MHB Is my vertical component calculation correct?

AI Thread Summary
The discussion revolves around calculating the vertical component of the force exerted by a pin on a beam connected to a cable and a weight. Two methods were proposed, but each yielded different results for the vertical force at point D. The first method incorrectly accounted for the direction of the forces, particularly the tension in the cable. The second method raised questions about the origin of certain terms in the equations used. Overall, there is confusion regarding the correct approach to determine the vertical component accurately.
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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?
 

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Rido12 said:
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. :eek:

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)
 
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