题目内容
(13分)如图所示,两根直木棍AB和CD相互平行,斜靠在竖直墙壁上固定不动,一根水泥圆筒与两根直木棍的滑动摩擦因数μ,它从木棍的上部匀加速滑下,加速度为a,若保持两木棍倾角α不变,将两棍间的距离减小后固定不动,仍将水泥圆筒放在两木棍上部, 为了分析其加速度变化,有同学这样分析:每根直木棍对圆筒的摩擦力和每根直木棍对圆筒支持力满足f1=μFN1,两根直木棍对圆筒的摩擦力的合力为f,两根直木棍对圆筒支持力的合力为FN,它们也满足f=μFN,而FN=mgcosα不变,故f也不变。所以加速度a不变,你认为这位同学的做法正确吗?如不正解指出它的错误原因,并提供正确的解法。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001234898306.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001234898306.png)
不正确,解析如下。
试题分析:在垂直于木板方向研究,木棍给圆柱两个支持力
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123598366.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123614391.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123629497.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123598366.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123614391.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001236921042.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123598366.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123614391.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123598366.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123614391.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000123770538.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001237853803.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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