例2-2已知:图示平面共点力系; 求:此力系的合力 解:用解析法 FR=2F.=F cos30-F, cos60-F cos 45+ F 45=129 3N FR=2F=FSin 30+ F sin 60'-F sin 45-F sin 45=112 3N F2+F2=171.3N R V F F cOS 0.7548 300 F COS B==0.6556 00N 6=40.99°,B=49.01求：此力系的合力. 解：用解析法 =  = 1 cos30 − 2 cos60 − 3 cos45 + 4 cos45 =129.3N     FRx Fix F F F F =  = 1 sin 3 0 + 2 sin 6 0 − 3 sin 4 5 − 4 sin 4 5 =112.3N     FRy Fiy F F F F 171.3N 2 2 FR = FRx + FRy = cos = = 0.7548 R Rx F F θ cos = = 0.6556 R Ry F F β   θ = 40.99 , β = 49.01 例2-2 已知：图示平面共点力系；