x y θ = 31.8 = 48.0 N. a 3.00 m/s

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Ross Neil Caldwell
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1 4.5.IDENTIY: Vecor addiion. SET UP: Use a coordinae sse where he dog A. The forces are skeched in igure 4.5. EXECUTE: + -ais is in he direcion of, A he force applied b =+ 70 N, = 0 A B B A = cos60.0 = (300 N) cos60.0 =+ 150 N B = sin 60.0 = (300 N)sin 60.0 =+ 60 N B igure 4.5a R = + A B R = A + B =+ 70 N N =+ 40 N R = A + B = N =+ 60 N R = R + R igure 4.5b R = (40 N) + (60 N) = 494 N R anθ = = R θ = 31.8 EVALUATE: The forces us be added as vecors. The agniude of he resulan force is less han he su of he agniudes of he wo forces and depends on he angle beween he wo forces. 4.9.IDENTIY: Appl = a o he bo. SET UP: Le + be he direcion of he force and acceleraion. = 48.0 N. EXECUTE: 48.0 N = a gives = = = 16.0 kg. a 3.00 /s EVALUATE: The verical forces su o zero and here is no oion in ha direcion IDENTIY and SET UP: Use Newon s second law in coponen for (Eq.4.8) o calculae he acceleraion produced b he force. Use consan acceleraion equaions o calculae he effec of he acceleraion on he oion. EXECUTE: (a) During his ie inerval he acceleraion is consan and equal o 0.50 N 1.56 /s a = = = kg We can use he consan acceleraion kineaic equaions fro Chaper = v0+ a 0 (1.56 /s )(.00 s), = + so he puck is a = 3.1. v = v0 + a = 0 + (1.56 /s )(.00 s) = 3.13 /s. (b) In he ie inerval fro =.00 s o 5.00 s he force has been reoved so he acceleraion is zero. The speed sas consan a v = 3.1 /s. The disance he puck rels is 0 = v0 = (3.1 /s)(5.00 s.00 s) = A he end of he inerval i is a = = 1.5.

2 In he ie inerval fro = 5.00 s o 7.00 s he acceleraion is again a = 1.56 /s. A he sar of his inerval v 0 = 3.1 /s and 0 = 1.5. = v + a = (3.1 /s)(.00 s) + (1.56 /s )(.00 s) = = Therefore, a = 7.00 s he puck is a = = = 1.9. v v a = 0 + = 3.1 /s + (1.56 /s )(.00 s) = 6.4 /s EVALUATE: The acceleraion sas he puck gains 1.56 /s of veloci for ever second he force acs. The force acs a oal of 4.00 s so he final veloci is (1.56 /s)(4.0 s) = 6.4 /s. 4.. IDENTIY: = a refers o forces ha all ac on one objec. The hird law refers o forces ha a pair of objecs eer on each oher. SET UP: An objec is in equilibriu if he vecor su of all he forces on i is zero. A hird law pair of forces he he sae agniude regardless of he oion of eiher objec. EXECUTE: (a) he earh (gri) (b) 4 N; he book (c) no, hese wo forces are eered on he sae objec (d) 4 N; he earh; he book; upward (e) 4 N, he hand; he book; downward (f) second (The wo forces are eered on he sae objec and his objec has zero acceleraion.) (g) hird (The forces are beween a pair of objecs.) (h) No. There is a ne upward force on he book equal o 1 N. (i) No. The force eered on he book b our hand is 5 N, upward. The force eered on he book b he earh is 4 N, downward. (j) Yes. These forces for a hird-law pair and are equal in agniude and opposie in direcion. (k) Yes. These forces for a hird-law pair and are equal in agniude and opposie in direcion. (l) One, onl he gri force. () No. There is a ne downward force of 5 N eered on he book. EVALUATE: Newon s second and hird laws give copleenar inforaion abou he forces ha ac. 4.9.IDENTIY: Since he observer in he rain sees he ball hang oionless, he ball us he he sae acceleraion as he rain car. B Newon s second law, here us be a ne force on he ball in he sae direcion as is acceleraion. SET UP: The forces on he ball are gri, which is w, downward, and he ension T in he sring, which is direced along he sring. EXECUTE: (a) The acceleraion of he rain is zero, so he acceleraion of he ball is zero. There is no ne horizonal force on he ball and he sring us hang vericall. The free-bod diagra is skeched in igure 4.9a. (b) The rain has a consan acceleraion direced eas so he ball us he a consan easward acceleraion. There us be a ne horizonal force on he ball, direced o he eas. This ne force us coe fro an easward coponen of T and he ball hangs wih he sring displaced wes of verical. The free-bod diagra is skeched in igure 4.9b. EVALUATE: When he oion of an objec is described in an inerial frae, here us be a ne force in he direcion of he acceleraion. igure 4.9

