# Physics Riddle

1. Aug 12, 2014

### micromass

Staff Emeritus
What will happen to the balance:

2. Aug 12, 2014

### Nathanael

I assume the balls are equal volume?

I think the balance will tip to the right, only because the string on ping pong ball is pulling up on the container. If the ping pong ball were held in place (so it didn't float up) from the outside (like on the right side) then my answer would be that the scale will not move.

My reasoning is that the difference in weight (between the balls) is irrelevant because the excess weight of the steel ball is counteracted by the string holding it up. Both of the balls only push down on the water with the bouyant force proportional to their volume, so their mass is irrelevant.

But since there is a force of tension upwards on the left container, it must push down on the scale less, thus the scale tips to the right.

3. Aug 12, 2014

### Staff: Mentor

Ain't saying. :tongue:

(Good one, micro! )

4. Aug 12, 2014

Staff Emeritus
The steel ball will rust.

5. Aug 12, 2014

### TurtleMeister

I'll give it a try. I'm not certain of my answer though. Assuming we're supposed to say what will happen after insertion of the balls: The scale will tilt down on the side of the ping pong ball. (I haven't read the spoiler)

6. Aug 12, 2014

### AlephZero

The weight of both containers is the same. The depth of fluid in both containers is the same, so the hydrostatic pressure acting on both scale pans is the same. The only difference is the upwards force from the string on the left hand pan.

7. Aug 12, 2014

### TurtleMeister

I forgot to give the reason for my answer.

The balance scale measures a difference in mass. After insertion of the balls, the mass on the left plate has changed but the mass on the right plate has not changed (the steel ball is supported by the external hanger). Buoyancy is irrelevant because it's the same for both.

8. Aug 12, 2014

### ZetaOfThree

At first I thought the scale tips to the left, but now I think it tips to the right. Assume the containers have the same mass so forget about them. Also assume the balls have the same volume. The plate on the left feels the weight of the ping pong ball plus the water. The plate on the right feels the weight of the water plus the buoyant force from the steel ball (the rest of the force is taken care of by the string so that it stays put in the water). The buoyant force from the steel ball is the weight of a ball of the same volume as the steel ball, but made of water. So since a ball made of water is heavier than the ping pong ball, the force on the right plate is greater, and the scale tips to the right.

Last edited: Aug 12, 2014
9. Aug 12, 2014

### Staff: Mentor

Way to go Zeta!

There apparently are a lot of people here who think they can pick themselves up by their bootstraps.

10. Aug 12, 2014

### davenn

LOL ... I would have been another

Dave

11. Aug 12, 2014

Staff Emeritus
Good job, Zeta!

12. Aug 13, 2014

### rcgldr

It might have been interesting to also include a cup filled with water and no ball in it, then ask to compare the weights of all 3 cups, but that might have given away the answer. It could also be noted that the cup with the ping pong ball has lower average density, and no external forces (other than gravity) are exerted onto the ping pong ball.

13. Aug 13, 2014

### BobKat

It boils down to this question: Does the upward force caused by the air inside the ping pong ball counterbalance the added weight of the ping pong ball itself and the string which holds it. Both balls displace the same amount of water (presumably) and the ball on the right adds no weight to the right side. I suspect the scale stays even, because the air lifts the ball and string - BUT the string is attached to the bottom of the glass, so I don't think it can lift the glass by its own bootstrap (string).

14. Aug 13, 2014

### BobKat

Guess I don't understand how the steel ball contributes the weight of its volume in water to the weight on the right...

15. Aug 13, 2014

### OmCheeto

I just did the experiment. Zeta is correct. Although, I replaced the two balls with a plastic tool pick container, and the string with a rubber band. 9 oz vs 12 oz.

Buoyancy problems always have me scratching my head, and mopping up in the kitchen afterwards. :tongue:

16. Aug 13, 2014

### OmCheeto

Replace the steel ball on the right, with a ping pong ball. Instead of a string, smoosh the ping pong ball down with your fingers. I actually did this, but my ping pong ball had too little volume to indicate a weight change on my cheesy little 16 ounce dollar store diet scale. Hence my toothpick container upgrade.

Last edited: Aug 13, 2014
17. Aug 13, 2014

### rcgldr

Gravity exerts a downwards force onto the steel ball equal to the steel balls weight, which in turn exerts some of it's weight onto the water, and some onto the wire connected to the steel ball. This force is opposed by an upwards buoyant force exerted by what would the the weight of the water displaced by the steel ball. The remaining component of downwards force is opposed by the external (to the cup) upwards force from the wire connected to the steel ball. So the net downwards force exerted by the steel ball onto the water is equal in magnitude to the upwards buoyant force (Newton third law pair of forces), which is equal to what would be the weight of water displaced by the steel ball.

Last edited: Aug 13, 2014
18. Aug 13, 2014

### BruceW

you can think of it like a variation of energy problem, too.
If the right side moves down by $\delta y$ (i.e. positive $\delta y$ means right side goes down), then the change in GPE on the right is equal to $- (V_{water} + V_{ball}) \rho_{water} g \delta y$ This is because all the water moves down by $\delta y$ plus some small amount to account for the fact that the steel ball has moved upwards relative to it. The GPE of the steel ball itself does not change. Also, the left side will move upwards, so the change in GPE on the left side is: $V_{water} \rho_{water} g \delta y + \rho_{pingpong} V_{ball} g \delta y$ This is simply because everything on the left side moves upwards by $\delta y$ (including the ping pong ball). So, the total change in GPE is $(\rho_{pingpong}-\rho_{water}) V_{ball} g \delta y$ Which is negative, since the density of the ping pong ball is less than that of water, so in conclusion, if the right side moves down and left moves up, the total GPE will decrease, so that is what will happen.

19. Aug 13, 2014

### Staff: Mentor

For those still struggling:
If the string pulling up on the steel ball supported the full weight of the ball, then both sides would be balanced. But it doesn't. (The buoyant force partially supports the ball.) Thus the right side is heavier.

20. Aug 13, 2014

### chingel

I think it is easiest to think of it in terms of external forces on the containers, since their weights are the same. The pressure of the water depends only the the height of the water level, which is the same for both of them, but the left container has an additional string pulling it up.
I think we have to be careful not to say that the steel ball's weight somehow channels through the water, because the water pressure is the same in both containers.