Here is the set up. I have a balance and two
identical beakers, which I fill with exactly the same amount of water, except in one of
the beakers there is a submerged ping pong ball tethered to the base of the beaker. And
in the other there is an identically sized ball but made of acrylic. So it is denser
than water and it is supported by a thread. Now I want you to make a prediction. If I
were to release this balance and allow it to rotate freely, which way do you think it
would tip, toward the acrylic ball, toward the ping pong ball or would it remain perfectly
balanced? You can make your selection by clicking on one of the on screen annotations or clicking
a link in the description.>>If they are both displacing the same amount,
like, I think it won’t rotate. So, yes, it will not rotate.>>Stay balanced.>>The ping pong ball does have mass, albeit
not very much. I would predict that this end will sink.>>I almost feel like this end would lift,
because it is lighter than … but it wouldn’t be pulling, because it is tethered to the
lever. I don’t know.>>I am thinking it would go this way. I am
thinking because of the tension that is pulling this up.>>So I guess if your summing the forces,
this one is more. So then this is going to rise. No… All right. So I just switched
my mind. I am going to go with this one is going to rise.>>But, I mean, it seems like this shouldn’t
be contributing any force to this side, because it is suspended and the forces are balanced.>>That means you would have more of a normal
force on this side and it would tip that way.>>It will tip down on the ping pong ball
side?>>Yeah, definitely the ping pong ball side.>>Yeah. That is so weird.>>So now you think it is going to up on this
side and down on this side.>>Are you ready? Three, two, one.

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Dennis Veasley

98 thoughts on “Beaker Ball Balance Problem”

  1. B ; I am almost certain the correct answer is B, but let me know if I am wrong. It is easiest to think of removing the water in both cases. You can do this because there is the same amount of water in both beakers. Now you have a ball on each scale and… the acrylic ball in heavier than the ping pong ball so the scale will drop toward the acrylic ball, right? WRONG!!!! You have to take into account the force that he is pushing downward on the ping pong ball to keep it under water! This force + the weight of the ping pong ball should definitely be heavier that just the weight of the acrylic ball. So the scale will tip downward on the side of the ping pong ball; B. Furthermore the force he is pushing downward will not be cancelled out because it is an external force. We could cancel out the tension on the string produced by the buoyancy of the ping pong ball because it was an internal force within the system. BAM

  2. none will move, because the forces are the same, since both are submerged in water, and both are the same depth.

  3. the ping pong ball is the one that resists submersion so it would actually create more weight. The heavy is only creating displacement.

  4. Let
    d1: density of the ping pong ball
    d2: density of the water
    V: volume of each ball
    p: total weight of the water in each beaker and the beaker itself
    The force that pulls the first beaker down is gravity
    F1= p+d1*V
    For the second one, there are two forces that pull it down: gravity (acts on the water and the beaker only) and the reaction of the acrylic ball. The reaction of the ball is equal to the buoyancy by the water on the ball. Therefor, we can find the sum of the forces that pull it down.
    F2= p+d2*V
    Now, let's compare F1 and F2
    F1 – F2 = (p+d1*V)-(p+d2*V) = V*(d1-d2)
    we know that d1 < d2. Therefor, F1 – F2 <0
    => F1 < F2
    So the answer is "towards the acrylic ball"

  5. Ping pong rise …..I think ping pong will rise because there's air inside and that air is trying to up that makes it lighter ( so I think )

  6. Don't get too hung up forces, etc. Just focus the average density inside the beaker. The ping pong ball represents a density 'void'. The acrylic ball is equivalent to neutral buoyancy, no density void.

  7. Balanced forces are cancelled no real weight is added other then the displacement forces assuming both are the same size the scale shouldn't move

  8. Can I just know which university you went to, because in my uivsersity where I studied engineering, there were maybe two girls out of 20 students!

  9. Acrylic would weigh more by the weight of the water displaced by the ball. The ping pong ball and string have been added to the mass of the water, a very small increase in weight. They could be placed outside on the scale and the new weight of water plus ball and string would stay the same because the submerged ping pong ball is tied to the bottom of the beaker, canceling the effect of displacement. The acrylic is supported from off the scale allowing the scale to weigh the displacement. It's exactly the same as pouring in the same volume of water as the volume of the ball. That additional volume of water weigh's a lot more than a piece of string and a ping pong ball… Not to mention the blob of glue holding the ping pong ball's string to the bottom.

  10. i can't believe people find it this hard. this is essentially freshman physics — just write out the balance of all the forces

  11. Since the mass of the water is the same, wouldn't that be cancelled out, not part of the equation. So in essence, wouldn't this be no different then attaching a helium balloon with string to the left side of the scale, and nothing to the other side?

