Preview, Review Summary - University Physics I Dr. Hulan E. Jack Jr U 7. Newton's Laws of Motion

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1 56 Preview, Review Summry - University Physics I U 7. Newtn's Lws f Mtin Objectives T lern vi experiment nd thery 1. fr bth Liner nd Rttinl mtins 2. the nture f rces nd Trques 3. the nture nd rles f mss nd mment f inerti. 4. t drw nd use ree Bdy Digrms (BD). THE EVENT Hit the stick. D Hit the Stick I nd Hit the Stick II, belw - experience nd discver frce, trque, mss nd mment f inerti (See Sectin 7.3). Overview Our humn mechnicl experiences primrily invlve interctins sscited with mtin, mre specificlly, chnging mtin. In wlking nd running, we strt frm rest, speed up then slw dwn nd stp. We trip nd try t stp ur fll. We hit nd push t strt r stp the mtin f bjects. We thrw nd ctch. We re lwys experiencing net externl frces nd trques. When we re still, if we re cmfrtble then ur thughts turn t ther things. If we re nt cmfrtble, then ur thughts becme cncerned with wht net externl frces nd trques we cn pply t chnge the situtin. Becuse f this we humns re very strngly Newtn s 2 nd Lw cretures. llwing this led, here we will crry ut the fllwing scheme. irst we will briefly stte nd discuss in their usul rder. Then we will cver ech in detil in reverse rder. T prperly use Newtn s 2 nd lw requires the 3 rd Lw Pint Bdies nd Extended Bdies. We need t immeditely distinguish between types f bdies. 1. the pint bdy It hs n sptil extent. Its length lng ll three sptil directins is zer (L=0, fr ll directins). 2. the extended bdy This hs length greter thn zer ( L>0) lng t lest ne sptil dimensin. Then, there re tw types f extended bdies, 2. rigid bdies These re defined s thse bdies fr which the reltin between pints n, r within, them re invrint. Such bdies cnnt defrm in ny wy - cn nt stretch, sher, r chnge shpe r vlume. 2b. nn-rigid bdies These bdies re ble t defrm Initil Sttement f.

2 57 Newtn s 1 st Lw f Mtin: A bdy subjected t n net externl frce mintins cnstnt mtin - it mves lng stright line t cnstnt speed. This gives trnsltinl equilibrium, ls clled equilibrium f the 1st kind ; Σ = 0. An extended bdy subjected t n net externl trque remins in cnstnt rttinl mtin - it rttes cnstnt ngulr velcity. This gives rttinl equilibrium, ls clled equilibrium f the 2 nd kind; Σt = 0. Newtn s 2 nd Lw f Mtin r Trnsltin: A bdy f mss m subjected t net externl frce experiences n ccelertin given s Σ = m. r Rttin: An extended bdy f Mment f Inerti I but the xis thrugh O subjected t net externl trque but xis O experiences n ngulr ccelertin given s Στ = I α. Newtn s 3 rd Lw f Mtin: r every ctin, there is n equl nd ppsite rectin. 7.2 Newtn s 3rd Lw f Mtin: r every ctin, there is n equl nd ppsite rectin. This llws us t drw free bdy digrms f bdies which re criticl fr slving mny prblems. But, t first sight this is mysterius sttement. The mystery, hwever, clers under clse exmintin. It is sttement f pint f view - wh sees wht. The pictures belw illustrte the mening f this sttement. The Actin-Rectin Pirs Anntted re the key. Let s lk t the T 3 's. The weights pint f view is tht the jint pulls it up. But, the jint s pint f view is tht the weight is pulling it dwn. Similrly fr the Ceiling CEILING T 1 Jint w Hnging Weight T 1 T 2 T 2 Brek ll physicl cnnectins nd T 3 replce them with T 3 ctin-rectin pirs w w Actin-Rectin Pirs T T - 2 JOINT pulls STRING 2 DOWN JOINT pulls STRING 1 DOWN 1 T 1 STRING 2 pulls JOINT UP STRING 1 pulls JOINT UP θ 1 θ 2 -T 3 T 3 W T 2 - WEIGHT pulls JOINT DOWN JOINT pulls STRING 3 UP USING NEWTON'S 3rd LAW O MOTION, REPLACE ALL THE BROKEN CONNECTIONS WITH ACTION_REACTION PAIRS Actin-Rectin Pirs Anntted T1 ree Bdy Digrm (BD) f the Jint ree Bdy Digrm (BD) f Weight T 1 T 2 T 2 T 3 w w T 3 ree Bdy Digrms

