(12分)如图所示，质量为25 kg的小孩坐在5 kg的雪橇上，大人用与水平方向成37°斜向上的大小为100 N的拉力拉雪橇，使雪橇沿水平地面做匀速运动，(sin 37°＝0.6，cos 37°＝0.8,g=10m/s2)
求：(1)雪橇对地面的压力大小；
(2)雪橇与水平地面间的动摩擦因数的大小．
240N 1/3
题目分析：（1）以小孩和雪橇整体为研究对象，总重力G=30×10N=300N
分析整体的受力情况如图，由于雪橇沿水平地面做匀速直线运动，则有
N+Fsin37°=G得N=G-Fsin37°=300N-100N×0.6=240N
由牛顿第三定律求得雪橇对水平地面的压力大小N′=N=340N，方向竖直向下．
（2）雪橇所受的滑动摩擦力大小为f=Fcos37°=100×0.8N=80N由f=μN得 μ=
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