# Multiple choice buoyancy problem

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1. Oct 14, 2015

### 1729

1. The problem statement, all variables and given/known data
A wooden cube is floating in water. The upper face of the cube meets the surface of a pool of water when a block with mass 0.200 kg is placed on top of the cube. When you remove the block, there's a 2 cm difference in height between the upper face and the water surface.

What's the length of the cube's edge?

(A) 9 cm
(B) 10 cm (this is the correct answer)
(C) 18 cm
(D) this problem is unsolvable without knowing the density of the wood used

2. Relevant equations
$$F=mg \\ B=\rho g V$$ The length of a cube's edge is the cube root of its volume

3. The attempt at a solution
Notice that the volume of the cube is equal to the volume of the water displaced in the first picture.
Since forces are at rest, they will sum to zero newton. (Newton's second law)
$$\rho_{water}V_{cube}g=g(0.200\mathrm{\ kg}+\rho_{wood}V_{cube})\\ \Leftrightarrow V_{cube}=\frac{0.200 \mathrm{\ kg}}{\rho_{water}-\rho_{wood}}$$ Since the density of wood is unknown, I would suppose (D) is the correct answer.

Last edited: Oct 14, 2015
2. Oct 14, 2015

### SteamKing

Staff Emeritus
You haven't stated for what question the choices A-D are the answers.

3. Oct 14, 2015

### 1729

4. Oct 14, 2015

### Krylov

I checked and (B) is right as far as I can see.

You are only using one of the two situations in the figure to derive an equation. You may obtain a second equation by using the other situation that is displayed. From this system of equations you will see that the density of wood is not needed. (In these equations, it may also help to express the volume of the cube directly in terms of the length of its edges.)

5. Oct 14, 2015

### haruspex

Further to Krylov's post:
Because you obtained an equation in which the cube's dimensions depend on the wood's density, you assumed answer D. But it turns out that there is enough information here to find the wood's density.

6. Oct 15, 2015

### 1729

Thank you all for your insightful replies. This has tremendously helped me develop my intuition in buoyancy problems.

In the right figure, notice that the forces are at rest and thus, by Newton's second law: $$\sum F_y = 0$$ The only considerable forces are gravity and the buoyant force. $$\Leftrightarrow m_{cube}=\rho_{wood}s^3=\rho_{water}(s^3-0.02\mathrm{\ m} \cdot s^2)$$ By solving the system of equations: $$s^3=\frac{0.200 \mathrm{\ kg}}{\rho_{water}-\rho_{wood}}\\\rho_{wood}s^3=\rho_{water}(s^3-0.02\mathrm{\ m} \cdot s^2)$$ the desired result follows: $$s=0.10 \mathrm{\ m} \land \rho_{wood}=800 \mathrm{\ \frac{kg}{m^3}}$$
Therefore, (B) is the correct answer.

7. Oct 15, 2015

### Krylov

Well done, and thank you for letting us know that you solved it, that is always nice.