Re: [問題] 一題物理
棍子外露在碗外的長度X, 棍子與水平線夾角t, 棍子在碗內受力N1, 棍子在碗緣受力N2,
木棒重W
N1/sint = N2/sin(90-2t) = W/sin(90+t)
N1/sint = N2/cos2t = W/cost -----(A)
令棍子和碗內交點為支點
故 W[(L+X)/2]cost = N2*L -----(B)
利用A式N2/cos2t = W/cost, 化簡B式
得1–tan^2 t = (L+X)/2L ,再利用等腰三角形(腰r , 底L , 夾角t) 求tan t
2–(4r^2/L^2) = (L+X)/2L
得X= 3L-(8r^2)/L
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