Lets see how good is Veeky Forums at solving problems

Right. The steel ball displaces the water which increases the volume in the right beaker which means an increased mass on the right side.

The answer hasn't changed since 2 days ago: it tips right.

On the left, you add the weight of a ping-pong ball and string (i.e. not much).

On the right, you add the weight of the volume of water displaced by the ball.

If the water wasn't there, the tension on the string would equal the weight of the ball. By submerging the ball in water, you're reducing the tension by the ball's buoyancy (equal to the weight of water displaced by the ball, independent of the ball's density so long as the ball is dense enough not to float). By Newton's third law, the upward force on the ball equals the downward force on the water.

Is this the new "50-50 either it happens or it doesn't"?

> On the right side you have the mass of the water only
If you think that dunking an object in the water won't change the total force, you should try it.

Place a glass of water on a weighing scale, and lower any more-dense-than-water object into it.

Even a ghetto scale will work; it doesn't need to be calibrated. E.g. a spring or piece of elastic, so long as it's not fully compressed/stretched by the weight of the glass.

No, it's an explanation of why it tips right.

But it doesn't make any sense.

First of all, why do you add the weight of the volume of water displaced by the ball on the right side, but not on the left? Second of all, submerging the ball in water does not reduce the tension in the string-it creates the tension to balance the upward buoyant force of the water. With no water, there's no tension. If you drained the water and you cut the string, would the ball fly into the air?

Its Newton's third law senpaitachi

On the left the buoyant force is cancelled by the string inside the cup, negligible mass added

On the right the buoyant force is exerted upwards on the steel ball (and also downwards on the jar) equal to the volume of water displaced

The right is greater than the left so it tips right

but the steel ball is supported by the force of tension in the spring so there is no more bouyant force on the steel ball than on the ping pong ball

but a scale measures force not mass.

>On the right the buoyant force is exerted upwards on the steel ball (and also downwards on the jar) equal to the volume of water displaced

I agree. And the same thing happens on the left side.

>On the left the buoyant force is cancelled by the string inside the cup,

Again, I agree. But the string has tension at both ends. Just like the tension at the top pulls down on the ball, the tension at the bottom also pulls up on the beaker. So the downward force acting on the scale is (Buoyant force)-(Tension) (in addition to the weight of everything).

how much does each ball weigh? You only state that their volumes are equivalent.

>So the downward force acting on the scale is (Buoyant force)-(Tension) (in addition to the weight of everything).

That's zero though

Right
You've added the equivalent weight of a ball made of water to the right and a ball made of air to the left

> First of all, why do you add the weight of the volume of water displaced by the ball on the right side, but not on the left?
On the left, the force is internal, as the string is tied to the glass.

On the right, it's external (gravity). Gravity pulls the ball on the right down. But the ball isn't accelerating, partly due to tension on the string, partly due to buoyancy. The part that's due to buoyancy has an equal and opposite force which adds to the downward force on the right of the scale.

> Second of all, submerging the ball in water does not reduce the tension in the string
Submerging the ball on the right reduces the tension.

If the ball on the right was a ping-pong ball, it would float; there would be no tension. If the ball on the right was made of ice, it would float, but only barely (9/10 submerged). There would still be no tension. So clearly its entire mass would be supported by the water, meaning that it adds to the weight on the right. Right? You understand that much?

Now increase the density to 1.01x that of water, so that the ball just starts to sink, until the string stops it. At which point, most of its mass is still supported by the water with a small fraction supported by the string.

If you can't understand this from a theoretical perspective, just try it. You don't need a lab-grade scale, just something sensitive enough to observe that dunking something in the water increases the down-force.

> how much does each ball weigh? You only state that their volumes are equivalent.
From the diagram, it's implied that the ball on the left is less dense than water while the ball on the right is more dense than water.

That's all that matters (assuming that the string on each side has negligible mass and volume).

>If the water wasn't there, the tension on the string would equal the weight of the ball. By submerging the ball in water, you're reducing the tension by the ball's buoyancy (equal to the weight of water displaced by the ball, independent of the ball's density so long as the ball is dense enough not to float). By Newton's third law, the upward force on the ball equals the downward force on the water.

Hang on, are you talking about the right side here? If so that makes sense and I feel silly.

In , the last paragraph is a follow-on to the one before, i.e. it's referring to the right-hand side.

The left-hand side is easy: you've added the mass of a ping-poing ball.

The right-hand side is what confuses people, because the mechanism isn't obvious. Clearly you aren't adding the entire weight of the right-hand ball, due to the string. But nor are you adding nothing; buoyancy doesn't magically disappear once the ball is heavy enough to sink, it just "caps" at the weight of the equivalent volume water.

