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如图所示，质量为1kg的物块放在倾角为37°的固定斜面上，若用一平行于斜面的力F推物块，物块和斜面保持静止，斜面与物块之间的最大静摩擦因数μ=0.5，重力加速度g取10m/s2，则外力F的大小可能为（　　）

A．1N

B．6N

C．9N

D．12N

答案

物体重力的沿斜面向下的分力为：

G1=mgsin37°=10×0.6=6N；

垂直于斜面的分力为：

G2=mgcos37°=8N；

则最大静摩擦力为：

Fm=μFN=μG2=0.5×8=4N

若推力向上，由共点力的平衡条件可知，推力最小值为等于重力沿斜面向下的分力；故F=6N；此时静摩擦力为零；

增大推力后，摩擦力变大，当达到最大静摩擦力时，推力最大为F′m=4+6=10N；

而当推力向下时，最大静摩擦力小于重力的分力，物体不可能处于静止状态；

故外力最大为10N，最小为6N；

故BC正确；

故选：BC．

填空题

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单项选择题

对于低余额提醒业务，描述正确的是（）。

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B、一般情况下无需发送，等有客户反馈相关问题时才有必要发送

C、不能在短信中显示客户卡的余额和卡号（后四位）

D、以上说法都不对
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