正交分解法
一、正交分解法
把力按相互垂直的两个方向分解叫正交分解
![](https://wkrtcs.bdimg.com/rtcs/image?w=90.65&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"90.65","h":"50.23","dataType":"gif","c":"word/media/image1.gif"})
二、用力的正交分解求多个力的合力
1、建立直角坐标系（让尽量多的力在坐标轴上）
2、正交分解各力（将各力分解到两个坐标轴上）
3、分别求出x 轴和y 轴上各力的合力：
![](https://wkrtcs.bdimg.com/rtcs/image?w=147.58&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"147.58","h":"24.56","dataType":"gif","c":"word/media/image3.gif"})
![](https://wkrtcs.bdimg.com/rtcs/image?w=150.93&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"150.93","h":"25.67","dataType":"gif","c":"word/media/image4.gif"})
4、求出FX 和 Fy 的合力，
即为多个力的合力
大小：![](https://wkrtcs.bdimg.com/rtcs/image?w=167.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_7","w":"167.00","h":"43.00","dataType":"wmf","c":"word/media/image5.png","imgOriW":334,"imgOriH":86})
方向：![](https://wkrtcs.bdimg.com/rtcs/image?w=145.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_7","w":"145.00","h":"49.00","dataType":"wmf","c":"word/media/image7.png","imgOriW":290,"imgOriH":98})
![](https://wkrtcs.bdimg.com/rtcs/image?w=122.67&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"122.67","h":"76.00","dataType":"gif","c":"word/media/image8.gif"})
三、用力的正交分解求解物体平衡问题
1、画出物体的受力图。
2、建立直角坐标系。
3、正交分解各力。（将各力分解到两个坐标轴上）
4、物体平衡时各方向上合力为零,分别写出x 方向 和y 方向方程。
![](https://wkrtcs.bdimg.com/rtcs/image?w=172.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"172.00","h":"24.00","dataType":"gif","c":"word/media/image9.gif"})
![](https://wkrtcs.bdimg.com/rtcs/image?w=174.67&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"174.67","h":"25.33","dataType":"gif","c":"word/media/image10.gif"})
5、根据方程求解。
例题2：如图所示，质量为m的物体放在粗糙水平面上，它与水平面间的滑动摩擦因数为μ，在与水平面成θ角的斜向上的拉力F作用下匀速向右运动。求拉力F的大小。
![](https://wkrtcs.bdimg.com/rtcs/image?w=197.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"197.33","h":"94.67","dataType":"gif","c":"word/media/image11.gif"})
例题2：如图所示，质量为m的物体放在粗糙水平面上，它与水平面间的滑动摩擦因数为μ，在与水平面成θ角的斜向上的拉力F作用下匀速向右运动。求拉力F的大小。
![](https://wkrtcs.bdimg.com/rtcs/image?w=165.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"165.33","h":"113.33","dataType":"gif","c":"word/media/image13.gif"})
∵物体匀速运动，合外力为零
由x方向合外力为零，有：
![](https://wkrtcs.bdimg.com/rtcs/image?w=100.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"100.00","h":"21.33","dataType":"gif","c":"word/media/image14.gif"})
由y方向合外力为零，有：
![](https://wkrtcs.bdimg.com/rtcs/image?w=124.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"124.00","h":"21.33","dataType":"gif","c":"word/media/image15.gif"})
解得：![](https://wkrtcs.bdimg.com/rtcs/image?w=176.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_7","w":"176.00","h":"35.00","dataType":"wmf","c":"word/media/image16.png","imgOriW":352,"imgOriH":70})
例题3：如图所示，质量为m的物体在倾角为θ的粗糙斜面下匀速下滑，求物体与斜面间的滑动摩擦因数。
