T1cos30°+T2cos45°=0
√3T1+√2T2=0 (1)
T1sin30°+T2sin45°=600
T1+√2T2=1200 (2)
(2)-(1)
(1-√3)T1=1200
则 T1=1200/(1-√3)=-1200(1+√3)=-150(1+√3)
T2=...
取绝对值得
.
T1cos30°+T2cos45°=0
√3T1+√2T2=0 (1)
T1sin30°+T2sin45°=600
T1+√2T2=1200 (2)
(2)-(1)
(1-√3)T1=1200
则 T1=1200/(1-√3)=-1200(1+√3)=-150(1+√3)
T2=...
取绝对值得
.