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a=0,5; b=1 m; c=2,5
suma sił
-F1-F2+Ra+Rb=0
suma momentów
Ma: -F1a-F2(a+b)+Rb(a+b+c)=0
Rb=F1a+F2(a+b)/(a+b+c)
Rb= (3,5*0,5+3*1,5)/4=1,5625 kN
Ra=F1+F2-Rb= 3,5+3-1,5625=4,9375 kN
sprawdzenie
Ra+Rb= 1,5625+4,9375=6,5kN
F1+F2=6,5 kN