Calculating Moments Using the Method of Joints: A Tutorial

[chetzread]

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

 

[chetzread]

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

 

[billy_joule]

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

 

[chetzread]

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?

 

[billy_joule]

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

 

[chetzread]

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' ?

 

[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.

 

[chetzread]

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??

 

[chetzread]

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?

 

[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.

 

[billy_joule]

Edit:Double post.

 

[chetzread]

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?

 

[billy_joule]

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.

 

[chetzread]

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??

 

[billy_joule]

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.

 

[chetzread]

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?

 

[billy_joule]

See post #10

 

[chetzread]

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)

 

[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.

 

[chetzread]

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?

 

[chetzread]

So, the author purposely make Ay = 0?

 

[billy_joule]

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:

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.

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.

 

[chetzread]

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...

 

[billy_joule]

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

.
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...

 

[chetzread]

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 ?

 

[chetzread]

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 ?

 

[billy_joule]

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.

 

[chetzread]

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 ?

 

[chetzread]

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?

 

[billy_joule]

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.

Since Ay =0? Then Cy = 30? If viewing follow the author's working?

I'm not sure what you mean by that.

 

[chetzread]

It's right.
Your subsequent working must have had an error as your conclusion was incorrect.

Why not try again, be more careful with your algebra and see how far you get.I'm not sure what you mean by that.

If I consider the rod CB only, then my moment about B = Cy(8) =0... Then Cy =0... But in the author's working, the author found that Ay = 0
My answer is different with the authors ans

 

[billy_joule]

If I consider the rod CB only, then my moment about B = Cy(8) =0... Then Cy =0... But in the author's working, the author found that Ay = 0
My answer is different with the authors ans

You forgot to include Cx.

 

[chetzread]

You forgot to include Cx.

OK, I gt the answer... So for this type of problem which involve multiple force member, we should consider forces equilibrium in 1 force member only if we can't solve the forces using equilibrium at all 3 members ?

 

[chetzread]

I

OK, I gt the answer... So for this type of problem which involve multiple force member, we should consider forces equilibrium in 1 force member only if we can't solve the forces using equilibrium at all 3 members ?

In this case we only consider moment about B in a rod only? If we consider moment about B in 2 rods, we can't solve the question?

 

[billy_joule]

OK, I gt the answer... So for this type of problem which involve multiple force member, we should consider forces equilibrium in 1 force member only if we can't solve the forces using equilibrium at all 3 members ?

I'm not sure if there's a hard and fast rule.
Seeing the quickest and easiest approach will come with practice.
This particular problem could be solved much faster with the method of joints.