  12. I like all your videos but in this video you used the cheap trick of finishing the video at a key point. Also, you started publishing the videos first on the paid website and later on youtube.com…. I think it is time to unsubscribe. But I think one viewer doesn't matter so much to you. So I wish you all the best 🙂

  13. Thanks for excellent food for thought. I think it will tilt towards acrylic ball. The reason for this can be viewed after drawing FBD on both the glasses. ping pong ball is connected to the string that produces tesion in the string which will reduce the normal reaction on the left glass.However on the right glass normal reaction will be higher

  14. I have seen the explanation video. Can it go like this?
    Both balls have same dimensions i.e. same volume i.e.displace equal volume of water i.e. displace equal mass of water. Archimedes says "simply experience same buoyant force. Thus upward forces are balanced. 
    Now a bit childlike,
    we know that density=mass/volume i.e. mass=density x volume.
    for equal volumes if density is more mass(weight-g is same right?) is more. 
    Downward forces unbalanced. Acrylic side more down that side goes down.

  15. Question? If you have 2 neutrally buoyant balls filled with liquid floating in water, but one is hot and one is cold. Will one rise or sink?

  16. I think the scale will stay balanced ( C ). Here's why: Consider a situation where an external agent (ex. Derek's hand), keeps a ping pong ball fixed in place in each of the 2 beakers. The scale would have to stay balanced, because the same thing is happening on both sides. Now, swap out one ping pong ball for a new, same-sized mystery ball, and let Derek keep it in place without it moving. As far as the water is concerned, the new ball is the same as the old one; it's still fixed in the same place, and it still displaces the same amount of water. Now let's say the mystery ball happens to be blue and acrylic, and instead of Derek holding it in place, a suspension cable does that work for him. Compared to the forcibly submerged ping pong ball, the acrylic ball still displaces the same amount of water. So in both cases, both balls exert the same buoyant force on the water, so they push the water down the same amount, so the resultant forces on each side of the scale must be equal.
    Thanks for reading!

  17. The scale with the greater mass on it will fall.  Mass is equal to density times volume.  Both beakers have the same volume (mass of water) before and after the balls are placed in the water.  Both balls displace the same amount of water.  But the mass of the volume occupied by the balls is not equal.  The acrylic ball is heavier (denser) than water.  Therefore, the mass of the volume occupied by the supported acrylic ball is equal to the mass of the water displaced by the ball.  The ping pong ball is lighter (less dense) than water.  Therefore, the mass of the volume occupied by the tied down ping pong ball is equal to the mass of the ping pong ball.  Since the mass of the acrylic ball is greater than the mass of the of the ping pong ball, the density (mass) of the acrylic ball beaker is greater and thus falls.

  18. @Veritasium          you know , you often say you can click the on-screen link and they never work! These don't work either.

  19. You should always interview physics students. They give reasonable answers using the knowledge they learned from university, yet you can still correct misconceptions.

    The people on the street have no background nor application of the knowledge, and just guess.

  20. I mean what the heck!!! Are they Physics Students??? Even I can answer that using High School Physics & i'm not boasting……

  21. I think the acrylic ball beaker will sink because the string isn't holding all of its weight. The water is supporting some of its weight with the buoyancy force, so there's more force downward on the acrylic beaker, since the ping pong ball seems to contribute negligible force.

  22. B. As you are applying a force onto the pingpong ball to keep it submerged which also creates a pressure on the water and subsequently the beaker which therefore should create a higher force downwards and cause the balance to tip over with the ping pong being lower (i think)

  23. Well i got alternative and more efficient/easy explanation that since first you said that both balls have same size and the beaker contains same amount of water but the density of the Acyclic ball is more than water while with ping pong its opposite so now lets use some equations.
    D(density)= M/V
    V is constant since you said sizes of both the balls are same.
    which means D1(Density of ping pong ball)= M1(mass of ping pong ball)/V
    and D2(Density of Acyclic ball)= M2(mass of Acyclic ball)/V
    but D1<D2 (since D1<Dw(density of water) and D2>Dw(density of water)
    and then that means
    M1/V<M2/V
    since V is constant it leads upto
    M1<M2
    which means M2(Acyclic ball) displaces more mass/amount of water than M1(ping pong ball) 
    hence it drops on right side.

  24. Astounding how many intelligent persons cannot reason this one out. Do they think a ship requires cables to hold it on the surface?

