对F3应采用直接投影法 4 Fy=Fsin ncos p Fy=F sin y sin p 60yF2 2.5m F COSr F3 BC 42+3 F B sin AB 0.8944 42+32+2.5 cOSy=0.4472 3m CD D q BO 422=0.8 BD 3 COS BC 42+3 Fx= F sin y Cos=1500×0.8944×0.6=805N F sin y sin=-1500×0.8944×0.8=-1073N F= FcOS y=1500×0.4472=671Nφ γ 对 F3 应采用直接投影法 cos sin sin sin cos F F F F F F zyx ===0 4472 0 8944 4 3 2 5 4 3 2 2 2 2 2 cos . . AB . BC sin = = + + + = = 4 m 2. 5m 3m x y z F 1 F 2 F 3 0 60 AC D B 0 6 4 3 3 0 8 4 3 42 2 2 2 . BC BD cos . BC CD sin = + = = = + = = Fx = F sin cos =1500 0.8944 0.6 = 805 N Fy = − F sin sin = −1500 0.8944 0.8 = −1073 N Fz = F cos =1500 0.4472 = 671 N