题目内容
图中氢气球重力为10N,空气对它的浮力为16N。由于受到水平向左的风力的作用,使系气球的绳子与地面成60°,试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024265873451.png)
(1)绳子的拉力;
(2)水平风力的大小。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024265873451.png)
(1)绳子的拉力;
(2)水平风力的大小。
(1)4
N(2)2
N
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002426603344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002426603344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250024266342857.jpg)
试题分析:解析:对氢气球受力分析如图所示,将绳子的拉力正交分解,由平衡条件得
水平方向F风=Fcos60°
竖直方向F浮=Fsin60°+mg
由(1)(2)联立得:F=4
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002426837346.png)
F风=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002426837346.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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