Calculating Center of Gravity and Force for a Flagstaff

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Homework Help Overview

The discussion revolves around determining the center of gravity and the forces acting on a flagstaff that is 10 meters long. The original poster presents a scenario where the flagstaff is balanced at both ends, requiring different forces to maintain horizontal equilibrium. The problem involves concepts from mechanics, particularly moments and forces.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the setup of the problem, questioning the clarity of the original statement and the meaning of certain terms. There are discussions about drawing diagrams to visualize forces and moments acting on the flagstaff. Some participants suggest taking moments about a pivot point to eliminate unknown forces.

Discussion Status

The conversation is ongoing, with participants providing hints and guidance on how to approach the problem. There is an emphasis on the importance of clear diagrams and understanding the concept of moments. Multiple interpretations of the problem are being explored, and participants are encouraged to express their reasoning and confusion.

Contextual Notes

Some participants express difficulty in understanding the problem due to language barriers and unclear terminology. There are also mentions of the need to grasp fundamental concepts of moments and forces to progress in solving the problem.

  • #31
tiny-tim said:
erm … what happened to the 10 - x … and where did that 10F come from? :confused:

I don't know how to explain I don't think I understand some basics here.
why is the 1 one 0 + wX and the second one 10 - x ?

Is there another way to solve it ? =s

http://www2.ucdsb.on.ca/tiss/stretton/PhysicsFilm/ForceTorque/sld011.html
Found this. Could this problem be solved in a similar way ?
 
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  • #32
Newton86 said:
Iwhy is the 1 one 0 + wX and the second one 10 - x ?

Because first the cog is distance x from the wall,

but when you turn the flagpole round, the cog is distance (10 - x) from the wall! :smile:

So the second equation has (10 - x)W where the first equation has xW. smile:

(and no, that picture doesn't help, because it shows two known forces at the same time, and this problem only has one at a time … nice website though! :smile:)
 
  • #33
That make sense :)
1: 0 + xW -10F
2: (10-x)W + 40F then ?
 
  • #34
Newton86 said:
… then ?

erm … to find x, you eliminate F.

no? :smile:
 
  • #35
So the equations are correct now ?
Its Wx we need to find ? Do i solve it by using both equations ? never been good at equations either.
 
  • #36
Multiply one equation by a number so that the same amount of F is in both equations … then just subtract the equations … that gets rid of (eliminates) F! :smile:
 
  • #37
Found another (easy) way to solve it :)

thin end 1/5* 10= 2
thick end 4/5*10= 8
So the gravity of force is 2m from the thick end and 8m from the thin end.
So now you can reveal your way that Iv tryed to understand for 4 days now :)
 
  • #38
Newton86 said:
Found another (easy) way to solve it :)

thin end 1/5* 10= 2
thick end 4/5*10= 8

hmm … but where did you get 1/5 and 4/5 from?
 
  • #39
tiny-tim said:
hmm … but where did you get 1/5 and 4/5 from?

Forces. 4+1 = 5F
4/5 in the thick end
1/5 in the thin end :blushing:

It is correct isn't it ?
 
  • #40
Yes, of course it's correct! :smile:

But which equations did you get it from?

You say …
Newton86 said:
4+1 = 5F

(I assume you mean 4F + F = 5F)

… but these forces are not applied at the same time, so what is your justification for adding them? :confused:

(… and what would you do if the wall was not at the end of the flagpole, but was 1m from the end, either on one or both occasions? :wink: )
 
  • #41
I don't see why I cant:confused:
And I don't know about that but I need to get the task done so I need to do the way I can :wink: But I would like to see your ways of doing it caus its probably a better and "safer" way.
 
  • #42
:smile: ok …
Newton86 said:
That make sense :)
1: 0 + xW -10F
2: (10-x)W + 40F then ?

Multiply the first equation by 4:

4xW = 40F​

Combine it with the second equation:

(10-x)W = 40F​

and you get:
4xW = (10-x)W

so 5xW = 10W

so x = 2 :smile:

(suppose the wall was 1m from the end on both occasions … then it would be 4(x-1)W = (9-x)W, so 5xW = 13W, so x = 2.6)

What worries me is why you couldn't see that :frown: :confused:

You have to be able to solve these equations.
 
  • #43
Yeah I should =| But I believe I learned a bit from it.
Thanks for your time:smile:
 

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