题目内容
森林发生火灾,我军某部接到命令,派直升机前去灭火。但火场离水源较远,直升机需沿水平方向匀速飞往水源取水灭火,悬挂着m=500 kg空箱的悬索与竖直方向的夹角θ1=45°.直升机取水后飞往火场,加速度沿水平方向,大小稳定在a=1.5 m/s2时,悬索与竖直方向的夹角θ2=14°,如果空气阻力大小不变,且忽略悬索的质量,试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016136114214.png)
水箱中水的质量M.
(重力加速度g取10 m/s2;sin 14°≈0.242;cos 14°≈0.970)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016136114214.png)
水箱中水的质量M.
(重力加速度g取10 m/s2;sin 14°≈0.242;cos 14°≈0.970)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001613626664.png)
试题分析:直升机取水过程,对水箱受力分析,如图所示,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016136422114.jpg)
直升机沿水平方向匀速飞往水源,水箱受力平衡,由共点力平衡条件得:
在水平方向上:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001613658688.png)
在竖直方向上:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001613689785.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001613704725.png)
直升机取水后飞往火场过程,做匀加速直线运动,对水箱受力分析,如图所示
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016137201766.jpg)
由牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016137361098.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016137671087.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250016137671155.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001613782859.png)
故水箱中水的质量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001613626664.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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