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BEcos={1.5,-3,3}/|{1.5,-3,3}|={1/3,-2/3,2/3}
BDcos={1.5,-3,-3}/|{1.5,-3,-3}|={1/3,-2/3,-(2/3)}
CFcos={-3,-6,2}/|{-3,-6,2}|={-3/7,-6/7,2/7}
M=Tbe*BEcos×AB+Tbd*BDcos×AB+Tcr*CFcos×AC={-780-2FA+2FB,0,-1170+FA+FB}=0
解得:
Tbe=390
Tbd=780
Tbe*BEcos(i)+Tbd*BDcos(i)+Tcr*CFcos(i)+Ai=0,i=1,2,3
解得:
Ax=A1=-195
Ay=A2=1170
Az=A3=130
BDcos={1.5,-3,-3}/|{1.5,-3,-3}|={1/3,-2/3,-(2/3)}
CFcos={-3,-6,2}/|{-3,-6,2}|={-3/7,-6/7,2/7}
M=Tbe*BEcos×AB+Tbd*BDcos×AB+Tcr*CFcos×AC={-780-2FA+2FB,0,-1170+FA+FB}=0
解得:
Tbe=390
Tbd=780
Tbe*BEcos(i)+Tbd*BDcos(i)+Tcr*CFcos(i)+Ai=0,i=1,2,3
解得:
Ax=A1=-195
Ay=A2=1170
Az=A3=130
追问
看不懂啊,能解释一下吗?那两个公式怎么用?我想理解了自己做。
追答
先求力的方向余弦,由力和力臂矢量叉乘得力矩矢量和,平衡则M=0,并可解得弹力,再由三个方向力平衡求A的支反力。
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