3 58 ceiling-jint views fr the T 1 pir nd the T 2 pir. In the middle picture the pirs re given their true nmes (e.g., T 1,-T 1 ). In the ther digrms the - is left ut becuse the directin n the digrm tkes cre f the Newtn s 2nd Lw f Mtin: Get smth flt wd r plstic ruler tht is unifrm lng its length. Unifrm lng its length mens if yu were cut it int mny pieces prllel t the lterl xis, the pieces wuld ll lk like. ind it nd mrk the lterl line where the ruler blnces n yur finger s shwn in the picture t the left. Lter we will find tht this blnce pint is clled the lngitudinl center f mss. lngitudinl xis lter l xis unifrm ruler ruler yur finger ruler yur finger Then ly it n smth flt level surfce, like very, very, very clen wd, plstic r glss tble tp, desk tp r flr. It is very imprtnt t limit the effects f frictin s much s pssible. Clen ll surfces with mmni, rther thn sp. Sp leves residue tht cn be hrd t get ff. Then plish with dry clen clth r hndy wipe. With yur finger, hit the ruler t its blnce pint with quick, shrp blw. Then hit it gin with the sme frce but t tw different distnces, r 1 nd r 2, frm the blnce pint. Ntice the mtin f the ruler s it mves lng the smth flt level surfce. The figures () nd (b), belw, illustrte the situtin. The right, ig. (b), hs lrger frce, but the sme tw vlues f r 1 nd r 2. Hit the Stick I liner ccelertin liner ccelertin r1 α 1 ngulr ccelertin θ 1 r 1 ngulr ccelertin ngulr displcement θ 2 r 2 α 2 r 2 () (b)

4 59 Hit the Stick II Nw let s tke tw equl msses f cly nd stick them n the ruler nd hit the ruler gin s shwn belw. liner ccelertin liner ccelertin liner ccelertin r 1 α1 ngulr ccelertin θ 1 r1 α 1 ngulr ccelertin θ 1 r 1 α 1 ngulr ccelertin θ 1 ngulr displcement α 2 θ 2 α 2 θ 2 α 2 θ 2 r 2 r 2 r 2 () (b) (c) Observtins nd Reflectins: We bserve tw mjr effects. irst, the frce cting prllel t the lterl xis cuses the ruler t mve lng stright line prllel t the lterl xis - trnsltinl mtin. When the frce cts t the center f mss, this is the nly effect. Secndly, when the frce cts t lngitudinl distnce, displcement perpendiculr t the frce, frm the center f mss, the ruler hs the sme trnsltinl mtin, but nw it ls rttes. As this lngitudinl distnce, displcement perpendiculr t the frce, increses (r 2 > r 1 ), rttinl mtin increses, but the trnsltinl mtin stys the sme. This new entity tht cuses the rttin is clled trque.(sectin 7.3.3) Wht did yu bserve? 1. Wht type f mtin did frces ffect? 2. Wht type f mtin did trques ffect? 3. Wht hppens when the fixed frce remins fixed s the mss deceses? Increses? 4. Wht hppens s the frce n fixed mss increses? Decreses? 5. Wht hppens when the frce nd the bdy remin fixed s the displcement f its pplictin, r, increse? Decreses? 6. Wht hppens when the trque nd the mss f the bdy remins fixed while the cly pieces re mved t greter displcement frm the center?

5 60 But wht re these tw physicl entities? Wht re they? Wht d they relly d? We will discver tht frces re simple entities. They re push r pull tht cn be reduced t 1-dimensinl vectr. Trque, n the ther hnd, re much mre cmplicted in tht they require full 3-dimensins t describe r represent them. Let s explre these issues, first fr frces, then fr trques rces nd Newtn s 2 nd Lw fr Trnsltinl : A bdy f mss m subjected t net externl frce experiences n ccelertin given s The bve figure depicts the behvir f tw bdies f different msses subjected t the sme net externl frce. They strt frm rest, v = 0. The bdies re shwn t their respective displcements t equl time intervls. The clck bve the bdy t ech psitin registers the time. The frce, ccelertin nd velcity vectrs re n ech the bdy t ech psitin. Everything, displcements nd vectrs, is t scle. Crefully nte the sizes f the displcement, frce, ccelertin nd velcity t ech psitin. The bve picture grphiclly illustrtes the =m reltinship. If either the resisting mss (inerti) m -> m/2, r the cusing frces -> 2, the resulting respnse, the ccelertin, dubles, nd s n.

