题目内容
如图所示,重G=100N的木块放在倾角θ=20°的斜面上静止不动,现用平行于斜面底边、沿水平方向的外力F拉木块,则F为多少时,可使木块沿斜面匀速滑下,已知木块与斜面间动摩擦因数μ=0.5,取sin20°=0.34,cos20°=0.94.
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110952405712435.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110952405712435.png)
物体在斜面上的受力如图所示,根据共点力的平衡条件得:
滑动摩擦力:f=
=μGcos30°
代入数据得:F=32.4N;
答:F为32.4N时,可使木块沿斜面匀速滑下.
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110952407114807.png)
滑动摩擦力:f=
F2+(Gsinθ)2 |
代入数据得:F=32.4N;
答:F为32.4N时,可使木块沿斜面匀速滑下.
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110952407114807.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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