Math question. Non school related, for once.

[IEatChildren]

This is about leverage and balance, for a personal building project.

Let's say I have a lever, and the fulcrum is 2/3 of the way so that one side of the bar is half the length of the other side.
If I put 10 lbs on the short side, how much weight do I need on the other side to have perfect balance? Assuming of course that the weight of the bar itself is negligible.

What if the bar was 3 times longer on one side than on the other?

 

[Legendary]

This is a question of torques - how far away is the weight put onto the bar from the fulcrum?

Edit: Anyway, to answer the question - the Weight times the distance to the fulcrum on one side has to equal the weight times the distance to the fulcrum on the other side. So if 10 pounds is on one side that's 1 m away, then there has to be 5 pounds on the other side that is 2 meters away.

 

[IEatChildren]

Assuming they're at the ends. One side is half the length of the other. If 10 lbs is on the short end, how much needs to be on the other end?

 

[Legendary]

See the edit.

 

[GroundZero]

Originally posted by: Legendary
This is a question of torques - how far away is the weight put onto the bar from the fulcrum?

Edit: Anyway, to answer the question - the Weight times the distance to the fulcrum on one side has to equal the weight times the distance to the fulcrum on the other side. So if 10 pounds is on one side that's 1 m away, then there has to be 5 pounds on the other side that is 2 meters away.

yep

 

[Ionizer86]

Fr=Ia=T
Force times radius = inertial times angular acceleration = torque

 

[IEatChildren]

So the weight necessary to balance it is directly proportional to the relative distance between the weights and the fulcrum, correct?