题目内容
(14分)如图16所示,光滑匀质圆球的直径d=40 cm,质量为M=20 kg,悬线长L=30 cm,正方形物块A的厚度b=10 cm,质量为m=2 kg,物块A与墙之间的动摩擦因数μ=0.2.现将物块A轻放于球和墙之间后放手,取g=10 m/s2,求:
(1)墙对A的摩擦力为多大?
(2)施加一个与墙面平行的外力于物体A上,使物体A在未脱离圆球前贴着墙
沿水平方向做加速度a=5 m/s2的匀加速直线运动,那么这个外力的大小和方
向如何?
(1)墙对A的摩擦力为多大?
(2)施加一个与墙面平行的外力于物体A上,使物体A在未脱离圆球前贴着墙
沿水平方向做加速度a=5 m/s2的匀加速直线运动,那么这个外力的大小和方
向如何?
(1)20 N (2)20
N,与水平方向成arctan
角斜向上
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141428346322.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141428486338.png)
(1)球与物体A的受力图分别如图所示
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241414285803891.jpg)
对球有:tanθ=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141428627385.png)
FN=Mgtanθ=150 N
对物体A有:Ff=μFN球=μFN=30 N>mg
所以A物体处于静止状态,静摩擦力Ff=mg=20 N.
(2)A沿水平方向运动,面对着墙看,作出A物体在竖直平面内的受力图如图所示,有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241414287671768.jpg)
Fsinα=mg
Fcosα-μFN=ma
解出:F=20
N,α=arctan
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241414285803891.jpg)
对球有:tanθ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141428627385.png)
FN=Mgtanθ=150 N
对物体A有:Ff=μFN球=μFN=30 N>mg
所以A物体处于静止状态,静摩擦力Ff=mg=20 N.
(2)A沿水平方向运动,面对着墙看,作出A物体在竖直平面内的受力图如图所示,有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241414287671768.jpg)
Fsinα=mg
Fcosα-μFN=ma
解出:F=20
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141428346322.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141428486338.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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