Example question given -
"Find Reactions at A. Draw a free body diagram and the positions of all forces, check for equlibrium conditions."

I've attached the diagram given, and a diagram with the forces which I think are at A (F1 & F2).

I don't know what the forces will be so help to find this would be appreciated.

Yes, you got them at the right place but in the wrong direction. What you drew are the resultant forces on A, not the reaction forces at A as asked. Can you try it again?

To find F2, resolve the 9 kN force, since there will then be only one horizontal force, that one will be F2.

For F1, you can apply the same logic as to the previous question, the one about a mass supported by two supports A and B. The mass here is the vertical component of the 9 kN force. Part is supported by the 3 kN force, and the other part by the fixed point A. Can you find it now?

Ok there's 2 diagrams I've added, the first is an edit of the original with different directions for the forces at A.

The second has added forces, would these be here and be equal to B?
So where ? = 3kN

If so this would leave 3kN (9kn - (A+B)=3kN) for the forces F1 & F2, or am I on the wrong lines?

Cheers.

Ok, in the first diagram, the upward force at A is about 4.79 kN. (the vertical component of 9 kN is given by (9 sin 60 = 7.79 kN, then removing the 3 kN gives 4.79 kN at A for the system to be in equilibrium)

For the second one, I'm a having some trouble understanding the question... If the forces are indeed there, the would be more force by the lower arrow than by the upper arrow for the rod/plank to remain in equilibrium.

The second was just a guess, assuming forces would be here (its also not to scale so the forces aren't represented by arrow length in the example).

I'm still struggling to get my head around how to work this one out and finding all the forces acting on the diagram.

The first diagram was good (perhaps instead of the curved force, you can put a vertical force, acting upwards. The curve is more used to describe a moment). The horizontal force is given by the horizontal component of 9 kN, given by 9 cos (60) = 4.5 kN.

Ah I get you now, so the diagram should look like one of these? Does it matter where the verticle pointing arrown is placed? There are two examples attached.

Right, better not make them 'cross' each other though. My teacher told me that this is a sort of convention, and it's better so as not to mess things up.

Well done! :)