the weight of each pulloy bo tho samo, and P bo equal to tho weight of all the inorablo pulloys (» in number), the mechanical advantage in - (n+l)(2»-l). n ] II Inclined Plane. 1. What weight is that which it would require the same nxertion to lift, as to draw a weight of 4 Iba. up a plane inclined at an angle of 30° to the horizon] 2. The power acting along the plane, if its length be S2 in. and height 8 in., find the mechanical advantage. 3. P acting along tho plane, if /•=9 lV>s., find FT when the height of the plane is 3 in. and the base 4 in. 4. An inclined plane rises 3 ft. 6 in. for every 5 ft. of length; if W=2QQ, find P, supposing it to act along the plane. 5. What force is necessary to support a weight of 60 lbs. ou a plane inclined at an angle of 60° to the horizon, the force acting horizontally 1 6. The power acting horizontally, if W=\'2 lbs., and the base be to the length as 4 is to 5, find P. 7. The power being horizontal, if IT =48 lbs., and the base be to the height as 24 is to 7, find P and R. 8. The power acting horizontally, if ^ = 2 lbs., and P= 1 lb., find W and the inclination of the plane. ' 9. The length of an inclined plane is 5 ft., and the height .'i ft. Find into what two parts a weight of 104 lbs. must be divided, so that one part hanging over the top of the plane may balance the other resting on the plane. It r |e 127 10. Whon a certain inclined piano ABC, whose length in AB, IS placed on AC as base, a power of 3 lbs. can support u weight of 5 lbs. ; find the weight which the same powpr could support if the plane wore placed on BC ;.\3 base, AC Xmng then the height of the plane, the power in both cases acting along the plane. 11. If the power which will support a weight, when acting along the plane, be half that which will do so acting hori- zontally, find the inclination of the plane. 1 2. In the inclined plane, the power acting parallel to the piano, determine the relation between P and IF. Deduce the same also from the triangle of forces. A string, inclined upwards at an angle of 30° to a plane whose inclination to the horizon is 30°, is just strong enough to support a weight W. Shew that if the string bo parallel to the plane, it will be strong enough to sustain a weight 2 W 13. When a given weight is sustained on a given inclined plane by a force in a given direction, find the pressure on the plane. Given the weight, and the magnitude and direction of the sustaining force, find the angle of the plane. 14. Find the angle at which a given force must act in order that it may just support a given weight on a given inclined plane. 15. If ^ be the pressure on the plane when the power acts horizontally, and R' when it acts parallel to the plane, shew that W = RR'. 16. The inclination of a plane is 30°; a particle is placed at the middle point of the plane, and is kept at rest by a string passing through a groove in the plane, and attached 128 to the opposite extremity of the base; shew that the tension of the string is equal to the weight of the particle. 1 7. A power P acting along a plane can support W, and, acting horizontally, can support W ; shew that 18. On an inclined plane the pressure, force and weight are as the numbers 4, 5, 7 ; find the inclination of the plane to the horizon, and the inclination of the force's direction to the plane. 19. A weight W is sustained on an inclined plane by P three forces, each equal to — , one acting vertically upwards, o another parallel to the horizon, and a third lying between these and parallel to the plane ; find the plane's inclination. 20. A weight W would be suj)ported by a power P acting horizontally, or by a power Q acting parallel to the plane ; shew that 1 _ i _1_ 21. A weight W is sustained on an inclined plane by a force P acting by means of a wheel and axle, placed at i\w top, in such a manner that the string attached to the weight is parallel to the plane. Given R and r, the radii of the wheel and axle, find the inclination of the plane. 22. A weight on an inclined plane may be raised cither by a force along the plane, or by means of a system of pulleys of the first kind, with two movable pulleys, which will cause the power to act at an angle of 60° to the plane. Determine the means to be employed with most advantage. 23. A force which will just support a weight of 120 lbs. by means of three pulleys of the third system, is applied in the most advantageous manner to support a weight W on 129 a plane inclined to the horizon at an angle whose tangent IS i; find the magnitude of iV. 24. Two weights sustain each other on two inclined planes I'Hvmg a common altitude, by means of a string, parts of which are parallel to each of the planes; compare the pres- sures on the planes. 25. A weight w is supported on an inclined plane by two I'orces, each equal to ^, one of which acts parallel to the hase. Shew that equilibrium may be possible when the inclination of the plane is not greater than 2 tari' (-) n 1 • . . ^n ' ' being a positive integer. 26 A triangle ABC is placed with its plane vertical and Its side £0 in contact with the line of greatest slope of an inclined plane; shew that if prevented from sliding, it will topple over, if the angle of the inclined plane be greater than ° tan .!« + c cos B c sin B 27. An inclined plane being free to move about its lower edge, and a weight being supported on it by a string which being parallel to the plane, passes through a small smooth nng attached to the top of the plane, and tlien passing over a smooth fixed pulley supports a weigh^ at its other end • shew that when there is equilibrium, p = d cot a, where d is the distance of the first weight from tlie foot of the plane p the perpendicular from the foot on the inclined portion of the string, and a the inclination of the plane to tlie horizon. 28. Two weights sustain each other upon two inclined planes having a common altitude by means of a string which is attached to each; find their position, taking into account the weight of the string, which is supposed to be uniform. m 130 29. A uniform beam of given length and weight, rests with one end on a given inclined plane, and the other attached to a string AFP passing over a pulley at F given in position. Having given the weight P attached to the other end of the string, find the position in which the beam rests. 30. A smooth cylinder is supported on an inclined plane with its axis horizontiU, by means of a string which, passing over the np|x;r surfiice of the cylinder, has one end attached to a fixed point and the other to a weight W which hangs freely; if a l»e the inclination of the plane to the horizon, and the inclination to the vertioil of that part of the string which is fastened to the fixed jioint, the weight of the cylinder is 2 WcoaiO si n {j - a) sin a being measured from the upper part of the vertical through the point to which the string is attached. The Screw. 1. What must be the length of a lever at the extremity of which a pressure of 1 lb. will support a weight of 1000 lbs. on a screw, the distance between two contiguous threads threads being | inch 1 2. The tangent of the angle of a screw is |, the radius of the cylinder 4 in., and the length of the power arm 2 ft. ; find the ratio of IF to P. 3. Supposing that 5J turns cause the head of a screw to advance | in., what power applietl at the extremity of an arm 18 in. long will be required to prod\»ce a pressui-e of 1000 lbs. upon the head of the screw ] 4. What force must be exerted to sustain a ton weight on a sa-ew, the thread of which makes 158 turns in the 3W to I of an ^■e of piglit |i the 131 course of 12 in., and which is acted on by an arm 6 ft. long 1 5. Find tho inclination to the horizon of the thread of a screw which, witli a pressure of 5 lbs. acting at an arm 2 ft. long, can support a weight of 300 lbs. on a cylinder of 3 in. I'adius. 6. The circumference of the circhi corresponding to the point of api)Iication of P is 6 ft. ; find how many turns the screw must make on a cylinder 2 ft. long, in order that W may be equal to 144 P. ' 7. The intei-val Ijctween the threads of a screw being T^ in., and the diameter of the cylinder | in., find the length of tlie thread in 14 revoUitions. 8. If a screw make m turns in a cylinder a foot long, and the length of the power arm be n ft., find the mechanical advantage. 9. The length of the jiower arm is n times the radius of the cylinder ; find the pitch of the sci-ew that tlie mechanical advantage may Ix; n. 10. The angle of a screw is 30°, and the length of the power arm is n times the radius of the cylinder ; find the mechanical advantage. 11. Shew that the mechanical advantage in the case of the screw is the same as that obtained by a horizontal power acting through a lever on a weight supported on an inclined plane, the arm of the power of the lever being to the arm of the weight as the power arm of tho screw is to the radius of the cylinder, and tho iuclination of the plane being the same as the pitch of the screw. 12. Prove that the power is to the whole pressure on th(( thread in a ratio compounded of the ratio of the interval between the thremls to the length of the thi'ead in one turn of th(3 screw, and of the ratio of the radius of the screw to the radius of the circle described by the power. 132 i|i5 '■ Virtual Velocities. 