Physics for Scientists and Engineers I

Transcription

1 Phscs for Scentsts nd Engneers I PHY 48, Secton 4 Dr. Betr Roldán Cuen Unverst of Centrl Flord, Phscs Deprtment, Orlndo, FL Chpter - Introducton I. Generl II. Interntonl Sstem of Unts III. Converson of unts IV. Dmensonl Anlss V. Prolem Solvng Strteges

2 I. Oectves of Phscs - Fnd the lmted numer of fundmentl lws tht govern nturl phenomen. - Use these lws to develop theores tht cn predct the results of future eperments. -Epress the lws n the lnguge of mthemtcs. - Phscs s dvded nto s mor res:. Clsscl Mechncs PHY48. Reltvt 3. Thermodnmcs 4. Electromgnetsm PHY49 5. Optcs PHY49 6. Quntum Mechncs II. Interntonl Sstem of Unts POWER PREFIX ABBREVIATION QUANTITY Length Tme UNIT NAME meter second UNIT SYMBOL m s pet ter gg meg P T G M Mss logrm g 3 lo Speed m/s hecto h Accelerton m/s de d Force Newton N - dec D Pressure Energ Power Temperture Pscl Joule Wtt Kelvn P N/m J Nm W J/s K cent mll mcro nno pco c m µ n p -5 femto f

3 III. Converson of unts Chn-ln converson method: The orgnl dt re multpled successvel converson fctors wrtten s unt. Unts cn e treted le lgerc qunttes tht cn cncel ech other out. Emple: 36 feet/h m/s feet h m 36.7 m/ s h 36s 3.8 feet IV. Dmensonl Anlss Dmenson of quntt: ndctes the tpe of quntt t s; length [L], mss [M], tme [T] Dmensonl consstenc: oth sdes of the equton must hve the sme dmensons. Emple: +v t+t / [ ] [ L] Note: There re no dmensons for the constnt / [ L] [ ] [ ] [ L] T + T T [ ] [ T ] [ L] + [ L] [ L] L + + Sgnfcnt fgure one tht s rell nown. Zeros m or m not e sgnfcnt: E: - Those used to poston the decml pont re not sgnfcnt. - To remove mgut, use scentfc notton..56 m/s hs 3 sgnfcnt fgures, decml plces..56 m/s hs 3 sgnfcnt fgures nd 6 decml plces.. m hs 3 sgnfcnt fgures. 5 m s mguous.5 3 fgures, fg., fgs. Order of mgntude the power of tht pples. 3

4 V. Prolem solvng tctcs Epln the prolem wth our own words. Me good pcture descrng the prolem. Wrte down the gven dt wth ther unts. Convert ll dt nto S.I. sstem. Identf the unnowns. Fnd the connectons etween the unnowns nd the dt. Wrte the phscl equtons tht cn e ppled to the prolem. Solve those equtons. Alws nclude unts for ever quntt. Crr the unts through the entre clculton. Chec f the vlues otned re resonle order of mgntude nd unts. Chpter - Vectors I. Defnton II. Arthmetc opertons nvolvng vectors A Addton nd sutrcton - Grphcl method - Anltcl method Vector components B Multplcton 4

5 Revew of ngle reference sstem 9º º<θ <9º 9º<θ <8º 8º θ θ º Orgn of ngle reference sstem 8º<θ 3 <7º 7º θ 3 θ 4 7º<θ 4 <36º Angle orgn Θ 4 3º-6º I. Defnton Vector quntt: quntt wth mgntude nd drecton. It cn e represented vector. Emples: dsplcement, veloct, ccelerton. Dsplcement does not descre the oect s pth. Sme dsplcement Sclr quntt: quntt wth mgntude, no drecton. Emples: temperture, pressure 5

6 II. + + Arthmetc opertons nvolvng vectors Vector ddton: - Geometrcl method Rules: s + commuttve lw 3. s c + + c ssoctve lw 3. Vector sutrcton: d Vector component: proecton of the vector on n s. cosθ snθ 3.4 Sclr components of tnθ Vector mgntude Vector drecton 6

7 Unt vector: Vector wth mgntude. No dmensons, no unts. ˆ, ˆ, ˆ unt vectors n postve drecton of,, es ˆ + ˆ 3.6 Vector component Vector ddton: - Anltcl method: ddng vectors components. r + + ˆ + ˆ Vectors & Phscs: -The reltonshps mong vectors do not depend on the locton of the orgn of the coordnte sstem or on the orentton of the es. - The lws of phscs re ndependent of the choce of coordnte sstem. θ θ ' + + ' + ' 3.8 Multplng vectors: - Vector sclr: f s - Vector vector: Sclr product sclr quntt dot product cos

