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