例1.已知:F、a、L求:A端约束反力 解 F A 解法I 0l B M B 2mA=0: MA-FsinQL=0 . MA=Fsin a L 2Fx=0: FAx-Fcos &=0 FAx=Fcos a 2Fy=0: FAy-Fsina=0 . FAY=Fsina 解法工 A FA=F M B ∑m=0:MA- Sina L=0 MA =SIna L例1. 已知：F、α、L 解： 求：A端约束反力 解法Ⅰ A α L B F A B α F A B α F MA FAy FAx 解法Ⅱ ∑mA = 0 : ∑F = 0 : y ∑F = 0 : x MA －FsinαL = 0 ∴ MA = FsinαL FAx－Fcosα= 0 ∴ FAx = Fcosα FAy－Fsinα= 0 ∴ FAy = Fsinα FA MA FA = F ∑m = 0 : MA －FsinαL = 0 ∴ MA = FsinαL