8 8 9 cos cos Rule: 3. cos9 cos Multplng vectors: - Vector vector Vector product vector sn ˆ ˆ ˆ c c + cross product Mgntude Angle etween two vectors: cosϕ 3. Rule: 9 sn sn Drecton rght hnd rule contnng plne to perpendculr c, Plce nd tl to tl wthout lterng ther orenttons. c wll e long lne perpendculr to the plne tht contns nd where the meet. 3 Sweep nto through the smllest ngle etween them. Vector product

9 9 Rght-hnded coordnte sstem Left-hnded coordnte sstem sn

10 ˆ ˆ P: If B s dded to C 3 + 4, the result s vector n the postve drecton of the s, wth mgntude equl to tht of C. Wht s the mgntude of B? Method Method Isosceles trngle B + C B + 3ˆ + 4 ˆ D D ˆ θ C D C tnθ 3/ 4 θ 36.9 B + 3ˆ + 4 ˆ 5 ˆ B 3ˆ + ˆ D B θ B / θ sn B Dsn 3. D B/ B P: A fre nt goes through three dsplcements long level ground: d for.4m SW, d.5m E, d 3.6m t 6º North of Est. Let the postve drecton e Est nd the postve drecton e North. Wht re the nd components of d, d nd d 3? Wht re the nd the components, the mgntude nd the drecton of the nt s net dsplcement? c If the nt s to return drectl to the strtng pont, how fr nd n wht drecton should t move? d4 d + d.8ˆ.8 ˆ +.5ˆ.ˆ.8 ˆ m N d.4cos45.8m D d4 + d3.ˆ.8 ˆ +.3ˆ +.5 ˆ.5ˆ +.4 ˆ m D d m E.4sn 45.8 D m d.5m d 45º d d 4 d 3 d d3.6cos6.3m d3.6sn 6.5m.4 θ tn North of Est c Return vector negtve of net dsplcement, D.57m, drected 5º South of West P d 5 ˆ 6ˆ 4ˆ + d ˆ ˆ 3ˆ + + d 4ˆ 3 ˆ ˆ r d d + d3? Angle etween r nd +? c Component of d long d? d Component of d perpendculr to d nd n plne of d, d? r d d + d 4ˆ + 5 ˆ 6ˆ ˆ + ˆ + 3ˆ + 4ˆ + 3 ˆ + ˆ 9ˆ + 6 ˆ 7ˆ 3 r ˆ 7 r cosθ 7 θ cos 3.88 r m d perp d d c d d dd cosθ cosθ d d θ d // d d d// d cosθ d 3.m d d 3.74 d m d d d d d// + dperp d perp m d m P3 If d ˆ 4ˆ 3ˆ + d 5ˆ ˆ ˆ + d + d d 4d? d + d contned n d, d plne d 4d 4 d d 4 perpendculr to d, d plne perpendculr to cos9 4 Tp: Thn efore clculte!!!

11 P4: Vectors A nd B le n n plne. A hs mgntude 8. nd ngle 3º; B hs components B -7.7, B -9.. Wht re the ngles etween the negtve drecton of the s nd the drecton of A, the drecton of AB, c the drecton of AB+3? ˆ A 3º Angle etween nd A B Angle, A B C ngle ˆ, ˆ ecuse C perpendculr plne A, B 9 c Drecton A B + 3ˆ D E B + 3ˆ 7.7ˆ 9. ˆ + 3ˆ ˆ D A E ˆ ˆ 8.39ˆ ˆ ˆ 3 D ˆ D ˆ 8.39ˆ ˆ ˆ 5.4 ˆ 5.4 cos D θ θ 99 D 97.6 P5: A wheel wth rdus of 45 cm rolls wthout sleepng long horontl floor. At tme t the dot P pnted on the rm of the wheel s t the pont of contct etween the wheel nd the floor. At lter tme t, the wheel hs rolled through one-hlf of revoluton. Wht re the mgntude nd the ngle reltve to the floor of the dsplcement P durng ths ntervl? Vertcl dsplcement: R. 9m Horontl dsplcement: πr. 4m r.4m ˆ +.9m ˆ r m R tnθ θ 3.5 πr d P6: Vector hs mgntude of 5. m nd s drected Est. Vector hs mgntude of 4. m nd s drected 35º West of North. Wht re the mgntude nd drecton of +?. Wht re the mgntude nd drecton of -?. c Drw vector dgrm for ech comnton. W - - N 5º + S E 5ˆ ˆ 4sn cos35 ˆ.9ˆ ˆ +.7ˆ ˆ m 3.8 tnθ θ ˆ ˆ m 3.8 tnθ θ or North of West