q q HA MB VA VB
Transkrypt
q q HA MB VA VB
3U]\NáDG5DPD]DPNQL WD]HFLJLHP :\]QDF]\üUHDNFMH l q ql l l l 5R]ZL]DQLH 8NáDGVLáRWU]\PDQ\SRXZROQLHQLX]ZL ]yZSU]HGVWDZLRQ\MHVWQDU\VXQNXSRQL*HM 4 3 2 q ql 1 A HA B VA VB x MB 2EOLF]P\ UHDNFM +A MHG\Q SR]LRP Z\NRU]\VWXMF UyZQDQLH ∑P ix = 0 . Przyjmuje ono i SRVWDü+A - ql = 0 ⇒ HA = ql. 'R Z\]QDF]HQLD SR]RVWDá\FK UHDNFML NRQLHF]QH MHVW UR]G]LHOHQLH SRGSyU $ L % : roz- ZD*DQ\P XNáDG]LH HOHPHQW\ UDP\ WZRU] REZyG ]DPNQL W\ 3RZRGXMH WR *H GOD Z\RGU EQLHQLD GZX UR]áF]Q\FK F] FL QLH Z\VWDUF]\ UR]G]LHOHQLH MHGQHJR SRáF]HQLD 3RG]LHOP\XNáDGZSU]HJXEDFKL=DQLPZSURZDG]LP\VLá\Z]DMHPQHJRRGG]LDá\ZDQLDZ W\FK SXQNWDFK SU]HDQDOL]XMP\ HOHPHQW\ L 6 WR SU W\ QLHREFL*RQH ]DNRF]RQH SU]HJXEDPL = ZDUXQNX LFK UyZQRZDJL Z\QLND *H UHDNFMH QD NRFDFK PXV] PLHü NLHUXQHN RVLSU WyZ3URZDG]LWRGRXNáDGXVLáSU]HGVWDZLRQHJRQDSRQL*V]\PU\VXQNX I S2 S2 2 S1 q ql S1 ql MB VA VB x ∑M Z równania I i2 = 0 PR*QD Z\]QDF]\ü QLH]QDQ\ PRPHQW UHDNF\MQ\ 0B (w punkcie 2 i SU]HFLQDM VL OLQLH G]LDáDQLD QLH]QDQHM UHDNFML 9B i niewiadomych S1 i S2, co eliminuje je z równania): ∑ M = 0 ⇒ -MB –ql l = 0 ⇒ MB = – ql. I i2 i =QDMF 0B PR*HP\ WHUD] ] UyZQD ∑M iA =0 i i ∑M iB = 0 REOLF]\ü UHDNFMH SLRQRZH i 5yZQDQLDWHSU]\MPXMSRVWDü VB 2l - MB+ ql l – ql l/2 = 0 -VA 2l + ql 3/2l + ql l – MB = 0, co daje VB = – 3/4 ql i VA =7/4 ql =HVWDZLHQLHREOLF]RQ\FKZLHONRFLSU]HGVWDZLDU\VXQHN q ql ql 3/4 ql 7/4 ql 2 ql2