Why is this force considered in the moment equation?

[Jarvis88]

Homework Statement

After reviewing a quiz I had in Statics, I realized I did not understand why a certain force was considered in the moments equation. If a force goes directly through the point of the moment, it is not considered when solving for a force using moments.

The question is asking us to use the method of sections by splitting the truss from section 1-1 and find the force in member BC.

Homework Equations

ΣM_H=0
By the way, I substituted kips for k.

The Attempt at a Solution

ΣM_E=0 → -A_y(60 ft) +(40k)(45 ft)+ (40k)(30 ft)+ (40k)(15 ft)-(30k)(20 ft)=0.
A_y=50k.

The equation I would have used to find F_BC: ΣM_H=0 → -50k(15 ft) + F_BC(20 ft)=0.
The equation the professor used to get the correct answer for the force along member BC: ΣM_H=0 → (-30k)(20 ft) -50k(15 ft) + F_BC(20 ft)=0

Why is the 30k force considered when it points directly into H? I understand it being used to find the moment about E, but not for the moment about H. Also, why is the lever arm distance 20 ft when the force is right at H?

 

[TomHart]

I can't see how the professor came up with the equation that he/she did. Your equation looks correct to me.

 

[NascentOxygen]

The equation I would have used to find F_BC: ΣM_H=0 →

You have not taken into account the horizontal component of F_A.

 

[TomHart]

You have not taken into account the horizontal component of FA.

I looked at this for, not a long time, but a fair amount of time, considering, "Is there any other force that I am not seeing?" And still I missed it. :(

 

[CWatters]

Hint: It's NOT the 30kips force at H they are using in the moments equation. My guess is you hadn't drawn a FBD of the bridge showing all the forces acting on it.