题目内容
(9分)如图,将质量m=0.1kg的圆环套在固定的水平直杆上。环的直径略大于杆的截面直径。环与杆间动摩擦因数
=0.8。对环施加一位于竖直平面内斜向上,与杆夹角
=53°的拉力F,使圆环以a=4.4m/s2的加速度沿杆运动,求F的大小。(取
=0.8,
=0.6,g=10m/s2)。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004073202021.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000407258301.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000407274297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000407289515.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000407305544.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004073202021.jpg)
9N
试题分析:令Fsin53°=mg,F=1.25N,
当F<1.25N时,杆对环的弹力向上,
由牛顿第二定律
Fcosq-
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000407258301.png)
FN+Fsinq=mg,
解得F=1N,
当F>1.25N时,杆对环的弹力向下,
由牛顿第二定律
Fcosq-
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000407258301.png)
Fsinq=mg+FN,
解得F=9N。
![](http://thumb2018.1010pic.com/images/loading.gif)
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