1. In the first system of pulleys, if there bo five strings at the lower block, and IF rise through 6 in., find how much P will descend. 2. In the second system of pulleys, the distance of the highest pulley from the fixed end of the string which passes round it is 16 ft.; find the greatest height through which the weight can be raised, there being four pulleys. 3. In the first system of pulleys, if P descend through 1 ft., W will rise through - inches, where 7i is the number of pulleys on the lower block. 4. In the first system of pulleys, if P descend through 12 feet while W rise through 2 feet, find the number of strings at the lower block. 5. Assuming the statical principles demonstrated in i)re- vious chaptei*s, shew that the principle of virtual velocities holds on the inclined plane when the power acts horizontally, and when it acts at a given angle to the plane. G. Employ the principle of virtual velocities to determine the relation between P and W on the inclined plane when the power acts hoi'izontally, and when it acts at a given angle to the plane. 7. If two forces P, Q, acting at a point, have a resultant H, and the point I'eceive a displacement so that p, q, r are the virtual velocities, of these forces, prove Pp+Qq = Rr. 8, Investigate, by means of the principle of virtual velo- cities, the conditions of equilibrium on the common wheel and axle, and on the differential axle. In converting the former into the latter by increasing the diameter of part of the axle, how far may the increase be continued without loss of advantage 1 Examine the effect of coiling the rope over both axles in the same direction. 133 9. A weight is supported on an inclined plane by a power acting through a wheel and axle placed at the top of the plane, the string attached to the weight being parallel to the i)lane and passing over the axle. Apply the principle of virtual velocities to determine the relation between P and W. 10. Apply the principle of virtual velocities to the fol- lowing : In the screw, if the arm of the power be six times the radius of tJie cylinder, and the thread be wound at an angle of 30° to the axis, what power will sustain a weight of three tons 1 11. By the principle of virtual velocities find the relation between P and IF on a single movable pulley, the parts of the string not being parallel. 12. Apply the principle of virtiial velocities to determine the ratio of the power to the weight, when the weight slides along a smooth vertical rod, and is attached by an inex- tensible string to a point in the rod, while the power acts horizontally at the middle point of the .string. bant are Work. 1. "What is the distinction between the " mechanical effi- ciency of !iu agent," and the " mechanical advantage of a machine]" A weight is i-aised through 100 ft. ten times in an hour by a force of 1 lb. acting thi'ough a machine whose advan- tage is 33 J find the equivalent '• horse-power." 2. What should be the horse-power of a steam engine capable of dragging GOO tons of stone per day of 10 hours up a smooth inclined plane of length 100 yards, and incli- nation 30° ] What if friction acts, coeflicient of friction being .6 ? 3. A well forty feet deep, six feet diameter, and half full of water, is to be emptied by an endless chain of buckets, w m if: 111 ii 134 each of which is one foot deep and two feet diameter ; find the work to be done, and shew that a ten-horse-power engine would do it in little more than three minutes, disregard- ing the weight of the buckets, friction and resistances, and noting that a cubic foot of water weighs 1000 ounces. Friction. 1. Find the coefficient of friction if a weight just rest on a rough plane inclined to the horizon at an angle of 30°. 2. If the height of a rough inclined plane be to the length iis 3 is to 5, and a weight of 15 lbs. be supported by friction alone, find the force of friction in lbs, 3. A body placed on a horizontal plane requires a hori- zontal foi'ce equal to its own weight to overcome the fric- tion ; supposing the plane gradually tilted, find at what angle the body will begin to slide. 4. A weiglit of 10 lbs. just rests on a rough plane inclined at an angle of 45" to the horizon ; find the pressure at right angles to the plane, and the force of friction exerted. 5. A force of 3 lbs. can, when acting along a rough inclined plane, just support a weight of 10 lbs., while a force of G lbs. would be necessary if the plane were smooth ; find the resultant pressure of the rough plane on the weight. G. A body is supported on a rough inclined plane by a force acting at a given angle to the incline. Shew that the reaction will be greatest when the body is just on the point of moving down the plane, and least when just on the point of moving up. 7. A body is just kept by fi-iction from sliding down a rough plane inclined at an angle of 30° to the horizon ; shew that no force acting along the plane would pull the body upwai'ds unless it exceeded the weight of the body. 135 8. A weight of 10 oz. is just supported on a rough plane whose inclination is 75° by a power of 5 oz, acting parallel to the plane j find the inclination of the plane on which the weight would just rest of itself. 9. A uniform ladder 10 ft. long rests \^•ith one end against a smooth vertical wall and the other on the ground, the coefficient of friction being | ; find how high a man whose weight is four times that of the ladder may rise be- fore it begins to slip, the foot of the ladder being 6 ft. from the wall. 10. A beam rests with one end on the ground, and the other in contact with a vertical wall. Having given the coefficients of friction for the wall and the ground, and the distances of the centre of gravity of the beam from the ends, determine the limiting inclination of the beam to the horizon. 11. A cylinder whose altitude is double the diameter of the base is placed on an inclined plane ; find what must be the coefficient of friction so that it may be just on the point of sliding and overturning. 12. A right cu'cular cylinder of radius one-sixth of its length, standing on a rough inclined plane, will topple over just as it begins to slide ; find the coefficient of friction. 13. A given force P, acting parallel to the horizon, just sustains a body of given weight IF, on a rough inclined plane, the angle of which is 0. The same body will just rest without support on a plane of tbe same matex'ial, the inclination of which is a. Determine 0. 14. A uniform I'od is held at a given inclination to a rough horizontal table by a string attached to one of its extremities, the other extremity resting upon the table ; find the least angle at which the string can be inclined to the horizon without causing the rod to slide along the table. 136 M •!!■ ■ 'SIR I 15. Find the work done by a traction of 20 lbs., acting at an angle of 30° to the horizon on a body which moves along a rough level plane through a distance of 100 yards. Dcterniino the coefficient and the amount of fiiction in this case, the weight of the body being 50 lbs. 16. A botly is kept at rest on a given inclined plane by a force making a given angle with the plane ; shew that the reaction of the plane when it is smooth is a harmonic mean between the greatest and least reactions when it is rough. 17. Two inclined planes are joined at their vertex, their angles of inclination being 30° and 60° respectively ; find the relation between their coefficients of friction, that two equal weights of the same substance may be at rest when supported on them by a string passing over the vex'tex. 18. Two inclined planes of the same altitude, one of which is twice as long as the other, the angles of inclination being complementary, are joined at the top ; find what must be the relation between the coefficients of friction for the planes, so that two equal heavy particles, placed one on each plane, and connected by a cord passing over a smooth pulley at the top, may be just in equilibrium. 19. 0^ is the vertical radius of a rough quadrant A B ; P and Q are two weights connected by a string, and P hangs freely along A 0, while Q rests on the convex side of the quadrant. Prove that the difference between the greatest and least values of P consistent with equilibrium is 2 fi Q cos a, a being the angle A Q, and fi the coefficient of friction. 20. A body resting on a rough plane of inclination a, is just on the point of moving up the plane imder the influ- ence of a horicontal force equal to its own weight ; the plane is then tilted until the body is on the point of sliding down, the power remaining horizontal and having the same mag- nitude ; find the inclination of the plane when this takes place. 137 21. Two uniform rods, each of length 2 a, are fiustoncd together so as to form two sides of a square. One of tlieni rests on the top of a rough peg. Shew that the distance f>t" the point of contact from the middle point of the rod is ^(l-(i), fi being the coefficient of friction. 22. An isosceles wooden trestle, consisting of two legs liinged at the top, standing on a rough floor, will support a weiglit only when the legs include an angle less than 30". It is made to support a given weight on the same floor when the legs include an angle of 60° by connecting the feet by a string ; determine the limiting values of the tf ii- sion of the string and the thrust of a leg, neglecting the weight of the trestle. 