Hallo Kemal,
FH = 3,454 kN ; FN = 21,801 kN , Reibungskraft Fr wegdenken :-)
FH = FG · sin(w) G1
FN = FG · cos(w) G2
G1 / G2 :
FH / FN = sin(w) / cos(w) = tan(w) ≈ 0,1584 → w ≈ 9°
α in G1 :
FG = FH / sin(w) ≈ 22,08 kN
m = FG / g ≈ 22,08 * 1000N / (9,81 N/kg) ≈ 2250,8 kg
Gruß Wolfgang