  25. I'm going to take a stab at explaining it before watching the other video.
    The ping-pong ball, due to being filled with air, basically has the effect of removing its own volume's worth of water from the beaker.
    The acrylic ball, though it is independently suspended, is still also suspended in the water. Imagine if you were holding the string of the acrylic ball and you lowered it into the water. You would feel the decrease in weight.
    Therefore, my prediction is the ping-pong side goes up.

  26. On the left we have m_beaker + m_water + m_pingpong + m_tether.  We don't know the mass of the tether.  It is possible for the tether to have almost no mass if the string is somehow glued or affixed to the bottom of the beaker.  It is also possible that a weight was used.  I admit it would be sneaky to add an unknown tether weight to the ping pong ball side, but I think this what he did.  The mass of this weight would be more than the mass of displaced water minus the mass of of the ping pong ball.  m_tether > m_displaced – m_pingpong.  The total on the left is therefore greater than m_beaker + m_water + m_displaced.

    On the right we have m_beaker + m_water + mass_acrylic – F_tether / g.  Subtracting the tether is the key point.  Notice that the tether of the acrylic ball is not on the balance, to the force on the tether is subtracted.  The force of the tether is (m_acrylic – mass_displaced)*g.  Combining give the total mass to be: m_beaker + m_water + mass_displaced.

    Looking at the equations the left is heavier and the ping pong ball side will fall.  This is actually pretty obvious if you assume that an unknown weight was used to tether the ping pong ball.  He could have used the core of a neutron star or other obscenely massive object.  If the outcome is knowable and one side has an unbounded large mass, that side is heavier.

    Of course if my assumption is invalid and the ping pong ball is tethered with glue, the opposite will happen and the ping pong ball will rise.

  27. Physics problem, my favorite 😀 

    So, we've got two beakers filled with the same amount of water. let's say that one of these has a weight of X newtons. Let's say the white ball's weight is W and the blue one's is B, I'll say right now their densities won't matter, you'll understand why soon. 

    So, on the left we have X + W weighing down on that scale. Now, you may think the tethering of the ball means it isn't helping push down, but that's part of the internal system. It's like saying your weight changes based on whether your arms are up or down (but not moving), it's still the same.

    Now, on the right, that's a bit different. There's always the weight X acting on the scale on the right, but the weight of the blue ball isn't all acting on the scale. There's an upwards force by the spring, S, to hold the blue ball up. So, the force acting on the right is equal to X + B – S. 

    I could stop right here and say that it depends on whether the string force is high enough to make it "lighter" than the other side, and that if there is enough force that
    (B – S) < W, then the left side will go down, if (B – S) > W, it will stay balanced, and if (B-S) > W, the right side will go down, but we can actually answer this question!

    Let's call the water's density 1000 kg/m^3. The density of a ping pong ball is around 8.4E-5 kg/m^3. The density of acrylic (likely not the same as what the ball is, but for the problem's sake let's say it is) is 1.18E-3kg/m^3.

    They appear to be roughly the same volume, which we'll call V. so, now we can say that the force on the left side is X + (8.4E-5 x V).

    For the right side, again, it's more complicated. We know that the buoyant force (the force that will be added to the apparent weight of the scale) is equal to the weight force of the volume of the liquid that's been displaced. That volume is V, and that liquid's density is 1000 kg/m^3, meaning our force would be equal to 1000V. This may look like a large number, but remember V is in terms of meters, and the ball is very small, it's actually a reasonable number. You may be thinking that I didn't mention this force in the left side, that's because that's an internal force, it will not effect the perceived weight on the scale. 

    So, that means the right side's weight force would be equal to X + 1000V. To solve which side is heavier, let's look at the two sides:

    X + 8.4E-5*V ( < or = or > ) X + 1000V
    subtract the X
    8.4E-5V (< or = or >) 1000V
    divide V
    8.4E-5 (< or = or >) 1000
    well, clearly the right is heavier, 1000 is significantly bigger. 

    Therefore, we can come to the conclusion that the right side will go down, until the forces become equal again. To go the extra mile, I'll just say that the scales will stop when the volume of water displaced by the blue ball is 8.4E-8 times the original volume, so it'd BARELY be touching.

    If this doesn't really make sense to you, instead of an acrylic ball and a ping pong ball, let's say the left had a fake hand with an incredibly light material in it, and you held your hand in the right side in the same place, and your hand's attachment to your arm was the spring, wouldn't it make sense for the side with your hand in the water to go down? 

    I will post this before watching the new video with the answer. If I'm wrong, I will not delete this, it'll be a testament to my failure at a physics problem (:

  28. I believe (in a theoretical environment) that they would stay balanced.
    Both balls displace the same amount of water and should not add to the weight.

