Solution manual for statics and mechanics of materials 1st edition by goodno

Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.2-1   8 G  3.439 10 ft lbf  s Part (a) rplane  rearth  5280 rplane  3965.629mi  23 mearth = 4.0925 ×10 ⋅slug ft mi mearth  mplane rplane  g  32.174 sec ft 24 Part (b) 35000 ft mearth = 5.9726⋅10 kg Fplane  G rearth  3959 mi  155141.821lbf  Wplane  mplane g  1.555  10  lbf  3 mplane  4.8331 10  slug Fplane  155.1 kip Wplane  155.5 kip © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.2-2 G  6.673  10  11 m kg s 24 mearth  5.9726 10 kg 22 mmoon  7.3477 10 kg RmoonE  378000km 27 mjupiter  1.8983 10 kg RjupiterE  588.5  10 km 23 mmars  6.4174 10 kg RmarsE  55.7 10 km 26 msaturn  5.6836 10 kg ( a) FmoonE  G RsaturnE  1.195  10 km mearth mmoon RmoonE ( c) FmarsE  G mearth mmars 17  2.05  10  kN ( b) FjupiterE  G mearth mjupiter RjupiterE 13  8.244  10  kN ( d) FsaturnE  G RmarsE mearth msaturn RsaturnE 17 2 15  2.185  10  kN 14  1.586  10  kN 15 ( a) FmoonE  2.05  10  kN ( b ) FjupiterE  2.185  10  kN The moon has the largest gravitational interaction with earth 13 ( c) FmarsE  8.244  10  kN 14 ( d ) FsaturnE  1.586  10  kN © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.2-3  23slug WmanE  180lbf mearth  4.092  10  3   8 G  3.439  10 ft 4 lbf  s 21 d  238.9  10  mi mmoon  5.035  10  slug rearth  3959 mi rmoon  1080 mi g  32.174 mearth mman rearth s mman  5.6 slug Fearth  G ft  180.4  lbf Fmoon  G mman mmoon rmoon  29.8 lbf Fmoon  29.8 lbf â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.2-4 m1  50kg m2  150kg R1  50km R2  25km 24 mearth  5.9726 10 kg rearth  F1E  G F2E  G mearth m1 R1  rearth mearth m2 R2  rearth  0.483  kN  1.461  kN 12742  km  6371 km m1  g  0.49 kN m2  g  1.471  kN Mass has the larger gravitational force â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-1 V   10     15  dircos  V V   0.555     0.832  nV  V V   0.555     0.832  nV  nV  i  nV  j nV   10  V  V  nV     15  n V  ( 0.555 )  i  ( 0.832 )  j Direction cosines Vx  10 Vy  15 θx  acos( l)  56.31  deg l  Vx V  0.555 m  Vy V θy  acos( m)  33.69  deg  0.832 n  cos( 90deg)  θz  acos( n )  90 deg © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-2 V1   10     15  V2       12  Vs  V1  V2   18    3 2 Vd  V1  V2     27  Vs  18.248 ns  Parallelogram Law Vs Vs  Vd  27.074  0.986     0.164  nd  15 θ1x  atan   56.31  deg  10  10 θ1y  atan   33.69  deg  15  12 θ2x  atan   56.31  deg 8 θ2y  atan   33.69  deg  12  Vd Vd   0.074  < unit vectors    0.997  α  θ1y  θ2y  67.38  deg θsx  atan   9.462  deg 18   θ1x  θ2x  112.62 deg sin( α) Vsmag   V2  18.248 sin θ1x  θsx   Vs  Vsmag n s   18    3 θdy  atan   4.236  deg 27   sin( α) Vdmag   V2  27.074 sin θ1y  θdy   2 Vd  Vdmag n d     27  © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Triangle Law αs  θ1y  θ2y  67.38  deg Vsmag   V1 2    V2 2  2   V1  V2  cos αs  18.248  αd  180deg  θ1y  θ2y  112.62 deg Vdmag   V1 2   V2 2  2   V1  V2  cos αd  27.074 © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-3 V1   10     15  V2  V1  18.028      12  V2  14.422 V1  V2  32.45 Vs  V1  V2   2    3  Vs  3.606 Vd  V1  V2   18     27  Vd  32.45 V1  V2  3.606 10 θ1x  atan   33.69  deg  15  15 θ1y  atan   56.31  deg  10  θ2x  atan   33.69  deg 12   12 θ2y  atan   56.31  deg 8 Sum and difference vectors have equal/opposite lines of action (collinear, see fig.'s below) so sum/difference of magnitudes are equal to sum/diff of magnitudes of each vector Unit vectors along sum and difference vectors ns  nd  Vs Vs Vd Vd    0.555     0.832   0.555     0.832 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-4 Vector sum 4  Vs   8    7  Unit vector along V1  0.635  n   0.773      dot product of Vs along n1 is V1  2.314  V1  Vs n  n   2.817      vector difference gives V2  6.314  V2  Vs  V1   5.183      V1 V2 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-5   5  V1    V2   Vy     14   3    V1  V2  35   Vy  42 Orthogonal so 35   Vy  42 = or  Vy  35   Vy  42 = solve Vy  Check 5 7          3   14  10 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-6 distance vector from C to A 0  d CA     4  5  5   9  m    0   distance vector from B to C 5  d BC    m    4  unit vector along CA force vector F  0.453  n CA    0.815    d CA  0.362   8.148  F  ( 18N)  n CA   14.667  N    6.519  d CA distance vector from B to D 5 d BD    m    4  11 © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-7 From Prob 1.3-6 l  8.15 N F  0.453  n CA   0.815     0.362   0.453 m   8.15  