Question

如圖10所示，在光滑的桌面上疊放著一質(zhì)量為m_a=2.0 kg的薄木板A和質(zhì)量為m_b=3.0 kg的金屬塊B.A的長度l=2.0 m.B上有輕線繞過定滑輪與質(zhì)量為m_c=1.0 kg的物塊C相連.B與A之間的動(dòng)摩擦因數(shù)μ=0.10,最大靜摩擦力可視為等于滑動(dòng)摩擦力.忽略滑輪質(zhì)量及與軸間的摩擦.起始時(shí)令各物體都處于靜止?fàn)顟B(tài)，繩被拉直，B位于A的左端（如圖），然后放手，求經(jīng)過多長時(shí)間后B從A的右端脫離（設(shè)A的右端距滑輪足夠遠(yuǎn)）？（取g=10 m/s²）

圖10

Answer 1

解析：分別隔離A、B、C進(jìn)行受力分析，并利用牛頓第二定律，得

對(duì)Aμm_Bg=m_Aa_A                                                                                               ①

對(duì)B F-μm_Bg=m_Ba_B                                                                                            ②

對(duì)C m_Cg-F=m_Ca_C                                                                                             ③

其中，F(xiàn)為B、C間繩子張力大小，a_B=a_C，                                        ④

又a_Bt²-a_At²=l                                                              ⑤

聯(lián)立以上各式可得：t=4.0 s.                                                      ⑥

答案：4.0 s