题目内容
(13分)如图所示,用完全相同的、劲度系数均为k的轻弹簧A、B、C将两个质量均为m的小球连接并悬挂起来,两小球均处于静止状态,弹簧A与竖直方向的夹角为30°,弹簧C水平,已知重力加速度为g,试求出轻弹簧A、B、C各自的伸长量。(所有弹簧形变均在弹性范围内)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250005238546020.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250005238546020.jpg)
xA=
,xB=
,xC=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523901745.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523870735.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523885741.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523901745.png)
试题分析:将两小球看作一个整体,对整体受力分析,可知整体受到重力2mg、弹簧A和C的拉力FA、FC的作用,受力如下图甲所示,根据共点力平衡条件有:FA=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523901766.png)
根据胡克定律有:FA=kxA,FC=kxC
联立以上各式解得弹簧A、C的伸长量分别为:xA=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523870735.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523901745.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250005239632912.jpg)
对B、C间的小球进行受力分析,其受力如图乙所示,根据平衡条件有:FB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523979795.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523995676.png)
根据胡克定律有:FB=kxB,解得弹簧B的伸长量为:xB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000523885741.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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