Analytical Mechanics 7th ed

【实例简介】
Analytical Mechanics 7th ed [SOLUTIONS MANUAL] - G. Fowles, G. Cassiday (2005) WW.pdf
1.3 cos 6= B(a)(a)+(2a2a)+(0)(3a)5a2 AB √5a2√l4a 14 6 ≈53° )A=i+j+k gond -√:+)+kk=√ B=i+j: face diagonal B=|Bl=√2 )C=AxB=11 d )cosB A- B 0∴=90 AB L.5 B AC sina C=Csin o A·C= Ac cose C.=CcOs6日 dx式 A X 一Cx+ A t ABA A+-B×A 6-2()72(m)+()=in+m+ky j2B+k6 70=AB=(q)(q)+(3)(-q)+(1)(2)=q2-3q+ (q-2)(q-1)=0,q=1or2 1.8 +B=(+B),(i+B)=+B2+2B 小+=2+82+2AB Since d B= ARcos≤AB,A+剧≤|A+ IA+BI A ,剧=14Bcs6l-|4os414l 官 Bcos≤B A Bcos Show Ax((×C)=(4C)B-(不B)C ×B.B,B=(4C+C,+AC:)B-(1,,+B+4E)C CCC =(4,B,C+A, BC +A B,C: -A, B, C:-A,B, C-A, BC)i +(A, B, C+A,B, C,+A B, C:-A, BC-A, B, C, -A.B.C)3 +(A, BC+A, B C, +A.B. C: -,,:-4 B,C:-A B c:k (A, B, C,+A, B, C: -A,B, C, -4,B,C)i +(A, B,C,+A, B, C 2-A,B, Cr-AB.c)j +(A,B. C:+A, B, Cr-AB, C-A, B,C i (A, BCr-A, B, Cr-AB. C+AB,C =+j(4 B, C 2-A,B C,-1, B, C+ A.B.C B, C-B C, B,C-B. C. BC -B, C-+(B C-A,B,C: -A, B C:+A,B C, 风 dsin e A=2 5y+(B-x)=xy+1'B-ry=ABsin O A=A×B k .b B (B×C)=41.B,B=.B,a B-(Ax C. C. C.C. C. C I12 A=(Ax, Ay, Az ,B=(0, By. )and C =(0. Cy, Cz) Cz is the perpendicular distance between the plane A. B and its opposite. li =BxC is directed along the x-axis since the vectors B. c are in the y z plane. Brd=BC is the area of the parallelogram formed by the vectors B. C. Multiply that area times the height of plane A, B= Ax to get the volume of the parallopiped V=A, ,A,B,C:=A(Bre a3 For rotation about the z axis SIn sin For rotation about the y axis ↑l sin b k.沪 sin COS sin 0( cosφsinφ cosTcoφ coso sinφ-sin T COS Sin COS sin e cos p sin 8 sin cos 0 1.14 2.氵=cos30° j·i=sin30 k.1=0 ? /3 sin 30 j·=c0s30° k·j=0 k·k √3+ √3 0‖13 A=3232+1.598-k 1.15 1. Rotate thru about z-axis d=45 2. Rotate thru 0 about x-axis 0=45 3. Rotatc thru y about z-axis y=45 3,z 0 01 100 2 R=0 0 √2√2 0 2√2 + 2222√22 R(, 0,0)=R,RoRA 十 2 √222√2 or o=(y0.pB Condition is;x"=R℃ where x'=o and r Since·i=l we have:v+β2+a2=l After a lot of algebra: a B=- y 116Ⅴ=vz=ct =y+一h=C+n at I =√ bc and a=c÷+cr cos 6 va bc√2c2 =45 1.17 v((=-ibosinor)j2b@ cos(or) 1=(bo'sin'of+4b 2cos'cor)=b(+3 cus'or' (O=-ib@ cos of-j2b@ sin or l=bo2(1+3m2on)5 2b0 可=bo 1.18 v((=ibo cos aot-jbosin at+k2ct ibo sin (t-jba cos af +k2. =(b t+b0 cos at+ 4 b2a)2+4 19=e+p bke (F-p)4+()+2:0)n=b( e+ bck k(k2-c2e2+262cke φ 4c2k2 kk+c cOS中= a constant (k2+c2)(k2+c)(k2+c 1.0a=(R-R)4e+(2R+R9) d==bore,+2ce =(b23+4c2 121F()=i(1-e") F(0=ike"*+jke Ike+k'e Trajectory: k= 56 52 1234567890 62 26200 02 0 06 0.8 1.22卩=+psin+e eiybo sin 5 1+cos(4an) o sin(4or F=eba cos -cos( 4ct D-eobo"sin(4or) 8 bol cos-cos 4ot+=sin'4or 8 Path is sinusoidal oscillation about the equator 123p…=y ivy 2v·d=21 8 124 F(× (×a)+F·(下×) v(vxa)+i l 0+F[0+(5×a) F·(×a)=F( =ⅵand=a, S CT 1.26Forl.14. be)3coso1· Sin at+b2u3 Sinl(t· cOS(r+4c2 bo cos at+ba sinat +4c 4c1 b+4cr b2o2+4c2 16c 60+4c b2k( k2-c)e+2b2c'ke For I15 C be k+c b2e2(k2+c2)-b2k2e2(k2+ bce"(k 1.27ν=v,a=v+—n ×a 9 【实例截图】
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