Well?

but the left side is heavier than the right side....

50%. it either happens or it doesnt

correct

The question contains insufficient information.

But based upon reasonable assumptions, it will tip to the right.

The left side gets an additional downward force equal to the weight of the ping-pong ball and string.

The right side gets an additional downward force equal to the weight of the volume of water displaced by the ball and string.

If the situation on the right side isn't clear, consider a ball with the same density of water, so that the ball is almost light enough to float (i.e. the string is slack). Clearly, you've added the weight of such a ball to the right-hand side.

Increasing the density slightly so that the string just becomes taut isn't going to suddenly remove that weight, while increasing the density further (increasing the tension on the string) won't change anything.

Or to view it another way:

Allowing a ping-pong ball and string to float on the water on the right-hand side will result in the scale balancing due to symmetry (tying the string to the bottom would change nothing).

Increasing the ball's density so that it starts to sink (while the string remains slack) will clearly tip the scale to the right. Increasing the density more will tip the scale more, up to the point where the string becomes taut, at which point further increases in density will have no effect.

This

gtfo, Bayesian statistifaggot pls

>insufficient information
it's incomplete, but not insufficient
>pingpong ball lighter in weight than steel ball
common knowledge
>principle of buoyancy, Archimedes 212BC
common knowledge
>forces in equilibrium, Newton AD1666
common knowledge

Anyone have or know where to find more of these basic physics sort of problems? It'd really help me study for my babby physics class.

...

>I am a 20 yo undergrad

The beaker on the right reduces tension in the string which increases the weight of the beaker. The water partially supports the steel ball.

>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.

Right side goes down.

>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.

>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.

>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.

vid related:
youtube.com/watch?v=stRPiifxQnM

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.

better question. What happens if you freeze both?

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.

Both sides see an increase in downward force.

LHS: weight of ping-pong ball and string

RHS: reaction to pressure on steel ball and submerged portion of string

Force in RHS is greater, as water has a greater density than the ping-pong ball. Ergo right side drops.

>weight and rope tension cancel out
wat

>more mass on the left side
wat

>it'll just fall on her
wat

>the left side is heavier
wat

>weight of the ball on the right is irrelevant
true, but irrelevant

>The water partially supports the steel ball.
>No, it does not.
Lrn2buoyancy fgt pls

I was making a joke that the system isn't supported by anything, it's just floating in the air.

This answer seems correct to me. But I'm a layperson. Can someone explain why this is wrong to me if it is?

The right side sinks. Some of the tension on the right string is reduced by the displaced water, adding more weight from the heavier ball.

>The buoyancy force isn't cancelled.

Yes, it is.

The tension from the string cancels the bouyant force, as you said here:

> you're left with only the normals for buoyancy and tension. The net effect is equal to the weight of the ping-pong ball.

I was using the term "Cancels" in the algebraic sense.

> the right side adds buoyancy or the weight of a "water" ball

The ball on the right isn't bouyant, it sinks.

The water displacement on the left side and right side are equal (assuming equal volume for both spheres, of course), and thus the bouyancy forces cancel eachother out.

The ball on the right side adds no net downward force because it is not physically connected to the scale, and since the ball on the right side is heavier than water, and not connected to the right side of the scale, it applies no net force on the right side of the scale.

The only difference between the forces acting on the left and right side of the scale is the added weight of the ping pong ball and string on the left side.

Hence, the left side of the scale is heavier, and lowers.

>>The water partially supports the steel ball.
>>No, it does not.
>Lrn2buoyancy fgt pls

Then why steel ball sink in water if water supports it?

This.

I don't understand why you faggots are even arguing after this answer

Alright guys, look.

(Assuming that the balls on either side are of the same volume)

The water displacement on either side is the same. (and thus cancels eachother out)

The ammount of water on either side is the same. (and thus cancels eachother out)

The only difference in this equation is the fact that the ball on the left is physically attached to the scale, while the ball on the right is not.

Hence, the scale on the left contains Water, a string, and a ping pong ball.

While the scale on the right contains only water.

Hence, the scale on the left is heavier, and lowers because it has a greater mass.

This.
Assuming string has zero mass, buoyancy is F, garvity force of ping pong ball and steel ball are Gp and Gs, tension of string on right side is T, then
T + F = Gs, T = Gs - F
Extra downforce on the left: Gp
Extra downforce on the right: Gs - T = F
Since F > Gp, right side goes down.
I wonder why no one use fomula ITT

>buoyancy cancelled by string
Okay
>metal ball isn't part of the system
Okay

But then how come when you actually do the experiment, the side with the suspended metal ball goes down? Does the scale displace more easily than the water?

>Hence, the left side of the scale is heavier, and lowers.
>Hence, the scale on the left is heavier, and lowers because it has a greater mass.
If you actually do it, the right side lowers.

ikr

basic fluid mechanics and a tad of common sense are all it takes

> I don't understand why you faggots are even arguing after this answer
Because it's WRONG. Right side lowers.

See for an explanation with more detail than should be needed. Well, two explanations, really. But as you appear to be pathologically stupid, I'll provide a third:

Submerging a ball via an external force (whether by gravity or a rigid rod) adds the weight of the displaced volume. While submerging a ball by tying it to the base of the jar adds the weight of the ball.

If you just dropped the steel ball in there so it sank all the way to the bottom, the right side would lower, right? Surely you understand that much?

Well, if you tie a string to it so that it submerges without hitting the bottom, you've reduced the force by the tension in the string.

And the tension in the string is (as should be obvious to anyone not you) less than the ball's weight. If the ball wasn't resting in the water, the tension would equal the weight. But resting the ball in the water reduces the tension due to buoyancy (which doesn't depend upon density, only volume). Which means that the presence of the ball is pushing the right side down by the same force (Newton's third law).

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