题目内容
如图所示,某同学在地面上拉着一个质量为m=30kg的箱子匀速前进,已知箱与地面间的动摩擦因数为μ=0.5,拉力F与水平面夹角为θ=45°,g=10N/kg.求:绳子的拉力F为多少?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241655180553035.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241655180553035.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824165518070576.png)
试题分析:如图所示,某同学在地面上拉着一个质量为m=30kg的箱子匀速前进,已知箱与地面间的动摩擦因数为μ=0.5,拉力F与水平面夹角为θ=45°,g=10m/s2.求:绳子的拉力F为多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241655187263038.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241655187881091.png)
解:箱子受力如图所示 (1分)
由平衡条件得:Fsinθ+FN-G=0 (1分)
Fcosθ-f=0 (1分)
f=μFN (1分)
G=mg (1分)
整理得F=μmg/(cosθ+μsinθ) (1分)
代入数据得F=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824165518070576.png)
点评:本题关键是对物体受力分析,然后根据共点力平衡条件并运用正交分解法列平衡方程求解.
![](http://thumb2018.1010pic.com/images/loading.gif)
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