23. A uniform rod of weight TTx'ests with one end agi'.inst a rough vertical plane, and with the other end attached to a string which passes over a smooth pulley veitically ab()\»' the former end, and supports a weight P. Find the limit- ing position of equilibrium, and shew "^hat equilibrium will be impossible unless P be not less than IF cos e, tan e being the coefficient of friction. IS iu- ^ne ra, tea 24. A uniform heavy rod having one extremity attache provided 2 m seconds be the unit of time, and c the unit of distance. 12. Find the numerical value of the -ation when in half a second a velocity is produced whicn would carry a l)ody over four feet in every quarter of a second. 13. A body moving with uniform acceleration describes 40 ft. in the half second which follows the first second of its motion ; find the acceleration. 14. "Hiscuss the whole motion of a particle which has n Telocity of 100 at a cei-taln point, and suffers a i*etardatiou of 10 during each unit of time. 15. A particle moving with a velocity 5 ft. per second is uniformly i-etarded, and after 30 seconds is found to be moving with the same velocity in the opposite direction ; find the amount of the retardation, and express it in yards per minute. 16. A body moving from rest is observed to move over a feet and h feet respectively in two consecutive seconds; shew that the acceleration is 6 - a, and find the time from rest. Ids 141 17. Uniform accelonition in defined as tliiit wliicli grner- iites equal velocities in ecjual times; would it he correct to defmo it ns that wliich generates ecjual velocities while the body moves through equal spaces ? 18. Shew that in the formula 8= ut - \ft''i s will not re- present tlie space described when the time elapsed is greater u than /■ 19. In the formula s = ut - J/l!^ discuss the motion as / increases, and explain the two values of t for a given value of «. 20. If w be the initial velocity, / the acceleration, v the velocity at time t, and s the space described, shew that V - u 2 « V + u t. in 21. A particle projected in the direction of a uniform accelemtion, describes P and Q feet in the ;>th and qth seconds respectively. Find the magnitude of the accelera- tion, and the veloci* ' of projection. 22. If u be the initial velocity of a particle, and v its velocity after a time t, the motion being uniformly accele- rated, find by general reasoning through what space the particle must move from rest to acquire the velocity v - u. 23. A particle moving with a cei-tain velocity is observed n seconds afterwards to be moving with the same velocity in the opposite direction ; shew that the space passed over in the interval is half that which would have been described if the particle had moved unifomily. 24. A body describes 72 ft. while its velocity increases from 16 to 20 ft. per second ; find the whole time of motion and the acceleration. 25. The velocity of a uniformly accelerated particle on passing a given point i* is 10 feet per second ; at a distance of 15 feet from Pit is 20 feet per second ; find its accelera- tion — also its distance from P when at rest. m 142 26. Two bodies uniformly accelerated in passing over the same space have their velocities increiised from « to 6 and from u to V respectively j compare their accelerations. 27. Having given the velocities at two points in the path of a particle whose motion is uniforndy accelerated, / being the acceleration, determino the distance between them. 28. A pai'ticle is moving at a certain instant at the mte of 10 ft. per second; after 4 seconds it is moving at the rate of GOO yards per minute, and after 1 seconds more at the rate of one mile in 48 seconds. Sh(!W that, so far as your information goes, the particle may be moving with uniformly accelerated motion, and assuming that this is the ciuse, find the numerical value of the acceleration. 29. A stone is droj>ped into a well, and after 2 seconds is heard to strike the water ; required the depth to the surface of the water, the velocity of sound being 11 32 ft. per second, and the uniform acceleration of the stone's motion 30 ft. per second. 30. A body starts from rest under a uniform acceleration, but at the commencement of each successive second the accel- eration is decreased in a geometrical pro))ortion (y- .^) ; shew that the space described in n seconds =(2n~ 3+.,';j2s. where s is the space described in the first second. Parallelogram of Velocities. 1 . A i>article is moving with such a velocity, and in such a direction, that the resolved parts of its velocity in the direcuion of two lines perpendicular to each other are respt ^- tively 3 and 4 ; determine the direction and velocity of the piirticle's motion. 2. If t>, y' be two component velocities of a particle, and a the angle between their directions, the resultant velocity \H i/ (v^ + v''^ + 2vv' cos «). 143 3. A particle is proceecling with unifomi velocity along a certain straight line, and anotlier velocity is communicated to it snch that it .aoves along a lino making 60"^ with its former direction and with imaltered velocity. Determine the magnitiule and direction of the communicatetl velocity. per such the kpt 0- the and IcitN 4. If velocities ha sejiarately communicated to a particle on the circumference of a circle, such that they would respectively cause the particle to move to the circumference along chords at right angles to euch other in the Siime time, shew that when the vehxnties are connnunicat(Hl to it simul- ttuu'ously, the time of its re^iching the circumference will be unaltered. 5. If a particle \\nn a velocity of 20 where a day and a mile ai"e tlu* units of time and space, and a velocity jMU'pen- dicular to the former of 2 1*^ when the units ai-e a mii\ute and a foot, find the resultant velocity when an hour and a yard are the units. 6. A boat is rowwl across a river (2 miles wide) in a direc- tion making an angle of G0° with the bank and against the sti-eam. The boat travels at the rate of 5 miles per hour, and the river flows at tiie r.ite of thi-ee miles i)er hour. Find at what point of the opposite Ixink the boiit will land. 7. In the preceding problem, at what angle to thi; liank should the boat be rowed in order to land at the opi>osite point of the river 1 8. If a \n^ the distj\nce lietween two moving jioints at any time, K the i-esultaut of their velocities com 1 lined, ii, v the resolved parts of V in and perpendicular to the direction of a, shew that their distance when nearest each other is ^r and the tinie of arrivinir at this distance is ait 9. Two vessels ai-e sailing with velocities V and V in dire<;tions which make an angle a. with each other j if U- Ivii ' •i- ;■ If! 144 the angle between the line joining them and the direction in which a gim must be fired from one of them in order to strike the other, find for any given position of the vessels, and shew that its greatest value is given by the equatioii V sin ^ = {V + V'^-2VV' cos a)*, where v is the velocity of the shot, supposed uniform. 10. If a point be situated at the intersection of the pei*- pendiculars drawn from the angular points of a triangle to the sides respectively opposite to them, and have component velocities represented in magnitude and direction by its distances from the angular points of the triangle, prove that its resultant velocity Avill tend to the centre of the circle about the triangle, and will be represented by twice the dis- tance of the point from the centre. CHAPTER IT. Mass. 1. If a Itoily weighing 30 Ihs. l)e moved by a constant force whicli generates in it in a second a velocity of 50 feet per second, fiml wliat weight the force woukl statically support. 2. A body weighing n lbs. is moved by a constant force which generates in the body in one second a velocity of a foot per second ; find the weight which the force could Kui)port. .3. Shew that a force 100 times the difference in the \voie 3, find the unit of mass. 0. If P = ?h/ be the relation between the pressure, mass and acceleration, fiuil the unit of time, when that of space is two feet, and that of pn>ssure the weight of a iinit of mass. 7. In the equation W=mg, what must be the relation between the units of time and space in order that the unit of mass may be the mass of a unit of weight? 8. A pressure P ])roduces an accelerating effect f on a mass m; determine the relation Ix-tweeii P, m and /, the unit of pressure being 1 lb., the unit of mass the mass of a cubic foot of water, and the ludt of acceleration the accelc- latioii produced by gravity. K >m.v1 fli^ i ■ : 140 9. In what time will a force which can support a weight of 5 lbs. move a mass of 10 lbs. weight through a space of 50 ft. on a smooth horizontal plane '? 10. Find in what time a force which would support ii weight of 4 lbs., would move a weight of 9 lbs. through 4li ft. along a smooth horizontal plane ; and find the velocity acquii-ed. 11. A mass of 5 lbs. is ol>served to move from rest through a distance of 10 feet in one second, and a mass of 6 lb.s. through a distance of 25 feet in two seconds ; compai'e the forces acting to produce these motions. 1 2. Find how far a force which would support a weight of n lbs., would move a weight of m lbs. in t seconds ; and iind the velocity acquired. 