题目内容
如图3-4-3所示,一根重8 N的均匀直棒AB,A端用细绳吊在固定点O上,今用水平力F=6 N作用于B端,细绳与竖直方向的夹角α=37°.则棒静止,问水平力F与重力G的合力大小和方向怎样?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204232781456.jpg)
图3-4-3
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204232781456.jpg)
图3-4-3
由直角三角形勾股定理可得:
F合=
N="10" N
合力的方向与竖直方向的夹角为α
tanα=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120423340753.png)
查三角形函数表得α=37°.
F合=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204232931222.png)
合力的方向与竖直方向的夹角为α
tanα=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120423340753.png)
查三角形函数表得α=37°.
均匀直棒受三个力作用而静止.由三力汇交原理可知,三个力作用线延长线相交于O′点,水平力F=6 N,重力G作用点为棒中点,且方向竖直向下,这二力互相垂直,依力的平行四边形定则可作出图.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204233711356.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204233711356.jpg)
![](http://thumb2018.1010pic.com/images/loading.gif)
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