6 Wht but when m-> 2m, r -> /2? 61 The bve picture grphiclly illustrtes the directinl nture f =m. Nte the directinl reltinship between the cusing frce nd the resulting respnse, the ccelertin, A frce,, is simply push r pull. It is directinl, s it is vectr. It ffects trnsltinl (liner) mtin. All liner mtin cn be reduced t mtin n ne-dimensinl stright r curved line. Sme curved lines, s yu will lern in yur lter trvels in clculus, cn be represented in curviliner crdinte systems. r exmple, n the sphericl erth, line f cnstnt lngitude, r ny circle n the sphere tht hs the sme rdius s the sphere, is gedesic stright line. rces in Eucliden system cn be represented s ne-dimensinl vectr by using n pprprite crdinte system. In the picture, the vectr A is 2-D in the x-y crdinte system, but 1-D in the x -y system. Inerti is the tendency f bdy t resist chnge in its liner mtin. The mss, given the symbl m, is mesure f inerti. Mss hs n directinl effects. It is sclr. Mss is fundmentl dimensin Memry Aids The Tringle = m, = /m, m = /. y A x y A x A mnemnic = m -----> m (sy it)! Cuse nd Effect In the Newtnin view, whenever there is n ccelertin there must be net externl frce ext cusing it. Tht is, frces cuse ccelertins. Accelertin is respnse t frce. Accelertins d nt cuse frces!!! Exmples: 1 A bdy f mss m = 4 kg experiencing net externl frce = 24 N Est. Wht is its ccelertin?

7 62 = m. S, = /m = 24 N Est / 4 kg = (24/6) m/s 2 Est = 4 m/s 2 Est 2 A bdy f mss m = 5 kg is underging n ccelertin = 10 m/s 2 45 Suth f Est. Wht is the cusing frce? =m, = 5 kg x 10 m/s 2 45 Suth f Est = (5x10) (kg m/s 2 ) 45 Suth f Est = 50 N 45 Suth f Est Exercise 1. Get severl bjects f widely vrying msses (bserved by wy f their weight). Plce them n smth tble r flr. Using but the sme frce, hit ech bject in mny different directins. Recrd the munt f frce nd its directin, the mss f the bject nd its respnse fr ech hit. Levels f rces: A very smll frce might be ne where nly the fingers f the hnd mve s hrd s yu cn - the hnd itself des nt mve t the wrist. A lrger frce might be wrist slp - the whle hnd mves s hrd s yu cn t the wrist. The next level f frce might be where the frerm mves t the elbw s hrd s yu cn - n mvement f the upper rm except fr rttin. In ech cse s hrd s yu cn will be mre reprducible thn t d it sftly, then little hrder, nd s n. Als shw pictures f sme f yur exmples. Smple Dt Tble Object nd its mss rce Respnse Mgnitude Directin Mgnitude Directin cn f sup 16 Hrd t the left quickly mved 12" t the left 2. Clculte the ccelertin when bdy f mss m = 2 kg is subjected t frce = 10 N 45 N f E Trques nd Newtn s 2 nd Lw fr Rttin Trques but the xis thrugh the pint, t Trque cuses chnges in rttinl mtin. As we hve seen in the previus chpter n mtin, rttinl mtin is full three-dimensinl entity. When the bserved mtin is in plne, the mtin is referenced s being rund n xis perpendiculr t tht plne. Hence rttinl mtin is described by the vectr r crss prduct. Trque cuses the fully 3 dimensinl rttinl mtin described by the vectr prduct, then trque itself is full 3-Dimensinl entity described by the vectr prduct. It is defined s fllws. t = r x *t *= t = r (sinθ) = r,

8 perpendiculr frce = sinθ. 63 = (rsinθ) = r, perpendiculr distnce r = rsinθ, The trque vectr, t, is perpendiculr t the plne f the pper nd psses thugh pint. = sinθ is the cmpnent f the frce tht is perpendiculr t the vectr r. r = rsinθ is the shrtest nd perpendiculr distnce frm the line f ctin f the frce t the pint O cunter clckwise (ccw). Sme exmples fllw. The pint O is ften clled the fulcrum. The distnce r is ften clled the lever rm r the mment rm. Trques but the xis O cn tend mke bdy rtte but tht xis clckwise (cw) r τ = 2.5 Nm ccw τ = 2.5 Nm cc τ = 4.33 Nm cc τ = 4.33 Nm ccw Exercise: Experience Trque