13. Two bodies urged from rest by the same uniform force describe the same space, the one in half the time the other does ; compare their final velocities, and their mo- menta. 14. A body weighing 10 lbs. is falling under the action of gravity, and is being pressed vertically downwards with a pressure of 1 lb. by another boily, such as the hand ; detev- jnine its velocity at the end of 3 seconds from rest. 15. If a weight of 8 lbs. be placed on a horizontiil plane which is nuule to descend vertically with an acceleration of 12 ft. per second, find the pi-essui-e on the plane. 16. If a weight of n lbs. be placed on a horizontal plane which is made to ascend vei'tically with an acceleration / find the pressure on the plane. 17. If a force of a lbs. protluce in a cubic foot of matter in one second a velocity of n ft. per second, find the specific gravity of the matter. .^^ 147 1. State Newton's three laws of motion, and apply them to a general explanation of the following cases : (1) A heavy particle moving along a smooth Iiorizontal plane, gravity always acting pei'pendicular to the plane. (2) A heavy particle moving down smooth inclined planes of different inclinations. (3) Different heavy particles moving down the same inclined plane. 2. Particles are projected simultaneously from a point with equal velocities, and at tlie same elevation, and are subject to the action of gravity; prove that at any subse equal; then ifyandy"' be the expressions for the accelerating effects of the two forces, referred to the same unit of time, 5. A particle falls through the same distance at two dif- ferent places on the earth's surface, and it is observed that the time of falling is t" less, and the velocity acquired m ft. greater at one place than at the other. If g, g' be the accele 'ations of gravity at the two places, shew that gg' — —. G. A constant force (y') acts ui)on a body from rest during 3 seconds, and then ceases. In the next 3 seconds it is found that the body describes 180 ft. Find both the velocity (v) of the body at the end of the 2nd second of its motion, and the numerical values of the acceleratijig force (1) wlien a second, (2) wlion a minute, is taken as the unit of time. 7. A body is projected vertically ui)wards with a velocity which will carry it to a height 1g\ find after what interval the body Avill be descending with the velocity g. 8. A point moving with uniform acceleration describes 20 ft. in the half second which elapses after the first second of its motion; compare its acceleration y with that of a fall- ing heavy particle, and give its numerical measure, taking a minute as the unit of time, and a mile as that of space. 9. A body falling in vacuo under the action of gravity is okserved to fall through 14 4 -9 ft. and 177*1 ft. in two suc- cessive seconds; determine the accelerating force of gravity, and the time from the beginning of the motion. 10. A stone is thrown vertically upwards, and is observed to be at a height of twenty feet after thi-ee seconds. How nnich higher will it ascend 1 11. A particle is projecti'd vertically upwards, from a height of 2r)7'G ft., with a velocity of OG'G ft.; find the time of its reaching the ground, and the momentum with which it impinges. U9 12. A pai'ticle projected from the deck of a vessel which is moving uniformly at the rate of 14 miles per hour, rises to a height above the deck of Gi'i feet; find how far the vessel will have gone before the particle again comes to the level of the deck. 'ibes lond ai- ls i-cd )W a ine Ich I 13. Two balls are projected at the same instant towards (^ach other, from the extremities of a vertical line, each with the velocity that would be acquired in falling down it. Where will they meet? 1 4. At the same instant one body is dropped from a given height, and another is started vertically upwards from the ground with just sufficient velocity to attain that height; compare the time they take to meet with the time in which the first would have fallen to the groinid. 15. A particle is let fall from a tower 200 ft. high. Shew that the velocity with which another must be projected vertically downwards from the same point, two seconds afterwai'ds, so as to reach the ground at the same instant, is 105 '3 ft. per second, nearly. IG. A ball weighing 30 lbs. is projected along a horizontal surface with an initial velocity of 50 ft. per second. Its motion is uniformly retarded, and it is reduced to rest at a distance of 250 feet from the point of projection. What is the value of the retarding moving force? 1 7. A balloon ascends with a uniformly accelei'ated velocity, so that a weight of 1 lb. produces on the hand of the aeronaut sustaining it a downward pre o that which 17 oz. would produce at the eax'th's surface. Ij'iud the height which the balloon will have attained in one minute from the time of starting, not taking into account the variation of the accelerating effect of the earth's attraction. 18. A particle falls from rest under the action of gravity, and the whole time of falling is divided into any number of intervals, during each of which it falls through the same Br |i> ,■■(■ ■ I* 1 '\ ii 150 ttistance; show that tlie accelerations of the motion during t he with and jtth of these intervals are as \/m - \/m - 1 : i/ M - \/n - 1. 19. A. particle moving under the action of a uniform force in a straight line is observed at a point B; two seconds pre- viously it was seen to be at A, where AB= 10 feet, and was then moving with a velocity 6, but in which direction was not observed. Shew that its motion may have arisen from a foi'ce in direction AB which would produce an acceleration 11, or from one in direction BA giving an acceleration 1; and the particle either was at B 2\x seconds previously, or will be there again after 8 seconds. 20. A particle moves from rest at A, under the action of a constant force f, towards B; when it arrives at B, the force instantly changes direction, and again when it reaches A, and so on. Shew that when it passes thi-ough B the «th time its (velocity )2 will be / J^ {2« - 1 - ( - 1)"}, and that the particle then comes to rest at a distance AB |2u + l-(-l)"}4r from^. Motion on Inclined Planes. 1. Find the velocity acquired by a particle in sliding down a plane of length 4 ft., inclined at an angle of 30° to the horizon. 2. Shew how to place a plane of given length in order that a body may acquire a given velocity by falling down it. 3. A heavy particle slides down a smooth plane of given height ; prove that the time of its descent varies as the secant of the inclination of the plane to the vertical. 4. If a particle be projected vertically upwards at a height of a feet, with the velocity which it would have acquired in falling down a smooth plane of length 21, inclined to the ^mbMI 151 vertical at an angle of 45°, find the height to which it will I'isa, the time when it will return to the point of projection, and when it will reach the earth. If a were very large compared with the earth's radius, for what reason would your result not apply 1 5. A particle being projected down an inclined plane with the velocity which would be acquired in falling down its perpendicular height, the time of descent is found to be that of falling down the height. Find the plane's inclination. 6. If a heavy particle be projected with a velocity of 10 feet per second up an inclined plane, which rises at the rate of 1 vertical in 3 horizontal, through what space will it have moved at the end of 5 seconds ] 7. A heavy particle is projected directly up an inclined plane (inclination a) with velocity u, and is attached to the point of projection by an inextensible string, whose length is half the distance a free particle would ascend ; detei'mine the time which elapses before the particle returns to the point of projection. 8. If two heavy particles commence to descend along the sides of a triangle, whose plane is vertical and base horizon- tal, at the same instant that a third falls freely from the vertex, find geometrically the position of the former particles when the third has reached the base. 9. The time of descent of a particle down any chord of a vertical circle fi'om the highest or to the lowest point is the siinie. The velocity acquired in falling to the end of the chord in either case varies as the length of the chord. 10. The time of descent from a given point to the centre of a circle vertically below it is the same as that to the cir- cumference down a tangent. 152 11. Find by a goometrical construction or otlmrwiso the line of quickest clescont, (i.) From a givon straight lino to a given jtoint. (ii.) From a given point witliin a given circle to the circle, (iii.) From a given circle to a given point within it. (iv.) From a given circle to a given straight line. 12. APB, AQG are two circles with their centres in the same vertical line AJiC, anil touching each otlier at their highest points. If AFQ, Apq be any two chords, the times of descent down PQ, pfj, from rest at P and p, ai'C c(pial. 13. Two bodies, A and £, descend from the same extre- mity of the vertical diameter of a circle, one down the diameter, the otlier down the chord of 30''. Find the ratio of A to Ji when their centre of gravity moves along the chord of 120°. 14. AB is the vertical diameter of a circle ; through .