题目内容
(12分)如图所示,某人用轻绳牵住一只质量
=0.6kg的氢气球,因受水平风力的作用,系氢气球的轻绳与水平方向成
角。已知空气对气球的浮力为15N,人的质量
,且人受的浮力忽略不计(sin370=0.6,cos370=0.8)。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023422212965.jpg)
求:(1)水平风力的大小
(2)人对地面的压力大小
(3)若水平风力增强,人对地面的压力如何变化?(要求说明理由)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342174337.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342190378.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342205670.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023422212965.jpg)
求:(1)水平风力的大小
(2)人对地面的压力大小
(3)若水平风力增强,人对地面的压力如何变化?(要求说明理由)
12N 491N 不变
试题分析:(1)对氢气球进行受力分析,设氢气球受绳子拉力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342236304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342252421.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342283762.png)
由共点力平衡条件:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342299936.png)
计算可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342314649.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342330535.png)
(2)对人进行受力分析,设人受地面的支持力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342346357.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342377871.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002342408527.png)
(3)若风力增强,人对地面的压力不变(2分)
视气球及人为一整体可知,其竖直方向上的受力情况没改变,故地面的支持力不变(2分)
![](http://thumb2018.1010pic.com/images/loading.gif)
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