9 64 As yu mve the ld, the sck r keys, recrd the distnce x nd yur bservtins - wht d yu feel. Hw hrd r esy is it fr yu t supprt this situtin. Use severl scks f different msses. Smple Dt Tble Ld - Item Ld Weight x Trque yur experience Ptt Sck 5 p 1 ft 5p x 1f = 5 pf esy t supprt Ptt sck 5 p 3 ft 5p x 3f = 15 pf lmst brke my wrist - culdn t hndle it! Newtn's 2nd Lw fr Rttin St O = I O α τ= Σr i i α - i i m i m i+1 r r m i+1 m i m i-1 -(i;i+1) = m pushes m m i m i-1 i i+1 (i;i+1) m pulls m frwrd i+1 i i -(i-1;i) m pushes mbck i-1 i bck (i-1;i) m ipulls m i-1 frwrd i (i;i+1)-(i-1;i) = m ii = mr iα i Let s imgine rd s shwn bve. We cn brek the rd int mss elements m i. The mss elements interct thrugh intertmic frces s illustrted in the encircled inset. Nw let s lk t the tp mss element m i tht is distnce r io bve the pint f rttin O. This element is cted n by the net effect frce i nd underges the respnse ccelertin i = αr io, where α is the ngulr ccelertin f the whle rd. In rigid bdy every piece f the bdy hs the sme ngulr mtin (See Sec ). S, the mss elements m i is cted n by the trque τ i but the pint O given by τ i = i r i.

10 Using Newtn s 2 nd Lw fr Trnsltin = m, this becmes τ i = m i i r i. 65 Since the reltinship between liner ccelertin nd ngulr ccelertin α is = αr, we get τ i = m i α r i r i = m i r i 2 α. Summing ver ll f the mss elements gives Σ τ i = Σ m i r i 2 α = ( Σ m i r i 2 ) α. By Newtn s 3 rd Lw, this sum f internl trques must equl the net externl trque Στ ext = xr. S, Στ ext = ( Σ m i r i 2 ) α = I α, where I O is clled the mment f inerti but the pint O, r but the xis perpendiculr t the plne thrugh O, nd is defined s I = Σ m i r i 2. Nte tht I O depends n the msses nd their distributin. Nw let s lk t the mment f inerti. 7.4 Mment f Inerti but. I = S m i r i 2. [ML 2 ] I O depends the msses nd their distributin but the xis thrugh O. It hs the physicl dimensins [ML 2 ] nd IS unit kg m 2. Nw let s lk t sme very simple exmples Exmples: Exercises: 1. Verify the bve vlues fr I. 2. Shw tht if r dubles then I increses by fur times. If r triples, I becmes 9 times lrger.

11 Experience Mment f Inerti I Smple Dt Tble 1. Unscrew brm stick frm brm. 2. ind the center f mss f the stick by finding the pint t which it blnces. 3. Mrk this pint. Mrk ff 15 cm, 30 cm nd 45 cm n bth sides f the stick. 4. Tie equl weights t the tw 15 cm mrks ne n ech side f the center. The weights might be tw 5 p sck f pttes, tw 1 p bxes f sugr. 5. Hld the lded stick try gently rtting by few degrees. Recrd yur feelings. 6. Mve the weights t the tw 30 cm mrks. D 5 gin. 7. Mve the weights t the tw 45 cm mrks. D 5 gin. 8. Use tw new weights tht re equl t ech ther but different frm the first set. D 5 t 7 gin. Item nd weight psitin I = Em i r i 2 cmment ptt sck 5 p = 2.27 kg sugr 1 p= kg 30 cm =0.3m 15 cm = 0.15m 2x(2.27kgx(0.3m) 2 = very hrd t rtte kg m 2 2x(0.453kgx(0.15m) 2 very, very esy t rtte = kg m Mments f Inerti f sme cmmn bdies but their center f mss UNIORM ROD UNIORM HOOP ll mss n uter rim UNIORM DISK UNIORM SPHERE L L/2 R R R In the bve picture the mss M nd mment f inerti I O They re ll given in terms f their mss nd effective rdius, R, nd L/2 fr the rd. I f the disk nd sphere re given in terms f mss density ρ. r generliztin, I cn be written s I = βmr 2, where β is mss distributin prmeter. The vlue f β pprches 0 s the mss becmes centered t the xis O. The mximum vlue is β =1. The vlues f β fr the bve bdies re: β = 1/3 fr rd when L/2 is treted s R, β = 1 fr hp, β = ½ fr disk, nd β = 2/5 fr sphere.

