两个物体a,b质量分别为m1,m2,靠在一起放在光滑的水平面
\u4e24\u4e2a\u7269\u4f53A\u548cB,\u8d28\u91cf\u5206\u522b\u4e3am1\u548cm2,\u4e92\u76f8\u63a5\u89e6\u653e\u5728\u5149\u6ed1\u6c34\u5e73\u9762\u4e0a\uff0c\u5bf9\u7269\u4f53\u65bd\u4e00\u6c34\u5e73\u63a8\u529bF\uff0c\u5219A\u5bf9B\u7684\u529b\u4e3aA\u548cB\uff0c\u8fd8\u6709AB\u6574\u4f53\u5177\u6709\u76f8\u540c\u7684\u52a0\u901f\u5ea6\uff0c\u5f97\u65b9\u7a0b\uff1aF=(M1+M2)a F'=M2a \u4e24\u5f0f\u76f8\u9664\u5373\u5f97\u7ed3\u679c \u9009C \u505a\u6b64\u7c7b\u53d7\u529b\u5206\u6790\u9898\u65f6\u8981\u7075\u6d3b\u8fd0\u7528\u6574\u4f53\u5206\u6790\u4e0e\u4e2a\u4f53\u5206\u6790\u65b9\u6cd5
\u5982\u679c\u7b97A \u5e94\u8be5\u8fd9\u6837\u5217\u5f0f\uff1aF-F'=M1a \u56e0\u4e3aA\u53d7\u5230\u4e24\u4e2a\u529b \u6574\u4f53\u7684\u548cB\u5bf9A\u7684\u5440 \u505a\u6b64\u7c7b\u9898\u65f6\u8981\u6ce8\u610f\u753b\u53d7\u529b\u5206\u6790\u56fe
\uff081\uff09\u636e\u9898\u610f\u6709\uff1aN1=0\uff0cf1=0
\u56e0\u6b64\u6709\uff1aNcos\u03b8=m1g [1] , F-Nsin\u03b8=m1a [2]
\u7531[1]\u5f0f\u5f97A\u3001B\u95f4\u76f8\u4e92\u4f5c\u7528\u529b\u4e3a\uff1aN=m1g/cos\u03b8
\uff082\uff09N2=m2g+Ncos\u03b8 [3] , f2=\u03bcN2 [4]
\u5c06[1]\u3001[3]\u4ee3\u5165[4]\u5f0f\u5f97\uff1a f2=\u03bc(m1+ m2)g
\u53d6A\u3001B\u7ec4\u6210\u7684\u7cfb\u7edf\uff0c\u6709\uff1aF-f2=(m1+ m2)a [5]
\u7531[1]\u3001[2]\u3001[5]\u5f0f\u89e3\u5f97\uff1aF=m1g(m1+ m2)(tg\u03b8-\u03bc)/m2
\u6545A\u3001B\u4e0d\u53d1\u751f\u76f8\u5bf9\u6ed1\u52a8\u65f6F\u7684\u53d6\u503c\u8303\u56f4\u4e3a\uff1a0\uff1cF\u2264m1g(m1+ m2)(tg\u03b8-\u03bc)/m2
a=（F1-F2）/(m1+m2)
然后再单独对B分析，B的加速度等于整体的加速度
由牛二得
F-F2=m2a
A对B的力
F=m2a+F2
=m2（F1-F2）/(m1+m2) +F2
分别受到水平推力F1,F2,
F1>F2
A施於B的作用力为
F=(F1-F2)m2/(m1+m2)
m1 m2当在光滑的水平面时就忽略了阻力了！！水平推力？？方向呢？？？是否一致的？？当一致时就是F1减去F2了！！相向时就是F1加上F2 了！！题都没说清~~
F2
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