Calculating Moments Using the Method of Joints: A Tutorial

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Discussion Overview

The discussion revolves around the calculation of moments using the method of joints in a static equilibrium problem. Participants explore the application of free body diagrams (FBDs) and the relevance of specific forces in determining moments about points B and C. The conversation includes aspects of homework-related problem-solving and technical reasoning.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the validity of the author's approach to calculating moments about point B, specifically regarding the exclusion of forces Cx and Cy.
  • Others argue that only forces acting on the boom should be included in the FBD for that member, leading to the conclusion that Cx and Cy do not affect the moment about B.
  • There is a repeated inquiry into the definition of the "boom" and its role in the problem, with some participants seeking clarification on why certain points are included or excluded in calculations.
  • Some participants propose that the method used by the author is a quicker way to show that Ay equals zero, while others express confusion over why this method was chosen over their own calculations.
  • There are discussions about the implications of taking moments about different points and how it affects the equations derived from the problem.
  • Participants express uncertainty about the correctness of their own equations compared to the author's, with some suggesting that their approach yields different results.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to take moments about points B and C. Multiple competing views remain regarding the relevance of certain forces and the methods used to derive the equations.

Contextual Notes

There are unresolved questions about the assumptions made in the calculations, particularly concerning the definitions of the boom and the forces acting on it. The discussion highlights the complexity of static equilibrium problems and the various methods that can be employed to analyze them.

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Homework Statement


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 ?

Homework Equations

The Attempt at a Solution

 

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because the moment Cy(0.8) +Ay(0.8) can cause the member to turn about B , right ?
 
Cy and Cx don't act on the boom so don't appear in the fbd for that member.
 
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billy_joule said:
Cy and Cx don't act on the boom so don't appear in the fbd for that member.
What do you mean by boom?
 
Read your first attachment.
The fbd for the boom appears at the bottom of your 3rd attachment.
 
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billy_joule said:
Cy and Cx don't act on the boom so don't appear in the fbd for that member.
OK, saw it.. Why they don't act on the 'boom' ?
 
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.
 
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billy_joule said:
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.
What is the boom actually??
 
billy_joule said:
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.
Then, why is point C given in the problem?
 
  • #10
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.
 
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  • #11
Edit:Double post.
 
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  • #12
billy_joule said:
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.
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
chetzread said:
We need to consider moment at C due to all members in the problem, right?
Your second attachment takes moments about C for the entire structure.

So, we need to consider the forces at C, right?
The question asks you to find the reaction forces, so yeah, we need to consider the forces at C, and find them.
 
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  • #14
billy_joule said:
Your second attachment takes moments about C for the entire structure.The question asks you to find the reaction forces, so yeah, we need to consider the forces at C, and find them.
No, I am asking why for moment about B, ' we don't have to consider force Cx and Cy??
 
  • #15
chetzread said:
No, I am asking why for moment about B, ' we don't have to consider force Cx and Cy??
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
billy_joule said:
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.
What is the boom here?
 
  • #17
See post #10
 
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  • #18
billy_joule said:
See post #10
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
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.
 
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  • #20
billy_joule said:
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.
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
So, the author purposely make Ay = 0?
 
  • #22
chetzread said:
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?
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:

billy_joule said:
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.

Finding the quickest, easiest way will come with practice.
chetzread said:
So, the author purposely make Ay = 0?

No, we don't get to choose what the reaction forces are! We need to find them using math and physics.
 
  • #23
billy_joule said:
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.
I still don't understand why the author choose to take moment about C at one boom only...
 
  • #24
Because it's the fastest way to find Ay.
Did you try this?;
billy_joule said:
.
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...
 
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  • #25
billy_joule said:
Because it's the fastest way to find Ay.
Did you try this?;
i will end up getting Cy(0.8) +Ay(0.8) + Cx(0.6) = 0
Cy(0.8) +Ay(0.8) -40(0.6) = 0 , where Cy = Ay = 15N ?
Which is not same with the author where Ay = 0 ...Is it wrong ?
 
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  • #26
billy_joule said:
Because it's the fastest way to find Ay.
Did you try this?;
so , how does the author conclude that Ay = 0 in this case ?
 
  • #27
chetzread said:
i will end up getting Cy(0.8) +Ay(0.8) + Cx(0.6) = 0
Cy(0.8) +Ay(0.8) -40(0.6) = 0 , where Cy = Ay = 15N ?
Which is not same with the author where Ay = 0 ...Is it wrong ?
Something went wrong with your algebra.
Show all your working.

I've had another look and taking moments about B for the entire structure offers no new information, all it says is that Ay + Cy = 30, which we already know from the vertical forces equilibrium equation. I apologise for saying it could be used to find Ay!

The question can't be solved without doing a fbd of either one of the rod or boom. We can take moments about B for the rod to find Cy but is faster to do what the author did.
 
  • #28
billy_joule said:
Something went wrong with your algebra.
Show all your working.

I've had another look and taking moments about B for the entire structure offers no new
information, all it says is that Ay + Cy = 30, which we already know from the vertical forces equilibrium equation. I apologise for saying it could be used to find Ay!
The question can't be solved without doing a fbd of either one of the rod or boom. We can
take moments about B for the rod to find Cy
but is faster to do what the author did.
So, my moment about B = Cy(0.8) +Ay(0.8)
+ Cx(0.6) = 0
is wrong? Or it's not suitable to solve this question ?
 
  • #29
billy_joule said:
Something went wrong with your algebra.
Show all your working.

I've had another look and taking moments about B for the entire structure offers no new information, all it says is that Ay + Cy = 30, which we already know from the vertical forces equilibrium equation. I apologise for saying it could be used to find Ay!

The question can't be solved without doing a fbd of either one of the rod or boom. We can take moments about B for the rod to find Cy but is faster to do what the author did.
Since Ay =0? Then Cy = 30? If viewing follow the author's working?
 
  • #30
chetzread said:
So, my moment about B = Cy(0.8) +Ay(0.8)
+ Cx(0.6) = 0
is wrong?
It's right.
Your subsequent working must have had an error as your conclusion was incorrect.
Or it's not suitable to solve this question ?
Why not try again, be more careful with your algebra and see how far you get.

chetzread said:
Since Ay =0? Then Cy = 30? If viewing follow the author's working?
I'm not sure what you mean by that.
 

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