12 67 Mment f Inerti R r L/2 = M = Rd Sphere Disk Hp 1.00 ' m 0.10 ' kg ' kg*m^ ' kg*m^ ' kg*m^ ' kg*m^ ' kg ' kg ' m 0.10 ' kg ' kg ' kg ' m 0.10 ' kg ' kg ' kg Exercises: 1. Verify in detil, including units, vlues f I in the bve tble. 2. Shw tht fr disk f the sme mteril if the rdius R is reduced t ½ f its riginl vlue tht the mment f inerti I O becmes 1/8 f its riginl vlue. If R is reduce t 1/3 it riginl vlue, I becmes 1/27 s lrge. Hw nd why might this be f interest in hrd drives in prtble cmputers? 7.5 1st Lw: A bdy subjected t n net externl frce mintins cnstnt mtins - mves lng stright line t cnstnt speed. This gives trnsltinl equilibrium, ls clled equilibrium f the 1st kind ; Σ = 0. A bdy subjected t n net externl trques remins in cnstnt rttinl mtin - mves with cnstnt ngulr velcity. This gives rttinl equilibrium, ls clled equilibrium f the 2 nd kind; Σt 0 = 0. We cn nly discver the mening f these ides thrugh Newtn s 2 nd Lw f Mtin urther, we cn slve nly the simplest definitinl prblems withut Newtn s 3 rd Lw f Mtin. S we must mke jurney thrugh ll three lws befre we cn mke prcticl uses f frces nd trques. 7.6 Sme Definitins nd Ntins rce [ML/T 2 ] - is push r pull. It's vectr. The dimensins ML/T 2 fllws frm =m. Weight W is the frce due t grvity nd lwys cts dwnwrd. It defines dwn. W = mg, where g is the dwnwrd ccelertin due t grvity. g = 32 f/s 2 = 9.8 m/s 2 n the surfce f the erth. Newtn <-> Pund Cnversin W = mg ; S 1kg x 9.8 m/s 2 = 9.81 N. Thus 1 kg weighs 9.8 N. But, n erth, 1 kg mss is mesured t weigh p. 1 kg p 1N= 1 N( ) (2.06) = p, thus 1p= N N kg

13 Prblem fr Trnsltin. The prblem n the left is fr sttic equilibrium. The ne n the right is the sme system but nw in n ccelerting elevtr. Crefully ntice the very smll, but physiclly very significnt differences between the tw. Stte the ill in the Principles Detils +^Σ v =0; T 1 sinθ 1 +T 2 sinθ 2 - T 3 = 0 verticl directin +>Σ h = 0; -T 1 csθ 1 + T 2 csθ 2 = 0 hrizntl directin +^Σ v =0; T 3 - mg = 0 verticl directin After sme lgebr we get T 1 = mgcs2 2 /sin(θ 1 +θ 2 ) T 2 = mgcs2 1 /sin(θ 1 +θ 2 ). NOTES: 1. Ntice the lyut. Next t ech ree Bdy Digrm(BD), Stte the Principles, then fill in the detils pprprite t tht bdy. This lyut helps yu t keep trck f things. r ech frce n the BD there shuld be ne entry in the detils Stte the ill in the Principles Detils +^Σ v =m jint ; T 1 sinθ 1 +T 2 sinθ 2 -T 3 = 0 m jint = 0 becuse m jint =0. +>Σ h = 0; -T 1 csθ 1 + T 2 csθ 2 = 0 n ccelertin in the hrizntl directin +^Σ v =m; T 3 - mg = m verticl directin After sme lgebr we get T 1 = m(g+) csθ 2 /sin(θ 1 +θ 2 ) T 2 = m(g+)csθ 1 /sin(θ 1 +θ 2 ). NOTES- Cntinued 2. Alwys be sure t put yur ssumed + directin in the Principles nd stick t it in the Detils. 3. The grey writing is the sme fr bth.. Ntice tht the nly difference between the tw prblems is the ccelertin f the elevtr, Prble m Newtn s Lws f Mtin fr Rttin Thred is wrpped rund the rim f spl. The spl hs mss M, rdius R nd mment f inerti but O, I O = $MR 2, where $ is cnstnt. The xis f rttin, O, is fixed. A mss m is ttched t the end f the thred s shwn in the figure. ind the ngulr ccelertin, ", in terms f M, m nd R. The thred des nt slip..

