题目内容
如图所示,整个装置处于平衡状态,左右两段轻线与天花板所成的角度分别为45°和60°,中间一段轻线水平,则悬于轻线上两个物体的质量m1 m2(填大于、等于或小于);质量之比m1:m2= 。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300507615911.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300507615911.png)
小于 ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300507921660.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300507921660.png)
分析:分别对两节点受力分析,用两物体的重力表示出绳上的拉力F,一根绳上的拉力大小是相同的,利用这个关系比较两个物体的质量关系.
解答:解:分别对两节点受力分析如图:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082413005088626692.png)
由平衡条件得:F′=m1g;F″=m2g
图1:由三角函数关系得:tan45°=
,解得:T=m1g①
图2:由三角函数关系得:tan60°=
,解得:T=
m2g②
联立①②得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130051260611.png)
故答案为:小于,![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130051260611.png)
解答:解:分别对两节点受力分析如图:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082413005088626692.png)
由平衡条件得:F′=m1g;F″=m2g
图1:由三角函数关系得:tan45°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050902587.png)
图2:由三角函数关系得:tan60°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050933616.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130051073419.png)
联立①②得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130051260611.png)
故答案为:小于,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130051260611.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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