(完整word版)理论力学答案(谢传峰版)

FA1—5aN

W

W

:F126F21.63F2B

60oFCD

解:2-6a1

1F3FF1FiF2Fi,

F3

FRFi3Fj，MA F BM0NDaGlcosFlcos0A AOB与Oyz夹角为,绳与轴Ox的平行线夹角为，已知F62FF42FF112FFpsin450N0

FfN2MfS22pDfS2M0N1p(1fS2)0 M0 FaPcosaPsina0NB

M0

FaPcosaPsina0NAFBfs2FNBfs2fs123

FA2M(方向如图所示)F32(M2qa)(受压)x(1）

FT FNP（2)FsfsFN,FNF0

FFF2P0ND

FFSDFSE1FFmin{4P,4P,4fsP,4fsP}4P

RR31fs13fs R，Stan12

S

C

S

F

B

rAOA，rBOB，rCO1CrDO1D，rBrC,rDrE zC

alcoszCa2lsinsin

PzCP(

lsin)0lsin0arcsin(2a)13zARzCRlcoszCR2coslsinsin

PzCP(

coslsin)0coslsin0zAzBasinzCzAzBacosl2asinlacosPzCF2l(PacosF2acos)0F2k(2asina)[2Pcoska(2sincos)]a0a(2sincos)FAxxA00可得：FAx0（1)y21yC1yA1yC1yA(FAyF11F2F31M)yA0

F1y1F2y2F3y3M0

(2） y3yC3 y2(MA2F1F23F3M)0对任意0可得：MA7(kNm)，逆时针方向。FBrBcos450MF2y2cos1500F3y30

rB62 (6FBM93F23F3)0FAxxAF1x1F2x2cos12000

（2) (3):y2y31yC1yA

1y21yA(FAy1F33F21M)yA0 x2xC6rDa,rK rErDa P1F1 rErDarDrEaF1rDcos1200P2rDcos1500P2rKcos12000(F123P2)a0P23F1对任意0可得： 6（受压）.rDa,rK rErDa rC0F1rDcos600P3rEcos1500P3rKcos12000(F123P3)a0 6（受拉)。VV弹VFdV0有:sin0和kb(1cos)2F0即：0和cos12F也就是：y0和y2F(kbF)两个平衡位置。d2V(kb2F)bcoskb2cos2d2V2Fb0，即y0时是不稳定平衡。当cos12F时，d2V4F(kbF) (1）VmgzCmg[(Rr)cos4cos()]Vmg[(Rr)cos4cos(Rr)]dVmg(Rr)[4sin(Rr)sin] d2Vmg(Rr)[4(Rr)cos(Rr)cos]1－3llk4lkcos2cos2 ay2lksin83lk 1－6cosyan,所以：aanv 1－7

1－10

F 1－11xvA

（a) xRx2F

sinR,cosx2R4x,y0FmRx5,FNmgmRx51－13由此可得：vcosve，OC杆的角速度为ve,OAl，所以vcos当450时，OC杆上C点速度的大小为：vCaavcos45av1－15cos30vMva2ve2v220.529m/s1－17 vevaR，vavrR,1OeC由上式解得:1ae322R121－19 vMvAva2ve2b22m/s,vrveb2m/s,1vA2242rad/sata aat167215.23m/s21aa10.16rad/s2tvrBr2rvatanve2rtan60023r，vraeOM2r02r2aeM

1－21 其中：vavB,veRB,vA， aanvr1.78m/s2rFO

Fraan

由此可求出：vavevcoscos

因为aanvavaataantanvsin. F

dtddt

RFRmartmdvrFecosdRdtdtvrdvrR22r2

vrR2RO2F

以为tandy,yc，dyc,因此tanc2tanP（c)FPP11P1xP1c3 2F

dpxmbFfFf(mm)gdpxmb(mm)aFfFf(mm)g由此解得平台的加速度为：am2bfg（方向向左)F

2－4 2－5 mdvFmgFNvrdmln(qvrfv)fln(m0qt)c u

dLO[J(2ar)r2] 2－14 ,aAdvA

F（2)vr4gR,ve1gR215F

F

（b)其中相对加速度为已知量，arae473g,F94mg,FN3.6267mg2－18 （a)

1111而2arccos

3333由此解得：mvrcosvrcos m,tanh

T10,T21m0r221mva2W12mgnh 将vrcos代入上式，可求得：v LOMOdtddt drd 3－3 可得：vcos300vl， vBvBCva23lvAvDcos300，vD43l3－4 ABvA3rad/svB15rad/sOB4vB0,OB0ABvA10OBaB302(瞬时针） 3－7

IIanBAa1acosaOaOaAaOaOtaOtr2IIasin,asin r1r22(r1r2)aOacosa1, OOacosa1 3－9 vvaRrRrvORvRaOvOvRaRRrRranBOa

aBa2a2

aCaOaCOaCn0aCxaOaCO0aCyaCO3－10

vB4m/s

aBatBanaAatBAaBanvB8m/s2BvBatB aC(an)2(aCB)282m/s23－13 vA1rad/s aBaBA0.8m/s2aaaearaK

aa2m/s2u

vOvave3u,vrve2u

vBvOvBOvByvOvBOcos3023uu

vBv2v213u (a）vnu2，Oar3u2,aa2u2,圆盘的角加速度为：art3u2

aBaOatBOanBO （b) aBxancos300atBOsin300aByaOansin300atBOcos300r3u2由此可得：aB37uvevatan300

vCDve3rvayveyvryveyO2ABCBC1,vr(2

aAaaaCaetCar 3－16(d）取BC杆为动系,套筒A为动点,BC1,vr(2

(见3－15d）BC为BC杆的角加速度.aCaCOaCOa'ra'K aC432r,BC323－17 ABvAvAAP2RABABvA0cos45aByaAansin4503vAaBa2a210vAvCavCevCr vDavDevDrvDaxvDevDrsin300

vDayvDrcos3003raCaCaaCeaCraCKaDaDaaDeaDraDKaDaxaDrsin300aDKcos30092raDayaDeaDrcos300aDKsin300332r3－21 maCFcosFSaCFr(rcosr0)m(r22)，FNmgFsin3－22 1ml2Flcos1mR2FR

(b）

(c)FSfFNf(m1m2)g

F由(a）(b)（c)（d)(e）联立可解得：3－29

AB,vCl, 212

32m1g3－33 vCvAlABLAm2vCl1m2l2AB212m2vCl1m2l2ABlI212(c）vA2I3－34 LA(BC)mvC3l1ml2BCvCvCvCClAClBC 11ml25ml22lI

l1ml21ml21ml2212

B I1ml21ml2lI联立求解（a）(b）可得AC6I2.5rad/s23－35碰撞前，弹簧有静变形(m1m3)vm32gh k，将

k,则

F

AC3vg f()asingcos3vgtan*a a2g2,3vga2g23vg03v28l(ga2g2)FCImaCmABlMCI1ml2ABmglcosFCIlMCI0 4－7 FCIa由此求得:aA1.25g,aC0.625g FCIymCaCy

即：af(gasin300) F4－8 FInm22rFItm2rMI1m(2r)22rMFr0，4－9 FI1mOAl,FI2m(ABlOAl),MI11ml2OA,MI21ml2AB, FI1lFI23l