Help! distance from pivot x to balance beam?

1. Dec 26, 2008

Richardos

Help! ive spent 2 days now trying to do this now and it driving me mad.

http://www.emvino.com/pivot.jpg [Broken]

Given the following diagram, what distance from the pivot (x) is required to balance the beam?

any help would be highly appreciated.

Thanks

Last edited by a moderator: May 3, 2017
2. Dec 26, 2008

sai_2008

2.5 m

3. Dec 26, 2008

Domnu

Hello Sai_2008,

Please help Richardos with his question, but do not answer it for him, as this doesn't help anybody with the problem. Show exactly how you obtained this answer, so that everyone else with the same problem can see how you did it, should they have the same question.

Thanks very much,

Domnu

My Solution
We need

$$\sum F_{left} x_{left} = \sum F_{right} x_{right}$$

where $$F_{i}$$ denotes each force at a particular point (on the left or right) and $$x_{i}$$ denotes the distance away from the pivot at a particular point. What do we get from here?

4. Dec 26, 2008

Domnu

Oh, btw sai_2008, I'm not sure, but I think I know who you are? I just PMed you.. check please =]

5. Dec 26, 2008

sai_2008

Sorry...

The force moments about the pivot must be equal on both sides (Exactly as domnu said in the equation below) for the balance to occur.

i.e., 10k x 5 = 20k X 2.5

sai

6. Dec 26, 2008

NoMoreExams

Just a word of advice, don't use x or X (especially not both in one line) to denote multiplication. I would use "*" for that if you don't want to parse it in TeX