MHB Is my vertical component calculation correct?

[Dethrone]

In the structure shown, a cable is attached to the 50 kN weight and to the beam A-D at point B. If the horizontal uniform beam weighs 8 kN/m, determine the following:
(a) The horizontal and vertical component of the force that the pin at D exerts on the beam A-D.
(b) The force in the cable.
...

View attachment 3383

I just want to focus on getting the vertical component right...and I have developed two methods, each of which return a different answer.
1) Taking moments about A (i know doing it about B is simpler):
$$\sum M= T_y -40(2.5)-(T-50)(3)+D_y(5)$$
$$\sum F_y = T_y-40-50+D_y =0$$
$$=T \sin36.9-90+D_y$$

Now solving the two equations, I get $D_y = 61.98 \text{kN}$2) Starting with this line...
$$\sum M_B = -40(1.5)-50(2)-(50-T)(2)+4(V)=0$$

Any of them correct?

 

[I like Serena]

In the structure shown, a cable is attached to the 50 kN weight and to the beam A-D at point B. If the horizontal uniform beam weighs 8 kN/m, determine the following:
(a) The horizontal and vertical component of the force that the pin at D exerts on the beam A-D.
(b) The force in the cable.
...

https://www.physicsforums.com/attachments/3383

I just want to focus on getting the vertical component right...and I have developed two methods, each of which return a different answer.
1) Taking moments about A (i know doing it about B is simpler):
$$\sum M= T_y -40(2.5)-(T-50)(3)+D_y(5)$$
$$\sum F_y = T_y-40-50+D_y =0$$
$$=T \sin36.9-90+D_y$$

Now solving the two equations, I get $D_y = 61.98 \text{kN}$2) Starting with this line...
$$\sum M_B = -40(1.5)-50(2)-(50-T)(2)+4(V)=0$$

Any of them correct?

Doesn't look like it. (Doh)

In the first method, you have a force of $(T-50)$ that would act in a clockwise fashion (negative).
However, $T$ is upward while $50\text{ kN}$ is downward.
So the force should be $(50-T)$ instead.

Furthermore, in the sum of the vertical forces I seem to be missing a $T$ term.

In the second method I'm assuming that with $V$ you mean the same thing as $D_y$.
Anyway, here you do have the direction of the $(50-T)$ correct!
But... where is the $-50(2)$ term coming from? (Wondering)