I saw this demonstration on Veritasium’s YouTube channel (see Beaker Ball Balance Problem video). It is really simple to run as long as you have a good balance. The result is not obvious and so the reason for the result requires a little analysis. Therefore, I think it would be good to write it up and provide a more detailed explanation of the demonstration complete with free body diagrams.
- Three identical cups
- Two ping pong balls, one filled with sand of something to weigh it down (actually any two balls with the same diameter as long as at only one of them floats)
- A stand
- Place identical volumes of water in two cups but leave enough space for at the top of the cups so that it will not overflow when you place the balls in the cups.’
- lock the balance so that it is level for steps 3-7.
- Take the empty ping-pong ball and tape some string to it and then tape the string to the base of the empty cup. Use as little string as possible. When you pour the water in, the ball has to be fully submerged.
- Pour the water from one of the cups into the one with the taped ball.
- Place the two cups with water (one with a ball attached) on either side of the balance.
- Suspend the heavy ball from the stand into the second cup such that it is fully submerged.
- Poll your students to see which way they think the balance will tip when it is released.
- Unlock the balance and observe which way the balance tips (it should go up on the side with the ball taped to the bottom of the cup)
We start by looking at the setup (see figure below). The two cups have identical volumes of water in them and both have a submerged ball with identical volumes which, therefore, displace identical volumes of water. Therefore, the depth of water in each cup is the same. For the purposes of this explanation light ball is denoted as ball (1) and the heavy suspended ball is ball (2).
We now examine the forces acting each ball (see figure below). Ball (1) has weight down (W1), buoyancy force up (FB), and the tension force in the string acting down (T1). The forces sum to zero as the system is in equilibrium. Ball (2) has the same set of forces acting on it. However, for ball (2) the tension force (T2) is acting up as the ball is denser than water such that the buoyancy force is less than the weight. Hence the ball must be suspended from above. The only important thing here is that the balls displace the same volume of water such that the water levels in each cup are identical.
We now turn our attention to the forces acting on the cups (see figure below).
For cup (1) there is the hydrostatic pressure force acting down, PA, that depends only on the cup geometry and the depth of water in the cup. There is also the weight of the cup (Wcup), the tension on the string acting up (T1) and the force due to the scale Fscale(1). Therefore,
Fscale(1) = PA+Wcup-T1 (1)
The forces acting on cup (2) are the hydrostatic pressure force acting down, PA, the weight of the cup (Wcup), and the force due to the scale Fscale(1). Therefore,
Fscale(2) = PA+Wcup (2)
As the cups are identical and the water depth in is the same in each then both PA and Wcup are the same in equations (1) and (2). Therefore, subtracting (1) from (2) gives
Fscale(2) – Fscale(1)= T1
Fscale(2) > Fscale(1)
and the balance goes down on the side with the suspended ball.
In fact, you could run this test with any two sized balls as long as:
- One ball floats and is tethered to the base of a cup
- One ball does not float and is suspended from above
- The water level in the two identical cups is the same.
An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (firstname.lastname@example.org). I also welcome comments (through the comments section or via email) on improving the demonstrations.