Answer
1253.553Kn\approx1254Kn
or 1.3Kn
Work Step by Step
We have 3 Forces , and the inclination is 0, i.e no need to take components, However for 2.5 m inclined distance we will take the component.
Now the first Force 600N would be negative as it is CLOCKWISE, M1= -600(1m)=600Nm
Now for 300N , it is in clockwise, direction as well, so take it -ve as well, but as far as distance is concerned , we would be having distance as 2.5 Sin 45
i.e M2= -300 * (2.5 Sin 45) = -530.33Nm
now for 500N force, it is Anticlockwise , taken +ve, the distance are measure from O i.e (1m + 2m + 2.5 Cos 45m)
therefore M3= 500 * (1+2+2.5 Cos45) = 2383.883
Thus Adding M1, M2 and M3 , we get
Mf= -600-530.33-2383.883= 1253.553 or 1254 Nm or 1.3 KN
