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Solution Manual for Engineering Mechanics: Dynamics, 2nd Edition

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha Answers to Selected Even-Numbered Problems Please note that answers are not provided for the following type of even-numbered problems:  Concept Problems.  Computer Problems.  Design Problems. For Concept and Computer problems, please consult the solutions manual. Chapter 1 1.2 .r B=A / ` D 3:883 ft 1.4 E r B=A ˇ ˇ xy system D .4:000 O { 1:000 O | / ft E r B=A ˇ ˇ pq system D .3:313 O u p 2:455 O u q / ft ˇ ˇ E r B=A ˇ ˇ xy system D ˇ ˇ E r B=A ˇ ˇ pq system D 4:123 ft 1.6 E v D .79:68 O { C 7:190 O | / ft = s 1.8  D 3:229 rad  D 0:087 31 rad 1.10 v r D 504:6 ft = s and v  D 353:3 ft = s 1.12  D 101:1 ı 1.14 x P 2 D 0:7679 ft and y P 2 D 5:330 ft 1.16 E v A D .20:02 O { C 12:81 O | / ft = s and E a A D . 1:414 O { C 4:243 O | / ft = s 2 1.18 r D 26:36 mm 1.20 ŒI xx ç D ŒI yy ç D ŒI ´´ ç D ML 2 Units of I xx , I yy , and I ´´ in the SI system: kg  m 2 ; Units of I xx , I yy , and I ´´ in the U.S. Customary system: slug  ft 2 D lb  s 2  ft. 1.22 Concept problem. 1.24 the units of E are kg = m  s 2  . Chapter 2 2.2 Concept Problem 2.4 Concept Problem 1 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.6  E r 1 D 8:747 O u r m and  E r 2 D 13:73 O u r m E v avg  1 D 5:467 O u r m = s and E v avg  2 D 8:579 O u r m = s 2.8 E v D . 154:0 O { C 266:7 O | / ft = s 2.10 E v D .145:4 O { C 69:81 O | / ft = s 2.12  E v 1 D .0:4623 O { C 0:08155 O | / m = s  E v 2 D .0:002550 O { C 0:0004499 O | / m = s 2.14  1 D 12:21 ı and  2 D 17:30 ı  1 D 102:2 ı and  2 D 72:70 ı 2.16 .x/ D cos 1  3 C 4x p 10 C 24x C 16x 2  2.18 E a avg D . 0:04588 O { C 3:886 O | / ft = s 2 E a avg E a.5 s / D . 0:001154 O { 0:003175 O | / ft = s 2 2.20  E r D E 0 E v avg D E 0 d D 1:394 ft v avg D 0:3485 ft = s 2.22 Computer Problem 2.24 v max D 2v 0 D 58:67 ft = s and v min D 0 ft = s y v min D 0 ft and y v max D 2R D 2:300 ft E a v min D v 2 0 R O | D .748:2 ft = s 2 / O | and E a v max D v 2 0 R O | D . 748:2 ft = s 2 / O | 2.26 R x D 8v 2 0 a 3 y 2 C 4a 2  2 R y D 4v 2 0 a 2 y y 2 C 4a 2  2 2.28 E v D .203:2 O { C 117:3 O | / ft = s E a D . 27:53 O { C 47:69 O | / ft = s 2 2.30 v max D 80:82 ft = s 2.32 E v D .51:32 O { 1:046 O | / ft = s E a D .0:3873 O { C 19:00 O | / ft = s 2 2 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.34 v D ! p d 2 x 2 s 1 C h 2  x d 2 x 3 d 4  2 v D 0:5236 ft = s ; v D 0:4554 ft = s ; and v D 0 2.36 E v A D .72:00 O { C 96:00 O | / ft = s E a A D .460:8 O { 345:6 O | / ft = s 2 2.38 Computer Problem 2.40 t braking D 7:854 s 2.42 j a j max D 3:600 m = s 2 t j a max j  1 D 3:927 s and t j a max j  2 D 11:78 s s j a max j  1 D 12:84 m and s j a max j  2 D 128:5 m 2.44 Largest distance traveled in 1 s is d D 0:2667 m, corresponding to a D ˇ 1 p t 2.46 v.0/ D 0:080 00 ft = s 2.48 v 0 D 10:10 m = s 2.50 a c D 22:36 g 2.52 t stop D 2:880 s 2.54  D 98:10 s 1 2.56 P x D ˚ 7:000 sin Œ.1:000 rad = s /tç C 10:50 cos Œ.0:5000 rad = s /tç 10:50 m = s x D ˚ 7:000 7:000 cos Œ.1:000 rad = s /tç C 21:00 sin Œ.0:5000 rad = s /tç .10:50 s 1 /t m = s 2 2.58 t stop D 0:2233 s 2.60 v term D 4:998 m = s 2.62 j v j max D 1:128 m = s s j v j max D 0:1250 m and s j v j max D 0:1250 m 2.64 v.t/ D mg C d  1 e C d t=m  v term D mg C d 2.66 v f D 3:563 m = s 2.68 Computer Problem 3 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.70 s wet D 26:31 m .s wet s dry / s dry .100 % / D 65:21 % 2.72 P ./ D ˙ r P  2 0 C 2 g L cos  cos  0  2.74 P  min D 4:930 rad = s 2.76 P x D ˙ s v 2 0 C 2  g C kL 0 m  .x x 0 / k m x 2 x 2 0  2.78 t x max D 0:2433 s 2.80 (a) P r D p 2G .m A C m B / r r 0 r rr 0 (b) (i) P r D 5:980  10 5 ft = s and (ii) P r ! 1 2.82 Concept Problem 2.84 d D 186:4 ft 2.86 ˛ s D a p r C hv 2 p 2 r 3 ˛ s ˇ ˇ r D r 1 D 1:442 rad = s 2 and ˛ s ˇ ˇ r D r 2 D 3:631 rad = s 2 2.88 v D v 0 C a c .t t 0 / s D s 0 C v 0 .t t 0 / C 1 2 a c .t t 0 / 2 2.90 v min D 113:5 ft = s 2.92 t i D 3:497 s E v B D .33:33 O { 34:31 O | / m = s 2.94 R D 5:047 m 2.96 t flight D 6:263 s E r land D .281:8 O { 4:500 O | / m 2.98 d D 24:09 m 2.100 (a)  R max D  4 rad D 45 ı (b) R max D 233:5 % of the actual maximum range (c) t R max D 119:5 s 2.102  1 D 1 2 sin 1 .gR=v 2 0 / and  2 D 90 ı 1 2 sin 1 .gR=v 2 0 /  1 D 24:86 ı and  2 D 65:14 ı 4 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.104 .v 0 / min D 50:75 ft = s 2.106 E v initial D .5:920 O { C 7:786 O | / m = s : dy dx ˇ ˇ ˇ ˇ x D x B D 0:2244 2.108 h max D v 2 0 sin 2 ˇ 2g cos  2.110 Computer Problem 2.112 t f R max D 5:211 s R max D 437:1 ft and  R max D 45:66 ı 2.114 t max D 3:000 s H max D 49:21 m percent increase in height with no air resistance D 2:363 % 2.116 Computer Problem 2.118 (a) E a  E b D 1  6 C 2  3 C 3  0 D 0 (b) E a   E a  E b  D  84 O { 42 O | C 0 O k  (c) they are the same 2.120 (a) E ! 1 D 104:7 O k  rad = s ; E ! 2 D 104:7 O {  rad = s and E ! 3 D 104:7 O |  rad = s (b) E ! ` D 100 3 p 3  O { C O | C O k  rad = s D 60:46  O { C O | C O k  rad = s 2.122 Derivation, so no answer needed. 