题目内容
![](http://thumb.1010pic.com/pic3/upload/images/201202/16/33ee47f6.png)
分析:A、B始终处于静止状态,分别对带电前后进行受力分析,判断绳子拉力的变化.
解答:解:带电前:对B有:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/df3d95fa.png)
T2=GB
对AB组成的整体有:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/4db176f9.png)
T1=GA+GB
带电后:对B有:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/f55f5a3a.png)
=F电+GB
对整体:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/88ffb4c0.png)
T′1=GA+GB
综上所述:T1=
,T2<
故选:C
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/df3d95fa.png)
T2=GB
对AB组成的整体有:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/4db176f9.png)
T1=GA+GB
带电后:对B有:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/f55f5a3a.png)
T | ′ 2 |
对整体:
![](http://thumb.1010pic.com/pic3/upload/images/201211/67/88ffb4c0.png)
T′1=GA+GB
综上所述:T1=
T | ′ 1 |
T | ′ 2 |
故选:C
点评:处于受力平衡状态的物体在求解或进行大小变化的判断时,只要选好研究对象进行正确的受力分析,应用平衡条件判断即可.
![](http://thumb2018.1010pic.com/images/loading.gif)
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