^1, the highest point, any choi-d AC is di-awn, and through C, a tangent meeting the tangent at B in the point T. Shew that the time for a body's sliding down CT oc -r;^ • 15. If SP, SQ be the lines of longest and quickest descent from a point S to points on the circumference of a circle standing in the same vertical plane with S, and SP, SQ meet the circle again in p, q, prove that the sum of the squares of the times down SP, SQ is eijual to that for those down Sp, Sq. 16. A particle slides down a rough plane inclined at an angle a to the horizon ; find the acceleration, n being the coefficient of dynamical friction. 17. A body is projected up a rough inclined plane with the velocity 2gf ; the inclination of the plane to the horizon is 30°, and the coefficient of friction is equal to tan 15°. Find the distance along the plane which the body will describe. M 153 18. A lioavy body is projectoil up a rouj^'li [tlan»< inclin('(l lit an angle of 60'' to the horizon, with tho velocity which it would have acqiiinMl in falling freely through a space of 12 ft;et, and just reaches tho top of the i)lano ; find tho altitude? of tho plane, the coefficient of dynamical friction being — : • 19. A P, AQ are two inclined planes, of wliich A P is rough (/<-tan J*AQ) and AQ smooth, .1/^ lying above AQ ; shew that if bodies descend from rest at P and Q, they will arrive; at .1, (1) in the same time if PQ bo pori)endicular to AQ, (2) with the same velocity if PQ bo perpendicular to A P. 20. A body is projected up a rougli inclined plane, the inclination of the plane to the horizon 1 icing a, and the co- (-'fficiont of friction tan e ; if in be the time of ascending and ti the time of descending, shew that -«) (m\^ sin (a - 1 n' sin (a + { Projectiles. 1. A body is pi-ojected with a velocity 3^ at an inclination of 45° to the horizon ; determine the range. 2. What angle of projection gives the greatest range theoretically? Why does experiment contradict this result? 3. If the initial velocity of a projectile be given, the hori- zontal range is the same, whether the angle of projection be -r- + a or — - a. Compare also the times of flight. 4. Point oiit what parts of Newton's second law of motion are illustrated in the motion of a common projectile. Shew that the sum of the greatest heights in the two paths which have the same range is half the greatest range, the velocity of projection being given. IM Hj f). A body is projoctfid witli a vertical velocity (Ifi"), and a horizontal velocity (8); prove that its distance from the point of projection at the end of one second is one foot. G. A particle is projected with a horizontal velocity 12*7, and niov»)s under the action of gravity. Find the position of the particle, and the direction and magnitude of its velocity after live seconds, pointing out clearly the dynamical prin- ciples of which you assume the truth. 7. The greatest elevation of a particle projected with a velocity of 80 feet per second is 50 feet; ftnd the direction of projection. 8. Given two straight lines representing in magnitude and direction the horizontal and vertical velocities of a projectih' at the point of projection, obtain a geometrical construction for the vertical velocity at a jjoint where the direction of motion makes a given angle with the direction of projection. 9. During the motion of a heavy particle projected in vacuo with a velocity v and at an angle a (> 30°) to the horizon, the times when the whole velocity is double the vertical velocity are - (sin a ± — ::^ COS a). Interpret this result when a <[ 30°. 10. Having given the velocities at two points of the path of a projectile, find the difference of their altitudes above a horizontal plane. 11. If two particles be projrc Ifvl the same direction with dit' line joining them moves para o the d i point in , pi that the tion ui projection. 12. A body is projected vertically p wards from a point A with a given velocity (u); find the direction (a) in whicl' another body must be projected at the same instant with given velocity (v) from a point B in the same horizontal li with ^, so as to strike the first body. 155 13. From tlie top of a towor two hodios iiro projected with the same j^iven velocity (v), ut different given angles of eleva- tion («, fi), and they strike the horizon at the same place ; find the height of the tower. 14. A ball is fired from the ground at an angle of 4.')'', so as just to pass over a wall 10 feet high at a distance of 2G0 feet. Shew that it will strike the ground 1()'4 feet from the wall. 15. A particle is projectoil from the top of a towor 120 ft'ct in height with the velocity which would be acquired by !i particle falling freely 12^ feet under the action of gravity, 3 and in a direction inclined at an angle sin* - to the vertical. [u its descent it just passes over a wall (perpendicular to the plane of projection) 30 feet in height. Sliow that the particle will strike the ground 6 feet from the foot of the wall. IG. A plane is inclined at an angle of 45° to the horizon, and from the foot of it a body is projected upwai'ds along the plane, and reaches the top with one-fifth of its original velocity (v); where will it strike the ground? 17. Swift of foot was Hiawatha; He could shoot an arrow from him, And run forward with such fleetness That the arrow fell behind him ! Strong of arm was Hiawatha ; He could shoot ten arrows upward. Shoot them with such strength and swiftness. That the tenth had left the bowstring Ere the first to earth had fallen. Neglecting the resistance of the air, taking g = 32, suppos- ing one second to elapse between the discharge of each of the ten arrows, and making Hiawatha shoot at his longest range, shew that he must have been able to run at least at the rate of 70 miles an hour. 15G 18. In the motion of a projectile, prove that the direction of motion at any time t, when the horizontal and vertical velocities are u and v respectively, will intersect the plane It at a distance ^ (J f - from the point of projection. 19. A body is projected horizontally with a velocity 4'/ from a point whose height above the ground is 10(/; find the direction of motion (1) when it has fallen half avuv to the ground, (2) when half the whole time of falling h;is elapsed. 20. If a body be pi'ojected at an angle a to the horizon, with the velocity due to gravity in 1", its direction is inclined \\.t an angle — to the horizon at the time tan — , and at an anple at the time cot — . -'2 2 21. If a be the angle of pi'ojection of a projectile, t the time which elapses before the body strikes the ground, prove that at the time — — — , the angle which the direction of motion makes with the direction of projection is equal to a. 22. In the path of a projectile, the velocities are y„ v.^, at two points where the directions of motion are inclined to t^ich other at an angle

k«tMM« 18T as sniootl, ladined planes) from tl.e highest point to the onrnniference; prove that they all reach the circun.f<.rence in tlie same ti>ne; and if each particle, after descrihin. its chord move freely to the ground, they will all strike^he ground with the same velocity. 20. In tho motion of a projectile, shew that the tan^..time .n QR^ ...., prove that tlJ tangents of the angles made with the horizon by the direc- tions of motion at the points F, Q, A>, .... ,,« i^, arithmetic ])rogression. 2-. In the path of a connnon projectile, if the tangent be drawn to cut the horizontal and vertical lines passing througl> tue lughes point, one of the points of section m^ves with uniform velocity, and the other is uniformly accelerated. 28. TJu.e heavy bodies, P, Q, li, are projected in a vertical plane at tho same t.me from a point at angles of elevation a, ,^y, res,)c.ctively, and so that the direction PR is alwavs ^.rtica ami Qli horizontal; shew that the ratio of tho I at which the distances FR and QR change is n sec Y sin {a - y) u cos a ~ V cos /? where a, v are the velocities of projection of F and Q respec- 29. Find the range of a projectile fired horizontally on a I'laue mclmed at an angle of 10= to the horizon. 30. The range of a projectile on a plane inclined to the 158 31. Shfiw that the direction of projection which gives the greatest range upon any plane through the point of projec- tion, bisects the angle between the vertical and the plane on which the range is measured. 32. A particle projected at an angle a to the horizon and with velocity v, impinges perpendicularly on an inclined plane drawn through the point of projection at an angle /i 2 J)'* to the horizon; shew that the range on the plane = cos a cos (a - j3) sin 2^ .7 33. If a body fall down a plane of inclination a, uiul another be pi'ojected at the same instant from the starting point horizontally along the plane with velocity v, find the distance D between the two bodies (1) after a given time t, { 2) after the first body has descended through a given space n. 34. A ball is fired at a given angle of elevation, and with a given charge, from the mast head of a ship sailing uniformly ; prove that the distance of the bail from the mast on reaching the level of the deck is always the same; and shew that, tlie angle of elevation being a, this distance is 2-/ ( \ «2 sill 2a J' where v is the velocity of discharge, and h the height of tltc mast above the deck. 35. Two projectiles are shot from the same jwint, at the same instant, with the sanie velocity, and at angles wliicli are complementary to each other; prove that the horizontal distance between them is always equal to the vei-tical distance : and that they will be moving in parallel directions after a time equal to the ai'ithmetic mean between their times of flight. 