![](https://wkrtcs.bdimg.com/rtcs/image?w=269.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"269.33","h":"105.33","dataType":"gif","c":"word/media/image17.gif"})
解析：
![](https://wkrtcs.bdimg.com/rtcs/image?w=222.67&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"222.67","h":"129.33","dataType":"gif","c":"word/media/image19.gif"})
∵物体匀速运动，合外力为零
由x方向合外力为零，有：
![](https://wkrtcs.bdimg.com/rtcs/image?w=105.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"105.33","h":"21.33","dataType":"gif","c":"word/media/image20.gif"})
由y方向合外力为零，有：
![](https://wkrtcs.bdimg.com/rtcs/image?w=88.87&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"88.87","h":"21.21","dataType":"gif","c":"word/media/image21.gif"})
解得：![](https://wkrtcs.bdimg.com/rtcs/image?w=169.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_7","w":"169.00","h":"51.00","dataType":"wmf","c":"word/media/image22.png","imgOriW":338,"imgOriH":102})
练习一：如图所示，质量为m的光滑小球放在倾角为θ的斜面上被挡板挡住，求斜面对小球的弹力及挡板对小球的弹力。
![](https://wkrtcs.bdimg.com/rtcs/image?w=161.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"161.33","h":"112.00","dataType":"gif","c":"word/media/image23.gif"})
练习二：如图所示，质量为m的物体在与竖直方向成θ角的恒力F作用下沿粗糙墙面向上匀速运动，求物体与墙壁间的动摩擦因数。
![](https://wkrtcs.bdimg.com/rtcs/image?w=125.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"125.33","h":"122.67","dataType":"gif","c":"word/media/image24.gif"})
正交分解法巩固：
1、如图，物体重力为10N，AO绳与顶板间的夹角为45º，BO绳水平，试用计算法求出AO绳和BO绳所受拉力的大小。
FAOY=FAOcos45=G
FAOX=FBO=G
![](https://wkrtcs.bdimg.com/rtcs/image?w=228.00&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"228.00","h":"162.67","dataType":"gif","c":"word/media/image25.gif"})
2、如图，氢气球被水平吹来的风吹成图示的情形，若测得绳子与水平面的夹角为37˚，已知气球受到空气的浮力为15N，忽略氢气球的重力，求：
①氢气球受到的水平风力多大？
②绳子对氢气球的拉力多大？ 风
![](https://wkrtcs.bdimg.com/rtcs/image?w=390.67&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"390.67","h":"221.33","dataType":"gif","c":"word/media/image26.gif"})
3、如图，物体A的质量为m，斜面倾角α，A与斜面间的动摩擦因数为μ，斜面固定，现有一个水平力F作用在A上，当F多大时，物体A恰能沿斜面匀速向上运动？
解：FN=Fsinα+Gcosα
Fcosα=Gsinα+Ff
Ff=μFN
![](https://wkrtcs.bdimg.com/rtcs/image?w=257.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"257.33","h":"110.67","dataType":"gif","c":"word/media/image27.gif"})
![](https://wkrtcs.bdimg.com/rtcs/image?w=221.33&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"221.33","h":"168.00","dataType":"gif","c":"word/media/image28.gif"})
![](https://wkrtcs.bdimg.com/rtcs/image?w=426.67&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","r":"r_4","w":"426.67","h":"85.33","dataType":"gif","c":"word/media/image29.gif"})
用力的正交分解求解物体平衡问题
1、画出物体的受力图。
2、建立直角坐标系。
3、正交分解各力。（将各力分解到两个坐标轴上）
4、物体平衡时各方向上合力为零,分别写出x 方向 和y 方向方程。
![](https://wkrtcs.bdimg.com/rtcs/image?w=174.67&md5sum=31d2a738973b9beca50ebd8fcd8c1fc1&sign=3dbc5d3092&rtcs_flag=2&rtcs_ver=3.3&l=webapp&bucketNum=48&ipr={"t":"img","w":"174.67","h":"25.33","dataType":"gif","c":"word/media/image30.gif"})
5、根据方程求解。
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