(1)取ABC杆为隔离体:
ΣMA =0, FDx.6cm -F.4cm =0
FDx =(2/3)F,(向左)
.
DE为二力杆, 力作用线通过DE连线,
|FDy|/|FDx|=6/8,
即: FDy =(6/8)|FDx| =(6/8)|(2/3)F| =F/2,(向上)
.
ΣFy =0, FAy +FDy -F =0
FAy +F/2 -F =0
FAy =F/2 =200N/2 =100N(向上)
.
ΣFx =0, FAx -FDx =0
FAx -(2/3)F =0
FAx =(2/3)F =(2/3).200N ≈133N(向右)
.
(2)取结构整体为受力分析对象:
ΣFx =0, FAx -FEx =0
133N -FEx =0
FEx =133N(向左)
.
ΣFy =0, FAy +FEy -F =0,
100N +FEy -200N =0,
FEy =100N(向上)
兄弟,答案不对啊
怎么答案两个都是167n
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