Calculating Center of Gravity and Force for a Flagstaff

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SUMMARY

This discussion focuses on calculating the center of gravity and the forces acting on a flagstaff of length 10 meters and mass 250 kg. The problem involves determining the vertical force F needed to maintain horizontal balance when the flagstaff is supported at one end, and a horizontal force of 4F when supported at the other end. The solution requires understanding moments and forces, leading to the conclusion that the center of gravity is located at 2 meters from the thick end and 8 meters from the thin end.

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  • #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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