basic physics - principle of moments


This question is so easy I'm embarrassed to ask but i don't know the answer :)

A uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark. What is the value of X

I presume this means the pivot is at 40 cm and the mass is placed at the 10cm mark so its perpendicular distance from pivot is 30cm

but i don't know how to take into account the pivot isn't in the middle and hence the weight of the metre rule on either side.

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On one side of balance point you have 60 gm of stick.  
The CM of this mass is 30 cm from the balance point

30cm*60g = 20cm*40g + 30cm*Xgm

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andiejeAuthor Commented:
How should i have known to take the centre of mass of that weight as the perpendicular distance?

My initial thought was that for each cm there is 1gm of weight so you would add all of these up to get 40 on one side and 60 on the other and that would give you the force from the balance itself.

I am presuming CM means centre of mass. I havent come across that term yet i the book i'm reading
CM does mean "center of mass".

The essential simplifying assumption is to find the centers of mass for the two
sides of the stick.  

This is probably the first problem you've seen where the mass of the balance
comes into play.

Always make a drawing.  If the balance point is in the center, the mass of the
stick drops out.  Otherwise, you have to find the moment for each side.

For a moment, you need a mass and a length.  You could break it up into
60 x (1 gm segments) on one side and 40 x (1 gm segments) on the other.

But center of mass is an important and powerful concept.
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