36. A man st.anding on the edge of a ravine fires at a uiark on the opposite bank in the siime horizontal plane as his eye. His rifle is properly sighted for tlie distance, but he liolds it in such a way that the plane through the line ol' i 159 sight and the axis of the barrel is inclined to the vertical at an angle /5. The ball strikes the opi>osite bank which is vertical in n seconds. Shew that it strikes at a distance 'n?g sin' — below the mark, and - li^g sin /? to the right or left, the resistance of the air being neglected. 37, The l^aiTcl of a rifle sighted to hit the centre of the bull's-eye, which is at the same height as the muzzle, and distant a yards from it, would be inclined at an elevation a to the horizon. Prove that if the rifle Ije wrongly sighted, so that the elevation is a + 0,0 being small compared A,vith a, the target will lie hit at a height : — . above the centre of the bull's-eye. COS* a 38. In the motion of a projectile shew that the focus ot the pai-al^/ia is above or below the horizon, according as the angle of projection is greater or less than that which pro- duces the greatest range. Shew tliat the angles of projection which cause the foci to be at equal distances above and below the lioi-izon an* complementary. 39. In the two trajectories of a coiumon pixijectile, which for a given velocity of projection have the same i-ange, the sum of the gi'eatest heights is double the greatest height for the maximum range; and in this latter case the focus lies in the same hoi'izontal plane as the point of pi-ojection. 40. In the motion of an ordinary projectile, shew that the distance of the focus of the parabola from the point of pro- jection is inde[>endent of the angle of projection. Fidlinfj Bodies Connected by a Siring. 1. Twelve lioumls weight is ao distributetl at the extremi- ties of a string piussing over a pulley, that one end descends thi-ough 7 ft. in as many seconds. Wluit weight Ls at each end of the string 1 ii' 160 2. A weight of 10 lbs. is attached to one end of a string ; find the weight which must be attiiched to the other, iii order that when the system is suspended from a fixed puUoy, the acceleration may be half that of gravity. 3. Two weights connected by a string passing over a pulley are in motion, and it is observed that the space through which the lieavier descendfj in the nth second froiii the commencement of the motion is a feet ; compare the weights. 4. In Atwood's machine, if the weight of each box be 31 J 07.., and an oz. weight be placed upon cue, find the velocity generated in 5". If after a certain time tlie oz. weight be removed, and the space described during the next second bo 4 ft., find the time from rest, and the whole space described (y = 32). 5. Two weights, P and (?, of 10 and 6 lbs. respectively, are suspended by a string over a smooth pulley. If after 1 2 seconds of motion P suddenly loses 7 lbs., find the time that will elapse before P returns to the point where it was thus diminished ; and describe the subsequent motion, supposing P to take up again the weight it had lost. What would have been the difference in the motion if instead of P lo.sing 7 lbs., Q had taken up 16 lbs. ; this latter weight being supposed previously at rest 'i 6. Two weights of 14 oz. and 13 oz. are attached by means of a string passing over a smooth pulley, and are allowed to move from rest for 3 seconds, when 9 oz. of the former become detached ; shew that the remaining weights are momentarily reduced to rest in a quarter of a second. 7. P pulls Q over a smooth pulley ; and Q in ascending, as it passes a certain point A, catches and carries with it u certain additional weight, and on its descent the additioiiiil weight is again deposited at A. Supposing no impulse to take place when the weight is so caiight up, and that Q> iu this manner oscillates through an eijual space on either sith; of A, find the additional weight. 161 8. Two unequal weights connected by a string which passes over a smooth fixed pxilley, are in motion ; shew that if at any instant the string breaks, when the ascending motion of one of them ceases, the other will have descended through three times the space through which the fonner has ascended in the interval. II 0. A weight Q is drawn along a smooth horizontal table by a weight /"hanging vertically; find (1) the acceleration of P, (2) the vertical and horizontal accelerations of the centre of gravity of P and Q. 10. Two bodies, P and Q, are connected by an inextensible string which passes over a smooth fixed pulley ; find the tension of the string. 11. State Avhat parts of the laws of motion are illustrated and confirmed by At wood's machine. Prove that the tension of the string remains always the same during the motion. 1 2. In Atwood's machine the tension of the string is the harmonic mean of the moving effects of gravity on the two bodies. 13. In Atwood's machine shew how the proportionality of the acceleration to the moving force on the same mass may be tested. A weight of 10 lbs. hanging vertically, draws another of 7 lbs. down a smooth plane, inclined at an angle of 30° to the horizon. Find the vertical velocity of the latter at the end of 10 seconds from rest, and the tension of the string. 14. A uniform string hangs at rest over a smooth peg. Half the string on one side of the peg is cut off; shew that the pressure on the peg is instantaneously reduced to two- thirds its pi'evious amount. 15. A body P descending vertically draws another body Q up the inclined plane formed by the upper surface of a k:: I 162 right-angled wedge which rests on a smooth horizontal table ; find the force necessary to prevent the wedge from slidinj; along the table. 16. In Atwood's machine a weight P ( - W+l p), startincf from rest, draws up a weight Q { = np), and at the end of successive intervals of time, each equal to t, P gains, and Q loses a weight p; shew that the velocities at the ends of these successive intervals ai'e as the squares of the natural numbers, and that the space described in time nt n 27»» + 1 „ = - . or. 6 2n+l "^ 17. Two bodies, whose weights ai-e P and Q, hang from the extremities of a cord ptissing over a smooth jx;g ; if at the end of each second from the beginning of motion P be suddenly diminished and Q suddenly increased by -th of n their original difference, shew that their velocity will be zero at the end of n + 1 seconds 18. Two equal botlies connected by a string are placed upon two jJanes which are inclined at angles a, fl, to the Iiorizon, and have a common altitude. Prove that the acceleration of their centre of gi*avity is gr sin I (a -i3) cos ■' ^ (« + /?). 19. A fine inelastic thread is loaded with n equal particles at equal distances c fi-om one another ; the thread is stretched and placed on a smooth horizont c 5. As 4 to 3. C. - • - p, taking a unit of length to represent a (( (/ unit of force. CnAFfER II. — 2. Comploto parallelogram ; then prop, is evident since diagonals bisect one another. 4. 5 lbs. ,'i. V 3 x less force. fi. l/ 3 , and acta perp. to 1. 7. Use first fornnilaH in S 29 ; 4, 150^, 60". 8. Evident on completing parallulognim, and •Irawing diagonal. 9. 120"; use second formulas, § 29. 10. Fees. -1 7 . . . . . 7 in st. line. 11. Cos -: 12 _ 14. As 2 : l/ 6 i.e. the angle whose cosine is - - V'"ir+1;§29. 15 12 Aa 12. 5,5v/3. 13. i l/"6 :2:l/3'-l. 16. § 28, and Euc, Ek. I., Prop. 20. 17. In a line with the reversed force, and double it. 18. The sum or difference of the forces. 19. Components make angles (iO', 30°, with original force. 20. § 18, pressure on peg bisects angle between forces; resolving along this, pressure - 2 x force x cos (i angle between forces). 21. Press, on top peg = ?<» V 3 ; press, on 10 side peg8 = — :(|/ 3 _ i). 22. 0. 23. Apply two equal and opposite forces along direction of given force ; this ecjuivalent system can at once be resolved in the given directions. 24. Kesolve the forces along and perp. to 1, say; result, is v 3, and makes angle 30° with 3. 25. 2 V2, and bisects angle between 4 and 5. 26. 8 K 4 + 2 /"2, and makes with 8 the angle tani (1 + l/Y). 27. 180°. 28. The pt. is the bisection of the arc at the l)ase of the segment. 29. We have only three indep. relations, viz., the first «[«. of § 29 and {Q, R) + (K, P) + (P, Q) - 180°. Angles being ^iven, relative magnitudes of fees, are known. 30. Use prob. 2, and iudirect proof. 31. is the bisection of the line joining the bisec- tions of AB, CD; and PO : OQ :: I : 3. 32. Indirect proof. 33. a ]> 90°, P >• /?. A figure makes second part evident. 34. Each diag. is equivalent to two sides, and, taking away the common side, o\)\). sides remain. 35. Diag. through C of parall'". of which A C, liV are sides. 36. Let i?be bisection of BC ; then AB is result, of AC, CB; 2 AEol AB, AC; and 3 AD of AB, 2 AE. 38. Bisect AD'm E; join EO (centre), and draw BEC perp. to EO. 39. On BC desc. seg. BOC cont*. angle = 180 - A, and on AC seg. AOC cent*. angle= 180 -B. is pt., § 29. 