Question

slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular solid. The volume of the slab is 20 m3 and the surface area of the top is 14 m2. The density of ice is 900 kg/m3 and sea water is 1030 kg/m3. No need to show work. a) Calculate the percentage of the volume of the ice that is submerged. b) Calculate the height, in meters, of the portion of the ice that is submerged. c) Calculate the buoyant force acting on the ice. d) Assume a polar bear has a mass of 400 kg. Calculate the maximum number of polar bears that could be supported by the slab without the slab sinking below the surface of the water.

Answer 1

Answer:

a) $\%V = 87.36\,\%$, b) $x = 1.248\,m$, c) $F_{B} = 176488.341\,N$, d) Six polar bears.

Explanation:

a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:

$\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0$

The height of the ice submerged is:

$\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V$

$x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}$

$x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}$

$x = 1.248\,m$

The percentage of the volume of the ice that is submerged is:

$\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%$

$\%V = 87.36\,\%$

b) The height of the portion of the ice that is submerged is:

$x = 1.248\,m$

c) The buoyant force acting on the ice is:

$F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F_{B} = 176488.341\,N$

d) The new system is modelled after the Archimedes' Principle and Newton's Laws:

$\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0$

The number of polar bear is cleared in the equation:

$n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V$

$n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}$

$n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}$

$n = 6.5$

The maximum number of polar bears that slab could support is 6.