Question

a 10cm tube floats in water with height of 4cm remaining above the surface.what is the density of materials from which the cube is made?

Answer 1

A 10 cm cube floats in water with height of 4 cm remaining above the surface. What is the density of materials from which the cube is made?

Solution.

$$\rho_{w} = 1000 \frac{kg}{m^{3}}, H = 10cm = 0.1m, h = 4cm = 0.04m;$$

Newton's second law in vector form:

$$m \vec{a} = \vec{B} + m \vec{g}.$$

A cube is at rest, then:

$$\vec{a} = 0.$$ $$0 = \vec{B} + m \vec{g}.$$

Projection on Y:

$$0 = B - mg;$$ $$B = mg.$$

$B$ – a buoyancy force.

$m$ – a mass of a cube.

$$m = \rho_{c} V_{c};$$

$\rho_{c}$ – the density of a cube.

$V_{c}$ – a volume of a cube.

$$V _ {c} = H ^ {3}.$$ $$B = \rho_ {w} V g;$$

$\rho_w$ – a density of a water;

$V$ – a part of volume of a cube below water surface.

$$V = S (H - h);$$

$h$ - the height of the cube above the surface;

$S$ - the area of the face of the cube.

$$S = H ^ {2};$$ $$V = H ^ {2} (H - h);$$ $$\rho_ {w} V g = m g;$$ $$\rho_ {w} V = m;$$ $$\rho_ {w} V = \rho_ {c} V _ {c};$$ $$\rho_ {w} H ^ {2} (H - h) = \rho_ {c} H ^ {3};$$ $$\rho_ {c} = \frac {\rho_ {w} (H - h)}{H}.$$

The density of the cube is:

$$\rho_ {c} = \frac {1 0 0 0 \cdot (0 . 1 - 0 . 0 4)}{0 . 1} = 6 0 0 \left(\frac {k g}{m ^ {3}}\right).$$

Answer: The density of the cube is $\rho_{c} = 600\frac{kg}{m^{3}}$.