质量m=2kg的均匀绳,长L=1m,两端分别连接重物A和B,ma=8kg,mb=5kg,今在B端施以大小为F=180N的竖直拉力,物体向上运动,求张力T(X)。
正确答案:
对整体进行受力分析,加速度向上为,根据牛顿第二定律有:
![](https://img.274949.com/Images/2020-04/04/58vfteyc.jpg)
对进行受力分析,根据牛顿第二定律有:
![](https://img.274949.com/Images/2020-04/04/5we62yoa.jpg)
对一小段绳子受力如图,根据牛顿第二定律得:
![](https://img.274949.com/Images/2020-04/04/kaysqr26.jpg)
![](https://img.274949.com/Images/2020-04/04/58vfteyc.jpg)
对进行受力分析,根据牛顿第二定律有:
![](https://img.274949.com/Images/2020-04/04/5we62yoa.jpg)
对一小段绳子受力如图,根据牛顿第二定律得:
![](https://img.274949.com/Images/2020-04/04/kaysqr26.jpg)
![](https://img.274949.com/Images/2020-04/04/3lk6pg0g.jpg)
答案解析:有
![](/editor/images/201705/qrcode_for_gh_573e2a458573_258.jpg)
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质量m=2kg的均匀绳,长L=1m,两端分别连接重物A和B,ma=8kg,mb=5kg,今在B端施以大小为F=180N的竖直拉力,物体向上运动,求张力T(X)。
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