Ph 08

Zusatzaufgaben

Cork in the water


Q: A small piece of cork (density, ρ = 200 kg/m3) is held 0.7 m at the bottom of a bucket of water (ρ = 1000 kg/m3) by a piece of string. The cork has a volume of 3 cm3.

(a) What is the tension in the string?


(b) Once the cork rises to the top, how much of it stays above the surface?

Solution:

The first part of the question asks about a force, namely the tension in the rope. There's usually one sure-fire method that will get you through most force questions: Newton's Second Law.

Start off with an FBD (free-body-diagram) and add up those forces:

Figure 2 There are three forces which act on the cork:
  • The buoyant force, FB which acts upward.
  • The weight of the cork, W.
  • The tension of the rope, which pulls downward, T (ropes can only pull, they never push.)

Since the cork is going nowhere, the acceleration is zero

Eqn 2

I arranged it such that T was by itself, since that, after all, is what we're attempting to find. Now, plugging in the formulas for the other two forces, : FB = ρf·Vf ·g, and W = ρcork·Vcork ·g.

Heavy Concept
The equation for buoyant force should look familiar, but I still want to explain it a bit:
  • ρf - this is the density of the fluid
  • Vf - this is the volume of the displaced fluid. I always equate this with the volume of the submerged portion of the object. In this case, the entire cork is under water, so Vf is equal to the entire volume of the cork.

As for the equation for weight, you might be more familiar with W = mg. But since we're dealing with densities and volumes in the question, we can take advantage of the fact that m = ρ·V. That's why W = ρ·V·g!

Eqn 3

Now, Vf = Vcork (see the note on the right). We're given the volume of the cork (3 cm3), but we cannot stick "3" into the equation since it's the wrong units.

Be careful when converting the units for area and volume. 3 cm3 does not equal 0.03 m3! You need to move the decimal point back two times for each dimension. Or, to think of it another way:
3 cm3 = 3 (cm)·(cm)·(cm)
=3 (0.01 m)·(0.01 m·(0.01 m)
= 3 · 10-6 m3

Okay, time to plug some numbers in:

Eq. 4
Okay, now the cork has made it to the top and is just sitting there. Take a peek at the FBD on the cork as it is floating:
Fig. 4 The only two forces acting on the cork is its weight, W and the buoyant force, FB.

Note that, unlike in part (a), the Vf does not equal Vcork! . Go back up and read the HEAVY CONCEPT above.

We want to find how much of the cork is above the water, but we have Vf, which is the amount of cork below the water. No problem.

Eqn. 10
Finally, the volume of the cork above the water is:

3 · 10-6 m3 - 6 · 10-7 m3 = 2.4 · 10-7 m3
QED
Quelle:www.servtech.com/~wkimler/problems/ncalc/thermo/fluids/cork/cork.htm