题目内容
一种测定风作用力的仪器原理如图所示。它的细长丝线一端固定于O点,另一端悬挂着一个质量为m="1" kg的金属球。无风时,丝线自然下垂;当受到沿水平方向吹来的风的作用时,丝线将偏离竖直方向一定角度θ,风力越大,偏角越大。若某时刻丝线与竖直方向的夹角θ=37°,试求此时金属球所受风力的大小。(取g="10" m/s2,已知sin37°=0.6、cos37°=0.8)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241328438333484.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241328438333484.jpg)
7.5 N
:小球的受力分析如图所示,则平衡时有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241328438642381.png)
联立可得:
代入数据得:F=1×10×0.75 N="7.5" N
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241328438642381.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132843927688.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132843958574.png)
联立可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132844036700.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目