第五章单元测试
The block weighs W, and an external force F acts on the block with the action line outside of the angle of friction, as shown in the figure. Given θ=25°, angle of friction φm=20°, F=W, then the block is at rest.（ ）
As shown in the figure, split with a steel wedge and the angle of friction between the contact surfaces is φm. In order to make the wedge do not slip out after splitting, the angle θ should be （ ）.- In the case of friction, the angle between the total reaction force and the normal reaction force is called the angle of friction. （ ）
As shown in the figure, given W=100kN, P=80kN, and the coefficient of static friction is fs=0.2, then the block will move upward. （ ）.
Rectangular cabinet as shown in the figure weights G, and center of gravity C is in its geometric center. The coefficient of static friction is fs between the cabinet and the ground. A horizontal force F is applied to the cabinet on right, then the minimum value of F required to make the cabinet be in equilibrium is （ ）.
A:错 B:对
答案:对
A:θ≤φm B:θ≤2φm C:θ>φm D:θ>2φm
A:对 B:错
A:对 B:错
A:min{2Gfs, Ga/h} B:max{2Gfs, Ga/h} C:max{Gfs, Ga/2h} D:min{Gfs, Ga/2h}
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