F   14.67  N    6.52  14.67N F  0.815 12 F  18 N n  6.52 N F 0.362 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-8 2 l m n =1 l  m m  n n  so l  n m  n l F  F  m    n  F  1     1 F or F  F  ( i  j  k) If we use F = 18 N from Prob 1.3-6 F  18N 13 F     10.39       10.39  N        10.39  18 N â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-9 F1x  ( 13lbf )  cos( 30deg)  11.26  lbf F1y  ( 13 lbf )  sin( 30deg)  6.5 lbf F2x  ( 15lbf )  cos( 45deg)  10.61  lbf F2y  ( 15lbf )  sin( 45deg)  10.61  lbf 14 © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-10  60  F   120  kN    30  F  137.477 kN 60 θx  acos  F   64.1 deg   120  θy  acos   150.8  deg  F  30 θz  acos  F   77.4 deg   Also use unit vector along F to find direction cosines  25  nF    50   deg   F  12.5  F 15 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-11  cos( 30deg)    11.258   lbf F1  13lbf       sin( 30deg)   6.5  Part (a)  cos( 45deg)    10.607   lbf F2  15lbf       sin( 45deg)   10.607   21.865   lbf    4.107  Sum Fs  F1  F2  Difference  0.652   lbf Fd  F1  F2     17.107  Unit vectors ns  Fs Fs   0.983     0.185  Fs  22.247 lbf Fd  17.119 lbf nd  Fd Fd   0.038     0.999  Part (b) rOP   10in     12in  rOP  15.62  in n OP  rOP rOP 16   0.64    0.768 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-12 α  22deg β  40deg Sum and difference vectors  cos( β)    114.907  N F2  150N      sin( β)   96.418   cos( α)    114.971  N F1  124N      sin( α)   46.451   0.1  N Fs  F1  F2     142.9  Fd  F1  F2  Fs  142.9 N Part (a) - vector sum using parallelogram law Fsmag   F1 2   F2 2  2  229.9  N    50  Fd  235.2 N F1  F2  cos( α  β)  142.9 N Part (b) - vector difference using triangle law  Fdx θd  acos  Fd Fdmag  17    12.26  deg  sin[ 180deg  ( α  β) ]  sin β  θd   ( 124N)  235 N © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.4-1 g  32.174 ft s lbf mUSCS  3500  108.8 slug g 18 mSI  14.5939 ( 108.8)  1588 kg â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.4-2 mc  ( 58  14)  kg  72 kg 72 14.5939  4.93 mc  4.93 slug Wc  mc g  158.7  lbf ( 4.93) ( 32.2) 158.746 19 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.4-3 h  2.37 in r  1.295  in V  3  π r  h  4.16 in L L  L 20 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.4-4 EAl  10600  ksi γAl  175  lbf ft Modulus E 10600  kip  4448.22  EAl  7.308  10 10  in   4.715  107 N  kip  N Pa  25.4 10  in 3 m 4  6.452  10 4.715  10 N m 4 6.452  10 EAl  7.308  10  kPa EAl  7.308  10  MPa 10  7.308  10 Pa m EAl  73.08  GPa Unit weight 175  lbf   4.44822   ft   778.438 N  lbf  N γAl  27.5 m r   in Mass of sphere mUSCS  γAl g 778.438 N  0.3048 m    ft   N γAl  2.75  10   0.02832  m 3 N  2.749  10  0.02832 m m kN m V   V  0.356  slug 3  π r  113.1  in 0.356  ( 14.5939 )  5.195 21 mSI  γAl g V 5.2 kg â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.4-5 W  15lbf W mUSCS   0.466  slug g 15 32.2 W mSI   6.8 kg g 15 ( 4.44822 )  0.466 9.807 22 6.804 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.5-1 αa  5deg αb  10deg     esa    eca    eta  cos αa  0.996 tan αa  0.087 αd  20deg   0.087   180  αa  0.087  rad sin αa  0.087 αc  15deg π   sin αa αa  sin αa αb  0.175  rad  0.127  % esc    ecc    etc  tan αc  0.268   ecb    etb  cos αb  0.985    0.254  % tan αa tan αb  0.176 αa  tan αa 20  αd  0.349  rad   cos αc  0.966 esb     91.24  % cos αa αa  cos αa αc  0.262  rad sin αc  0.259   sin αb  0.174   sin αc αc  sin αc  1.152  % esd    ecd    etd     72.897 % cos αc cos αd  0.94    2.295  % tan αc tan αd  0.364 αc  cos αc αc  tan αc 23    82.277 % cos αb  αb  cos αb   tan αb  αb  tan αb  1.017  %   0.349   180    sin αd  0.342 π    0.51 % sin αb  αb  sin αb    2.06 % sin αd  αd  sin αd    62.853 % cos αd  αd  cos αd   tan αd  αd  tan αd  4.095  % © 2019 Cengage Learning® All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.5-2 R( α)  sin(  α)  cos(  α) cos( α)    Rapprox( α)   cos(  α) error αa  1.28 %   (  α) ( ) 1 error αc  12.77  % 24 R( α)  Rapprox( α) R( α)  ( 1)   error αb  5.33 % error( α)    error αd 25.02 % â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.5-3 P  2.728 P  Q  10.66 ( P  Q)  113.6 Q  7.93 P  Q  5.20 P Q  0.344 P Q 2.28 25 â 2019 Cengage Learningđ All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.3-12 α  22deg β  40deg Sum and difference vectors... in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.2-3... in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 1st Edition by Goodno Full file at https://TestbankDirect.eu/ Problem 1.2-4