题目内容
如下图所示,A、B两物体叠放在水平地面上,已知A、B的质量分别为mA=10 kg,mB=20 kg,A、B之间,B与地面之间的动摩擦因数均为μ=0.5.一轻绳一端系住物体A,另一端系于墙上,绳与竖直方向的夹角为37°,今欲用外力将物体B匀速向右拉出,求所加水平力F的大小,并画出A、B的受力分析图.(取g=10 m/s2,sin 37°=0.6,cos 37°=0.8)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242144271732402.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242144271732402.jpg)
160 N
试题分析:A、B的受力分析如下图所示.
对A应用平衡条件
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824214427469987.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242144285921019.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242144297163594.jpg)
联立①、②两式可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242144297311237.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824214430995869.png)
对B用平衡条件
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242144317592426.png)
点评:本题是两个物体平衡问题,采用隔离法研究,关键是分析物体的受力情况,作出力图.
![](http://thumb2018.1010pic.com/images/loading.gif)
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