3 4.31.IDENTIY: Idenif he forces on he chair. The floor eers a noral force and a fricion force. SET UP: Le + be upward and le + be in he direcion of he oion of he chair. EXECUTE: (a) The free-bod diagra for he chair is given in igure (b) or he chair, a = 0 so = a gives n g sin 37 = 0 and n = 14 N. EVALUATE: n is larger han he weigh because has a downward coponen. igure IDENTIY: = a us be saisfied for each objec. Newon s hird law sas ha he force C on T ha he car eers on he ruck is equal in agniude and opposie in direcion o he force T on C ha he ruck eers on he car. SET UP: The onl horizonal force on he car is he force T on C eered b he ruck. The car eers a force C on T on he ruck. There is also a horizonal fricion force f ha he highwa surface eers on he ruck. Assue he sse is acceleraing o he righ in he free-bod diagras. EXECUTE: (a) The free-bod diagra for he car is skeched in igure 4.33a (b) The free-bod diagra for he ruck is skeched in igure 4.33b. (c) The fricion force f acceleraes he sse forward. The ires of he ruck push backwards on he highwa surface as he roae, so b Newon s hird law he roadwa pushes forward on he ires. EVALUATE: T on C and C on T each equal he ension T in he rope. Boh objecs he he sae acceleraion a. T = Ca and f T = Ta, so f = ( C + T) a. The acceleraion of he wo objecs is proporional o f. igure IDENTIY: We can appl consan acceleraion equaions o relae he kineaic variables and we can use Newon s second law o relae he forces and acceleraion. (a) SET UP: irs use he inforaion given abou he heigh of he jup o calculae he speed he has a he insan his fee lee he ground. Use a coordinae sse wih he + -ais upward and he origin a he posiion when his fee lee he ground. v = 0 (a he aiu heigh), v 0 =?, a = 9.80 /s, 0 =+ 1. v = v + a ( ) 0 0 EXECUTE: 0 = ( 0) = ( 9.80 /s )(1. ) = 4.85 /s v a

4 (b) SET UP: Now consider he acceleraion phase, fro when he sars o jup unil when his fee lee he ground. Use a coordinae sse where he + -ais is upward and he origin is a his posiion when he sars his jup. EXECUTE: Calculae he erage acceleraion: v v 4.89 /s 0 ( a) = = = 16. /s s 0 (c) SET UP: inall, find he erage upward force ha he ground us eer on hi o produce his erage upward acceleraion. (Don forge abou he downward force of gri.) The forces are skeched in igure EXECUTE: 890 N = w/ g = = 90.8 kg 9.80 /s = a igure 4.39 g = ( a) = ( g + ( a ) ) = 90.8 kg(9.80 /s /s ) = 360 N This is he erage force eered on hi b he ground. Bu b Newon s 3rd law, he erage force he eers on he ground is equal and opposie, so is 360 N, downward. EVALUATE: In order for hi o accelerae upward, he ground us eer an upward force greaer han his weigh IDENTIY: Appl = a o he elevaor o relae he forces on i o he acceleraion. (a) SET UP: The free-bod diagra for he elevaor is skeched in igure The ne force is T g (upward). igure 4.50 Take he + -direcion o be upward since ha is he direcion of he acceleraion. The aiu upward acceleraion is obained fro he aiu possible ension in he cables. EXECUTE: = a gives T g = a T g 8,000 N (00 kg)(9.80 /s ) a = = =.93 /s. 00 kg (b) Wha changes is he weigh g of he elevaor. T g 8,000 N (00 kg)(1.6 /s ) a = = = 11.1 /s. 00 kg EVALUATE: The cables can give he elevaor a greaer acceleraion on he oon since he downward force of gri is less here and he sae T hen gives a greaer ne force IDENTIY: Noe ha in his proble he ass of he rope is given, and ha i is no negligible copared o he oher asses. Appl = a o each objec o relae he forces o he acceleraion. (a) SET UP: The free-bod diagras for each block and for he rope are given in igure 4.54a.

5 igure 4.54a T is he ension a he op of he rope and T b is he ension a he boo of he rope. EXECUTE: (b) Trea he rope and he wo blocks ogeher as a single objec, wih ass = 6.00 kg kg kg = 15.0 kg. Take + upward, since he acceleraion is upward. The free-bod diagra is given in igure 4.54b. = a igure 4.54b g = a g a = 00 N (15.0 kg)(9.80 /s ) a = = 3.53 /s 15.0 kg (c) Consider he forces on he op block ( = 6.00 kg), since he ension a he op of he rope ( T ) will be one of hese forces. igure 4.54c = a g T = a T = ( g+ a) T = 00 N (6.00 kg)(9.80 /s /s ) = 10 N Alernaivel, can consider he forces on he cobined objec rope plus boo block ( = 9.00 kg): igure 4.54d = a T g = a T g a which checks = ( + ) = 9.00 kg(9.80 /s /s ) = 10 N, (d) One wa o do his is o consider he forces on he op half of he rope ( =.00 kg). Le T be he ension a he idpoin of he rope. = a T T g = a T T g a = ( + ) = 10 N.00 kg(9.80 /s /s ) = 93.3 N igure 4.54e

6 To check his answer we can alernaivel consider he forces on he boo half of he rope plus he lower block aken ogeher as a cobined objec ( =.00 kg kg = 7.00 kg): igure 4.54f = a T g = a T = ( g+ a) = 7.00 kg(9.80 /s /s ) = 93.3 N, which checks EVALUATE: The ension in he rope is no consan bu increases fro he boo of he rope o he op. The ension a he op of he rope us accelerae he rope as well he 5.00-kg block. The ension a he op of he rope is less han ; here us be a ne upward force on he 6.00-kg block IDENTIY: Calculae a fro a = d r / d. Then ne = a. SET UP: w= g EXECUTE: Differeniaing wice, he acceleraion of he helicoper as a funcion of ie is 3 ˆ a= (0.10 /s ) i (0.1 /s ) kˆ and a = 5. 0s, he acceleraion is ˆ a= (0.60 /s ) i (0.1 /s ) kˆ. The force is hen 5 w ( N) ˆ ˆ 4 (0.60 /s ) (0.1 /s ) ( N) ˆ 3 = a = a = ( N) ˆ g (9.80 /s ) i k = i k EVALUATE: The force and acceleraion are in he sae direcion. The are boh ie dependen.

1. VELOCITY AND ACCELERATION

1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under