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1、第三章相互作用 章末总结提高Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Sli
2、des for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for
3、 .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.C
4、opyright 2004-2011 Aspose Pty Ltd.v 方法专题讲解v1受力分析v1定义：受力分析指的是把研究对象在特 定的物理环境中所受到的所有力都找出来 ，并画出受力示意图Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright
5、 2004-2011 Aspose Pty Ltd.v2某个力是否存在的判断依据v(1)条件判断：根据力的产生条件是否满足 ，来判断物体是否受某个力的作用v(2)效果判断：根据力的效果是否得以体现 ，来判断物体是否受某个力的作用v(3)相互作用判断：利用力的作用的相互性 ，即施力物体同时也是受力物体，从一个 物体是否受到某个力作用来判断另一个物 体是否受到相应的反作用力的作用Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with As
6、pose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v3受力分析的一般步骤v(1)确定研究对象，并把研究对象从系统中 隔离出来；v(2)先分析物体所受重力，然后依次分析各 接触面上的弹力及摩擦力；v(3)最后分析其他力的情况；v(4)检查 所画受力图，是否“多力”或“漏力” Evaluation only.Evaluation only. Created with Aspose.Slides for .NET
7、3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v【例1】 跨过光滑定滑轮的轻绳，两端各 拴一个物体，如图1所示，物体A和B重均为 20 N，水平拉力F12 N，若A和B均处于静 止状态，试分析物体A和B的受力情况，画 出力的示意图，并计算各力的大小Evaluation only.Evaluation only. Creat
8、ed with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with
9、 Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v【答案】 F20 N，FN10 N，F(10 12)Nv【点拨】 正确对物体A、B进行受力分析 是解此类题 的关键对未知力F的大小和 方向，我们可以用“假设法”，设其朝某一 方向，然后利用平衡条件列方程求解，若 求出为正值，说明假设的方向是正确的， 若求出力为负值 ，说明其方向与假设方向 相反Evaluation only.Evaluation onl
10、y. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v巩固训练v1如图3所示，GA8 N，GB16 N，A、 B之间及B与水平地面之间的动摩擦因数均 为0.25，物块B在水平拉力F的作用下向 左匀速运动，求F的大小Evaluation only.Eva
11、luation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v解析：先画出A、B的受力分析图如图4所示v则对A有FTFfAFNAGA.v对B有FFTFfAFf地2GAFNB2GA (GAGB)10 N.v答案：10 NEvaluation
12、 only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v图解法解题v1图解法常用在研究分力的动态变 化时，通过作出平行四边形或矢量三角形，观察线段的长短变化分析其对应的力的大小变化，具有直观、简单的特点v2用图解法
13、解题的一般步骤v(1)根据对物体的受力分析，利用分解或合成作出平行四边形或矢量三角形；v(2)按照题意分析，确定分力的变化v3图解法所解题目的特点：物体受力中有一个力的大小和方向均不变，有一个分力的方向不变而大小变化，另一个力的大小和方向均变化Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-201
14、1 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v【例2】 如图5所示，重为G的物体放在倾 角为的光滑斜面上，被竖直放置的光滑木 板挡住，若将挡板逆时针转动逐渐放低 试分析球对挡板的压力和对斜面的压力如 何变化Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyrig
15、ht 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v【解析】 球的重力产生两 个效果：压紧斜面和挡板， 即重力可分解为垂直斜面的 力G1和垂直于挡板的作用力 G2.这两个分力分别跟球对斜 面的压力和球对挡板的压力 大小相等在挡板放低时， 重力G的大小和方向不变； G1的方向不变，G2的大小和 方向都在变，如图6所示， 由图知，G2先减小后增大， 最小值为Gsin.G1始终在减 小Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.
16、5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.v【答案】 对挡板的压力先减小后增大，对 斜面的压力减小v【点拨】 用图解法解题的关键是作好图并 找出各个力的变化关系Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Asp
