题目内容
如图所示,重为100N的物体受推力F作用贴于墙面静止,F与墙的夹角θ=600,墙对物体的最大静摩擦力为40N,要使物体保持静止,求推力F的大小范围。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241630134793930.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241630134793930.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824163013682777.png)
试题分析:由题意,对物体受力分析知
⑴若物体有下滑趋势,则静摩擦力向上,当静摩擦力达到最大时,F有最小值
如图所示,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241630138222130.jpg)
根据平衡条件有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241630138381002.png)
解得:Fmin=120N 2分
⑵若物体有上滑趋势,则静摩擦力向下,当静摩擦力达到最大时,F有最大值
如图所示,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241630140413010.jpg)
根据平衡条件有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824163014056980.png)
解得:Fmax=280N 2分
综上所述,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824163013682777.png)
点评:关键是对物体进行受力分析,将物体相对墙的运动趋势分情况讨论
![](http://thumb2018.1010pic.com/images/loading.gif)
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