    In a practical setting the ping-pong ball beaker will probably sink because of the added mass of the ping-pong ball unless the adhesion force of the water on the acrylic cancels the added weight out.

    In this setup it can go either way since the water is added after the ball is submerged and the beaker has been filled to produce a certain outcome.

  29. I just discovered this experiment/video.  My explanation seems a bit more straight forward.  The water and beakers are the same mass – it doesn't matter about the displacement of the water unless the water spills out – they are equal on both sides so they can be ignored. The blue ball is not interacting at all with the beaker/water It is not transferring it's mass to the beaker so it has no effect on the weight on that side of the balance.  – you could just leave that out of the setup without any change. The white ball has an upward buoyant force which is directly applied to the beaker making that side of the system weigh less.  Similar to weighing something and lifting some of the weight – the overall weight of the system goes down. Thoughts? (and don't be rude)

  30. I figured this out a few months ago. The right side will sink UNTIL the buoyant force become equal to the ping-pong-balls (and the tethers ofc) mass. But if they are the same from the beginning then they will be balanced.

  31. Hi Veritasium! Im your fun and I hope u will add RUSSIAN SUBTITLES because Im Russian. Of course I translate your videos by youself, but it's really hard!

  32. there is one more possibility , I can make the white ball(less denser) tilt down if mp*g > ma*g-T where mp(mass of white ball) , ma(mass of blue ball)

  33. Consider another experiment: A beaker filled with water is kept on a weighing machine. What happens if we submerge the ping pong ball with a light rod completely into the liquid? No liquid overflows. Will the reading change or remain same?

  34. The scale would tip towards the acrylic ball besause of the bouyancy of the acrylic ball. Because the water applies some force to the ball then, according to Newton's third law the ball is appllying an equal and opposite force to the water and the beaker. And since the ball is not floating that force is going to go to the balance thus causing it to tip towards the side of the blue ball. On the other hand the ping pong ball and the beaker are tied together so they act as a single mass.

  35. What about the displacement of the strings…?
    I want to say that the pingpong ball side will rise, because I see that there appears to be more string in that side, and that would translate to slightly more displacement on that side, assuming both balls displace the same amount of water on their own. Also there is less overall mass on that side as the pingpong ball is hollow.

  36. the blue ball has a buoyant force acting on it so it's causing an equal an opposite force on the water. I guess you have to keep that in consideration…

  37. L ( Magnitude of forces acting on left base) = || weight of beaker + weight of water + weight of pp ball ||
    R (Magnitude of forces acting on right base) = || weight of beaker + weight of water + (weight of ac ball – tension in string) ||

    L – R = || weight of pp ball – (weight of ac ball – tension in string) ||

    || weight of ac ball – tension in string || = || buoyant force || = || weight of water displaced by ball ||

    hence,
    L – R = || weight of pp ball – weight of water of the same volume as ball ||

    So it comes down to the difference between the weight of the ping pong ball on the left side and the weight of water of the same volume as the acrylic ball on the right…..

  38. Hi Veritasium! What would happen if the ping pong has the same density with water and is still in the middle of the water? Will the ping pong side sink?

  39. (Identical downward pressure forces from water on both beakers) + (Upward tensional force by string on left one) = (Tilt right)

  40. the ping bong ball will go down because it is attached to the beaker. the blue ball wich i cant spell is just suspended in the water

  41. Interesting. So a friend proposed a similar question but in a more simplistic fashion.

    "If you have an aquarium and large fish with weight scales on the left and right side, would that side get heavier as the fish swims to that side?"

  42. Balanced. Next. Oh, explanation… Force from fingers = buoyant force of ball = might as well be water alone. Acrylic sphere hanging = tension balances buoyant force = might as well be water alone. Therefore balanced. QED.

  43. Both beakers contain the same volume. the one on the left is a closed system, the buoyancy is balanced within the beaker so it has only the same effect as if the pingpong ball was just sitting beside the beaker on the balance. The right side, the water is providing some buoyancy to the ball, and the ball is part of an external system, so the right side will experience a downward force equal to some water of the volume of the ball. That's more than the weight of the pingpong ball. The right side will go down.

  44. Basically, ideally you are weighing the beaker and same volume (assume same mass) of water on each side (unless water is salted on one side for instance), so it should balance. Got anything harder?

  45. When you put something in water, the combined weight is just the addition of each weights, doesnt matter if it floats or sinks. What weight should we add if there is a string holding from out of the water? Not the weight of the object, just the weight of the water it displaces, which in this case is much heavier than the weight of pingpong ball, which is all we can add in the other case.

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