Calculating Center of Gravity and Force for a Flagstaff

[tiny-tim]

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!

 

[Newton86]

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 :)

 

[tiny-tim]

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?

 

[Newton86]

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

It is correct isn't it ?

 

[tiny-tim]

Yes, of course it's correct!

But which equations did you get it from?

You say …

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?

(… 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? )

 

[Newton86]

I don't see why I cant
And I don't know about that but I need to get the task done so I need to do the way I can But I would like to see your ways of doing it caus its probably a better and "safer" way.

 

[tiny-tim]

ok …

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 ​

(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

You have to be able to solve these equations.​

 

[Newton86]

Yeah I should =| But I believe I learned a bit from it.
Thanks for your time