14 Drw the ree Bdy Stte the ill in Digrms f the wheel Physicl the Detils nd the mss. 30 pints Principles 30 pints 10 pints 69 Στ O = I O α; TR = βmr 2 α, (1) +dwnσ = m ; mg - T = m, (2) n slipping = Rα. (3) THE PHYSICS IS DONE! THE MATHEMATICS b. ind " in terms f m, M, R nd g. 30 pints Dividing (1) by R gives T = βmrα. (4) Rewriting (2) gives mg - T = m. (5) Add (4) nd (5) t eliminte T gives : mg = MRα + m. Nw substitute (3) in the bve gives, mg = βmrα + mrα = (βm+m)r α. mg g S, α = = (6) ( βm + mr ) (1 + β( M / m)) R dividingthrughbym c. ind the tensin in the thred T nd the liner ccelertin f mss m. mg rm (4) we get T s T = βmrα = β( M / m) (7) (1 + ( β M )/ m ) = Rα = g (1 + β( M / m)) rm (3) we get s (8) Discuss the behvir f α, nd T s β nd [M/m] vry. 8. CONTACT (SURACE) RICTION

15 Cntct (Surfce) rictin is frce between tw surfces. As with ll frictin, it lwys cts ppsite t the directin f ctul r impending mtin, ppsite t v. See Dissiptive rces belw. Sttic rictin, s is the frictin frce cting befre mtin strts. Its vlue is equl t r less thn the ppsing ctive frces nd 0 < s < µ s N, where µ s is clled the cefficient f sttic frictin. N is the nrml frce, it's perpendiculr t the surfces in cntct. Kinetic rictin, k, tkes ver nce the surfce mve reltive t ech ther. k = µ k N, where µ k is the cefficient f kinetic fictin. µ k < µ s. Sme Apprximte vlues MATERIALS sttic µ s kinetic µ k steel n steel brss n steel glss n glss cpper n glss tefln n tefln rubber n dry cncrete rubber n wet cncrete The NORMAL is tht frce which is lwys perpendiculr t the surfces t the pint f cntct. 8.2 Exercises: Shw tht the ngle f the inclined plne θ is ls the ngle between the weight mg nd its cmpnent (mgcsθ) perpendiculr t the inclined plne s shwn in the digrm DO THE RICTION SLIDE (It s crny but it mkes pint) 1. irst, stnd nd put yur feet tgether. Lift yur weight ff yur right ft nd slide it t the frnt. Then lift yur weight ff yur left ft nd slide it t yur right ft. Weight ff yur right ft nd slide it t the bck. Weight ff yur left ft nd slide it t yur right ft. 2. Nw put little weight n yur right ft nd slide it t the frnt. A little weight n yur left ft, slide t yur right ft. Then slide bck s befre with little weight n the sliding ft. 3. D it ll gin with mre weight n the sliding ft. 4. inlly, with s much weight s yu cn put n t the sliding ft, slide, slide, slide. WHAT DO YOU OBSERVE? Use s mny ft-flr vritins s yu cn. t - Bre, cttn sck, nyln sck, wl sck, lether sled shes, rubber sles shes (snekers) lr - Smth wd, vinyl tile, linleum, cermic tile (ften in the bthrm), bth tube r shwer flr. SAMPLE DATA TABLE

16 t lr Weight n sliding ft Cmments 71 in cttn sck cermic tile light slides very,very esily in cttn sck cermic very hevy slides with sme difficulty ANOTHER ORM O DATA TABLE Cmbintins Weight n sliding ft Sliding the ft t lr light mderte hevy esy nt esy hrd didn t sneker tile x x sneker tile x x 8.2 LAWS O CONTACT RICTION s =< s MAX = µ s N > k = µ k N These re empiricl lws becuse they cme directly frm experience nd experiment. We d nt knw why they behve his wy. The nswers t tht questin is in the hnds f therist. S fr the explntins re very cmplex, cmplicted nd incmplete Exmple: inding m s A blck is t rest n rugh vrible ngle inclined plne. At sme criticl ngle, θ criticl, the blck just begins t slide dwn the plne. Shw tht the cefficient f sttic frictin µ s = tnθ criticl. This ngle is ften clled the ngle f repse. Drw ree Bdy Digrm f the Blck just s it strt t slide Stte the Physicl Principles Lws f frictin s >= s MAX = µ s N + Σ = m; sum f frces prllel t the plne =0. Dwn plne is Σ = 0; ill in the Detils Here we re cncerned with s MAX becuse it is just t the nset f mving. S, s = s MAX = µ s N (1) mgsin θ criticl - µ s N = 0 (2)