2.124 E v A D 10:00 O | m = s ; E v B D 10:00 O | m = s ; E v C D 10:00 O | m = s ; and E v D D 10:00 O | m = s 2.126 E ! wheel D 65:71 O k rad = s 2.128 jE a P j D 19;300 ft = s 2 Orientation of E a P from x axis D 110:0 ı . ccw / 2.130 v 0 D 263:7 ft = s and r D 6875 ft 2.132 P r D 1:204 m = s and P  D 0:8504 rad = s 2.134 E v D .0:037 70 O u r C 0:5341 O u  / m = s E a D 1678 O u r C 236:9 O u   m = s 2 2.136 Concept Problem 2.138 P r ˇ ˇ  D 90 ı D 8:825 ft = s and P  ˇ ˇ  D 90 ı D 0:7746 rad = s 5 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.140 E a P D . 252:9 O u C 399:8 O u B / ft = s 2 2.142 Concept Problem 2.144 Concept Problem 2.146  D 49:69 ft 2.148 P v D 3:213 m = s 2 2.150  Südkurve D 50:56 m  Nordkurve D 92:54 m 2.152  D 327:3 ft 2.154 d D 831:0 ft 2.156 E a D E 0 2.158 E a D . 39:40 O { C 42:88 O | / m = s 2 2.160 Concept Problem 2.162 ˇ ˇ E a ˇ ˇ D .33:69  10 3 cos / m = s 2 2.164 E a D .121:2 O { C 230:0 O | / ft = s 2 2.166 E a D 18:96 ft = s 2  O u t C  69:91 ft = s 2  0:2917 s 2  s  O u n 2.168 P v D 5:343 m = s 2 and  D 17:50 m 2.170  D 282:2 ft 2.172  min D 564:5 ft t f t 0 D 4:742 s 2.174 jE a j D 1   v 0 2v 0  s  s 4v 2 0  2 C  v 0 2v 0  s  2 2.176 d D 304:4 ft and t f D 3:106 s 2.178 Concept Problem 2.180 Concept Problem 2.182 P r D 684:1 ft = s and P  D 0:011 57 rad = s R r D 4:944 ft = s 2 and R  D 0:000 428 1 rad = s 2 2.184 v D 1:816 m = s ˇ ˇ E a ˇ ˇ D 4:573 m = s 2 6 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.186 R r D 1:599 m = s 2 2.188  0 D 0 r 0 D 0:014 86 m 2.190 E v D . 1:3 m = s / O u r C 0:22 s 1  r O u  E a D 0:048 40 s 2  r O u r 0:5720 m = s 2  O u  2.192 E v D .0:1500 O u r C 1:732 O u  / ft = s and E a D . 0:3638 O u r C 0:1125 O u  / ft = s 2 2.194 E r D 30:20 O u r ft E v D .3 O u r C 7:550 O u  / ft = s E a D . 1:888 O u r C 1:500 O u  / ft = s 2 2.196  D 5:602  m ! D 2630 rad = s D 25;120 rpm 2.198 v D 5490 ft = s ˇ ˇ E a ˇ ˇ D 1:005  10 6 ft = s 2 2.200 E v D v 0 p  2 C .r 0 C / 2 O u r C v 0 .r 0 C / p  2 C .r 0 C / 2 O u  E a D v 2 0 .r 0 C / 3  r 2 0 C 2r 0  C .1 C  2 / 2  2 O u r C v 2 0 ˚ .r 0 C / 2 C 2   2 C .r 0 C / 2  Œ 2 C .r 0 C / 2  2 O u  2.202 P  A D 0:001 157 rad = s and P  H D 4:287 rad = s R  A D 1:684  10 4 rad = s 2 and R  H D 0:075 47 rad = s 2 2.204 P E a D  « r 3r P  R  3 P r P  2  O u r C h r  «  P  3  C 3 R r P  C 3 P r R  i O u  2.206 R r D 5:781 ft = s 2 R  D 1:421 rad = s 2 2.208 r D q h 2 C .d C / 2 P r D v 0 h p h 2 C .d C / 2 and P  D v 0 .d C / h 2 C .d C / 2 R r D v 2 0  . C d / d 2 C h 2 C d  d 2 C h 2 C  2 C 2d  3=2 and R  D v 2 0  h d 2 C h 2  2  d 2 C h 2 C  2 C 2d  2 2.210 E v D 1:758 O | m = s and E a D 136:9g O | 2.212 v D 0:1185 ft = s and ˇ ˇ E a ˇ ˇ D 0:086 42 ft = s 2 7 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.214 a r D K 2 a b 2  1 r 2  2.216 Computer Problem 2.218 Concept Problem 2.220 v H=F D h 5:630  10 5 0:8091 cos  x 2 i m = s a H=F D 5:237 