BD是二力杆，从几何尺寸知，|Fby|=|Fbx|=|Fb|sin45
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取ABC杆为隔离体, FBx向左， FBy向下 :
ΣMa =0, |Fbx|(0.2m) +|Fby|(0.1m) -F(0.7m) =0
将|Fby|=|Fbx|代入上式得：3|Fbx| -7F =0
|Fbx| =|Fby| = (7/3)F
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ΣFy =0, Fay - |Fby| =0
Fay =(7/3)F（向上）
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ΣFx =0, Fax -|Fbx| +F =0
Fax -(7/3)F +F =0
Fax = (4/3)F
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