题目内容
如图所示,位于水平地面上的重为300N的木箱,在大小为100N、方向与水平方向成α=37°角的拉力作用下沿地面作匀速直线运动。(sin370=0.6,cos370=0.8)
求:(1)地面对木箱的支持力?(2)木块与地面之间的动摩擦因数?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241148528451799.jpg)
求:(1)地面对木箱的支持力?(2)木块与地面之间的动摩擦因数?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241148528451799.jpg)
(1)240N (2)0.33
(1)木箱受力如图所示,由于木箱做匀速运动,则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114852861896.gif)
Fsinα + FN = G
所以地面对木箱的支持力为:
FN = G-Fsinα =300N-100×0.6N = 240N
(2)由题意,有 Fcosα =Ff= 0
又Ff=μFN
所以![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114852877954.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114852861896.gif)
Fsinα + FN = G
所以地面对木箱的支持力为:
FN = G-Fsinα =300N-100×0.6N = 240N
(2)由题意,有 Fcosα =Ff= 0
又Ff=μFN
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114852877954.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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