题目内容
如图所示,均匀木板AB长12 m,重200 N,在距A端3 m处有一固定转动轴O,B端被绳拴住,绳与AB的夹角为30°,板AB水平.已知绳能承受的最大拉力为200 N,那么重为600 N的人在该板上安全行走,离A端的距离应在什么范围?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241148371672261.jpg)
2 m≤x≤3.5 m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241148371832547.jpg)
作出AB板的受力图
人在O轴左端x处,绳子拉直拉力为零.由力矩平衡可得: G人×x-G×
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114837198253.gif)
x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114837214518.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114837230405.gif)
人在O轴右端y处,绳子的拉力T="200" N,由力矩平衡得:Tsin30°×BO-G人y-G×
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114837198253.gif)
y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241148372611341.gif)
即离A端3.5 m.
所以人在板上安全行走距A端的距离范围为
2 m≤x≤3.5 m
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目