Question

如图所示，水平面上有一重40N的物体，受到F₁=12N和F₂=6N的水平力作用而保持静止．已知物体与水平面间的动摩擦因数为μ=0.2，求：                                                                                                    

（1）此时物体所受的摩擦力多大？                                                                         

（2）将F₁撤去后，物体所受的摩擦力为多大？                                                      

（3）将F₂撤去后，物体所受的摩擦力为多大？                                                      

                                                                            

                                                                                                                               

Answer 1

解：物体所受最大静摩擦力为：f_m=μG=8（N）

（1）由于F₁﹣F₂=6N＜f_m

所以物体处于静止状态，所受摩擦力为静摩擦力：

故有：f₁=F₁﹣F₂=6N，方向水平向右．

（2）因为F₂=6N＜f_m，物体保持静止，

故所受静摩擦力为：f₂=F₂=6N，方向水平向左

（3）因为F₁=12N＞f_m，所以物体相对水平面向左滑动，

故物体受的滑动摩擦力：f₃=μG=8N，方向水平向右．