Question

如圖所示，水平面上有一重40N的物體，受到F₁=12N和F₂=6N的水平力作用而保持靜止．已知物體與水平面間的動摩擦因數(shù)為μ=0.2，求：                                                                                                    

（1）此時物體所受的摩擦力多大？                                                                         

（2）將F₁撤去后，物體所受的摩擦力為多大？                                                      

（3）將F₂撤去后，物體所受的摩擦力為多大？                                                      

                                                                            

                                                                                                                               

Answer 1

解：物體所受最大靜摩擦力為：f_m=μG=8（N）

（1）由于F₁﹣F₂=6N＜f_m

所以物體處于靜止狀態(tài)，所受摩擦力為靜摩擦力：

故有：f₁=F₁﹣F₂=6N，方向水平向右．

（2）因為F₂=6N＜f_m，物體保持靜止，

故所受靜摩擦力為：f₂=F₂=6N，方向水平向左

（3）因為F₁=12N＞f_m，所以物體相對水平面向左滑動，

故物體受的滑動摩擦力：f₃=μG=8N，方向水平向右．