# Homework Help: Moment at different points

1. Sep 19, 2016

1. The problem statement, all variables and given/known data
For the moment about B , Cy(0.8) +Ay(0.8) + Cx(0.6) = 0 , right ?

Why the author only give ,-Ay(0.8) = 0

Is it wrong ?

2. Relevant equations

3. The attempt at a solution

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Last edited: Sep 19, 2016
2. Sep 19, 2016

because the moment Cy(0.8) +Ay(0.8) can cause the member to turn about B , right ?

3. Sep 19, 2016

### billy_joule

Cy and Cx don't act on the boom so don't appear in the fbd for that member.

4. Sep 19, 2016

What do you mean by boom?

5. Sep 19, 2016

### billy_joule

The fbd for the boom appears at the bottom of your 3rd attachment.

6. Sep 19, 2016

OK, saw it.. Why they don't act on the 'boom' ?

7. Sep 20, 2016

### billy_joule

Because point C isn't part of the boom.
would the forces you apply to your keyboard appear in a free body diagram of your neighbours kitchen sink?
A fbd of the boom should include only the forces acting on the boom. Those are the forces at A and B.

8. Sep 20, 2016

What is the boom actually??

9. Sep 20, 2016

Then, why is point C given in the problem?

10. Sep 20, 2016

### billy_joule

The rod starts at C and ends at B.
The boom starts at A and ends at B.
Call them member BC and member AB if you prefer.

11. Sep 20, 2016

### billy_joule

Edit:Double post.

12. Sep 20, 2016

We need to consider moment at C due to all members in the problem, right? So, we need to consider the forces at C, right?

13. Sep 20, 2016

### billy_joule

The question asks you to find the reaction forces, so yeah, we need to consider the forces at C, and find them.

14. Sep 20, 2016

No, I am asking why for moment about B, ' we don't have to consider force Cx and Cy??

15. Sep 20, 2016

### billy_joule

You'll have to be more specific.

The worked solution takes moments about B for the boom only.

Cx and Cy don't act on the boom so are not included in the free body diagram for the boom and so are not included in any subsequent equilibrium equations.

If we were to take moments about B for the rod, or for the entire structure then Cx and Cy would be relevant.

16. Sep 20, 2016

What is the boom here?

17. Sep 20, 2016

### billy_joule

See post #10

18. Sep 20, 2016

ok , why we have to consider 1 'boom' only during calculation of moment about B ?? for moment about C , we consider 2 boom , right ? That's why the Ax(0.6) and -30(0.8)

19. Sep 20, 2016

### billy_joule

It's just the way the author choose to do it.
It's a quick way to show Ay is zero.
It could be done in literally infinite different ways, we can take moments about any point in space for any or all static members and the answer is always zero. The trick is to find the most convenient point to make the math quick and easy.

Last edited: Sep 20, 2016
20. Sep 20, 2016

So, this is the special case? If not, then my moment about B , Cy(0.8) +Ay(0.8) + Cx(0.6) = 0 would be correct?

21. Sep 20, 2016

So, the author purposely make Ay = 0?

22. Sep 20, 2016

### billy_joule

What do you mean by special case?

Your equation is correct, and can be used to solve the question.
Why not try to solve it your way and see what happens?
Is your way or the authors way faster?

Hopefully you can now see why the author did what he or she did. It comes down to my earlier comment:

Finding the quickest, easiest way will come with practice.

No, we don't get to choose what the reaction forces are! We need to find them using math and physics.

23. Sep 20, 2016

I still don't understand why the author choose to take moment about C at one boom only.....

24. Sep 20, 2016

### billy_joule

Because it's the fastest way to find Ay.
Did you try this?;

25. Sep 20, 2016