houbu
幼苗
共回答了21个问题采纳率:95.2% 举报
(1)取十字杆(包括绳及滑轮)为受力分析对象:
ΣMD =0,(FAx)(10r) -P(7r) =0
(FAx)(5m) - 300N(3.5m) =0
FAx = 210N(向左)
ΣFx =0,FDx -FAx =0
FDx -210N =0
FDx =210N(向右)
ΣFy =0,FDy -P =0
FDy =300N(向上)
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(2)取滑轮H为受力分析对象,设滑轮轴心为H
滑轮绳张力TG =P =300N
ΣFx =0,FHx =TG =300N(向右)
ΣFy =0 FHy = P =300N(向上)
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(3)取EBH横杆为受力分析对象:
FHy的反作用力FH'y =300N(向下)
ΣMB =0,-(Tec)(4r.sin45) -(FH'y)(6r) =0
-(Tec)(4x0.707) -(300N)(6) =0
EC绳张力 Tec ≈636.5N
1年前
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