由前实验装置分析: F·L Fl 2LRF E S·△L S( bS(n,-no) 2R 又:S=xD2;F=Mg 4 令:N=△Mn=n1=no 则: 2LRMg 8MgLR b·-元D2·N 丌DbN 4( ) 2 ( ) 2 . 1 0 1 0 b S n n L R F n n R b S F L S L F L E − = − = = D bN Mg L R b D N L R Mg E 2 2 8 4 1 2 = = ; 4 1 2 S = D F = Mg N = n = n1 − n0 则: 又: 令: 由前实验装置分析: