杠杆滑轮综合
来源:学生作业帮 编辑:作业帮 分类:物理作业 时间:2024/09/03 09:12:44
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![](http://img.wesiedu.com/upload/8/57/857b000538f8e9a4bf31f92638f837d0.jpg)
![](http://img.wesiedu.com/upload/8/57/857b000538f8e9a4bf31f92638f837d0.jpg)
解题思路: (1)求解物体配重A受到的拉力,先要对配重A受到的拉力和小成进行受力分析,根据处于平衡状态的物体,所受的合力为零求解,配重A对地面的压力根据压强公式P=F/S得出的F=PS公式计算. (2)动滑轮D的重力和物体配重A对绳子的拉力共同作用的杠杆的左端,要求解小成在H点施加竖直向下的拉力为T2和动滑轮的重力需通过分析杠杆的两次平衡,根据杠杆的平衡条件列出方程,结合F1:F2=20:19的关系组成方程组进行求解.
解题过程:
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最终答案:略
解题过程:
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![](http://img.wesiedu.com/upload/8/57/857b000538f8e9a4bf31f92638f837d0.jpg)
最终答案:略