MUL 05 / 21

NOTE:

[Note: GERMAN versions of these Week 5/21 tasks are available HERE. I welcome suggestions on how to improve the translations!]

 In the printed version of this blog, Week 5 consists of the first 5 tasks shown in Week 2 of this blog. The tasks shown below feature as Week 21 in the book!

This week's tasks involve the notion of density, or a particular aspect of it, concerned with objects floating in water. Here the key idea is that, for solid objects made from a given material (in our case, a given type of wood), the amount (mass) of an object that is under water is proportional to its total amount. The underlying principle here, of course, is Archimedes' principle which states that "the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces". (from Wikipedia)

Monday: This involves two planks made from the same wood and both with rectangular cross-sections. Eight twentieths, or 4/5 of the 25mm thick plank will be underwater. So it sinks to a depth of 20mm.

Students might feel that they need to know how long a plank is and how wide, to determine how far it sinks. However, this information is not relevant if we assume that the plank is a simple, rectangular cuboid. In effect we have reduced the problem to just one dimension - the thickness of the planks.
Students might raise another interesting question - why does (or why do we assume that) the plank floats in the way shown, ie ‘flat’ in the water with the shortest edges vertical, rather than either the ‘width’ or the ‘length’ vertical and ⅘ under water?
Some students might feel that they don’t know enough physics to solve this task. This would be an understandable reaction, though it is possible to make progress without knowing much about floating and sinking. For example, suppose someone suggests the 25mm plank will sink to a depth of 13mm (8mm + 5mm). In that case, what would happen with a 12mm (ie 20mm – 8mm) thick plank? Could it sit on top of the water?

Tuesday: Here we compare the depth in water of a floating plank with that of a pole. We are only after a qualitative answer about the depth of the pole - we are not trying to find the actual depth. The given quantities, 20mm and 5mm, are relevant only in as much as they tell us that less than half of the floating plank is under water.
The task is quite subtle as the pole’s cross-section is circular rather than rectangular. This means there is not a simple, linear relation between the depth to which the pole is underwater and the amount that is under water.

You might want to focus students’ attention on the first of the three possible positions of the pole shown in the diagram. Here the depth below water and the depth above is the same as for the plank; however, the proportion of the amount of material below and above the water is different for the pole and the plank - because of the circular cross-section, a larger proportion of the pole is above water, so the pole in that position would sink lower, roughly as in the middle position.

Wednesday: Two thirds of the mass of the boat on the right is under water. So two thirds of the left-hand boat will be under water too. This means that two thirds of the area of the upside-down T-shaped cross-section will be under water. So the boat sinks to a depth of 1.5m rather than 2m: the lower ‘deck’ of the boat is just submerged, while the upper deck is just out of the water. Some students might discern this directly from the diagram as the lower deck is twice the size of the upper deck.

Some students might think the larger boat should sink further, but depth is not dependent on absolute size. The middle boat, below, is smaller than the one on the right but will behave like the one on the left.

Thursday: Here we have two identical beakers, with some sand in the bottom, floating in water. The first beaker sinks to a depth that happens to be three times the depth of the sand it contains. Does this mean the second beaker will also sink to three times the depth of its sand? This seems to be Tracy’s thinking. This works if we are willing to treat the beakers as weightless, which is the kind of thing we often do in the maths or physics classroom.
But if we do treat the beakers as weightless, then the sand would have a density of three times that of water. In reality dry sand is about 1.5 times as dense. While even wet, packed sand is only about twice as dense as water. Of course, we don’t expect many students (or the rest of us!) to know this!

 

So what happens if we don’t treat the beakers as weightless? As we don’t know their actual weight, we don’t know how much of the 6cm depth of displaced water supports beaker A and how much supports the 2cm dept of sand, and so we can’t tell how far beaker B will sink. However, we can determine that it will be more than 7cm. Why?
We can also determine whether it will be less than or more than 9cm, though this is more challenging.                                       Note: This might help! D = mS + B.

Friday: The situation here might appear to be multiplicative, but it turns out that the task can be solved additively. Beaker B has an extra 1cm of water compared to A. So when it floats in water, it displaces that extra amount of water. This means it will be 1cm lower in the water than A, so it sinks to a depth of 7cm.
Note: The extra glass that is under water for beaker B will also displace some water, so we are assuming that the glass sides of the beaker are thin enough for this to be negligible.

Some students might notice that the depth of A (6cm) in the tank is twice the depth of the water it contains (3cm) and wrongly conclude that beaker B will sink to a depth of 2×4cm = 8cm. However, this ×2 relationship would only hold if the beakers were weightless and the liquid in the beakers was not water but something twice as dense.

A more circuitous but perfectly valid argument would be to say that for beaker A, the 6cm depth of displaced water supports the weight of the beaker and the 3cm depth of water, so 6cm – 3cm = 3cm is needed to support the beaker. So a depth of 3cm + 4cm of displaced water is needed to support beaker B with its 4cm depth of water.
(Note that we can simply argue in terms of depth of water rather than volume, because the beaker has a constant cross-section.)

Extra day: Here's a variant on Friday's task.

This is a simpler variant of a task that has stayed with me since the early days of the CSMS project when two colleagues from the science wing (Michael Shayer and Hugh Wylam) introduced the team to this:
A rowing boat is floating in a swimming pool. A person in the boat throws a concrete slab overboard, into the pool. What happens to the height of the water in the pool?

I can't remember whether they created the task or sourced it from elsewhere. But we were all very excited and challenged by it. We were trying to develop a Piagetian diagnostic test on floating and sinking at the time.