题目内容
如图所示,重为G的均匀棒水平地搁在一个圆柱体B上.二者的接触点D离棒的左端距离是棒长的3/5,当圆柱体顺时针方向转动时,在棒的右端与它紧靠着的木板C恰能沿光滑竖直墙面匀速下滑,则木板C的重为 .若木棒与圆柱体间的动摩擦因素为μ,则棒与圆柱体间的滑动摩擦力为 .![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_ST/images0.png)
【答案】分析:(1)木板C沿光滑竖直墙面匀速下滑,其重力与所受的木板的滑动摩擦力大小相等.对木板而言,力矩平衡,由力矩平衡条件求解木板C对棒的摩擦力.
(2)对木棒竖直方向力平衡,可求出木板对木棒的弹力,由f=μN求解滑动摩擦力.
解答:解:因为棒均匀,令棒长为L,c对棒的摩擦力为f,则重力在
处,以D为支点棒满足力矩平衡,有:
=![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/2.png)
可解得![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/3.png)
以C为研究对象,因为C匀速下降,故C处于平衡状态,棒对C的摩擦力等于C的重力
根据牛顿第三定律可得:![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/4.png)
(2)以棒为研究对象,在竖直方向受重力、C对棒竖直向下的摩擦力f和B对D点竖直向上的弹力N作用,根据平衡可得:
G+f=N
所以N=G+f=![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/5.png)
所以棒与圆柱体间的滑动摩擦力f滑=μN=![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/6.png)
故答案为:
,![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/8.png)
点评:本题中木板C受力平衡,木棒不仅力平衡,力矩也平衡,根据力平衡条件和力矩平衡条件结合处理本题,分析受力情况是关键
(2)对木棒竖直方向力平衡,可求出木板对木棒的弹力,由f=μN求解滑动摩擦力.
解答:解:因为棒均匀,令棒长为L,c对棒的摩擦力为f,则重力在
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/2.png)
可解得
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/3.png)
以C为研究对象,因为C匀速下降,故C处于平衡状态,棒对C的摩擦力等于C的重力
根据牛顿第三定律可得:
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/4.png)
(2)以棒为研究对象,在竖直方向受重力、C对棒竖直向下的摩擦力f和B对D点竖直向上的弹力N作用,根据平衡可得:
G+f=N
所以N=G+f=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/5.png)
所以棒与圆柱体间的滑动摩擦力f滑=μN=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/6.png)
故答案为:
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195620777170170/SYS201310291956207771701010_DA/8.png)
点评:本题中木板C受力平衡,木棒不仅力平衡,力矩也平衡,根据力平衡条件和力矩平衡条件结合处理本题,分析受力情况是关键
![](http://thumb2018.1010pic.com/images/loading.gif)
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