40. By Geomet. OD : OE : K i 170 OF— cos A : cos B : coa C; but, since is at rest, OD : OE : 0/^= sin ^ : sin JJ : sin C ; .'. tan ^1 = tan 7i — tan C; and A - B — C. 4'2. According to order in which aides are taken. 43. Com- pf)nents make angles 22^": GTi" with side not rept*. a force, and — a * /l H — j-rr, a being a side. 1/2 Chapter III. — 1. An area; a sq. in. 4. See notes. 5. Second pt., § 35. 7. 12 lbs. 8. Take mts. abt. pt. of attachment ; W tension of string disappears, and force = — —. 9. Take mts. abt. A : 1/3 tension =:/»!/ 3. 10. Mts. a1)t. fixed point. 11. 5 or 2 lbs. accd. .IS fulcrum is J or ^ length of lever from end. 12. IG, and acts at same pt. as 7. 13. Mts. abt. fixed pt. are equal. 14. 5 : l/ 6 • 15. Mts. abt. fulcrum. 10. Incln. of AB to horizon — P.AB + Q.BC coa ABC tafi' J ^ . : :— 7 . 17. Incln. of BC to horizon Q.BOshiABC (P+Q). BC+P.AB con ABO 18. 45" = incln. of rope to hori- tani P.AB am ABC z(Ui. 19. W. 20. If «'- wt. of rod, 17= suspended wt. ; incln. of rod IV + 2 W 21. J length of rod from end. 22. '2a, to horizon -- tafi^ — 2 W 2h, lengths of longer and shorter arms ; w^, w^ their wta. ; required aw, - l)W„ wt. — -. 2o. Let O.Arbe perp. from cr. on AB, IK-wt. 26 ot rod, w= attached wt. Taking mts. abt. cr., reactions of circle 2 W. OiV BC BC disappear, and W. ON sin A =w. j^ AC, w= • OC . diam. AC AC 24. Let 7? :: reaction of wall, T--- tension of string, W- wt. of rod, ('. -. length of string, -- incln ot rod to wall ; taking mts. abt. lit. where rod touches wall, mid/... vt. of rod, and pt. of attachment of string to wall, T a cot 6 -- W. ma sin 9, T {ma - a cot 6) — B. ma cos 6, a ^^. ma sin t - /?. - . — ; whence eliminate T, R, W. sin orT)erp. to it. 25. BD = BC 20. If O i)e cr., dif. of mts. is expressed by 4 x area of OrQ. 28. h V 30 + 12 V 3 x strength of one man. 2't. Let AB = a, BC -- h, angle bet. rod and string = 0, wt. = W. Take mts. abt. A ; press. -2 IF cos. 0— W. y4«'''-//^ + 6l/8a2+62 30, \Yo a V 2 obtain, takin<.? mts. abt. cr. (1) P cos A + Q cos B + li cos C = 0, (2) P sin A t- &c. = ; whence reins. If B = C, B = - Q, k P~0; If ^ =: i? - C, P+ Q + B ■-- 0. CitArTEU IV — 2. ThepoKU, ofcr. is same whatever ])o dim. of forces; let thei.i be perp. to id. of triangle, and take mts. al)t. /iC. 4. Let fees, aci at all the angles; cr. is cr. of lig. ; now icmovo one in force by applying an equal and opposite force, .ami the posn. of the reqd. resultant is obtd. at dis. from cr. of fig. eipuil to J length of rad. side. 5. At dis. from cr. =—7. 9. 1 ft. from end. 10. ^4 n-l II. 20 11)3. ft. from end ; changes ^V f'- 12. 90 lbs. accd. as x,w, +. 13. Accd. as «;,+....< v\ +... +nw ' !«,+...<[ !« + .... 15. Bisection of line joining bisections of opp. sides. 16. It must pass through intersn. of horiz. line througli li, and vert, through cr. rod (Prob. 3, Ch. 11.) 17. §43, and that if two forces counter- l)alance, they are equal and opp. 19. When on side if canted to riglit, cr. of gr. is to left of pt. of contact, and tendency is to resume old posn. 20. Each particle is equivalent to two particles of half its wt. at ends of side, and their cr. of gr. is that of triangle. a , — a , - a 21. Dis. from each ^yMft' + c'-^ri-), &a 22. |/I9, "/is, -1/7. 6 6 23. If cr. of gr. of B and D be at E, tliat of A and C must be there. If neither that of li and D nor of A and C be at E, tliat of the four cannot l)e there. 24. Each particle of the second set may be broken into three ecpial ones in posns. of the particles at whose cr. of gr. it is. 25. For it is cr. of gr. of each set of four. 26. Draw two purp. lines through cr. of gr., and drop perps. on these from the par- ticles. These pe. ps. rept. components of forces, and alg. sum is zero. 27. If O be any p*^., G cr. of gr., A posn. of a particle (w), 2 (xu. OA'^) ■^OCP 2 (i«H2 {w.QA% and 2 {w. OA'-) is 1st. when OU \). 29. If Al>-~-\ AC, result, of OA, OC is 2 OD, and then result, of OB and 2 OD must trisect BD in G spy, and =3 OG. 30. In oas*; of a parallelogram. 31. Indirect proof. 33. Result, of forces passes through cr. of gr., i.e., forces are. ctjual. 36. iv'73, Sv/ l'^> I 37. I'ct 2 a = vert, angle, 9 = angle made l)y line with ))ase, j) = perp. from vertex on base ; then line = ^ p sin a 1 1 ) ;H T7.— , ?■• ^^- By dividing it into two F'^ts of tri- {. -i— I. • (« + «)) , ". on diags. ■.os.{d-a) cos indes we may find two lines on each of which it lies, and tiien- intersn." 39. 2 BC. 40. The intersn. of is cr. of gr., and diags. bisect one another ; .•, tig. is a parallelogram. 41. Sim.' to I'rob. 38. 42. Let K bo the l)isuotioii of JU). If Efj~h EF, cr. of gr. is on a line through L j)arallel to BI> ; if KM -i KE, it is on a line through J/ parallel to AG ; and these lines intersect in KF. 40. The or. of gr. of rods A li, BC, CA is cr. of circle in triangle formed I'y joining their binictions. Similarly all the way up ; .'. cr. of gr. is in lino joining cr. of this '•irele t<. tlu' vertex, and is one-third the way up, since that of each triangular side is so. 47. Mts. al)t. vertex. On line from vcvtex to hisection of l);ise, and dist. from vertex I of this bne. 48, /iJ> 2 I'D. 19 49. J rad. from centre. 50. Dist. from opp. eorncr ofsq. 52, Alt. of triangle '^ .1!^ x aide of sq. 53 ( ^U 2 13 sidi inpare the sectn. with a triangle perp. to edges, and pi'ssing throuul; pt. where sertii. cuts orifice. 54. yl />'(' triangh', O intersn. i>t perps. ; AO may be shewn twice Tterp. from cr. of circle on li(.'. 55. CA I 2a, BC --- 2h; dis =; i/o.*- h* + '2 It' It' COS ACB. 57. Cr. 172 of gr. of perimeter is that of wts. at D, E, F proportional to the sidea, .and hisoctor of FDH ilividos FE in ratio of AC to Ali. i)!). A E, F bisectns. of sides. PD = h AO, &c. Result, of 0.1, OB, OC = result, of 01), OE, 0/'= result, of i AO with OP + h BO with OP + hCO with OP. (30. ABC triangle; D its centreT Take GE — ^ CB, and DO = I DE. G is cr. of gr. Gl. Breali wt. a into h a a,t B, and h a at C, &c. 02. Mts. abt. A. Presoi. passes through pt. where dirn. of IF euts BC. 64. Mts. aht. vert. pi. through C. (50. Mts. abt. vert. pi. through pt. of suspension. 07. The three forces pass through same pt., and may easily be sliewii that angle made liy string with reaction of wall = 00', Lc:, by string with vert. = 30° 70. Shew 2 8in2 C =Ji^!LZli ,2sm'B = 'h^lzL^ 4a2 71. a = angle ma le by strings with one another ; 3 ^= angle made by string carrying wt. with rod ; W = wt. of rod ; w = wt. to string. Then 2 a sin {a + ji)=h sin n ; i« sin /3 (a - c) + (k' sin {'2a + 3). c + '2w cos a sin (a + jS] (ii + c) = o, '2', I80-C. A', B\ C" a are \M. reqd. 74. Sliew^> = • sin A cos [B-C], &c. Chapter V. — Levers. — 1. 1^ ft. from 3. 2. P, (?, 7? pass through a pt., and P, Q are perp. 3. w = suspended wi;., W = wt. of CD, CG a{W+'2Q)-h{P+Q+W-jR) G its cr. of cr. w= W. tan a. 4. ^ AG R from P, where 7? -- press, on jjeg, W — wt. of beam, 2a ^^ length of beam, 1) =:dis. of fulcrum fiom P. 7. Arms are equal. I're.ss. = 4 /¥. 8. Ineln. of JCtohoiiz. = 2 cos^ /' + v/j^ ^ + 2(2i rTllT 2 (2 W + W) • 9. As 1 : V^ ; 120°. 11. n, h the arms; a, /3 inclns. to horiz.; P, Q forces; . AG = P. GX + W. G- -. 22. If P be first power, 23. 25 lbs. 24. As — has greatest effect ^W-P 4 P cos 00' is secoml anil most advantageous tan a : tan «, «, a' being inclns. of pis. i ft wlien G - 0. 20. Let f! be cr. of gr. ; C.'ZJperp. to BC: BE- J {(t + r cos //;. 6'A' := ^j^ sin />, and just as it topples (1BIf cyl. : revolving vert, and lioriz. v + 2 IF cos^ ^ 6 = R cos a, 2 \V cos A sin \ ti = R sin a. A' being reaction of pi. Eliminate R. Srr . 10. nVs. 11. In each case = rad. of pr. cot a . • rad. of cyl. Virtual Vehtritim. — 1. ,30 in. W -_ 5 P, and W x its rise = P x its descent. 2. 1 ft. 4. 0. 5. T.et a displacement (//) be made along the plane. Now /* cos a =r IT .sin a; .'. Ph cos a = Wli sin « ; . •. P X displacement in tlim. of /' =^ It' x displacement in dim. of \V. Simly. for 2nd part. 0. 2nd juirt. Suppose a displacenieiit (//) along pi. ; then by r. /•., /' ■ // cos W — IT x /( sin >, nr /' cds B =^ \V sin o. 7. I., R make angles a, 0, 7 with A B. Then y = .1 B cos j, &c., and ei(. liolds if P. AB cos a + V. .1 // cor* /3 — /'. .4 /; (tos 7 ; if /' cos a •)■ V cos /3 = /? cos 7; §23. ». Until diain. is incd. three times. I'revious wt. is to wt. now supjiorted as sum of radii tn dif. "f radii. 9. j>=:displ'. /• r of/*; .■. that of ir in dim. of U'— ;>. - sin a, . ". /). P=^i>. - sin a. W. ^ R ' ' R 1 10. : ton. II. Make dis[)l'. by letting 0110 string remain 2 V 3 stationary, and other move jiarallel to itself. 12. Let angle made by P with string change from a to a - /•, wluTc k is so small that sin k = k, cos k — I nearly. II' = i tan a /*. m 174 nWt—l. ffV- 2. 9^\; with friction 9^ + 30 V^S 11 3. The work is that of raising J-JJii jr lbs. through 240 + 241J + 242§ + 4788 inches. Ans. 3.20 ... min. Fri'lion. — (Note; the coeflF. of fric. is usually denoted 1)y ft), 1. —=,§96. 2. Fric. = 15xsin(incl. of pl.) = 9. 3. Fric.=/t W. V 3 and W^fx W; .'. n = 1, and angle reqd. is 45°. 4. /[i = tan45° = l ; 10 ,— ,_ R = -71= = 5 V 2 , and fric. =nR — bV 2. 5. From second state- V 2 ment sin a = J, cos o = ^ ; and iZ = 10 x ^ = 8, 3 + /« i? = 10 x | = C, , — W cos (a + 6) . • . press. -= v//e^ + fx' K^ - 1/73. 