如图所示，静止在水平桌面上的木块，在水平方向受到推力F１、F２和摩擦力。已知 F１＝10N，F２＝2N，若撤去F１，木块在水平方向上受到的合力为（ ）
A．8N
B．2N
C．10N
D．0
答案：D
题目分析：木块开始在水平方向受三个力而平衡，则有
；物体处于静止状态，则说明物体受到的最大静摩擦力大于8N；撤去
后，外力为2N，故物体仍能处于平衡，故合力一定是零；
故选D
点评：本题的关键是判断撤掉一个力后物体是否滑动，从而根据受力确定最大静摩擦力的范围。
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