题目内容
如图,质量为m的A物放在质量为M倾角为θ光滑的斜面B上,斜面B置于光滑的水平地面上,若水平向左推B物,且使A、B相对静止,则推力大小为______;若水平向右推A物,且使A、B相对静止,则推力大小为______.
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102135391581110.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102135391581110.png)
若水平向左推B物,且使A、B相对静止,对A分析,![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102135392202575.png)
根据牛顿第二定律得,
A的加速度a=
=gtanθ.
因为整体具有相同的加速度,
则推力F1=(M+m)a=(M+m)gtanθ.
当水平向右推A物,且使A、B相对静止,
对A,根据牛顿第二定律得,![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102135401252755.png)
F-Nsinθ=ma,
Ncosθ=mg
则F-mgtanθ=ma
对整体分析F=(M+m)a,
联立两式解得F=
.
故答案为:(M+m)gtanθ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102135392202575.png)
根据牛顿第二定律得,
A的加速度a=
mgtanθ |
m |
因为整体具有相同的加速度,
则推力F1=(M+m)a=(M+m)gtanθ.
当水平向右推A物,且使A、B相对静止,
对A,根据牛顿第二定律得,
![](http://thumb.1010pic.com/pic2/upload/papers/20140610/201406102135401252755.png)
F-Nsinθ=ma,
Ncosθ=mg
则F-mgtanθ=ma
对整体分析F=(M+m)a,
联立两式解得F=
(M+m)mgtanθ |
M |
故答案为:(M+m)gtanθ,
(M+m)mgtanθ |
M |
![](http://thumb2018.1010pic.com/images/loading.gif)
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