sin  x 2  m = s 2 2.222 v A=B D 144:7 m = s 2.224 E v C =A D .6:857 O { C 30:27 O | / m = s : E a C =A D .1:376 O { 1:000 O | / m = s 2 : 2.226 .v P / avg D 17:30 ft = s 2.228 Computer Problem 2.230 v p max D 330:7 ft = s D 225:4 mph 2.232 E v B D 2:928 O { ft = s 2.234 v rain D 16:37 ft = s 2.236 E a B D L 1 R  cos  L 1 P  2 sin  C L 2 R  cos  L 2 P  2 sin   O { C L 1 R  sin  C L 1 P  2 cos  C L 2 R  sin  C L 2 P  2 cos   O | 2.238  D 63:57 ı 2.240  D 14:90 ı 2.242 E v B D 3:088 O | ft = s : E a B D 0:3192 O | ft = s 2 : 2.244 E v B D 2:000 O | m = s : 2.246 v Ay D 4:596 ft = s : 2.248 E a B D 11:10 O { m = s 2 : @  t d D 0:1644 s 2.250 a 0 D 0:9374 ft = s 2 2.252 P ` D dv 0 p h 2 C d 2 2.254 Component of E v C =B along BC D 0 8 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.256 r D q .d C  cos / 2 C .h C  sin / 2 P  D v 0 . C d cos  C h sin / .d C  cos / 2 C .h C  sin / 2 and P r D v 0 .h cos  d sin / p .d C  cos / 2 C .h C  sin / 2 2.258 t D 0:1556 s 2.260 R y e D 9:81 m = s 2 R y C ˇ ˇ t D 0:4391 s D 44:51 m = s 2 2.262 Concept Problem 2.264 Concept Problem 2.266 t D 83:82 s 2.268 E v C D . 6:500 O u R C 5:520 O u  C 5:300 O u ´ / ft = s E a C D . 0:6624 O u R 1:560 O u  / ft = s 2 2.270 r P .r R  sin  C 2 P r P  sin  C 2r P  P  cos / r P  sin .r R  C 2 P r P  r P  2 sin  cos / D 0; r P  sin . R r r P  2 r P  2 sin 2 / P r.r R  sin  C 2 P r P  sin  C 2r P  P  cos / D 0; P r.r R  C 2 P r P  r P  2 sin  cos / r P . R r r P  2 r P  2 sin 2 / D 0 2.272 E v D .0:04869 O u R C 7:961 O u  0:2999 O u ´ / m = s E a D . 79:79 O u R C 0:9738 O u  0:8120 O u ´ / m = s 2 2.274 ˇ ˇ E a  max ˇ ˇ D 12:36 ft = s 2 2.276 E v D df d´ P ´ O u r C K f .´/ O u  C P ´ O u ´ E a D  R ´ C P ´ 2 d 2 f d´ 2 K 2 f 3 .´/  O u R C R ´ O u ´ 2.278 a r D 0:1598 m = s 2 ; a  D 0:1884 m = s 2 ; and a  D 0:2232 m = s 2 2.280 E v P =B D .10:17 O { B C 10:33 O | B / m = s and v P =B D v P =A D 14:50 m = s E a P =B D .7:822 O { B 1:258 O | B / m = s 2 and ˇ ˇ E a P =B ˇ ˇ D ˇ ˇ E a P =A ˇ ˇ D 7:923 m = s 2 2.282 Concept Problem 2.284 Computer Problem 2.286 s D mg C 2 d  C d t C m  e C d m t 1  2.288 x D  L 0 C gm k  1 cos r k m t  2.290 v 0  p gR 9 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 2.292  D 240:5 ft 2.294 The car will lose contact with the ground. a c D 1:606 m = s 2 2.296 P r ˇ ˇ  D 180 ı D 0 P  ˇ ˇ  D 180 ı D 2:941 rad = s R r ˇ ˇ  D 180 ı D 40:85 m = s 2 and R  ˇ ˇ  D 180 ı D 0 2.298 P  D 0:053 91 rad = s R  D 0:2380 rad = s 2 2.300 v max D 6:020 ft = s 2.302 h max D 0:8898 ft 2.304 jE a A j D 18:65 m = s 2 2.306 ˇ ˇ E a ˇ ˇ D 172;700 ft = s 2 2.308 ˇ ˇ E a ˇ ˇ  min D 2:534 m = s 2 Chapter 3 3.2 Concept Problem 3.4 E a A D 2:147 O | ft = s 2 : 3.6 a max D 17:75 ft = s 2 3.8 a B max D 0:1650 m = s 2 D 0:5417 ft = s 2 3.10 R x C EA mL x D 0 3.12 a B max D 5:048 m = s 2 D 16:57 ft = s 2 3.14 d D 6:630 ft 3.16 d D 417:5 ft If an entire train weighing 30  10 6 lb was skidding to a stop, instead of a locomotive, we would have the same stopping distance because the weight does not appear in our solution. 