题5.3图 对物B(图(a)),有F1=Fn,且 FsB=s? FNB ΣX=0,Fmn-Fs-Fs-Fn=0 ΣY=0,FANB-PB-FM=0 解得Fmin=3200N 5.4已知P=400N,直径D =0.25m,欲转动棒料需力偶矩M= 15N m i 求棒料与V形铁间的摩擦系数 ∫s 解棒料受力如图,在临界状态 FsB= fsF FsA= 题5.4图 2X =0, FNA FsB-P sin45 =0 2Y=0, FNB- FsA-P cos45=0 ∑M(F)=0,(F+F)9-M=0 解得 ∫s=0.223