Homework Help: Why is this force considered in the moment equation?

1. Apr 1, 2017

Jarvis88

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
ΣMH=0
By the way, I substituted kips for k.

3. The attempt at a solution
ΣME=0 → -Ay(60 ft) +(40k)(45 ft)+ (40k)(30 ft)+ (40k)(15 ft)-(30k)(20 ft)=0.
Ay=50k.

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

Last edited: Apr 1, 2017
2. Apr 2, 2017

TomHart

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

3. Apr 2, 2017

Staff: Mentor

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

4. Apr 2, 2017

TomHart

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. :(

5. Apr 4, 2017

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.