# What is the distance at which the triangular UDL is acting on in this problem?

I'm working on the statics problem below:
I'm tasked to draw the shear and bending moment diagram for the frame. I'm trying to work out the reactions and I'm taking the moment about pin E.Here's my working so far:
$$\Sigma M_E=0$$ $$(10)(5)-100-(2*10)(\frac{1}{2}*10+5)+(10)(15)-100-(\frac{30}{\sqrt{2}})+(0.5*5*4)(\frac{1}{3}*5)+A_y(20)=0$$ My question here is, for the triangular UDL, is the distance one-third or two thirds? My lecturer did a similar problem and he said that the distance is two thirds but I don't quite get why it is two thirds. Someone help.

It is 2/3 and also if we assume the moment direction positive clockwise, to be consistent with the rest of your positive signs, then that vertical loading's moment is negative, the correct sign and factor is (-0.5*5*4) (2/3*5).

The reason the distance is 2/3 is because C.G. of a the vertical 5m load triangle is at 1/3 from its base, which here is 2/3s from E.