如圖6所示，質量為mB＝24kg的木板B放在水平地面上，質量為mA＝22kg的木箱A放在木板B上．一根輕繩一端...

問題詳情：

如圖6所示，質量為mB＝24 kg的木板B放在水平地面上，質量為mA＝22 kg的木箱A放在木板B上．一根輕繩一端拴在木箱上，另一端拴在天花板上，輕繩與水平方向的夾角為θ＝37°.已知木箱A與木板B之間的動摩擦因數μ1＝0.5.現用水平向右、大小為200 N的力F將木板B從木箱A下面勻速抽出(sin 37°＝0.6，cos 37°＝0.8，重力加速度g取10 m/s2)，則木板B與地面之間的動摩擦因數μ2的大小為                    (　　)

圖6

A．0.3                       B．0.4                 C．0.5                       D．0.6

【回答】

A

解析　對A受力分析如圖*所示，由題意得

FTcos θ＝Ff1                                                                                                                                                                                                                       ①

FN1＋FTsin θ＝mAg                                                                                                         ②

Ff1μ1FN1                                                                                                                                                                                                                             ③

由①②③得：FT＝100 N

對A、B整體受力分析如圖乙所示，由題意得

FTcos θ＋Ff2＝F                                                                                                      ④

FN2＋FTsin θ＝(mA＋mB)g                                                                                        ⑤

Ff2=μ2FN2                                                                                                                                                                                                                             ⑥

由④⑤⑥得：μ2＝0.3，故A選項正確．

知識點：共點力的平衡

題型：選擇題