如图6所示，质量为mB＝24kg的木板B放在水平地面上，质量为mA＝22kg的木箱A放在木板B上．一根轻绳一端...

问题详情：

如图6所示，质量为mB＝24 kg的木板B放在水平地面上，质量为mA＝22 kg的木箱A放在木板B上．一根轻绳一端拴在木箱上，另一端拴在天花板上，轻绳与水平方向的夹角为θ＝37°.已知木箱A与木板B之间的动摩擦因数μ1＝0.5.现用水平向右、大小为200 N的力F将木板B从木箱A下面匀速抽出(sin 37°＝0.6，cos 37°＝0.8，重力加速度g取10 m/s2)，则木板B与地面之间的动摩擦因数μ2的大小为                    (　　)

图6

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

【回答】

A

解析　对A受力分析如图*所示，由题意得

FTcos θ＝Ff1                                                                                                                                                                                                                       ①

FN1＋FTsin θ＝mAg                                                                                                         ②

Ff1μ1FN1                                                                                                                                                                                                                             ③

由①②③得：FT＝100 N

对A、B整体受力分析如图乙所示，由题意得

FTcos θ＋Ff2＝F                                                                                                      ④

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

Ff2=μ2FN2                                                                                                                                                                                                                             ⑥

由④⑤⑥得：μ2＝0.3，故A选项正确．

知识点：共点力的平衡

题型：选择题