da da dN=Fsin +(F+dF)sin F F da 2F sin -+ dF sin 2 da dN 2F一=F fdN fN=(F+df)escos d da F+-dF' =dF cos dF dF F dF ca fda fda F 2 所以紧边和松边的拉力比为:-=ea F1(1--)      Fd d F d dF d F d F dF d dN F = = = + = + + 2 2 2 sin 2 2 sin 2 ( )sin 2 sin dF d dF d F d fdN F dF = = = + − 2 cos 2 cos 2 ( )cos    fd F dF =    =   0 2 1 fd F dF F F f F F = 2 1 ln f e F F = 2 1 所以紧边和松边的拉力比为: ) 1 1 (1 1 1 2 1 f f e F e F F = F − F = F − = −