题目内容
一氢气球重10 N,所受空气的浮力为16 N,由于受到水平风力F的作用,使氢气球的绳子和地面成600角,如图1—11所示。由此可知,绳子的拉力为 N,水平风力F的的大小是 N。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241232508882178.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241232508882178.png)
4
,2
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
考点:
专题:共点力作用下物体平衡专题.
分析:对小球受力分析,受到重力、浮力、拉力和风力,根据平衡条件列式求解.
解答:解:氢气球受力如图.根据共点力平衡,运用正交分解得,F=Tsin30°
F浮=mg+Tcos30°,联立两式得,绳子拉力T=4
N,风力F=2
N.
故答案为:4
,2![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241232510601827.png)
点评:解决本题的关键知道气球处于平衡,合力等于零.运用正交分解时,知道合力等于零,在x轴方向和y轴方向合力都等于零.
专题:共点力作用下物体平衡专题.
分析:对小球受力分析,受到重力、浮力、拉力和风力,根据平衡条件列式求解.
解答:解:氢气球受力如图.根据共点力平衡,运用正交分解得,F=Tsin30°
F浮=mg+Tcos30°,联立两式得,绳子拉力T=4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
故答案为:4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123250919341.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241232510601827.png)
点评:解决本题的关键知道气球处于平衡,合力等于零.运用正交分解时,知道合力等于零,在x轴方向和y轴方向合力都等于零.
![](http://thumb2018.1010pic.com/images/loading.gif)
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