Question

傾角為37°的斜面體靠在固定的豎直擋板P的一側(cè)，一根輕繩跨過固定在斜面頂端的定滑輪，繩的一端與質(zhì)量為m_A=3kg的物塊A連接，另一端與質(zhì)量為m_B=1kg的物塊B連接。開始時(shí)，使A靜止于斜面上，B懸空，如圖所示�，F(xiàn)釋放A，A將在斜面上沿斜面勻加速下滑，求此過程中，擋板P對(duì)斜面體的作用力的大小。（所有接觸面產(chǎn)生的摩擦均忽略不計(jì)，sin37°=0.6，cos37°=0.8，g=10m/s²）

Answer 1

F = 4.8N

解析:

設(shè)繩中張力為T，斜面對(duì)A的支持你為N_A，A、B加速度大小為a，以A為研究對(duì)象，

由牛頓第二定律

m_Agsin37° －T =ma                                                                                                            ①2分

N_A = m_Agcos37°④                                                                                                                ②1分

以B為研究對(duì)象，由牛頓第二定律

T－m_Bg = m_Ba                                                                                                                    ③2分

聯(lián)立解得 a = 2m/s²  T = 12N  N_A= 24N

以斜面體為研究對(duì)象，受力分析后，在水平方向

F = N′_Asin37°－Tcos37°                                                                                                       ④2分

N_A = N′_A                                                                                                                                                                                                                                                                           

解得 F = 4.8N                                                                                                                      1分

（或以整體為研究對(duì)象，由牛頓第二定律得F = m_Aacos37°）=4.8N，則本式給4分，①③式各給2分，共8分）