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What will happen with this scale and submerged objects

by warfreak131
Tags: happen, objects, scale, submerged
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Sep15-13, 11:54 AM
P: 181
1. The problem statement, all variables and given/known data

Hello all, I saw a picture today of a physics problem which is really a head scratcher in my opinion. See the attached photo. You have two columns of water on a scale. In the left column, you have a ping pong ball, attached by a string to the bottom of the container. In the right column, you have a metal ball fully submerged, but being held up by an external support.

Both containers have an equal mass of water, both balls have an equal volume, and the strings have no mass.

Which way will the scale move, and why?

2. Relevant equations

3. The attempt at a solution

My gut instinct is to say that the scale will dip towards the right. But I don't know if this is mostly due to the buoyancy of the ping pong ball, the added weight of the metal ball, or some combination of roughly 50% of both.

As far as the ping pong ball is concerned:
I know that if you were to attach a helium balloon for example to the container, it will act to lift the container due to the buoyancy of helium in air. Likewise, a ping pong in water will have a buoyancy force which wants to lift the ball and pull the container up since its attached, but something about that scenario doesn't seem right, which is why I'm here. Thank you in advance!
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Sep15-13, 12:20 PM
P: 124
Well the metal ball isn't part of the system? It's weight is being held from the outside, so it doesn't play a role. Thus the scale should go left, since there is more weight.
Sep15-13, 12:22 PM
P: 181
Ok, I see, but the buoyancy of the ping pong ball doesn't play any role?

Sep15-13, 12:27 PM
P: 124
What will happen with this scale and submerged objects

I don't think the force upwards on the ball plays any role for the entire force on the scale. Thus it is just a manner of which container is heavier.
Sep22-13, 08:51 PM
P: 1
The scale should be lower on the right side.

Imagine if you removed the spring from the left beaker. The ball would float to the top and would (practically) not displace any water. If you weighed it before and after removing the spring it would be the same (ignoring the weight of the spring in both cases). Think of it as trying to make yourself weigh less by pulling up on your legs when you are standing on a scale (not going to happen, though I wish I could at the doctors). Now since it is attached the buoyancy of that volume of water has to go somewhere, and it is going to be stored in the spring since there is going to be excess buoyancy acting up on the ball than there would be if were floating on the surface.

As for the steel ball, the buoyancy force acting up on the ball is going to be equal to the actual volume it is displacing. If you were to have a scale under the mount holding the steel ball it will read less after you submerge the ball equal to weight of the volume of water it is displacing. So if you were to read the scale while it is submerged and add the weight of the displaced water it would be equal to the weight of the ball.

Basically the buoyancy of the ping pong ball acting down is far less than the steel ball. Even though the same amount of water is being displaced, the weight added by the buoyancy force of the steel ball is larger.

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