The beaker on the right reduces tension in the string which increases the weight of the beaker. The water partially supports the steel ball.
Well?
Josiah Cruz
Sebastian Gray
>pingpong ball lighter in weight than steel ball
The weight of the ball on the right is irrelevant, so long as it's heavy enough to sink (i.e. more dense than water).
It's stated that the ball on the right is submerged, which requires that it's more dense than water (however, it doesn't change the answer; it could be a ping-pong ball if the string was replaced by a rigid rod; what matters is that it's submerged by an external force).
We're left to assume that the one on the left is less dense than water, as the drawing shows it not resting on the bottom and it's referred to as a ping-pong ball (which tend to be less dense than water).
That's a reasonable assumption. Or at least it would be if this wasn't troll central.
Lucas Adams
Right side goes down.
Alexander Peterson
>The water partially supports the steel ball.
No, it does not.
The balls on the left and right side have the same volume, and thus displace an equal volume/mass of water.
The ball on the left side, reguardless of bouyancy, adds it's weight to the left side of the scale. (it's bouyancy force is canceled by the string holding it in place)
The ball on the right does NOT add it's weight to the right side, because it's weight is supported by the string and not the water.
So, the difference in mass between the two balances is the ping pong ball and the string holding it in place.
Hence, the left side has more mass than the right side, and thus the left side goes down.
Hudson Walker
>it could be a ping-pong ball if the string was replaced by a rigid rod
This is not correct, as if the right side had a ping pong ball submerged, and held in place by a rigid rod, then it's bouyancy would actually introduce a downward force on the right side of the balance.
The metal ball does NOT do this, because it is more dense than water and sinks, instead of floating.
Colton Barnes
>its bouyancy would actually introduce a downward force on the right side of the balance.
why does this not happen with the metal ball? If the water exerts an upward buoyant force on the ball, then the ball exerts the same amount of force in the opposite direction on the water, by Newton's Third law.
Logan Thompson
vid related:
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Nolan Gomez
I feel like the density of the ball on the right is isolated from the balance system so long as it doesn't move. So if it was held in place by a rigid wire instead of a flexible string it could be any density and it wouldn't matter.
Imagine if it was hollow and you filled it with mercury and then somehow sucked it back out.
Jaxson Hernandez
better question. What happens if you freeze both?
Gabriel Wood
The buoyancy force isn't cancelled.
Force diagram of left ball: Buoyancy = Tension + Weight(ping-pong)
or Weight(p) = Tension - Buoyancy
When you later draw the force diagram for the left beaker with the ball removed, you're left with only the normals for buoyancy and tension. The net effect is equal to the weight of the ping-pong ball.
On the right ball, the equation is: Buoyancy + Tension = Weight(steel)
or Buoyancy = Tension - Weight(st)
When you draw the force diagram from the right beaker with the ball removed, you're only left with the normal for buoyancy. Thus, while the left side adds the weight of the ping-pong ball, the right side adds buoyancy or the weight of a "water" ball, meaning the right side has a larger downward force and tips.