I'd like to see you draw a force diagram of that. See if you still believe yourself.

...

Left, steel is about 8 times as dense as water so 7/8 of ball weight is on left and 1/8 on right. Ping pong ball adds another small amount of weight to the left.

Water pressure doesn't make for neat force diagrams. Doesn't make it any less real.

You realise that this experiment is something which can easily be performed by anyone with access to a weighing scale, right?

I think he's trolling. I hope.

> hydrostatics
> literally fluid mechanics 101
Or you could just shut up and calculate

Is it allowed to post your homework if you write it on image with Chinese cartoon?

It tips to the left. The weight of the steel ball can be completely neglected because the weight is supported by the string. The weight of the ping pong ball is added to the weight of the water.

u gotta be feeling pretty stupid by now eh?

Why would I feel stupid for giving the correct answer?

...

The steel ball is not completely supported by the weight of the string because of its buoyancy

Right + weight of steel ball
Left + weight of ping pong ball

It tips to the right.

Wrong, all of the ball's weight is on the right

How? It's supported by the crane on the left

Draw free body diagrams for each side of the scale and each ball. Then write the relations between them.

It tips to the left.

I hope you're trolling.

t. I failed freshman physics

This is the 100% non troll answer. It tips to the ping pong ball side.

It's an easy force equillibrium problem. The balls are the same size, and thus displace the same amount of water. There is no hidden inequal buoyancy forces or tension forces here. The reason the balance dips to the ping pong ball side is because the ping pong ball is resting in/on the water. This side weighs more. The right side weighs less because the steel ball is not resting in/on the water, it is suspended by the string above.

Tldr: (x mass of water) + (mass of ping pong ball) > (x mass of water)

I fell for the b8 ik.

>The right side weighs less because the steel ball is not resting in/on the water

It is though. The weight steel ball is held up mostly by the string, but a little bit by the water as well.

Imagine a glass ball full of water, made so that it's only slightly more dense than water. Then imagine holding it from a piece of string, and lowering it into water, you'd feel it go from weighing its full amount to having the water support most of the weight.

That video was proven wrong forever ago

Show proof or you're a liar.

Post more physics riddles.

i dont have any heh
some fags at ck started to discuss this instead of what op asked because he used it as the image for his post

To the right. The key is that the ping pong ball must be less dense than water in order to float. Now, the net force on the left is just weight of the water plus weight of ping pong ball plus weight of the beaker. The net force on the right is the buoyancy force of the displaced water down, plus the mass of the water and beaker. Since the buoyancy force is equal to the mass of the displaced volume, and the ping pong ball is less dense than the the water it displaces the down force of the buoyancy force on the right is higher than the down force of the ping pong ball and so by newtons second law the right side accelerates down.

You can show that it's correct in the comfort of your own kitchen.

Seriously, if you're going to try to troll people with hand-waving and misdirection, it helps if you pick an experiment which most people can't perform at home.

Did the experiment, used a steel weight, volume 10 mililiters, hung it from a rope. the balance showed an increase in weight of 10,65 grams. tried it again with a typical pingpong ball. Used a weight to lower the pingpongball below the water, the pingpongball and rope added 3-6 grams of weight, the pingpongball has a larger volume than the weight, my guess is that the added weight in the case of the pingpong ball is the mass of the pingpongball itself and the change in weight in the case of the weight is due to the displacement of water reducing the amount of tension in the rope. therefore right side of the balance would go down.

the ping-pong ball pushed that water back

Is the crane submerged?

yeah and it would have a higher force due to the mass you dumbfuck

enough of this shit: youtube.com/watch?v=QD3hbVG1yxM

>enough
>posts wrong video
youtube.com/watch?v=stRPiifxQnM

Do the balls even really factor into the equation?

The left beaker has gravity down and tension up

The right beaker has gravity down

Obviously it would tilt right?

The left ball doesn't factor in (beyond its weight) since that's a closed system.

The right ball does, every object has buoyancy so some of the weight is supported by the water and thus the scale. Think of how stuff weighs less under water.

It remains unchanged. Both beakers have the same amount of water in them and neither is being weighed down by anything. If the ping-pong ball was being pushed into the water by a a hand or something, the buoyant force would make that beaker "heavier", but the buoyant force is balanced by the string tied to the bottom, so nothing happens.
The steel ball does literally nothing.

The steel ball has buoyancy you double nigger

He's right
I'm wrong. Forgot to consider the buoyant force the water exerts on the steel ball.

buoyancy is a reaction force, it doesn't really exist