17 72 sum f frces perpendiculr t N - mgcsθ criticl = 0 (3) the plne =0. Upwrd is + directin. Eq (2) gives N=mgcsθ criticl. (4) Substituting this int Eq(3) yields mgsinθ - µ s mgcsθ criticl = 0. (5) Slving Eq(5) nd cnceling mg gives sinθ criticl = µ s csθ criticl (6) Dividing bth sides by cs2 nd using the definitin tn2 = sinθ/csθ, it fllws tht µ s = sinθ criticl /csθ criticl = tnθ criticl Exercise: Mesure m s Get hrd cver bk, prtrctr, sme bjects like cins, tpe cssette cse, pieces f vrius kinds f pper nd clth mterils, fr ech ne piece the size t wrp rund, tpe t, r glue t, the tpe cssette cse nd nther piece lrge enugh t cver the surfce f the bk.. 1. Plce cin n the bk while it is in hrizntl psitin. Then rise the ngle f the bk until the cin just strts t slide. Mesure this ngle with the prtrctr nd recrd it. The tngent f this ngle is the cefficient f sttic frictin. 2. Next, cver the surfce f the bk with piece f smth ntebk pper. Mesure nd recrd the bve tw ngles fr the sliding cin n this pper. 3. Next, cver the surfce f the bk with piece f newspper. Mesure nd recrd the bve tw ngles fr the sliding cin n this pper. 4. Next, cver the surfce f the bk with piece f glssy pper. Mesure nd recrd the bve tw ngles fr the sliding cin n this pper. 5. Next, cver the surfce f the bk with piece f clth. Mesure nd recrd the bve tw ngles fr the sliding cin n this surfce. 6. Nw using tpe cssette cse, first bre, then cvered with the smth ntebk pper, then with newspper, then with the glssy pper, with the clth, mesure nd recrd the tw ngles fr these sliding dwn ll f the previus surfces n the bk. Yu my nt hve time fr ll f these vritins. But, d t lest dzen vritins nce ech. Als d t lest ne ten times nd clculte the verge nd stndrd devitin. SAMPLE DATA TABLE Item Surfce Angle Strt µ s Angle Stp µ k qurter(clen) newspper qurter (clen) ntebk cver Exmple: Cnnecting crd NOT prllel t the inclined plne The system s shwn hs W 1 = m 1 g n plne inclined t n ngle θ t the hrizntl. It is cnnected t weight W 2 = m 2 g by mssless inextensible rpe ging ver mssless frictinless pulley. The rpe t m 1 g mkes n ngle φ with the plne. The cefficients f sttic nd

18 73 kinetic frictin between m 1 g nd the plne re µ s nd µ k, respectively. The whle system hs n upwrd ccelertin. Shw tht the nrml, N, cting between the plne nd m 1 g is N = m 1 (g+)csθ + m 2 (g+)sinφ.. Sketch the ree Bdy Stte the Physicl ill in the Detils f Digrm f W1 Principle(s) r ech 20 pints 20 pints 30 pints Lws f rictin s =< smx = µ s N > k = µ k N (1) Express in terms f cmpnents perpendiculr nd prllel t the plne. The ctin is lng these directins. + Σ = m + Σ = m; 1 1 ; N - m 1 gcsθ - Tsinφ = m 1 csθ (2) - - W 1 sinθ + Tcsφ = m 1 sinθ (3) Sketch the ree Bdy Digrm (BD) f W pints The type f frictin is left pen here. If we re cncerned with the nset f mtin then we use smx = µ s N, nd k = µ k N fr kinetic. Σ h = 0; Trivil becuse there re n hrizntl frces. +^Σ v = m 2 ; T - m 2 g = -m 2 (4) b. rm the bve, slve fr the expressin fr the nrml cting between W 1 nd the plne. 10 pints Equtin (4) gives us the vlue f T s T = m 2 (g + ). Substituting this int Equtin (2) give N = m 1 gcs2 + TsinN + m 1 csθ = m 1 gcsθ+ m 2 sinφ(g + ) + m 1 csθ = (g + ) m 1 csθ+ (g + ) m 2 sinφ. c. Briefly discuss why the nrml behves this wy. 10 pints The vlue f the nrml depends n ll frces perpendiculr t the plne. The tensin T hs cmpnent perpendiculr t the plne 9 Centripetl rce Review Sectin 6 (Centripetl Accelertin) ll!!! In Newtnin Physics whenever there is n ccelertin, there must be cusing frce - Newtn s 2 nd Lw f Mtin, = /m. Tht is, the ccelertin is the respnse t frce - n ccelertin implies there must be frce. S the frce cusing the centripetl ccelertin c is clled the centripetl frce c nd when cting n bdy f mss m hs the vlue

