Topic: Puzzle: balls and beakers

[seidensticker]

I've seen people pull their hair out over this one; I hope you like it.

You have a balance scale (the kind that tips to the side that's heavier).  On both sides is a beaker half full of water.  They're identical except that the left side has a ping pong ball held totally underwater by a string glued to the bottom of the beaker.  On the right side is a lead ball also in the middle of the water.  It's suspended by a string from the arm of a small crane standing on the table (not on the balance).

All the obvious simplifications apply -- the balls are of the same volume, both balls are submerged and neither touches the glass, the strings and glue are weightless, the beakers and the amounts of water are identical, and so on.

The question is: which side goes down?  The hard question is: why?

This might be a puzzle for the podcast.

[JD]

I'd suspect it's neither.  The weight of the balls have no effect.

[azinyk]

A ping pong ball has a mass of about 2.8 grams and a volume of about 28.8 ml.  Assume water has a density of 1.0 kg/L

Then if the water level is at the 500 ml line, the plastic side has 500-28.8 = 471.2 ml of water.

The lead side has the same amount of water because the water line and the volume of the ball are the same.

The force pushing down the plastic side is simply it's weight, (471.2 + 2.8)*9.81 = 4.65 N

The force pushing down the lead side is equal to the weight of the water plus the buoyant force pushing up on the lead ball (by Newton's third law, an equal and opposite force pushes down on the beaker).  The weight of the water is 471.2*9.81 = 4.62 N.
By Archemedes' principle, the buoyant force on the lead ball is equal to the weight of the water it displaces, which is 28.8*9.81 = 0.28 N.  Thus the total force pushing the lead side down is 4.62+0.28 = 4.91 N.  

Thus the lead side goes down.  It should be noted that the tension in the string holding the lead weight is reduced by the amount of buoyant force pushing up on the lead ball.  You can ignore the mass of the plastic ball if you like, the effect will only be emphasized.  Actually, the problem works just as well without the plastic side, but that makes a nice red herring for people who work out that the volumes are equal, and then stop there.

Ashley Zinyk

[reed]

Thus the lead side goes down.

Agree.  Both sides have the weight of the water and the container.
The plastic side has the weight of the ping pong ball.
The lead side has the buoyant force.

ping pong ball:
mass = 2.7e-3 kg
radius = 2e-2 m
volume = 3.35e-5 m^3
weight = 0.027 N
buoyant force = 0.33 N

The buoyant force is about 0.3 N greater than the weight of the ping pong ball, so the lead side goes down.