题目内容
如图所示,质量为m的小物体静止在半径为R的半球体上,小物体与半球体间的动摩擦因数为μ,小物体与球心连线与水平地面的夹角为θ,则下列说法正确的是( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110946307884024.png)
A.小物体对半球体的压力大小为mgcosθ |
B.小物体对半球体的压力大小为mgtanθ |
C.小物体所受摩擦力大小为μmgsinθ |
D.小物体所受摩擦力大小为mgcosθ |
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110946307884024.png)
对小物块受力分析如图1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110946308354517.png)
将重力正交分解,如图2,
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/2014061109463094412140.png)
A、由于物体静止在半球上,处于平衡态,
沿半径方向列平衡方程:
N-mgsinθ=0…①
得:N=mgsinθ,故AB错误;
C、沿切向列平衡方程:
f-mgcosθ=0…②
解得:f=mgcosθ,故C错误D正确.
故选:D.
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110946308354517.png)
将重力正交分解,如图2,
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/2014061109463094412140.png)
A、由于物体静止在半球上,处于平衡态,
沿半径方向列平衡方程:
N-mgsinθ=0…①
得:N=mgsinθ,故AB错误;
C、沿切向列平衡方程:
f-mgcosθ=0…②
解得:f=mgcosθ,故C错误D正确.
故选:D.
![](http://thumb2018.1010pic.com/images/loading.gif)
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