Last example I've got here:

"Find Reactions at L and R where Forces A= 2.2kN, B=1.3kN, C=6kN"

I've attached the diagram given.

Also,
The FBD I've drawn is a trapezium, is this correct?

I guess you have to take it as a normal problem. Just be sure which values you take. All the forces are acting only to the vertical or horizontal.

A: 2.2 kN.

Force due to A on L is (800 * 3)(2.2)/(800*4) [clockwise moment]
You get this by taking R as pivot, the moment is (800*3)(2.2).
Then, to balance the force, L should exert the same moment: F.
F * (800 * 4) = (800 * 3)(2.2)
F = (800 * 3)(2.2)/(800 * 4)

B: 1.3 kN

Force due to B on L is (1385)(1.3)/(800 * 4)
You need to extrapolate an imaginary line along B so that B is perpendicular to R. It's like you have an imaginary circle around R (of any size you want) and you need to make the line of action of B a tangent of that circle.
This gives you the moment due to B, at R. Force that L must exert so that system is in equilibrium: F
F * (800 * 4) = (1385)(1.3)
F = (1385)(1.3)/(800 * 4)

C: 6 kN

This one is easy, you don't need to consider the whole structure, but make as if C was on a straight plank.

Add those up, and they will give you the total force on L.

Once done, use the same method for R.

I'm not too sure about B though (I haven't dealt with such a problem before. If someone here knows that it isn't the proper way to tackle B, I'll be most pleased to see the correct way.

Okay, with the methods you gave, I came to the forces:

A: 1.65
B: 0.56
C: 3.00

Total = 5.21kN
(This is at L)

Does this look right?

Yes, it's correct! :)

Right, so they are correct,

For R:
C is the same for both sides as its equal.
A I got 0.55kN

B is what I am stuck with.

Ok, I told you already that I was not sure for B, and following my present guess of how to solve it, you need to extrapolate the force of B, so that this time, the circle is around L, L being the pivot.

The anticlockwise moment given by B is (1.3*1385) = 1800.5 kNm.

The force that R has to exert is (1800.5/3200) = 0.5626 kN.

Now, here, if you can picture it, B is not 'pushing' on R but is trying to lift R in the air. Hence, the forces on R is less (or that force calculated above is negative).

It's just like something in the shape of an 'L'. If you move the vertical part to the left, you are lifting the end of the horizontal part. If you are moving the vertical part to the right, the force pushes the horizontal part downwards.

Thanks, well from that I have 4.11kN for R. I'll see what the rest of the class got for their results and compare.

You didn't take the negative of the force due to B. You have to remove 0.56 from the total of forces A and C.

This gives 2.99 kN