6. R= ^ when body cos ti-n sin ..,. , ircos(o + 0) is just sliding down, and= ^ — when body is just sliding cos Q + fi sin Q up, and when motion is about to take place fx is greatest. 7. /» =-7=; R= ircosSO'-z: W.}L^; fric. = fi R = \ W ; force V 3 2 reqd. = W cos 60° + ^ /^ = i W+\ W= W. \ ^V^ - V2 . , .. 1 + v^ ^^ , ami a=ttaii' ::_- v's - I V ^-\ Take vert, and horiz. components, and mts. abt. an end. 10. (i, I) distances from lower und upper end of cr. of gr. ; iucln. = tan' a - h u a' W sin 0- P cos i) ^ 11. i 12. 1. 13. tano = /t — r. 8. I.e. find ^. = - V^ 9. or. 12 ft. /t (a^-h) ircose + i-'sinf) P /I \ = a + tan' — • 14. tan' I - + 2 tan a I, where a is the inclina- W \n / ,- i/¥ tion of the rod. 15. 3000 V 3 ; — ^-; 10 / 3 . 16. PI. smooth, cos (9+ r) cos (9 + a) 1 reaction =- W ~ — ; pi. rough, = W ; — • , cos cos y 1 + ju tan W cos (0 + a) 1 , — , — 17. i«i +/«, 1/3=1/3 - 1. or W cos 18. /ii + 2/i., = l. \ - \x tan y 20. 90 -o. 22. Without fric. T^ with fric. , = -^ f — - - \ tan IS"") . Thrust of leg : 2 \r/ Q / 21/1 1/3" 23. cos U (0 + £) = — cos (, where is inclu. of string to vertical. 24. = angle l)et. rod and line of greatest slope, a — incln. of ring to hori?;, , then sin =- At cot o. 26. /n iraud W, wliere iriswt. c tan' u of cyl. 29. \- ir, a - (• tan' fi DYNAMICS. CHAFreR I._L 5 rah. perhr. east 4. A second. 5. ^J second ; 85J. (CO)-' 2. 38-8 .... 3. I n 60a " 6- -- ft. 7. 2-19 ft. 8. 38400. 10. If X be the acceln. when the units in 9. — / wh. M, <.. T are^expresse.1 are the units of space and time; then /_ ___, F.= L± . 12. .^j. 13. 64. j4 p^^^ ^^^^^ ^^ rest in 10 units of time, space descrd hprnw r^nn t+ l i>ath, and in 10 units of 'tiLe^st UsiiS^Ll'- I^:iZZ\t and space descrd. increase indefinitely. 15.400. \G. _ " + ^ \1. v=. V 2fs , and . • . equal vels. are not generated in equal spaces. 18. . represents dis. from initial pt., and after time " body is retrac- ing its path. 19. §18. 21. ^ " ^ /jL!ig2lLz:Cliizl' 22. i («'-«)<. §9. 24. 16": 1. ^'- ~^- 28. 6. 29. 56-85 ft the veb.. and dr tLromrt' f o." • . ' ^'"'"^^ ''"''^ proportional to 5. 14G7- 5 (i oo"f ' f^r "»t«rsn. represents resultant vel. ^i'< .'.^.... o. —J ml. below opjjo. i)t 7 'i'i° S' c r> ■ one b<>. ly to rest impressing on bcJtl a^4l. .Juaf and opp ?„T fi /A'l. ' " ''T'r ■"' '"■■"^' '^ *" ^*^«t by impreiiTK. v" K mi both • /^y^ the present dirn. of Ji^s motion, C the Ltorsn'! oftlie d/rns of motion at first ; then ^" = < ^^' ^ V"-2 V V cos a)i no.- , sin DBA ., -, and DBA IS kno\vu, Zf.^! C Ijeing so. p-q 25. 10; 5 ft. 2(;>- • l'*^;« ^•^'P'-cssed by JC; acceln. by. •J2'L'; • 1K~ m X .ii J; ,.,.., ,;, i,s the number of times 322 is contarned 'in iF- . . unit ot mass is nuiss of .32-2 lbs T W,^^ ff r.o V V ; second, if ,/=-3.> 7 , > /■'''.'•x ;'— ^'•i*'^ "f 0% lbs. 0. 8 P \L \\^h,~'c. ,"•' "'»ts of space and time ; .v = .32-2/2 8. y 11^.. province an acceln. m I'of/y feet per second ODamlsH equal to that of m x Ct2'5 lbs. But P produces on a mass equal to tliat of J* lbs. an acceln. " nearly. VI X 62' 5 10. i-tg-t^ = i9; ve\. = *sa-t. n m Ifi 11. 4:3. g-t. 13. 1:2; 2:1. 14. Hg. 15. 8- /■> 96 12. h--0-^\ m = 5 lbs. ntarly. 1. (1) By Ist law motn. i.s uniform and in a st. line. (2) By 2nd law, since moving force varies and mass is const., acceln. varies. (3) Acceln. same for all (2nd law). 2. By 2nd law. Fri'c RtctUumd Motion.— \. Constant. § 8, Cor. 2; § 9, Cor. 2. o .,- — o 2p^,v;=2p^: '-^--('•,-'-,)=/K-^)-A ka 'A. ^it 5. <=V— ! — \'-, i=-\/-2.,a-V2y'a: 6. VeL=40; 16 " ij ij acccl. = 20 and 72000. 7. 3". 8. 32:32-2; 2i;\. 9. 32-2; f)". 10. 26tS ([/ = 32). 11- 8"; lOlw. 12. 82,-^ ft. 13. ,"5 length ir.o of line from top. 14. 1 : 2. 16. — ~ lbs. 17. 3022-5 ft. / 2m. 18. Time through m"". .-^pace Motion on Inclined Plaueif. — 1. V^4//. 3. '•= 2(»t— 1) sm' (vel.) 2cj X length 4. iVY; Jsn^-: ; j 2{a + lV-2) + ^ (I ^ .'/ __^~_ seconds. Cravity would vary seusildy for dif. heights. IJ sin- a. [I 100 V^ 10 , 2"v/ — 50. l.enyth 2 V 10 5. sinM|.^2--l.) w' . ''/In of strine =^ -7—; ; time of ascent = — -. — ( 1 r= 1 ; of A[i sm a ij sin a\ y/ .» / 8. On perp. cm descent = • 1 : ; reqd. time : iu'iiiiig tlu' givuii line and tliu given circle at its higlicst jit. Line tliniuuh jit. of contact with given line and liighcHt I't. of given circle ii< rei|il. line. Make use of the follov.ing prohleni. I'J. P<^> = {AC — y\Jt) cos QAC; accel. down /\> — + '■')• 1<>- (f^i'i " — /' .'/ CO.S a) ;/. 17. Accel, down pi. =^ = t.^; acccl. down AP^=r (I sin a ■fi) Tin 10 doMn AP-- s. •-' AP cos -3)1 I 'J A<,> —^-.■[ ; down^H.> = J— .-^ which are equal it' A I' cos (a — n)r=Ai^, tion goi ;/ sin (o — t) A(JP = W . ij sin (0+ £) '20. Retardation going up — ■• ; acccl. coming down cos t cos t ProjcrtiU.i. — 1. O.v. 2. ^T. Resistance of air. 3. C'us 2 a:] — sin 2 a. o. \"ert. and and hi'iiz. distances are '(i and 'N ft. (i. It ha.s moved vertically downwaids 12^ , y>C' the horiz. and vert. vels. ; draw .^1 /> making given angle with AC, cutting ^if produced if necessary in J) : JiJ) is re(id. vert. vel. !). Pts. where it would occur are lielow horizoi ii. IH. 1 '-' .'/ 1 1. Horiz. dis. bet. them at tin le ' = (r' — /•) cos a ■ f ; vert. dis. = (/•' — r) sin a ■ f ; .'. tan (incln. of line joining tlieni to horiz.) — tan n. 12. 'lime of /-'"s Au . A /I coming over A = — ■ : tlien A is at lit. = *■ COS J / AB \^ I I ; and /)' is at lit. \r cos 0/ =- (• sui n r cos (I and these must be equal ; . •. sin a ^:::r cos j3 An / An r cos o 1 3. .', (/ /- — »• sin a • .'. \»' cos n/ where f = 16 r' sin (a+J) 14. 23. tan/inclu. of line joinini,' them 31 ■ i I 178 to horiz.) /) sin o — (' sill II — ;, ivnd is const. 24. (vel.)- =: t ./ r cos a — (; cos o , " cos n X rad. 25. Is equal to — - (tan W — tin ^). If < be time .'/ (' sin a — 7 / to /', T time m arcs, tangents of inclinations are — — '■ — > II cos a (• sin a — 7 (t-\-T) n sin a.7 1/ fi T) , , . , — , , &c. 27. A highest \}t, r cos I /' cos (I B, C pts. of section of tangent with horizontal and vertical lines, (•-' sill- II M pt. in horizon vertically Ijelow A ; C M = r sin a ' t II r cos II — i J l'> and C is uniformly acceld. ; B A = - ■ (c sin 1 — ijt), - .'/ 2 (■■-' and B moves uniformly. 29. tan 10' sec 10'. 30. In both .'/ 2 M- cos a sin ((» — (i) cases = • 31. Let o = inclination of plane to .7 cos- y horizon, ii = angle dim. of projection makes with plane ; then range If cos' a 90, 7— \ sin (a -f- 2 /3) — sin n L wh. is gst. wlien a + 2 j3 i or /3 = J (90 — a). 33. «<;('*/ " ' — 34. Let \r/ sill a t be time of reaching deck; then range on deck ^^ « eoj u.t, i being given by sin2(a4-0)— sin2a — A = y sm « • < — 4 y <■'• 37. 1 ruf" dist. is a — ^ — — r-— - — 2 cos^ (a -|- y) cos 2 a „ V ^. '^''' ■= a • W, approx. 40. Dis. = — ■ cos'^ a 2 2 7^ /__ (P-[-QY accel. of Q =■■ 10. If y be the tension Pij - T ^= accel. of P = T— O7 2 /'(^ —,.-.T= ~-fj. 13. KeepP + (^ constant and vary /* — Q, and observe distances for same times. 1.,'/ 7 ; 2yV li>3' 14. 4 P = mass of string. When part is cut oft 2 Pij— T T—P'i = accel, of longer part = that of shorter = — ; n^^ 179 4 P,i ■'■ ^ ~ .1 ^ -Ili'I l)r(;s.siiic' oil i,eu ~ ^ ^''.1 _ •< l^O - I,). r/,/ ,,„y „ _ /^ — (? sm tt -^ T^-fy ' « '"i"'-' i"^'li"- "f i-l, Hi. Th,. v.I.s. a.v -^ ' - ,,t, sT^Tr''''" '«■ ^'-^ -'••■ "t -. -'f gv. .. f , (sin „ --M ,^r, h„n., aeccl =J , (sin „ - si„ ^j (co. a + .:,. ^). •'' ,r ■ ^^'^^'1' tllc■IMltidu.sl)ec•<.nu:a.■nntin- '.'/. UOUS .strin;; (vtl, )■' wIk 11 //Ih I,,ivc.s --.= ,y " . i'l" tJ.^ •20 ,, = lit „f wcl.^e, /> = its l.M.c, ,. = "v„l. „f l,,|| V - V..I of -'^ ' , \i'it. \cl. nil ivacliuiK or„u,„l — ,/'2.ry i .-. ''-A/?' ^'"'^ toginini.l^ /-". Ho;v. .listanc -l^^crd. by hall == /:^' . ,":/==,, ., j- ^^ + .,-,-" ^ li' *• "l"!:,: • "i'l'' t" K"- ^-t'l. '•■ -.. If the balls be nf e.i lal mass, an.l elasticity pcifeet. c. ;io^ 7. j . 4. /;,, _ {A + i>') ,r = J ,r + y,.,,'^ to eliminate A and />' S. n = ^^'^''1 . _Ax-~Cv ^1 + /^ "-"T+C ' ^"^''■"•i"--ito.l and.. 9. Communicate.lto the earth. The" rise ('^v ^Oi,.iii,< i.f 1 i i -" 'H^, use .. .s Ixnnu' ht. (lesccnded. 10. --- , a being Ills, bet. LTII. Ills, , and 11 ;• I'n v<.!u II Ti 11 .• ' Ml ., aim H, I ^11. \eJs. II. \\,^\\ contmias to reI)ound Or/ /!?; and (1 (2- / o , seconds. ,4. Distances are a, 2n,^ 2rt(^ &c.; time:...;-", 1> ;-" o, . /^" „ ,. „ \'„ ' \i :. ' -' \; -' &<-■• ii). Jfaii'.esaie A" 'TV.' '■v., «'ii 29, sinL'O- dencc) = sin '20 V V li (l -{-,') ^^' 17. tan (angle of inci- 18. First angle of incidence = tan' l+e + , 22. Cull us. (iccnr at distci es. --- X Qee from another, and balls will come tog. at A wh u+r one en mult, of '. + V XOc 180 ^ mult, of Qcc. 25. tan »„ + ,== -^ „ tan « , ; r^^ ^ ^ — ,. ^ . , «. cos 9^ 1 sin 0, cos n + \ I" .sill H+l -= r, sin «, , ultly. •_'!». Vi'l. of angle tun* - - - - with dirn. of /I'.s motion. .SO. vl moves ofl' A (1 + ') p(ir\}. to line joining crs. with vel. n —-' - ; />' niovi'S along this line M'lth vel. -• T tl K R N n. ropp, fi.MiK t <(i.. I'lUNTKRS, roi.Hdnxr -iTni'DT, niiinvT' rlTi \v 6. ^ u w SI IMAGE EVALUATION TEST TARGET (MT-3) /. 1.0 I.I "IIIIIM IM • 1112 iu 2.2 1.8 L25 1.4 1.6 ^ 6" _ ► V] <^ '3 "^3 A ^/,. /A '/ Photographic Sciences Corporation 33 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4S03 W 4^.1 Q^ iB cot-f, Clark * ci. Lir^. toHjrit).