Good evening Atika! I’m not so sure about this one, but the basic idea is that the moment of a force is the force applied times the perpendicular distance from the pivot.

The pivot is at A, so the 10 N force directly at A does not affect the rotation in any way at all.

The distance from A to B is 0.25 m. The force is not directed in a perpendicular manner, but at an angle. This diagonal force applied can be broken down into its horizontal force and its vertical force. Only its vertical force, which is perpendicular to the 0.25 m length, will contribute to the moment (in an anticlockwise manner).

Moment due to this force at B
= 0.25 m x (30 sin 45) N
= ... Nm anticlockwise

For the force at C, I presume we need to calculate the “straight-line distance AC” and the perpendicular force is based on the angle 90 degrees from AC, so this force at C is likely to cause an anticlockwise moment about the pivot (but then this again depends on the calculation); we need to take the component which is perpendicular to this “straight-line distance AC”.

For the force at D, the straight line distance is 0.37 m. Of course the force is perpendicular to this line AD, so it will create a moment of 0.37 m x 14 N clockwise.

We combine these to obtain the overall resultant moment about A. This will have to depend on the force at C.

I am not so sure if these workings are correct though.