理论力学全套解疑05

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WFmax=fWFmax=fNWN=W5-1aQ5-1(b)5-1Nααsin0sin,0QWNWQNy−==−+=∑α5-1(c)ααcos0cos,0WNWNy==−=∑5-1(d)QNQNX==−=∑0,0fNF=fNFmax=fN5-2(a)W=1000(N)P=100(N)f=0.4FFmax5-25-2b030sin,0=+−=∑DPWNyN=W−Psin30°=950(N)0cos,0=−=∑FPXαF=Pcosα=86.6(N)FmaxFmax=fN=0.4950=380(N)F=86.6(N)Fmax=380(N)Wf5-3(a)(b)P1P2αP5-3PP1P25-3(a)5-3c(1)0sin,01=+−=∑αPWNyN=W−P1sinα(2)0cos,0max1=−=∑FPXαFmax=fN(3)ααsincos1ffW+=P5-3(b)5-3(d)(4)0sin,02=−−=∑WPNyαN=P2sinα+W(5)0cos,0max2=−=∑FPXαFmax=fN(6)ααsincos2ffW−=P5-3(a)121Pα5-41(a)xPx=Psinα5-41b5-41Q5-42(a)xααcossinQPX−=∑PsinαQcosα05-42(b)PsinαQcosα05-42(c)2Gα5-43(a)fP5-425-43BABBAAAGNBFBNAGNANBO5-43gO0)(=+−=∑RFFmABORFRFA5-43g12F=Fmax1F2F=Fmax=fN5-5(a)ABABP1B2lBf1BFmaxFF5-5b5-50,00,0=−==+=∑∑PNYFNXBA0sincos21,0)(=⋅⋅−=∑ααABNBPmABFααctg2,ctg2,PFPNPNAB−===F2BF=FmaxFmaxBBFmax5-5(c)0,00,0max=−==−=∑∑PNYFNXBA0sincos,0)(=⋅−⋅⋅=∑ααlNsPmABFBfNF=maxαtg,,max⋅⋅====flsfPFNPNABFmax5-5(d)AB0,00,0max=−==+=∑∑PNYFNXBA0sincos,0)(=⋅−⋅⋅=∑ααlNsPmABFBfNF=maxαtg,,,max⋅⋅−=−===flsfPFfPNPNABsFmaxFmax=fNFmaxFFMFPF5-65-6SNFRF=FmaxRRSR1S5-7(a)5-7Sαϕαϕααcos0cos,0SNSNY==−=∑)tg(costgmaxϕαϕ=⋅⋅==fSfNFFααsin0sin,0SFSFX==−=∑FFmaxαϕαcostgsinmaxSFSF==αϕtgαtgϕsinαtgϕ⋅⋅cosαF=FmaxSFFmax2S5-7(b)αϕαcos0cos,0SNSNY==−=∑αϕcostgmaxSF⋅=Fααsin0sin,0SFSFX==−=∑FFmaxαϕαcostgsinmaxSFSF==αϕtgαtgϕsinαtgϕcosαFFmaxFFmaxQϕ2°P(a)PQPθ25°P5-8PQSPQSθα21=12.5°5-8bS5-8cGα=30°5-9(a)fPmin5-95-9(b)FB0cos,00sin,0=−==−−=∑∑ααGNYGFPXBFB=fN)sincos(αα+=fGP1C25-9cFBAFA0,0)(=−=∑RFRFmBAOFRBFAR12P1(1)0sin,0=−−=∑αGFNXBA(2)0cos,0=−−=∑αGFNYAB0,0)(=−=∑RFRFmBAOF(R)(3)BFB=fNB(4)ffGGNA−−+=1)sin(cossinααα5-9(c)ffGGNNPNPXAAA−−+==′==′−=∑1)sin(cossin0,0ααα21(1)0sin,0=−−=∑αGFNXBA(2)0cos,0=−−=∑αGFNYAB(3)0,0)(=−=∑RFRFmBAOFAFA=fNA(4)fGNA−=1sinαfGNNPNPXAAA−==′==′−=∑1sin0,0αPPPminα=30°0)sin(cos−ααfGPPfGPP−==1sinminα

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