19 74 c = m c = mv 2 /R = mt 2 R. 9.1 Exmples: A mss swinging rund n string - such s swinging set f keys n key chin A stellite rbiting prent bdy -such s the mn rund the erth, ny plnet rund the sun A negtive chrge rbiting psitive chrge - such s n electrn rbiting prtn in the hydrgen tm Here v, v tngentil, nd v ORBIT re the sme thing. 9.2 Prblems 9.2.1`The Unbnked Highwy A cr f mss m trvels with velcity v rund curve n flt highwy. The curve hs f rdius R. Shw tht fr the cr nt t slip ff the rd the cefficient f sttic frictin µ s between the cr tires nd the rd must stisfy the cnditin, µ s >= v 2 /(gr), where g is the grvittinl intensity (ls clled, the ccelertin due t grvity). Sketch the ree Bdy Stte the Physicl ill in the Detils f Digrm f the cr Principle(s) r ech 20 pints 20 pints 30 pints v int the pper Cr heding int the pper W=mg N s 2 c = v/r Lws f rictin s =< smx = µ s N > k = µ k N (1) Here we re ging t ignre he fct tht the tw tires trvel in circle f different rdii. In mst prcticl situtins R is in the rnge f 1000m while the tires re but 2 m prt. This puts yur errr under 1%. But if yu re cncerned with differentil tire wer, yu cnnt neglect this difference. Newtn s 2 nd Lw +^Σ verticl = 0 N = mg (2) +>Σ hrizntl = m = m c = mv 2 /R s = mv 2 /R (3)

20 THE MATHEMATICS Cmbing Eq.(1), (3) then (2) gives mv 2 /R = s =< smx = µ s N = µ s mg (4) Cnceling m gives v 2 /R <= µ s g. Hence µ s > = v 2 /(gr). (5) 75 INTERPRETATION The result f Equtin (5) is tht the cr trvels in the circle f rdius R t the speed v withut sliding ff the rd s lng s µ s is equl t r greter thn v 2 /(gr). r vlues f µ s less thn this vlue the cr slides ff the rd seeking circulr pth f lrger R r must be slwed t smller velcity v The Bnked Highwy A cr f mss m trvels with velcity v rund curve n highwy. The curve hs f rdius R. The highwy curve is bnked n ngle θ B such tht n frictin is required t keep the cr n the curve. The frce tht keeps the cr n the curved pth is the hrizntl cmpnent f the nrml frce f the rd. Shw tht the bnk ngle θ B is tn θ B = v 2 /(gr). Sketch the ree Bdy Stte the Physicl ill in the Detils f Digrm f the cr Principle(s) r ech 20 pints 20 pints 30 pints N=Ncs v N θ B θ Β Cr heding int the pper θ B W=mg v int the pper N=Nsin h 2 = v/r c θ Β +>Σ hrizntl = m c = mv 2 /R; Nsinθ B = mv 2 /R, (1) +^Σ verticl = 0 ; Ncsθ B - mg = 0, (2) The cmpnents f N re s shwn by the tringle then slid ver t ct frm the rd surfce. THE MATHEMATICS: Eq. (2) becmes Ncsθ B = mg (3) Dividing (1) by (3) gives Nsinθ B /Ncsθ B = (mv 2 /R)/mg. Cnceling N nd m yields tnθ B = v 2 /(gr). (4) INTERPRETATION The result f Equtin (4) is tht the cr trvel in the circle f rdius R t the speed v withut sliding ff the rd s lng s tnθ B is equl t v 2 /(gr). r ther vlues f tnθ B cr slides up r dwn n the rd seeking different R r it velcity must chnge t sty n the initil circulr pth Dn t fll ff the tp f the lp-the-lp tp A cr is trveling thrugh lp-the-lp s shwn. If the cr is nt trveling fst enugh it breks cntct the lp t the tp. Shw tht the velcity f the cr,v, must stisfy the reltin CAR R

21 76 v 2 > gr. Sketch the ree Bdy Stte the Physicl ill in the Detils f Digrm f the cr Principle(s) r ech 20 pints 20 pints 30 pints N W=mg 2 =v/r c +9Σ hrizntl = m c = mv 2 /R; N + mg = mv 2 /R, (1) the directin f the ccelertin is chsen s + fr the cnvenience f keeping the right side f the equtin +. The cnditin fr stying in cntct with the N>0 S t the tp f the lp-the-lp N=mv 2 /R - mg > 0 (2) THE MATHEMATICS: 1. Eq. (2) fter dding mg t bth sides gives mv 2 /R > mg 2. Nw cncelling m nd crss multiplying by R, we get the desired result v 2 > gr. (3) INTERPRETATION This prblem pplies t clss f situtins. It pplies t ny situtin tht yu re upside dwn t the tp f verticl circulr pth. Sme exmples re: 1 If yu re flying upside dwn nd Eq. (3) is true, then yu d nt need yur set belt t keep yu in yur set. When v 2 < gr yu need yur set belt. 2 A mtrcycle trveling lp-the-lp. Cn yu think f ther situtins?