题目内容

倾角为37°的斜面体靠在固定的竖直挡板P的一侧,一根轻绳跨过固定在斜面顶端的定滑轮,绳的一端与质量为mA=3kg的物块A连接,另一端与质量为mB=1kg的物块B连接。开始时,使A静止于斜面上,B悬空,如图所示。现释放AA将在斜面上沿斜面匀加速下滑,求此过程中,挡板P对斜面体的作用力的大小。(所有接触面产生的摩擦均忽略不计,sin37°=0.6,cos37°=0.8,g=10m/s2
F = 4.8N
设绳中张力为T,斜面对A的支持你为NAAB加速度大小为a,以A为研究对象,
由牛顿第二定律
mAgsin37° -T =ma                                                                                                            ①2分
NA = mAgcos37°④                                                                                                                ②1分
B为研究对象,由牛顿第二定律
TmBg = mBa                                                                                                                    ③2分
联立解得a = 2m/s2 T =" 12N " NA= 24N
以斜面体为研究对象,受力分析后,在水平方向
F = N′Asin37°-Tcos37°                                                                                                       ④2分
NA = N′A                                                                                                                                                                                                                                                                           
解得F =" 4.8N                                                                                                                     " 1分
(或以整体为研究对象,由牛顿第二定律得F = mAacos37°)=4.8N,则本式给4分,①③式各给2分,共8分)
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