3.18 E a D  C d m v 2 g  O | : 3.20 . s / min D 0:3436 3.22 v i D 5:297 ft = s 10 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 3.24 Computer Problem 3.26 v i D 32:68  10 6 m = s 3.28 v max D g r m k 3.30 Spring compression D 0:7367 ft 3.32 k D 5:084  10 4 lb = ft 3.34 ı max D 0:1675 m .F s / max D 586:1 N 3.36 Computer Problem 3.38 x stop D 0:1597 m 3.40 k D 5:930 lb = ft 3.42 k D 9034 kg = s 2 3.44 v.0/ D s 2 k m  x 2 0 C 2L 0  L q x 2 0 C L 2  3.46 Concept Problem 3.48 Concept Problem 3.50 Concept Problem 3.52  D 1:401  10 4 ft 3.54 No value of ! c can be found that would cause the person to slide up the wall 3.56 .F OA / before release D mg cos  ; .F OA / after release D mg cos ; . % change in F OA /  D 30 ı D 25:00 % 3.58 jE a j D 3:620 g j F L j D 43;060 N 3.60 R x C k m x 1 r u p x 2 C y 2 ! D 0; R y C k m y 1 r u p x 2 C y 2 ! D 0 3.62 R r r P  2 C k m .r r u / g cos  D 0 and r R  C 2 P r P  C g sin  D 0 3.64 t s D 0:7466 s 3.66 y D m C d tan  0  e C d x=m 1  m 2 g 2C 2 d v 2 0 cos 2  0  e C d x=m 1  2 11 Last modified: August 16, 2012

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    Engineering Mechanics: Dynamics 2e Gray, Costanzo, Plesha 3.68 a  D 0 D a ´ ; a R D g. s C tan / 1  s tan  D 120:1 ft = s 2 ; F  D 0; F D  s mg cos   s sin  D .83:16 ft = s 2 / m; N D mg cos   s sin  D .92:40 ft = s 2 / m; v max D p g s  s C tan  1  s tan  D 363:4 ft = s D 247:8 mph 3.70 F D 5971 lb 3.72 .! c / max D 2:733 rad = s 3.74 (a) E a D .318:8  10 3 O u r C 603:0 O u  / m = s 2 (b) P D 5:898  10 6 N and R D 11:33  10 3 N 3.76 ! D 9:939 rad = s 3.78  B D 41:81 ı 3.80 Concept Problem 3.82 R r r P  2 C G m e r 2 D 0 and r R  C 2 P r P  D 0 3.84 ı min D r 5mgr k v B D p 2gr 3.86 Computer Problem 3.88 h max D 0:3618 ft 3.90 R x L 2 y 2  C R yxy C x P x 2 L 2 y 2  C P y 2 L 2 x 2  C 2xy P x P y L 2 x 2 y 2 C gx q L 2 x 2 y 2 D 0; R y L 2 x 2  C R xxy C y P x 2 L 2 y 2  C P y 2 L 2 x 2  C 2xy P x P y L 2 x 2 y 2 C gy q L 2 x 2 y 2 D 0 3.92 R  P  2 sin  cos  C .g=L/ sin  D 0 and R  sin  C 2 P  P  cos  D 0 3.94 R r r P  2 D 0 and r R  C 2 P r P  D 0 3.96 M ´ D 2m! 2 0 r q r 2 r 2 0 j v r j D ! 0 q d 2 C 2dr 0 v D ! 0 q 2d 2 C 4dr 0 C r 2 0 3.98 Concept Problem 12 Last modified: August 16, 2012

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